0-rank tensor vs vector in 1D The Next CEO of Stack OverflowWriting Breit-Pauli spin-spin-coupling Hamiltonian as a sum of irreducible spin tensor operatorsHow to prove this mathematical result about the Green's displacement tensor?Determinant of a mixed rank-2 tensorMetric Tensor as the Simplest Mathematical Object that can Define a Space (or Spacetime)Derivatives in Euler-Lagrange for fieldsWhat exactly is the Parity transformation? Parity in spherical coordinatesIt is correct to say that a tensor is simply a multidimensional array of related quantities? But what about a tensor as a transformation?Dirac's book *General Theory of Relativity*: Doesn't this show the partial derivative of the metric tensor is zero?Doubts on covariant and contravariant vectors and on double tensorsHow does the partial derivative of a tensor of rank $n$ creates a tensor of rank $n+1$? (cartesian coordinates)

Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact

Is it a bad idea to plug the other end of ESD strap to wall ground?

Is this a new Fibonacci Identity?

Early programmable calculators with RS-232

Is it possible to make a 9x9 table fit within the default margins?

My boss doesn't want me to have a side project

Is a distribution that is normal, but highly skewed, considered Gaussian?

Why doesn't Shulchan Aruch include the laws of destroying fruit trees?

Salesforce opportunity stages

logical reads on global temp table, but not on session-level temp table

Is it possible to create a QR code using text?

Strange use of "whether ... than ..." in official text

Is it correct to say moon starry nights?

Are British MPs missing the point, with these 'Indicative Votes'?

How can I separate the number from the unit in argument?

Could a dragon use its wings to swim?

Could you use a laser beam as a modulated carrier wave for radio signal?

Can a PhD from a non-TU9 German university become a professor in a TU9 university?

Why was Sir Cadogan fired?

Masking layers by a vector polygon layer in QGIS

Is it okay to majorly distort historical facts while writing a fiction story?

How to find if SQL server backup is encrypted with TDE without restoring the backup

Creating a script with console commands

Upgrading From a 9 Speed Sora Derailleur?



0-rank tensor vs vector in 1D



The Next CEO of Stack OverflowWriting Breit-Pauli spin-spin-coupling Hamiltonian as a sum of irreducible spin tensor operatorsHow to prove this mathematical result about the Green's displacement tensor?Determinant of a mixed rank-2 tensorMetric Tensor as the Simplest Mathematical Object that can Define a Space (or Spacetime)Derivatives in Euler-Lagrange for fieldsWhat exactly is the Parity transformation? Parity in spherical coordinatesIt is correct to say that a tensor is simply a multidimensional array of related quantities? But what about a tensor as a transformation?Dirac's book *General Theory of Relativity*: Doesn't this show the partial derivative of the metric tensor is zero?Doubts on covariant and contravariant vectors and on double tensorsHow does the partial derivative of a tensor of rank $n$ creates a tensor of rank $n+1$? (cartesian coordinates)










10












$begingroup$


What is the difference between zero-rank tensor $x$ (scalar) and vector $[x]$ in 1D?



As far as I understand tensor is anything which can be measured and different measures can be transformed into each other. That is, there are different basises for looking at one object.



Is length a scalar (zero rank tensor) or is it a 1D vector (rank 1 tensor)?



In books it is said that temperature, pressure and other "numbers" are 0rank tensors, they are invariant under transformations and posess no "direction" (that is there is no basis for them). But I'm cofused about units... I thinks of units as somekind of basis.



ex.:



  • physical parameter: writing pen's length

  • tensor: $l$

  • length in "inches basis": $[5.511811023622]$

  • length in "centimeters basis": $[14]$

  • transformation law: 1cm = 2.54inch

so $l$ is a scalar (0rank), but on the other hand it's a vector (1rank).



The same logic can be applied to mutate classical examples of 0rank tensors: pressure, temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.










share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise.
    $endgroup$
    – gented
    yesterday















10












$begingroup$


What is the difference between zero-rank tensor $x$ (scalar) and vector $[x]$ in 1D?



As far as I understand tensor is anything which can be measured and different measures can be transformed into each other. That is, there are different basises for looking at one object.



Is length a scalar (zero rank tensor) or is it a 1D vector (rank 1 tensor)?



In books it is said that temperature, pressure and other "numbers" are 0rank tensors, they are invariant under transformations and posess no "direction" (that is there is no basis for them). But I'm cofused about units... I thinks of units as somekind of basis.



ex.:



  • physical parameter: writing pen's length

  • tensor: $l$

  • length in "inches basis": $[5.511811023622]$

  • length in "centimeters basis": $[14]$

  • transformation law: 1cm = 2.54inch

so $l$ is a scalar (0rank), but on the other hand it's a vector (1rank).



The same logic can be applied to mutate classical examples of 0rank tensors: pressure, temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.










share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise.
    $endgroup$
    – gented
    yesterday













10












10








10


1



$begingroup$


What is the difference between zero-rank tensor $x$ (scalar) and vector $[x]$ in 1D?



As far as I understand tensor is anything which can be measured and different measures can be transformed into each other. That is, there are different basises for looking at one object.



Is length a scalar (zero rank tensor) or is it a 1D vector (rank 1 tensor)?



In books it is said that temperature, pressure and other "numbers" are 0rank tensors, they are invariant under transformations and posess no "direction" (that is there is no basis for them). But I'm cofused about units... I thinks of units as somekind of basis.



ex.:



  • physical parameter: writing pen's length

  • tensor: $l$

  • length in "inches basis": $[5.511811023622]$

  • length in "centimeters basis": $[14]$

  • transformation law: 1cm = 2.54inch

so $l$ is a scalar (0rank), but on the other hand it's a vector (1rank).



The same logic can be applied to mutate classical examples of 0rank tensors: pressure, temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.










share|cite|improve this question











$endgroup$




What is the difference between zero-rank tensor $x$ (scalar) and vector $[x]$ in 1D?



As far as I understand tensor is anything which can be measured and different measures can be transformed into each other. That is, there are different basises for looking at one object.



Is length a scalar (zero rank tensor) or is it a 1D vector (rank 1 tensor)?



In books it is said that temperature, pressure and other "numbers" are 0rank tensors, they are invariant under transformations and posess no "direction" (that is there is no basis for them). But I'm cofused about units... I thinks of units as somekind of basis.



ex.:



  • physical parameter: writing pen's length

  • tensor: $l$

  • length in "inches basis": $[5.511811023622]$

  • length in "centimeters basis": $[14]$

  • transformation law: 1cm = 2.54inch

so $l$ is a scalar (0rank), but on the other hand it's a vector (1rank).



The same logic can be applied to mutate classical examples of 0rank tensors: pressure, temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.







vectors coordinate-systems tensor-calculus linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







coobit

















asked 2 days ago









coobitcoobit

380111




380111







  • 3




    $begingroup$
    Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise.
    $endgroup$
    – gented
    yesterday












  • 3




    $begingroup$
    Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise.
    $endgroup$
    – gented
    yesterday







3




3




$begingroup$
Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise.
$endgroup$
– gented
yesterday




$begingroup$
Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise.
$endgroup$
– gented
yesterday










4 Answers
4






active

oldest

votes


















14












$begingroup$

“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.



