Confusion in Proof of Noether's theoremTransformation of $d^4x$ under translation disregarded?Noether's theorem under arbitrary coordinate transformationNoether's theorem: meaning of transformation of coordinatesNoether's theorem and translationsBig puzzle about Noether's theorem of coordinate transformation (spacetime symmetry)Global and local symmetries in Noether's theorem. And also Stress-Energy tensorsNoether's theorem with infinite parametersVariation of the Action under infinitesimal arbitrary transformations and Noether's TheoremNoether's theorem - Making a global symmetry local (via the SEM tensor)Transformation of coordinates in Noether's TheoremNoether's Theorem in Classical Field theory ConfusionNoether's theorem under arbitrary coordinate transformation
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Confusion in Proof of Noether's theorem
Transformation of $d^4x$ under translation disregarded?Noether's theorem under arbitrary coordinate transformationNoether's theorem: meaning of transformation of coordinatesNoether's theorem and translationsBig puzzle about Noether's theorem of coordinate transformation (spacetime symmetry)Global and local symmetries in Noether's theorem. And also Stress-Energy tensorsNoether's theorem with infinite parametersVariation of the Action under infinitesimal arbitrary transformations and Noether's TheoremNoether's theorem - Making a global symmetry local (via the SEM tensor)Transformation of coordinates in Noether's TheoremNoether's Theorem in Classical Field theory ConfusionNoether's theorem under arbitrary coordinate transformation
$begingroup$
This question is related to this Noether's theorem under arbitrary coordinate transformation and this Transformation of $d^4x$ under translation disregarded?
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x^prime$is given by
$$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^prime-int mathscrL(phi,partial_muphi)d^4x$$
For me this variation should always be zero since this is just a change of variables so there is nothing new.
Suppose that in one dimension flat space we have $phi(x)=sin(x)$ and our action is given by
$$S=intsin(x)dx$$
under a trasnlation $xmapsto x+a$ we should have $phi(x) mapsto phi'(x)=sin(x-a)$.
Under a change of coordinates $xmapsto y=x+a$, $phi(x) mapsto phi'(y)=sin(y-a)$ , since $x$ and $y$ are dummy index the action is the same in both case.
Now assuming we have $sqrt g= x^2$. Translating will give us an action
$$S=int x^2sin(x-a)dx$$
But change of variables will give us an action
$$S=int (y-a)^2sin(y-a)dx$$
witch is different.
Shouldn't the variation of the action be written like this $$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^-int mathscrL(phi,partial_muphi)d^4x$$
symmetry noethers-theorem classical-field-theory
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
$endgroup$
|
show 10 more comments
$begingroup$
This question is related to this Noether's theorem under arbitrary coordinate transformation and this Transformation of $d^4x$ under translation disregarded?
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x^prime$is given by
$$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^prime-int mathscrL(phi,partial_muphi)d^4x$$
For me this variation should always be zero since this is just a change of variables so there is nothing new.
Suppose that in one dimension flat space we have $phi(x)=sin(x)$ and our action is given by
$$S=intsin(x)dx$$
under a trasnlation $xmapsto x+a$ we should have $phi(x) mapsto phi'(x)=sin(x-a)$.
Under a change of coordinates $xmapsto y=x+a$, $phi(x) mapsto phi'(y)=sin(y-a)$ , since $x$ and $y$ are dummy index the action is the same in both case.
Now assuming we have $sqrt g= x^2$. Translating will give us an action
$$S=int x^2sin(x-a)dx$$
But change of variables will give us an action
$$S=int (y-a)^2sin(y-a)dx$$
witch is different.
Shouldn't the variation of the action be written like this $$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^-int mathscrL(phi,partial_muphi)d^4x$$
symmetry noethers-theorem classical-field-theory
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
$endgroup$
$begingroup$
What about the Jacobian arising in front of the measure when changing variables?
$endgroup$
– Valter Moretti
May 2 at 7:14
$begingroup$
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help.
$endgroup$
– Manvendra Somvanshi
May 2 at 7:16
$begingroup$
Don't we make a transformation of the fields themselves, i.e., $phi(x)tophi(x)+deltaphi(x)$ where the variation $deltaphi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $deltaphi(x)neqfracpartialphi(x)partial x^mudx^mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken.