The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.



But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.



Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.



Under any other transformation group, the distinction between scalars and vectors is similar.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
    $endgroup$
    – coobit
    2 days ago











  • $begingroup$
    @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
    $endgroup$
    – Chiral Anomaly
    2 days ago










  • $begingroup$
    Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
    $endgroup$
    – G. Smith
    2 days ago











  • $begingroup$
    @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
    $endgroup$
    – Carmeister
    2 days ago










  • $begingroup$
    I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
    $endgroup$
    – coobit
    yesterday


















5












$begingroup$

First of all, I'll constrain the discussion assuming:



1) Finite-dimensional vector spaces



2) Real Vector spaces



3) Talking just about contravariant tensors



4) Physics which use the standard notion of Spacetime



$$* * *$$



To answer your question I need to talk a little bit about Tensors.



I) The tensor object and pure mathematics:



The precise answer to the question "What is a tensor?" is, by far:




A tensor is a object of a vector space called Tensor Product.




In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.



I.1) What truly is a Vector?



First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:




A vector is a element of a algebraic structure called vector space.




So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.



I.1.1) Some facts about vectors



Consider then a vector formed by a linear combination of basis vectors:



$$mathbfv = sum_j = 1^n v^jmathbfe_j tag1$$



This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):




A set $mathcalS$ is a basis for a vector space $mathfrakV$ if:



1) the vectors of the set $mathcalS$ are linear independent



2) the vectors of the set $mathcalS$ spanned the vector space $mathfrakV$, i.e. $mathfrakV equiv span(mathcalS)$




So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.



Another fact is that you can change the basis $mathbfe_j$ to another set basis of basis $mathbfe'_j$. Well, when you do this the vector components suffer a change too. And then the components transforms like:



$$v'^k = sum^n_j=1M^k_jv^jtag2$$



but, of course, the vector object, remains the same:



$$mathbfv = sum_j = 1^n v^jmathbfe_j = sum_j = 1^n v'^jmathbfe'_j$$



So, a vector, truly, is a object of a vector space, which have the form of $(1)$ and their components transforms like $(2)$.



I.1.2) The "physicist way" of definition of a Tensor



When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:




A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:



$$T'^ij = sum^n_k=1sum^n_l=1 M^i_kM^j_l T^kl tag3$$




This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said).
Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $(1)$, which the basis vectors spans the entire $mathfrakV$) and their components have "transformation behaviour" like $(2)$. If you compare what I exposed about vector and $(3)$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing.
The fact is, the definition $(3)$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $mathbfT$.



I.2) What truly is a Tensor?



Well, you have the transformation of components of a tensor, i.e. $(3)$, well defined. But what about their "space" and "form" (something like $(1)$)?



So, the space is called tensor product of two vector spaces:




$$Votimes W tag4$$




The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $Votimes W$ and $mathfrakLin^2(V^* times W^*; mathbbK)$:




$$Votimes W cong mathfrakLin^2(V^* times W^*; mathbbK) tag5$$



where $mathfrakLin^2(V^* times W^*; mathbbK)$ is the dual vector space of all bilinear functionals.




So a tensor have the form:




$$mathbfT = sum^n_i=1sum^n_j=1 T^ij (mathbfe_iotimesmathbfe_j) tag6$$
And $mathbfT in Votimes W$.




Well, given the transformation rule $(3)$ the space, $(5)$, and form, $(6)$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $(6)$ the tensor basis $(mathbfe_iotimesmathbfe_j)$ spans $Votimes W$.



By virtue of the general construction of tensor product and the identification given by $(5)$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:




$$beginarrayrl
mathbfT :V^*times W^* &to mathbbK \
(mathbfv,mathbfw)&mapsto mathbfT(mathbfv,mathbfw)=: v^icdot_mathbbKw^j
endarray$$



Where the operation $cdot_mathbbK$ is the product defined in the field.




With this picture we say that a tensor like $(6)$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.



II) The tensor object and physics



The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics.
The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.



Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:




$$[mathbfT(x^k)] = sum^n_i=1sum^n_j=1 [T^ij(x^k)] ([mathbfe_i(x^k)]otimes[mathbfe_j(x^k)]) tag5$$
A tensor field is the object which attaches a tensor to every point p of the Manifold.




With the manifold theory, the transformation rule becomes:




$$[T'^ij(x^m)] = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l [T^kl(x^m)] equiv T'^ij = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l T^kl tag7$$




Notice that the partials are simply the transformation matrices $M$. The matrices $M$ are called the Jacobians transformation matrices and the matrices $M$ became these jacobians by virtue of Manifold theory.



In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.



III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?



So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $mathbbK$ and the other is a member of a vector space.
Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).




$$phi'= phi$$




$$* * *$$



[*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
    $endgroup$
    – coobit
    yesterday






  • 1




    $begingroup$
    Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
    $endgroup$
    – Brevan Ellefsen
    yesterday


















1












$begingroup$

I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.



You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.



Scalar field



We want to consider a space $M$ consisting of points $p$ that we give coordinates $x(p)$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $x'(p)$ but the underlying point $p$ is the same. A scalar function is any function that assigns to each $pin M$ real number:



$f:pmapsto f(p)$.



However, we often abuse notation and write the function as a function of the coordinates:



$f:xmapsto f(x)$ and $f':x'mapsto f'(x')$



the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $p$:



$f'(x'(p)) = f(x(p)) = f( Lambda^-1(x'(p))$



where $Lambda$ is the function that changes coordinates i.e. $xmapsto x' = Lambda (x)$.



Vector Field



There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $V_p$ at each point $pin M$ of the same dimension as $M$ (so 1 in our case). If then maps each point $p$ in $M$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $v^imapsto v^i' = J^i'_i v^i$ on the individual components of and vector $vin V_p$. Here, $J$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.



Example



Consider the scalar function $f(x)=e^x$ and the vector function $vecv(x)=e^xhatx$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $xmapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=e^x'/a$.



The vector function transforms to become $vecv'(x') = e^x'/a hatx$. However as noted above we must also transform the basis vector and the rule amounts to $hatx = a hatx'$ in this case. So the new vector function is $vecv'(x') = a e^x'/a hatx'$.



The case of length



For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $2$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $L(x)=x$ rather than the vector transformation law for $vecL=xhatx$ which would be different.




*The notations $hatx$ and $hatx'$ mean the basis vector appropriate for the $x$ and $x'$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
    $endgroup$
    – coobit
    yesterday










  • $begingroup$
    @coobit I have updated the answer. If it is still unclear tell me.
    $endgroup$
    – jacob1729
    yesterday










  • $begingroup$
    Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
    $endgroup$
    – coobit
    yesterday










  • $begingroup$
    But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
    $endgroup$
    – coobit
    yesterday











  • $begingroup$
    @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
    $endgroup$
    – jacob1729
    yesterday


















1












$begingroup$

You seem to confuse transformations with conversion of units. A transformation is a purely mathematical concept and as such has no notion of (physical) units. Physics gives rise to the concept of units which are used to describe physical quantities.