$endgroup$
– Dvij Mankad
May 2 at 7:17
$begingroup$
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action
$endgroup$
– amilton moreira
May 2 at 7:21
$begingroup$
@Valter Moretti $d^4x^prime=|J|d^4x$
$endgroup$
– amilton moreira
May 2 at 7:28
|
show 10 more comments
$begingroup$
This question is related to this Noether's theorem under arbitrary coordinate transformation and this Transformation of $d^4x$ under translation disregarded?
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x^prime$is given by
$$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^prime-int mathscrL(phi,partial_muphi)d^4x$$
For me this variation should always be zero since this is just a change of variables so there is nothing new.
Suppose that in one dimension flat space we have $phi(x)=sin(x)$ and our action is given by
$$S=intsin(x)dx$$
under a trasnlation $xmapsto x+a$ we should have $phi(x) mapsto phi'(x)=sin(x-a)$.
Under a change of coordinates $xmapsto y=x+a$, $phi(x) mapsto phi'(y)=sin(y-a)$ , since $x$ and $y$ are dummy index the action is the same in both case.
Now assuming we have $sqrt g= x^2$. Translating will give us an action
$$S=int x^2sin(x-a)dx$$
But change of variables will give us an action
$$S=int (y-a)^2sin(y-a)dx$$
witch is different.
Shouldn't the variation of the action be written like this $$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^-int mathscrL(phi,partial_muphi)d^4x$$
symmetry noethers-theorem classical-field-theory
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
$endgroup$
This question is related to this Noether's theorem under arbitrary coordinate transformation and this Transformation of $d^4x$ under translation disregarded?
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x^prime$is given by
$$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^prime-int mathscrL(phi,partial_muphi)d^4x$$
For me this variation should always be zero since this is just a change of variables so there is nothing new.
Suppose that in one dimension flat space we have $phi(x)=sin(x)$ and our action is given by
$$S=intsin(x)dx$$
under a trasnlation $xmapsto x+a$ we should have $phi(x) mapsto phi'(x)=sin(x-a)$.
Under a change of coordinates $xmapsto y=x+a$, $phi(x) mapsto phi'(y)=sin(y-a)$ , since $x$ and $y$ are dummy index the action is the same in both case.
Now assuming we have $sqrt g= x^2$. Translating will give us an action
$$S=int x^2sin(x-a)dx$$
But change of variables will give us an action
$$S=int (y-a)^2sin(y-a)dx$$
witch is different.
Shouldn't the variation of the action be written like this $$delta S=int mathscrL'(phi^prime,partial_muphi^prime)d^4x^-int mathscrL(phi,partial_muphi)d^4x$$
symmetry noethers-theorem classical-field-theory
symmetry noethers-theorem classical-field-theory
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
edited May 2 at 9:10
amilton moreira
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
asked May 2 at 6:58
amilton moreiraamilton moreira
377213
377213
$begingroup$
What about the Jacobian arising in front of the measure when changing variables?
$endgroup$
– Valter Moretti
May 2 at 7:14
$begingroup$
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help.
$endgroup$
– Manvendra Somvanshi
May 2 at 7:16
$begingroup$
Don't we make a transformation of the fields themselves, i.e., $phi(x)tophi(x)+deltaphi(x)$ where the variation $deltaphi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $deltaphi(x)neqfracpartialphi(x)partial x^mudx^mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken.
$endgroup$
– Dvij Mankad
May 2 at 7:17
$begingroup$
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action
$endgroup$
– amilton moreira
May 2 at 7:21
$begingroup$
@Valter Moretti $d^4x^prime=|J|d^4x$
$endgroup$
– amilton moreira
May 2 at 7:28
|
show 10 more comments
$begingroup$
What about the Jacobian arising in front of the measure when changing variables?
$endgroup$
– Valter Moretti
May 2 at 7:14
$begingroup$
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help.
$endgroup$
– Manvendra Somvanshi
May 2 at 7:16
$begingroup$
Don't we make a transformation of the fields themselves, i.e., $phi(x)tophi(x)+deltaphi(x)$ where the variation $deltaphi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $deltaphi(x)neqfracpartialphi(x)partial x^mudx^mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken.