Coming back to your example, let's consider a one-dimensional coordinate system and a pen therein. The pen's head is located at $vecx_head=0$ and its tail is located at $vecx_tail = t$. These two positions are (one-dimensional) vector quantities while the pen's length is a scalar quantity, defined by $L = ||vecx_tail - vecx_head||$ where $||cdot||$ is an appropriate metric with which the vector space is equipped (usually Euclidean metric in physics context but Minkowski metric is used as well). Now consider a reflection of the coordinate system about $vecx=0$. The tail position is going to change to $vecx'_tail = -t$ but the pen's length is still $L$, i.e. it is invariant under such transformations.



Now you brought in the concept of (physical) units but, as mentioned above, this is distinct from (mathematical) transformations. While you wrote "transformation law" the appropriate term would be "conversion law" and this also emphasizes the distinction between these two concepts. As you can observe from the previous paragraph, no notion of a unit was necessary to describe the transformation (reflection) of the coordinate system. Units on the other hand determine the value of $t$ (the pen's tail position) that we are going to obtain by our measurement. They ensure that you and I, if we measure the same pen, arrive at the same value for that position. A conversion of unit means that we change to a different ruler and hence obtain a different number (magnitude) for the position but this is all valid and compatible as long as we report this number together with its unit of measurement and we know the relevant conversion rules.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "151"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f469598%2f0-rank-tensor-vs-vector-in-1d%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    14












    $begingroup$

    “Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.



    The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.



    But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.



    Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.



    Under any other transformation group, the distinction between scalars and vectors is similar.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
      $endgroup$
      – coobit
      2 days ago











    • $begingroup$
      @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
      $endgroup$
      – Chiral Anomaly
      2 days ago










    • $begingroup$
      Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
      $endgroup$
      – G. Smith
      2 days ago











    • $begingroup$
      @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
      $endgroup$
      – Carmeister
      2 days ago










    • $begingroup$
      I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
      $endgroup$
      – coobit
      yesterday















    14












    $begingroup$

    “Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.



    The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.



    But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.



    Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.



    Under any other transformation group, the distinction between scalars and vectors is similar.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
      $endgroup$
      – coobit
      2 days ago











    • $begingroup$
      @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
      $endgroup$
      – Chiral Anomaly
      2 days ago










    • $begingroup$
      Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
      $endgroup$
      – G. Smith
      2 days ago











    • $begingroup$
      @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
      $endgroup$
      – Carmeister
      2 days ago










    • $begingroup$
      I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
      $endgroup$
      – coobit
      yesterday













    14












    14








    14





    $begingroup$

    “Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.



    The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.



    But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.



    Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.



    Under any other transformation group, the distinction between scalars and vectors is similar.






    share|cite|improve this answer











    $endgroup$



    “Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.



    The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.



    But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.



    Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.



    Under any other transformation group, the distinction between scalars and vectors is similar.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    G. SmithG. Smith

    10.3k11429




    10.3k11429











    • $begingroup$
      I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
      $endgroup$
      – coobit
      2 days ago











    • $begingroup$
      @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
      $endgroup$
      – Chiral Anomaly
      2 days ago










    • $begingroup$
      Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
      $endgroup$
      – G. Smith
      2 days ago











    • $begingroup$
      @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
      $endgroup$
      – Carmeister
      2 days ago










    • $begingroup$
      I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
      $endgroup$
      – coobit
      yesterday
















    • $begingroup$
      I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
      $endgroup$
      – coobit
      2 days ago











    • $begingroup$
      @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
      $endgroup$
      – Chiral Anomaly
      2 days ago










    • $begingroup$
      Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
      $endgroup$
      – G. Smith
      2 days ago











    • $begingroup$
      @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
      $endgroup$
      – Carmeister
      2 days ago










    • $begingroup$
      I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
      $endgroup$
      – coobit
      yesterday















    $begingroup$
    I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
    $endgroup$
    – coobit
    2 days ago





    $begingroup$
    I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
    $endgroup$
    – coobit
    2 days ago













    $begingroup$
    @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
    $endgroup$
    – Chiral Anomaly
    2 days ago




    $begingroup$
    @coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
    $endgroup$
    – Chiral Anomaly
    2 days ago












    $begingroup$
    Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
    $endgroup$
    – G. Smith
    2 days ago





    $begingroup$
    Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
    $endgroup$
    – G. Smith
    2 days ago













    $begingroup$
    @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
    $endgroup$
    – Carmeister
    2 days ago




    $begingroup$
    @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange
    $endgroup$
    – Carmeister
    2 days ago












    $begingroup$
    I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
    $endgroup$
    – coobit
    yesterday




    $begingroup$
    I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it?
    $endgroup$
    – coobit
    yesterday











    5












    $begingroup$

    First of all, I'll constrain the discussion assuming:



    1) Finite-dimensional vector spaces



    2) Real Vector spaces



    3) Talking just about contravariant tensors



    4) Physics which use the standard notion of Spacetime



    $$* * *$$



    To answer your question I need to talk a little bit about Tensors.



    I) The tensor object and pure mathematics:



    The precise answer to the question "What is a tensor?" is, by far:




    A tensor is a object of a vector space called Tensor Product.




    In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.



    I.1) What truly is a Vector?



    First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:




    A vector is a element of a algebraic structure called vector space.




    So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.



    I.1.1) Some facts about vectors



    Consider then a vector formed by a linear combination of basis vectors:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j tag1$$



    This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):




    A set $mathcalS$ is a basis for a vector space $mathfrakV$ if:



    1) the vectors of the set $mathcalS$ are linear independent



    2) the vectors of the set $mathcalS$ spanned the vector space $mathfrakV$, i.e. $mathfrakV equiv span(mathcalS)$




    So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.



    Another fact is that you can change the basis $mathbfe_j$ to another set basis of basis $mathbfe'_j$. Well, when you do this the vector components suffer a change too. And then the components transforms like:



    $$v'^k = sum^n_j=1M^k_jv^jtag2$$



    but, of course, the vector object, remains the same:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j = sum_j = 1^n v'^jmathbfe'_j$$



    So, a vector, truly, is a object of a vector space, which have the form of $(1)$ and their components transforms like $(2)$.



    I.1.2) The "physicist way" of definition of a Tensor



    When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:




    A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:



    $$T'^ij = sum^n_k=1sum^n_l=1 M^i_kM^j_l T^kl tag3$$




    This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said).
    Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $(1)$, which the basis vectors spans the entire $mathfrakV$) and their components have "transformation behaviour" like $(2)$. If you compare what I exposed about vector and $(3)$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing.
    The fact is, the definition $(3)$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $mathbfT$.



    I.2) What truly is a Tensor?



    Well, you have the transformation of components of a tensor, i.e. $(3)$, well defined. But what about their "space" and "form" (something like $(1)$)?