$endgroup$
– Dvij Mankad
May 2 at 7:17
$begingroup$
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action
$endgroup$
– amilton moreira
May 2 at 7:21
$begingroup$
@Valter Moretti $d^4x^prime=|J|d^4x$
$endgroup$
– amilton moreira
May 2 at 7:28
$begingroup$
What about the Jacobian arising in front of the measure when changing variables?
$endgroup$
– Valter Moretti
May 2 at 7:14
$begingroup$
What about the Jacobian arising in front of the measure when changing variables?
$endgroup$
– Valter Moretti
May 2 at 7:14
$begingroup$
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help.
$endgroup$
– Manvendra Somvanshi
May 2 at 7:16
$begingroup$
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help.
$endgroup$
– Manvendra Somvanshi
May 2 at 7:16
$begingroup$
Don't we make a transformation of the fields themselves, i.e., $phi(x)tophi(x)+deltaphi(x)$ where the variation $deltaphi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $deltaphi(x)neqfracpartialphi(x)partial x^mudx^mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken.
$endgroup$
– Dvij Mankad
May 2 at 7:17
$begingroup$
Don't we make a transformation of the fields themselves, i.e., $phi(x)tophi(x)+deltaphi(x)$ where the variation $deltaphi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $deltaphi(x)neqfracpartialphi(x)partial x^mudx^mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken.
$endgroup$
– Dvij Mankad
May 2 at 7:17
$begingroup$
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action
$endgroup$
– amilton moreira
May 2 at 7:21
$begingroup$
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action
$endgroup$
– amilton moreira
May 2 at 7:21
$begingroup$
@Valter Moretti $d^4x^prime=|J|d^4x$
$endgroup$
– amilton moreira
May 2 at 7:28
$begingroup$
@Valter Moretti $d^4x^prime=|J|d^4x$
$endgroup$
– amilton moreira
May 2 at 7:28
|
show 10 more comments
4 Answers
4
active
oldest
votes
$begingroup$
This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.
For example, consider translational symmetry. Sometimes this is written as
$$x to x' = x + a, quad phi to phi'(x') = phi(x).$$
However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $phi$ with a $phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have
$$int_-infty^infty x^2 f(x) , dx = int_-infty^infty (x+a)^2 f(x+a) , dx$$
even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant.
What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $phi$, and end up with the field $phi'$. Indeed in general, we have
$$int_-infty^infty x^2 f(x) , dx neq int_-infty^infty x^2 f(x+a) , dx.$$
You can derive the expression for $phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
$endgroup$
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
|
show 1 more comment
$begingroup$
In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$
$$ delta S~:=~S_V^prime[phi^prime]- S_V[phi]tagA$$
describes an active (as opposed to passive) infinitesimal transformation $$phi(x)tophi^prime(x^prime), tagB$$ $$ xto x^prime, tagC$$ of the action functional $$ S_V[phi]~=~int_V mathrmd^nx ~cal L(phi(x),partialphi(x),x).tagD$$
The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $phi$ variation does vanish cf. my Phys.SE answer here.
--
$^1$ The integration region $Vto V^prime$ is conventionally moved according to the horizontal variation (C).
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
$endgroup$
add a comment |
$begingroup$
Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $delta cal L=0$ for a coordinate transformation is a choice.
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
$endgroup$
add a comment |
$begingroup$
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x²$ is given by ...
I have here only one book where to search for Noether's theorem.
Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94))
$$mathscr L(x+a) equiv
mathscr L(phi_i(x+a), partial_muphi_i(x+a))$$
where $a$ may also depend on $x$. Variation of fields is written
$$deltaphi_i = delta a^mu(x),partial_muphi_i(x)$$
etc. No variation of coordinates in the action integral.
A final note may help you to clarify the matter. Variational
formulation in relativistic field theory is but an extension of the
one in ordinary mechanics. The analogy runs as follows:
$$matrix
hfill rm time & rightarrow & rm spacetime coordinates hfillcr
hfill rm lagrangian coordinates & rightarrow & rm fields. hfill cr$$
So the relevant variation is the one of fields, not of coordinates.
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.
For example, consider translational symmetry. Sometimes this is written as
$$x to x' = x + a, quad phi to phi'(x') = phi(x).$$
However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $phi$ with a $phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have
$$int_-infty^infty x^2 f(x) , dx = int_-infty^infty (x+a)^2 f(x+a) , dx$$
even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant.