    So, the space is called tensor product of two vector spaces:




    $$Votimes W tag4$$




    The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $Votimes W$ and $mathfrakLin^2(V^* times W^*; mathbbK)$:




    $$Votimes W cong mathfrakLin^2(V^* times W^*; mathbbK) tag5$$



    where $mathfrakLin^2(V^* times W^*; mathbbK)$ is the dual vector space of all bilinear functionals.




    So a tensor have the form:




    $$mathbfT = sum^n_i=1sum^n_j=1 T^ij (mathbfe_iotimesmathbfe_j) tag6$$
    And $mathbfT in Votimes W$.




    Well, given the transformation rule $(3)$ the space, $(5)$, and form, $(6)$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $(6)$ the tensor basis $(mathbfe_iotimesmathbfe_j)$ spans $Votimes W$.



    By virtue of the general construction of tensor product and the identification given by $(5)$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:




    $$beginarrayrl
    mathbfT :V^*times W^* &to mathbbK \
    (mathbfv,mathbfw)&mapsto mathbfT(mathbfv,mathbfw)=: v^icdot_mathbbKw^j
    endarray$$



    Where the operation $cdot_mathbbK$ is the product defined in the field.




    With this picture we say that a tensor like $(6)$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.



    II) The tensor object and physics



    The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics.
    The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.



    Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:




    $$[mathbfT(x^k)] = sum^n_i=1sum^n_j=1 [T^ij(x^k)] ([mathbfe_i(x^k)]otimes[mathbfe_j(x^k)]) tag5$$
    A tensor field is the object which attaches a tensor to every point p of the Manifold.




    With the manifold theory, the transformation rule becomes:




    $$[T'^ij(x^m)] = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l [T^kl(x^m)] equiv T'^ij = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l T^kl tag7$$




    Notice that the partials are simply the transformation matrices $M$. The matrices $M$ are called the Jacobians transformation matrices and the matrices $M$ became these jacobians by virtue of Manifold theory.



    In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.



    III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?



    So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $mathbbK$ and the other is a member of a vector space.
    Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).




    $$phi'= phi$$




    $$* * *$$



    [*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
      $endgroup$
      – coobit
      yesterday






    • 1




      $begingroup$
      Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
      $endgroup$
      – Brevan Ellefsen
      yesterday















    5












    $begingroup$

    First of all, I'll constrain the discussion assuming:



    1) Finite-dimensional vector spaces



    2) Real Vector spaces



    3) Talking just about contravariant tensors



    4) Physics which use the standard notion of Spacetime



    $$* * *$$



    To answer your question I need to talk a little bit about Tensors.



    I) The tensor object and pure mathematics:



    The precise answer to the question "What is a tensor?" is, by far:




    A tensor is a object of a vector space called Tensor Product.




    In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.



    I.1) What truly is a Vector?



    First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:




    A vector is a element of a algebraic structure called vector space.




    So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.



    I.1.1) Some facts about vectors



    Consider then a vector formed by a linear combination of basis vectors:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j tag1$$



    This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):




    A set $mathcalS$ is a basis for a vector space $mathfrakV$ if:



    1) the vectors of the set $mathcalS$ are linear independent



    2) the vectors of the set $mathcalS$ spanned the vector space $mathfrakV$, i.e. $mathfrakV equiv span(mathcalS)$




    So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.



    Another fact is that you can change the basis $mathbfe_j$ to another set basis of basis $mathbfe'_j$. Well, when you do this the vector components suffer a change too. And then the components transforms like:



    $$v'^k = sum^n_j=1M^k_jv^jtag2$$



    but, of course, the vector object, remains the same:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j = sum_j = 1^n v'^jmathbfe'_j$$



    So, a vector, truly, is a object of a vector space, which have the form of $(1)$ and their components transforms like $(2)$.



    I.1.2) The "physicist way" of definition of a Tensor



    When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:




    A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:



    $$T'^ij = sum^n_k=1sum^n_l=1 M^i_kM^j_l T^kl tag3$$




    This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said).
    Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $(1)$, which the basis vectors spans the entire $mathfrakV$) and their components have "transformation behaviour" like $(2)$. If you compare what I exposed about vector and $(3)$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing.
    The fact is, the definition $(3)$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $mathbfT$.



    I.2) What truly is a Tensor?



    Well, you have the transformation of components of a tensor, i.e. $(3)$, well defined. But what about their "space" and "form" (something like $(1)$)?



    So, the space is called tensor product of two vector spaces:




    $$Votimes W tag4$$




    The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $Votimes W$ and $mathfrakLin^2(V^* times W^*; mathbbK)$:




    $$Votimes W cong mathfrakLin^2(V^* times W^*; mathbbK) tag5$$



    where $mathfrakLin^2(V^* times W^*; mathbbK)$ is the dual vector space of all bilinear functionals.




    So a tensor have the form:




    $$mathbfT = sum^n_i=1sum^n_j=1 T^ij (mathbfe_iotimesmathbfe_j) tag6$$
    And $mathbfT in Votimes W$.




    Well, given the transformation rule $(3)$ the space, $(5)$, and form, $(6)$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $(6)$ the tensor basis $(mathbfe_iotimesmathbfe_j)$ spans $Votimes W$.



    By virtue of the general construction of tensor product and the identification given by $(5)$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:




    $$beginarrayrl
    mathbfT :V^*times W^* &to mathbbK \
    (mathbfv,mathbfw)&mapsto mathbfT(mathbfv,mathbfw)=: v^icdot_mathbbKw^j
    endarray$$



    Where the operation $cdot_mathbbK$ is the product defined in the field.




    With this picture we say that a tensor like $(6)$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.



    II) The tensor object and physics



    The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics.
    The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.



    Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:




    $$[mathbfT(x^k)] = sum^n_i=1sum^n_j=1 [T^ij(x^k)] ([mathbfe_i(x^k)]otimes[mathbfe_j(x^k)]) tag5$$
    A tensor field is the object which attaches a tensor to every point p of the Manifold.




    With the manifold theory, the transformation rule becomes:




    $$[T'^ij(x^m)] = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l [T^kl(x^m)] equiv T'^ij = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l T^kl tag7$$




    Notice that the partials are simply the transformation matrices $M$. The matrices $M$ are called the Jacobians transformation matrices and the matrices $M$ became these jacobians by virtue of Manifold theory.



    In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.



    III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?



    So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $mathbbK$ and the other is a member of a vector space.
    Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).




    $$phi'= phi$$




    $$* * *$$



    [*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
      $endgroup$
      – coobit
      yesterday






    • 1




      $begingroup$
      Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
      $endgroup$
      – Brevan Ellefsen
      yesterday













    5












    5








    5





    $begingroup$

    First of all, I'll constrain the discussion assuming:



    1) Finite-dimensional vector spaces



    2) Real Vector spaces



    3) Talking just about contravariant tensors



    4) Physics which use the standard notion of Spacetime



    $$* * *$$



    To answer your question I need to talk a little bit about Tensors.