What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $phi$, and end up with the field $phi'$. Indeed in general, we have
$$int_-infty^infty x^2 f(x) , dx neq int_-infty^infty x^2 f(x+a) , dx.$$
You can derive the expression for $phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
$endgroup$
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
|
show 1 more comment
$begingroup$
This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.
For example, consider translational symmetry. Sometimes this is written as
$$x to x' = x + a, quad phi to phi'(x') = phi(x).$$
However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $phi$ with a $phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have
$$int_-infty^infty x^2 f(x) , dx = int_-infty^infty (x+a)^2 f(x+a) , dx$$
even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant.
What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $phi$, and end up with the field $phi'$. Indeed in general, we have
$$int_-infty^infty x^2 f(x) , dx neq int_-infty^infty x^2 f(x+a) , dx.$$
You can derive the expression for $phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
$endgroup$
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
|
show 1 more comment
$begingroup$
This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.
For example, consider translational symmetry. Sometimes this is written as
$$x to x' = x + a, quad phi to phi'(x') = phi(x).$$
However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $phi$ with a $phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have
$$int_-infty^infty x^2 f(x) , dx = int_-infty^infty (x+a)^2 f(x+a) , dx$$
even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant.
What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $phi$, and end up with the field $phi'$. Indeed in general, we have
$$int_-infty^infty x^2 f(x) , dx neq int_-infty^infty x^2 f(x+a) , dx.$$
You can derive the expression for $phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
$endgroup$
This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking.
For example, consider translational symmetry. Sometimes this is written as
$$x to x' = x + a, quad phi to phi'(x') = phi(x).$$
However, you don't implement this symmetry transformation by substituting every $x$ with an $x'$ and every $phi$ with a $phi'$, because that's just a trivial change of variables that always leaves the action invariant, much like how a $u$-substitution always leaves an integral invariant. In other words, obviously we have
$$int_-infty^infty x^2 f(x) , dx = int_-infty^infty (x+a)^2 f(x+a) , dx$$
even though $x^2$, which stands in for the Lagrangian here, is not actually translationally invariant.
What they really mean is that we are considering a genuine, nontrivial transformation of the fields. We start with the field $phi$, and end up with the field $phi'$. Indeed in general, we have
$$int_-infty^infty x^2 f(x) , dx neq int_-infty^infty x^2 f(x+a) , dx.$$
You can derive the expression for $phi'$ by thinking "we map the point $x$ to $x'$" but you don't actually do that as well, because then you end up with a trivial transformation. Another way of saying this is that a symmetry means that a system stays the same if you change some things but not other things. In this case, we shifted the function $f$ relative to the fixed background $x^2$ in order to test translational symmetry, which didn’t hold. If you instead changed both, you wouldn't get any useful information.
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
edited May 2 at 15:05
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
answered May 2 at 8:40
knzhouknzhou
49.2k12136240
49.2k12136240
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
|
show 1 more comment
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
you are right. But my question is why the text books insist in this presentation?
$endgroup$
– amilton moreira
May 2 at 8:53
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
@amiltonmoreira Because they learned it from other people who also used this bad presentation. It's sad but true: there are about 50 things in the undergraduate physics curriculum that are almost always presented extremely poorly by tradition, and this is one of them. Everybody I know has had to get around each of these potholes themselves.
$endgroup$
– knzhou
May 2 at 8:55
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
could you give me a reference where this proof is made right?
$endgroup$
– amilton moreira
May 2 at 8:56
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@amiltonmoreira Sorry, I don't have one from a textbook; if anything they just get more and more sloppy the more advanced they get. But do look at the top answers in the questions you linked earlier, they're good.
$endgroup$
– knzhou
May 2 at 8:59
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
$begingroup$
@knzhou I think Srednicki's presentation in Chapter $22$ is pretty similar to what you are saying. He doesn't consider any coordinate changes at all and talks purely about true field transformations at each point with the coordinates untouched, i.e., $phi(x)tophi(x)+deltaphi(x)$. This seems to make it manifest that we are not doing some trivial transformations. Correct me if there is some subtle sloppiness I am missing.