    I) The tensor object and pure mathematics:



    The precise answer to the question "What is a tensor?" is, by far:




    A tensor is a object of a vector space called Tensor Product.




    In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.



    I.1) What truly is a Vector?



    First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:




    A vector is a element of a algebraic structure called vector space.




    So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.



    I.1.1) Some facts about vectors



    Consider then a vector formed by a linear combination of basis vectors:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j tag1$$



    This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):




    A set $mathcalS$ is a basis for a vector space $mathfrakV$ if:



    1) the vectors of the set $mathcalS$ are linear independent



    2) the vectors of the set $mathcalS$ spanned the vector space $mathfrakV$, i.e. $mathfrakV equiv span(mathcalS)$




    So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.



    Another fact is that you can change the basis $mathbfe_j$ to another set basis of basis $mathbfe'_j$. Well, when you do this the vector components suffer a change too. And then the components transforms like:



    $$v'^k = sum^n_j=1M^k_jv^jtag2$$



    but, of course, the vector object, remains the same:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j = sum_j = 1^n v'^jmathbfe'_j$$



    So, a vector, truly, is a object of a vector space, which have the form of $(1)$ and their components transforms like $(2)$.



    I.1.2) The "physicist way" of definition of a Tensor



    When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:




    A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:



    $$T'^ij = sum^n_k=1sum^n_l=1 M^i_kM^j_l T^kl tag3$$




    This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said).
    Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $(1)$, which the basis vectors spans the entire $mathfrakV$) and their components have "transformation behaviour" like $(2)$. If you compare what I exposed about vector and $(3)$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing.
    The fact is, the definition $(3)$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $mathbfT$.



    I.2) What truly is a Tensor?



    Well, you have the transformation of components of a tensor, i.e. $(3)$, well defined. But what about their "space" and "form" (something like $(1)$)?



    So, the space is called tensor product of two vector spaces:




    $$Votimes W tag4$$




    The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $Votimes W$ and $mathfrakLin^2(V^* times W^*; mathbbK)$:




    $$Votimes W cong mathfrakLin^2(V^* times W^*; mathbbK) tag5$$



    where $mathfrakLin^2(V^* times W^*; mathbbK)$ is the dual vector space of all bilinear functionals.




    So a tensor have the form:




    $$mathbfT = sum^n_i=1sum^n_j=1 T^ij (mathbfe_iotimesmathbfe_j) tag6$$
    And $mathbfT in Votimes W$.




    Well, given the transformation rule $(3)$ the space, $(5)$, and form, $(6)$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $(6)$ the tensor basis $(mathbfe_iotimesmathbfe_j)$ spans $Votimes W$.



    By virtue of the general construction of tensor product and the identification given by $(5)$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:




    $$beginarrayrl
    mathbfT :V^*times W^* &to mathbbK \
    (mathbfv,mathbfw)&mapsto mathbfT(mathbfv,mathbfw)=: v^icdot_mathbbKw^j
    endarray$$



    Where the operation $cdot_mathbbK$ is the product defined in the field.




    With this picture we say that a tensor like $(6)$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.



    II) The tensor object and physics



    The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics.
    The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.



    Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:




    $$[mathbfT(x^k)] = sum^n_i=1sum^n_j=1 [T^ij(x^k)] ([mathbfe_i(x^k)]otimes[mathbfe_j(x^k)]) tag5$$
    A tensor field is the object which attaches a tensor to every point p of the Manifold.




    With the manifold theory, the transformation rule becomes:




    $$[T'^ij(x^m)] = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l [T^kl(x^m)] equiv T'^ij = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l T^kl tag7$$




    Notice that the partials are simply the transformation matrices $M$. The matrices $M$ are called the Jacobians transformation matrices and the matrices $M$ became these jacobians by virtue of Manifold theory.



    In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.



    III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?



    So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $mathbbK$ and the other is a member of a vector space.
    Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).




    $$phi'= phi$$




    $$* * *$$



    [*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.






    share|cite|improve this answer











    $endgroup$



    First of all, I'll constrain the discussion assuming:



    1) Finite-dimensional vector spaces



    2) Real Vector spaces



    3) Talking just about contravariant tensors



    4) Physics which use the standard notion of Spacetime



    $$* * *$$



    To answer your question I need to talk a little bit about Tensors.



    I) The tensor object and pure mathematics:



    The precise answer to the question "What is a tensor?" is, by far:




    A tensor is a object of a vector space called Tensor Product.




    In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.



    I.1) What truly is a Vector?



    First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:




    A vector is a element of a algebraic structure called vector space.




    So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.



    I.1.1) Some facts about vectors



    Consider then a vector formed by a linear combination of basis vectors:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j tag1$$



    This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):




    A set $mathcalS$ is a basis for a vector space $mathfrakV$ if:



    1) the vectors of the set $mathcalS$ are linear independent



    2) the vectors of the set $mathcalS$ spanned the vector space $mathfrakV$, i.e. $mathfrakV equiv span(mathcalS)$




    So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.



    Another fact is that you can change the basis $mathbfe_j$ to another set basis of basis $mathbfe'_j$. Well, when you do this the vector components suffer a change too. And then the components transforms like:



    $$v'^k = sum^n_j=1M^k_jv^jtag2$$



    but, of course, the vector object, remains the same:



    $$mathbfv = sum_j = 1^n v^jmathbfe_j = sum_j = 1^n v'^jmathbfe'_j$$



    So, a vector, truly, is a object of a vector space, which have the form of $(1)$ and their components transforms like $(2)$.



    I.1.2) The "physicist way" of definition of a Tensor



    When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:




    A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:



    $$T'^ij = sum^n_k=1sum^n_l=1 M^i_kM^j_l T^kl tag3$$




    This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said).
    Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $(1)$, which the basis vectors spans the entire $mathfrakV$) and their components have "transformation behaviour" like $(2)$. If you compare what I exposed about vector and $(3)$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing.
    The fact is, the definition $(3)$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $mathbfT$.



    I.2) What truly is a Tensor?



    Well, you have the transformation of components of a tensor, i.e. $(3)$, well defined. But what about their "space" and "form" (something like $(1)$)?



    So, the space is called tensor product of two vector spaces:




    $$Votimes W tag4$$




    The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $Votimes W$ and $mathfrakLin^2(V^* times W^*; mathbbK)$:




    $$Votimes W cong mathfrakLin^2(V^* times W^*; mathbbK) tag5$$



    where $mathfrakLin^2(V^* times W^*; mathbbK)$ is the dual vector space of all bilinear functionals.




    So a tensor have the form:




    $$mathbfT = sum^n_i=1sum^n_j=1 T^ij (mathbfe_iotimesmathbfe_j) tag6$$
    And $mathbfT in Votimes W$.




    Well, given the transformation rule $(3)$ the space, $(5)$, and form, $(6)$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $(6)$ the tensor basis $(mathbfe_iotimesmathbfe_j)$ spans $Votimes W$.



    By virtue of the general construction of tensor product and the identification given by $(5)$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:




    $$beginarrayrl
    mathbfT :V^*times W^* &to mathbbK \
    (mathbfv,mathbfw)&mapsto mathbfT(mathbfv,mathbfw)=: v^icdot_mathbbKw^j
    endarray$$



    Where the operation $cdot_mathbbK$ is the product defined in the field.




    With this picture we say that a tensor like $(6)$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.



    II) The tensor object and physics



    The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics.
    The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.



    Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:




    $$[mathbfT(x^k)] = sum^n_i=1sum^n_j=1 [T^ij(x^k)] ([mathbfe_i(x^k)]otimes[mathbfe_j(x^k)]) tag5$$
    A tensor field is the object which attaches a tensor to every point p of the Manifold.




    With the manifold theory, the transformation rule becomes:




    $$[T'^ij(x^m)] = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l [T^kl(x^m)] equiv T'^ij = sum^n_k=1sum^n_l=1 fracpartial x'^ipartial x^kfracpartial x'^jpartial x^l T^kl tag7$$




    Notice that the partials are simply the transformation matrices $M$. The matrices $M$ are called the Jacobians transformation matrices and the matrices $M$ became these jacobians by virtue of Manifold theory.



    In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.



    III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?



    So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $mathbbK$ and the other is a member of a vector space.
    Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).




    $$phi'= phi$$




    $$* * *$$



    [*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered 2 days ago









    M.N.RaiaM.N.Raia

    560314




    560314











    • $begingroup$
      Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
      $endgroup$
      – coobit
      yesterday






    • 1




      $begingroup$
      Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
      $endgroup$
      – Brevan Ellefsen
      yesterday
















    • $begingroup$
      Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
      $endgroup$
      – coobit
      yesterday






    • 1




      $begingroup$
      Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
      $endgroup$
      – Brevan Ellefsen
      yesterday















    $begingroup$
    Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
    $endgroup$
    – coobit
    yesterday




    $begingroup$
    Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong?
    $endgroup$
    – coobit
    yesterday




    1




    1




    $begingroup$
    Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
    $endgroup$
    – Brevan Ellefsen
    yesterday




    $begingroup$
    Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post!
    $endgroup$
    – Brevan Ellefsen
    yesterday











    1












    $begingroup$

    I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.



    You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.



    Scalar field



    We want to consider a space $M$ consisting of points $p$ that we give coordinates $x(p)$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $x'(p)$ but the underlying point $p$ is the same. A scalar function is any function that assigns to each $pin M$ real number:



    $f:pmapsto f(p)$.



    However, we often abuse notation and write the function as a function of the coordinates:



    $f:xmapsto f(x)$ and $f':x'mapsto f'(x')$



    the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $p$:



    $f'(x'(p)) = f(x(p)) = f( Lambda^-1(x'(p))$



    where $Lambda$ is the function that changes coordinates i.e. $xmapsto x' = Lambda (x)$.



    Vector Field



    There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $V_p$ at each point $pin M$ of the same dimension as $M$ (so 1 in our case). If then maps each point $p$ in $M$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $v^imapsto v^i' = J^i'_i v^i$ on the individual components of and vector $vin V_p$. Here, $J$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.



    Example



    Consider the scalar function $f(x)=e^x$ and the vector function $vecv(x)=e^xhatx$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $xmapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=e^x'/a$.



    The vector function transforms to become $vecv'(x') = e^x'/a hatx$. However as noted above we must also transform the basis vector and the rule amounts to $hatx = a hatx'$ in this case. So the new vector function is $vecv'(x') = a e^x'/a hatx'$.



    The case of length



    For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $2$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $L(x)=x$ rather than the vector transformation law for $vecL=xhatx$ which would be different.




    *The notations $hatx$ and $hatx'$ mean the basis vector appropriate for the $x$ and $x'$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      @coobit I have updated the answer. If it is still unclear tell me.
      $endgroup$
      – jacob1729
      yesterday










    • $begingroup$
      Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
      $endgroup$
      – coobit
      yesterday











    • $begingroup$
      @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
      $endgroup$
      – jacob1729
      yesterday















    1












    $begingroup$

    I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.



    You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.



    Scalar field



    We want to consider a space $M$ consisting of points $p$ that we give coordinates $x(p)$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $x'(p)$ but the underlying point $p$ is the same. A scalar function is any function that assigns to each $pin M$ real number:



    $f:pmapsto f(p)$.



    However, we often abuse notation and write the function as a function of the coordinates:



    $f:xmapsto f(x)$ and $f':x'mapsto f'(x')$



    the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $p$:



    $f'(x'(p)) = f(x(p)) = f( Lambda^-1(x'(p))$



    where $Lambda$ is the function that changes coordinates i.e. $xmapsto x' = Lambda (x)$.



    Vector Field



    There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $V_p$ at each point $pin M$ of the same dimension as $M$ (so 1 in our case). If then maps each point $p$ in $M$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $v^imapsto v^i' = J^i'_i v^i$ on the individual components of and vector $vin V_p$. Here, $J$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.



    Example



    Consider the scalar function $f(x)=e^x$ and the vector function $vecv(x)=e^xhatx$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $xmapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=e^x'/a$.



    The vector function transforms to become $vecv'(x') = e^x'/a hatx$. However as noted above we must also transform the basis vector and the rule amounts to $hatx = a hatx'$ in this case. So the new vector function is $vecv'(x') = a e^x'/a hatx'$.



    The case of length



    For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $2$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $L(x)=x$ rather than the vector transformation law for $vecL=xhatx$ which would be different.




    *The notations $hatx$ and $hatx'$ mean the basis vector appropriate for the $x$ and $x'$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      @coobit I have updated the answer. If it is still unclear tell me.
      $endgroup$
      – jacob1729
      yesterday










    • $begingroup$
      Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
      $endgroup$
      – coobit
      yesterday











    • $begingroup$
      @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
      $endgroup$
      – jacob1729
      yesterday













    1












    1








    1





    $begingroup$

    I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.



    You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.



    Scalar field



    We want to consider a space $M$ consisting of points $p$ that we give coordinates $x(p)$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $x'(p)$ but the underlying point $p$ is the same. A scalar function is any function that assigns to each $pin M$ real number:



    $f:pmapsto f(p)$.



    However, we often abuse notation and write the function as a function of the coordinates:



    $f:xmapsto f(x)$ and $f':x'mapsto f'(x')$



    the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $p$:



    $f'(x'(p)) = f(x(p)) = f( Lambda^-1(x'(p))$



    where $Lambda$ is the function that changes coordinates i.e. $xmapsto x' = Lambda (x)$.



    Vector Field



    There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $V_p$ at each point $pin M$ of the same dimension as $M$ (so 1 in our case). If then maps each point $p$ in $M$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $v^imapsto v^i' = J^i'_i v^i$ on the individual components of and vector $vin V_p$. Here, $J$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.



    Example



    Consider the scalar function $f(x)=e^x$ and the vector function $vecv(x)=e^xhatx$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $xmapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=e^x'/a$.