$endgroup$
– Dvij Mankad
May 2 at 9:07
|
show 1 more comment
$begingroup$
In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$
$$ delta S~:=~S_V^prime[phi^prime]- S_V[phi]tagA$$
describes an active (as opposed to passive) infinitesimal transformation $$phi(x)tophi^prime(x^prime), tagB$$ $$ xto x^prime, tagC$$ of the action functional $$ S_V[phi]~=~int_V mathrmd^nx ~cal L(phi(x),partialphi(x),x).tagD$$
The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $phi$ variation does vanish cf. my Phys.SE answer here.
--
$^1$ The integration region $Vto V^prime$ is conventionally moved according to the horizontal variation (C).
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
$endgroup$
add a comment |
$begingroup$
In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$
$$ delta S~:=~S_V^prime[phi^prime]- S_V[phi]tagA$$
describes an active (as opposed to passive) infinitesimal transformation $$phi(x)tophi^prime(x^prime), tagB$$ $$ xto x^prime, tagC$$ of the action functional $$ S_V[phi]~=~int_V mathrmd^nx ~cal L(phi(x),partialphi(x),x).tagD$$
The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $phi$ variation does vanish cf. my Phys.SE answer here.
--
$^1$ The integration region $Vto V^prime$ is conventionally moved according to the horizontal variation (C).
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
$endgroup$
add a comment |
$begingroup$
In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$
$$ delta S~:=~S_V^prime[phi^prime]- S_V[phi]tagA$$
describes an active (as opposed to passive) infinitesimal transformation $$phi(x)tophi^prime(x^prime), tagB$$ $$ xto x^prime, tagC$$ of the action functional $$ S_V[phi]~=~int_V mathrmd^nx ~cal L(phi(x),partialphi(x),x).tagD$$
The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $phi$ variation does vanish cf. my Phys.SE answer here.
--
$^1$ The integration region $Vto V^prime$ is conventionally moved according to the horizontal variation (C).
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
$endgroup$
In most physics textbooks (including Noether's own seminal 1918 paper) on Noether's theorem, the infinitesimal variation$^1$
$$ delta S~:=~S_V^prime[phi^prime]- S_V[phi]tagA$$
describes an active (as opposed to passive) infinitesimal transformation $$phi(x)tophi^prime(x^prime), tagB$$ $$ xto x^prime, tagC$$ of the action functional $$ S_V[phi]~=~int_V mathrmd^nx ~cal L(phi(x),partialphi(x),x).tagD$$
The infinitesimal variation (A) needs not vanish in general. However, pure horizontal $x$-variation (C) without vertical $phi$ variation does vanish cf. my Phys.SE answer here.
--
$^1$ The integration region $Vto V^prime$ is conventionally moved according to the horizontal variation (C).
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
edited May 2 at 9:07
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
answered May 2 at 8:24
Qmechanic♦Qmechanic
109k122041262
109k122041262
add a comment |
add a comment |
$begingroup$
Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $delta cal L=0$ for a coordinate transformation is a choice.
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
$endgroup$
add a comment |
$begingroup$
Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $delta cal L=0$ for a coordinate transformation is a choice.
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
$endgroup$
add a comment |
$begingroup$
Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $delta cal L=0$ for a coordinate transformation is a choice.
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
$endgroup$
Noether's theorem states the conservation laws as a function of (the symmetries of) the Lagrangian. The lagrangian is chosen to reflect the symmetries of the physical system it describes. So whether $delta cal L=0$ for a coordinate transformation is a choice.
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
answered May 2 at 8:26
my2ctsmy2cts
6,3542722
6,3542722
add a comment |
add a comment |
$begingroup$
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x²$ is given by ...
I have here only one book where to search for Noether's theorem.
Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94))
$$mathscr L(x+a) equiv
mathscr L(phi_i(x+a), partial_muphi_i(x+a))$$
where $a$ may also depend on $x$. Variation of fields is written
$$deltaphi_i = delta a^mu(x),partial_muphi_i(x)$$
etc. No variation of coordinates in the action integral.
A final note may help you to clarify the matter. Variational
formulation in relativistic field theory is but an extension of the
one in ordinary mechanics. The analogy runs as follows:
$$matrix
hfill rm time & rightarrow & rm spacetime coordinates hfillcr
hfill rm lagrangian coordinates & rightarrow & rm fields. hfill cr$$
So the relevant variation is the one of fields, not of coordinates.
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
$endgroup$
add a comment |
$begingroup$
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x²$ is given by ...
I have here only one book where to search for Noether's theorem.
Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94))
$$mathscr L(x+a) equiv
mathscr L(phi_i(x+a), partial_muphi_i(x+a))$$
where $a$ may also depend on $x$. Variation of fields is written
$$deltaphi_i = delta a^mu(x),partial_muphi_i(x)$$
etc. No variation of coordinates in the action integral.
A final note may help you to clarify the matter. Variational
formulation in relativistic field theory is but an extension of the
one in ordinary mechanics. The analogy runs as follows:
$$matrix
hfill rm time & rightarrow & rm spacetime coordinates hfillcr
hfill rm lagrangian coordinates & rightarrow & rm fields. hfill cr$$
So the relevant variation is the one of fields, not of coordinates.
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
$endgroup$
add a comment |
$begingroup$
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x²$ is given by ...
I have here only one book where to search for Noether's theorem.
Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94))
$$mathscr L(x+a) equiv
mathscr L(phi_i(x+a), partial_muphi_i(x+a))$$
where $a$ may also depend on $x$. Variation of fields is written
$$deltaphi_i = delta a^mu(x),partial_muphi_i(x)$$
etc. No variation of coordinates in the action integral.
A final note may help you to clarify the matter. Variational
formulation in relativistic field theory is but an extension of the
one in ordinary mechanics. The analogy runs as follows:
$$matrix
hfill rm time & rightarrow & rm spacetime coordinates hfillcr
hfill rm lagrangian coordinates & rightarrow & rm fields. hfill cr$$
So the relevant variation is the one of fields, not of coordinates.
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
$endgroup$
To proof Noether's theorem every text book that i know assumes that the variation of the action under the trasformation $xmapsto x²$ is given by ...
I have here only one book where to search for Noether's theorem.
Itzykson-Zuber do not consider coordinate transformations. They write for space translations (eq. (1-94))
$$mathscr L(x+a) equiv
mathscr L(phi_i(x+a), partial_muphi_i(x+a))$$
where $a$ may also depend on $x$. Variation of fields is written
$$deltaphi_i = delta a^mu(x),partial_muphi_i(x)$$
etc. No variation of coordinates in the action integral.
A final note may help you to clarify the matter. Variational
formulation in relativistic field theory is but an extension of the
one in ordinary mechanics. The analogy runs as follows:
$$matrix
hfill rm time & rightarrow & rm spacetime coordinates hfillcr
hfill rm lagrangian coordinates & rightarrow & rm fields. hfill cr$$
So the relevant variation is the one of fields, not of coordinates.
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
answered May 2 at 14:05
Elio FabriElio Fabri
4,0671214
4,0671214
add a comment |
add a comment |
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$begingroup$
What about the Jacobian arising in front of the measure when changing variables?
$endgroup$
– Valter Moretti
May 2 at 7:14
$begingroup$
I don't think I understand the question. By varying the action and setting it to zero one arrives at Euler Lagrange equations. Noether's Theorem is derived by finding the variation in Lagrangian itself. Maybe a little clarification will help.
$endgroup$
– Manvendra Somvanshi
May 2 at 7:16
$begingroup$
Don't we make a transformation of the fields themselves, i.e., $phi(x)tophi(x)+deltaphi(x)$ where the variation $deltaphi(x)$ is an arbitrary variation and, thus, in my understanding, in general, $deltaphi(x)neqfracpartialphi(x)partial x^mudx^mu$. So, it is supposed to be a true change and not just a change of variables, at least, a priori. Correct me if I am mistaken.
$endgroup$
– Dvij Mankad
May 2 at 7:17
$begingroup$
yes that is what i think. but in proof Noether's theorem ,see for example Field Quantization-W. Greiner Eq. 2.38, 2.39, and 2.45. they insist in the variation of the action
$endgroup$
– amilton moreira
May 2 at 7:21
$begingroup$
@Valter Moretti $d^4x^prime=|J|d^4x$
$endgroup$
– amilton moreira
May 2 at 7:28