    The vector function transforms to become $vecv'(x') = e^x'/a hatx$. However as noted above we must also transform the basis vector and the rule amounts to $hatx = a hatx'$ in this case. So the new vector function is $vecv'(x') = a e^x'/a hatx'$.



    The case of length



    For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $2$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $L(x)=x$ rather than the vector transformation law for $vecL=xhatx$ which would be different.




    *The notations $hatx$ and $hatx'$ mean the basis vector appropriate for the $x$ and $x'$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.






    share|cite|improve this answer











    $endgroup$



    I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.



    You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.



    Scalar field



    We want to consider a space $M$ consisting of points $p$ that we give coordinates $x(p)$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $x'(p)$ but the underlying point $p$ is the same. A scalar function is any function that assigns to each $pin M$ real number:



    $f:pmapsto f(p)$.



    However, we often abuse notation and write the function as a function of the coordinates:



    $f:xmapsto f(x)$ and $f':x'mapsto f'(x')$



    the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $p$:



    $f'(x'(p)) = f(x(p)) = f( Lambda^-1(x'(p))$



    where $Lambda$ is the function that changes coordinates i.e. $xmapsto x' = Lambda (x)$.



    Vector Field



    There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $V_p$ at each point $pin M$ of the same dimension as $M$ (so 1 in our case). If then maps each point $p$ in $M$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $v^imapsto v^i' = J^i'_i v^i$ on the individual components of and vector $vin V_p$. Here, $J$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.



    Example



    Consider the scalar function $f(x)=e^x$ and the vector function $vecv(x)=e^xhatx$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $xmapsto x' = ax$. Then to give the same values $f(x)$ must change to become $f'(x')=e^x'/a$.



    The vector function transforms to become $vecv'(x') = e^x'/a hatx$. However as noted above we must also transform the basis vector and the rule amounts to $hatx = a hatx'$ in this case. So the new vector function is $vecv'(x') = a e^x'/a hatx'$.



    The case of length



    For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $2$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $L(x)=x$ rather than the vector transformation law for $vecL=xhatx$ which would be different.




    *The notations $hatx$ and $hatx'$ mean the basis vector appropriate for the $x$ and $x'$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered yesterday









    jacob1729jacob1729

    873414




    873414











    • $begingroup$
      Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      @coobit I have updated the answer. If it is still unclear tell me.
      $endgroup$
      – jacob1729
      yesterday










    • $begingroup$
      Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
      $endgroup$
      – coobit
      yesterday











    • $begingroup$
      @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
      $endgroup$
      – jacob1729
      yesterday
















    • $begingroup$
      Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      @coobit I have updated the answer. If it is still unclear tell me.
      $endgroup$
      – jacob1729
      yesterday










    • $begingroup$
      Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
      $endgroup$
      – coobit
      yesterday










    • $begingroup$
      But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
      $endgroup$
      – coobit
      yesterday











    • $begingroup$
      @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
      $endgroup$
      – jacob1729
      yesterday















    $begingroup$
    Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
    $endgroup$
    – coobit
    yesterday




    $begingroup$
    Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank?
    $endgroup$
    – coobit
    yesterday












    $begingroup$
    @coobit I have updated the answer. If it is still unclear tell me.
    $endgroup$
    – jacob1729
    yesterday




    $begingroup$
    @coobit I have updated the answer. If it is still unclear tell me.
    $endgroup$
    – jacob1729
    yesterday












    $begingroup$
    Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
    $endgroup$
    – coobit
    yesterday




    $begingroup$
    Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $pi$, $e$ or 0, -1 ect.
    $endgroup$
    – coobit
    yesterday












    $begingroup$
    But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
    $endgroup$
    – coobit
    yesterday





    $begingroup$
    But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind......
    $endgroup$
    – coobit
    yesterday













    $begingroup$
    @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
    $endgroup$
    – jacob1729
    yesterday




    $begingroup$
    @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings....
    $endgroup$
    – jacob1729
    yesterday











    1












    $begingroup$

    You seem to confuse transformations with conversion of units. A transformation is a purely mathematical concept and as such has no notion of (physical) units. Physics gives rise to the concept of units which are used to describe physical quantities.



    Coming back to your example, let's consider a one-dimensional coordinate system and a pen therein. The pen's head is located at $vecx_head=0$ and its tail is located at $vecx_tail = t$. These two positions are (one-dimensional) vector quantities while the pen's length is a scalar quantity, defined by $L = ||vecx_tail - vecx_head||$ where $||cdot||$ is an appropriate metric with which the vector space is equipped (usually Euclidean metric in physics context but Minkowski metric is used as well). Now consider a reflection of the coordinate system about $vecx=0$. The tail position is going to change to $vecx'_tail = -t$ but the pen's length is still $L$, i.e. it is invariant under such transformations.



    Now you brought in the concept of (physical) units but, as mentioned above, this is distinct from (mathematical) transformations. While you wrote "transformation law" the appropriate term would be "conversion law" and this also emphasizes the distinction between these two concepts. As you can observe from the previous paragraph, no notion of a unit was necessary to describe the transformation (reflection) of the coordinate system. Units on the other hand determine the value of $t$ (the pen's tail position) that we are going to obtain by our measurement. They ensure that you and I, if we measure the same pen, arrive at the same value for that position. A conversion of unit means that we change to a different ruler and hence obtain a different number (magnitude) for the position but this is all valid and compatible as long as we report this number together with its unit of measurement and we know the relevant conversion rules.






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      You seem to confuse transformations with conversion of units. A transformation is a purely mathematical concept and as such has no notion of (physical) units. Physics gives rise to the concept of units which are used to describe physical quantities.



      Coming back to your example, let's consider a one-dimensional coordinate system and a pen therein. The pen's head is located at $vecx_head=0$ and its tail is located at $vecx_tail = t$. These two positions are (one-dimensional) vector quantities while the pen's length is a scalar quantity, defined by $L = ||vecx_tail - vecx_head||$ where $||cdot||$ is an appropriate metric with which the vector space is equipped (usually Euclidean metric in physics context but Minkowski metric is used as well). Now consider a reflection of the coordinate system about $vecx=0$. The tail position is going to change to $vecx'_tail = -t$ but the pen's length is still $L$, i.e. it is invariant under such transformations.



      Now you brought in the concept of (physical) units but, as mentioned above, this is distinct from (mathematical) transformations. While you wrote "transformation law" the appropriate term would be "conversion law" and this also emphasizes the distinction between these two concepts. As you can observe from the previous paragraph, no notion of a unit was necessary to describe the transformation (reflection) of the coordinate system. Units on the other hand determine the value of $t$ (the pen's tail position) that we are going to obtain by our measurement. They ensure that you and I, if we measure the same pen, arrive at the same value for that position. A conversion of unit means that we change to a different ruler and hence obtain a different number (magnitude) for the position but this is all valid and compatible as long as we report this number together with its unit of measurement and we know the relevant conversion rules.






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        You seem to confuse transformations with conversion of units. A transformation is a purely mathematical concept and as such has no notion of (physical) units. Physics gives rise to the concept of units which are used to describe physical quantities.



        Coming back to your example, let's consider a one-dimensional coordinate system and a pen therein. The pen's head is located at $vecx_head=0$ and its tail is located at $vecx_tail = t$. These two positions are (one-dimensional) vector quantities while the pen's length is a scalar quantity, defined by $L = ||vecx_tail - vecx_head||$ where $||cdot||$ is an appropriate metric with which the vector space is equipped (usually Euclidean metric in physics context but Minkowski metric is used as well). Now consider a reflection of the coordinate system about $vecx=0$. The tail position is going to change to $vecx'_tail = -t$ but the pen's length is still $L$, i.e. it is invariant under such transformations.



        Now you brought in the concept of (physical) units but, as mentioned above, this is distinct from (mathematical) transformations. While you wrote "transformation law" the appropriate term would be "conversion law" and this also emphasizes the distinction between these two concepts. As you can observe from the previous paragraph, no notion of a unit was necessary to describe the transformation (reflection) of the coordinate system. Units on the other hand determine the value of $t$ (the pen's tail position) that we are going to obtain by our measurement. They ensure that you and I, if we measure the same pen, arrive at the same value for that position. A conversion of unit means that we change to a different ruler and hence obtain a different number (magnitude) for the position but this is all valid and compatible as long as we report this number together with its unit of measurement and we know the relevant conversion rules.






        share|cite|improve this answer











        $endgroup$



        You seem to confuse transformations with conversion of units. A transformation is a purely mathematical concept and as such has no notion of (physical) units. Physics gives rise to the concept of units which are used to describe physical quantities.



        Coming back to your example, let's consider a one-dimensional coordinate system and a pen therein. The pen's head is located at $vecx_head=0$ and its tail is located at $vecx_tail = t$. These two positions are (one-dimensional) vector quantities while the pen's length is a scalar quantity, defined by $L = ||vecx_tail - vecx_head||$ where $||cdot||$ is an appropriate metric with which the vector space is equipped (usually Euclidean metric in physics context but Minkowski metric is used as well). Now consider a reflection of the coordinate system about $vecx=0$. The tail position is going to change to $vecx'_tail = -t$ but the pen's length is still $L$, i.e. it is invariant under such transformations.



        Now you brought in the concept of (physical) units but, as mentioned above, this is distinct from (mathematical) transformations. While you wrote "transformation law" the appropriate term would be "conversion law" and this also emphasizes the distinction between these two concepts. As you can observe from the previous paragraph, no notion of a unit was necessary to describe the transformation (reflection) of the coordinate system. Units on the other hand determine the value of $t$ (the pen's tail position) that we are going to obtain by our measurement. They ensure that you and I, if we measure the same pen, arrive at the same value for that position. A conversion of unit means that we change to a different ruler and hence obtain a different number (magnitude) for the position but this is all valid and compatible as long as we report this number together with its unit of measurement and we know the relevant conversion rules.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 13 hours ago

























        answered yesterday









        a_guesta_guest

        1667




        1667



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Physics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f469598%2f0-rank-tensor-vs-vector-in-1d%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Wikipedia:Vital articles Мазмуну Biography - Өмүр баян Philosophy and psychology - Философия жана психология Religion - Дин Social sciences - Коомдук илимдер Language and literature - Тил жана адабият Science - Илим Technology - Технология Arts and recreation - Искусство жана эс алуу History and geography - Тарых жана география Навигация менюсу

            Club Baloncesto Breogán Índice Historia | Pavillón | Nome | O Breogán na cultura popular | Xogadores | Adestradores | Presidentes | Palmarés | Historial | Líderes | Notas | Véxase tamén | Menú de navegacióncbbreogan.galCadroGuía oficial da ACB 2009-10, páxina 201Guía oficial ACB 1992, páxina 183. Editorial DB.É de 6.500 espectadores sentados axeitándose á última normativa"Estudiantes Junior, entre as mellores canteiras"o orixinalHemeroteca El Mundo Deportivo, 16 setembro de 1970, páxina 12Historia do BreogánAlfredo Pérez, o último canoneiroHistoria C.B. BreogánHemeroteca de El Mundo DeportivoJimmy Wright, norteamericano do Breogán deixará Lugo por ameazas de morteResultados de Breogán en 1986-87Resultados de Breogán en 1990-91Ficha de Velimir Perasović en acb.comResultados de Breogán en 1994-95Breogán arrasa al Barça. "El Mundo Deportivo", 27 de setembro de 1999, páxina 58CB Breogán - FC BarcelonaA FEB invita a participar nunha nova Liga EuropeaCharlie Bell na prensa estatalMáximos anotadores 2005Tempada 2005-06 : Tódolos Xogadores da Xornada""Non quero pensar nunha man negra, mais pregúntome que está a pasar""o orixinalRaúl López, orgulloso dos xogadores, presume da boa saúde económica do BreogánJulio González confirma que cesa como presidente del BreogánHomenaxe a Lisardo GómezA tempada do rexurdimento celesteEntrevista a Lisardo GómezEl COB dinamita el Pazo para forzar el quinto (69-73)Cafés Candelas, patrocinador del CB Breogán"Suso Lázare, novo presidente do Breogán"o orixinalCafés Candelas Breogán firma el mayor triunfo de la historiaEl Breogán realizará 17 homenajes por su cincuenta aniversario"O Breogán honra ao seu fundador e primeiro presidente"o orixinalMiguel Giao recibiu a homenaxe do PazoHomenaxe aos primeiros gladiadores celestesO home que nos amosa como ver o Breo co corazónTita Franco será homenaxeada polos #50anosdeBreoJulio Vila recibirá unha homenaxe in memoriam polos #50anosdeBreo"O Breogán homenaxeará aos seus aboados máis veteráns"Pechada ovación a «Capi» Sanmartín e Ricardo «Corazón de González»Homenaxe por décadas de informaciónPaco García volve ao Pazo con motivo do 50 aniversario"Resultados y clasificaciones""O Cafés Candelas Breogán, campión da Copa Princesa""O Cafés Candelas Breogán, equipo ACB"C.B. Breogán"Proxecto social"o orixinal"Centros asociados"o orixinalFicha en imdb.comMario Camus trata la recuperación del amor en 'La vieja música', su última película"Páxina web oficial""Club Baloncesto Breogán""C. B. Breogán S.A.D."eehttp://www.fegaba.com

            What should I write in an apology letter, since I have decided not to join a company after accepting an offer letterShould I keep looking after accepting a job offer?What should I do when I've been verbally told I would get an offer letter, but still haven't gotten one after 4 weeks?Do I accept an offer from a company that I am not likely to join?New job hasn't confirmed starting date and I want to give current employer as much notice as possibleHow should I address my manager in my resignation letter?HR delayed background verification, now jobless as resignedNo email communication after accepting a formal written offer. How should I phrase the call?What should I do if after receiving a verbal offer letter I am informed that my written job offer is put on hold due to some internal issues?Should I inform the current employer that I am about to resign within 1-2 weeks since I have signed the offer letter and waiting for visa?What company will do, if I send their offer letter to another company