Why isn't scaling space and time considered the 11th dimension of the Galilean group?Can one derive Galilean transformations from the harmonic oscillator equations of motion and the relativity principle?Relationship bewtween the principle of Galilean Relativity and absolute timeWhy does galilean invariance imply that particles that start rest stay on the same line?Noether's Theorem and scale invarianceUnderstanding the physical operations corresponding to unit conversion and scale transformation of timeDoes the Newton's law break scale invariance?The Meaning of Newton's Second Law of Motion Being Invariant Under Certain TransformationsSilly question about Galilei GroupRelationship between the Galilei Group and the Phase Space
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Why isn't scaling space and time considered the 11th dimension of the Galilean group?
Can one derive Galilean transformations from the harmonic oscillator equations of motion and the relativity principle?Relationship bewtween the principle of Galilean Relativity and absolute timeWhy does galilean invariance imply that particles that start rest stay on the same line?Noether's Theorem and scale invarianceUnderstanding the physical operations corresponding to unit conversion and scale transformation of timeDoes the Newton's law break scale invariance?The Meaning of Newton's Second Law of Motion Being Invariant Under Certain TransformationsSilly question about Galilei GroupRelationship between the Galilei Group and the Phase Space
$begingroup$
Galilean transformations are said to have 10 degrees of freedom. Four for translation in space and time, three for rotation, and three for direction of the uniform motion.
If I scale space axis by $alpha$ and do the same with time axis, you can see that Newton's second law remains the same.
So why don't we consider scaling (of time and space) another type of Galilean transformation?
galilean-relativity scale-invariance
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
$endgroup$
add a comment |
$begingroup$
Galilean transformations are said to have 10 degrees of freedom. Four for translation in space and time, three for rotation, and three for direction of the uniform motion.
If I scale space axis by $alpha$ and do the same with time axis, you can see that Newton's second law remains the same.
So why don't we consider scaling (of time and space) another type of Galilean transformation?
galilean-relativity scale-invariance
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
$endgroup$
$begingroup$
I agree that "scale" is a badly neglected degree of physical freedom.
$endgroup$
– Steve
Jun 1 at 23:50
$begingroup$
@Steve, it is not a degree of freedom in classical mechanics.
$endgroup$
– Akerai
Jun 2 at 0:18
$begingroup$
@Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)?
$endgroup$
– Steve
Jun 2 at 0:38
1
$begingroup$
@Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims.
$endgroup$
– Thatpotatoisaspy
Jun 2 at 3:10
$begingroup$
Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114
$endgroup$
– AccidentalFourierTransform
Jun 2 at 14:13
add a comment |
$begingroup$
Galilean transformations are said to have 10 degrees of freedom. Four for translation in space and time, three for rotation, and three for direction of the uniform motion.
If I scale space axis by $alpha$ and do the same with time axis, you can see that Newton's second law remains the same.
So why don't we consider scaling (of time and space) another type of Galilean transformation?
galilean-relativity scale-invariance
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
$endgroup$
Galilean transformations are said to have 10 degrees of freedom. Four for translation in space and time, three for rotation, and three for direction of the uniform motion.
If I scale space axis by $alpha$ and do the same with time axis, you can see that Newton's second law remains the same.
So why don't we consider scaling (of time and space) another type of Galilean transformation?
galilean-relativity scale-invariance
galilean-relativity scale-invariance
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
edited Jun 2 at 5:09
Qmechanic♦
110k122101289
110k122101289
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
asked Jun 1 at 23:22
Shuheng ZhengShuheng Zheng
206210
206210
$begingroup$
I agree that "scale" is a badly neglected degree of physical freedom.
$endgroup$
– Steve
Jun 1 at 23:50
$begingroup$
@Steve, it is not a degree of freedom in classical mechanics.
$endgroup$
– Akerai
Jun 2 at 0:18
$begingroup$
@Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)?
$endgroup$
– Steve
Jun 2 at 0:38
1
$begingroup$
@Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims.
$endgroup$
– Thatpotatoisaspy
Jun 2 at 3:10
$begingroup$
Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114
$endgroup$
– AccidentalFourierTransform
Jun 2 at 14:13
add a comment |
$begingroup$
I agree that "scale" is a badly neglected degree of physical freedom.
$endgroup$
– Steve
Jun 1 at 23:50
$begingroup$
@Steve, it is not a degree of freedom in classical mechanics.
$endgroup$
– Akerai
Jun 2 at 0:18
$begingroup$
@Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)?
$endgroup$
– Steve
Jun 2 at 0:38
1
$begingroup$
@Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims.
$endgroup$
– Thatpotatoisaspy
Jun 2 at 3:10
$begingroup$
Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114
$endgroup$
– AccidentalFourierTransform
Jun 2 at 14:13
$begingroup$
I agree that "scale" is a badly neglected degree of physical freedom.
$endgroup$
– Steve
Jun 1 at 23:50
$begingroup$
I agree that "scale" is a badly neglected degree of physical freedom.
$endgroup$
– Steve
Jun 1 at 23:50
$begingroup$
@Steve, it is not a degree of freedom in classical mechanics.
$endgroup$
– Akerai
Jun 2 at 0:18
$begingroup$
@Steve, it is not a degree of freedom in classical mechanics.
$endgroup$
– Akerai
Jun 2 at 0:18
$begingroup$
@Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)?
$endgroup$
– Steve
Jun 2 at 0:38
$begingroup$
@Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)?
$endgroup$
– Steve
Jun 2 at 0:38
1
1
$begingroup$
@Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims.
$endgroup$
– Thatpotatoisaspy
Jun 2 at 3:10
$begingroup$
@Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims.
$endgroup$
– Thatpotatoisaspy
Jun 2 at 3:10
$begingroup$
Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114
$endgroup$
– AccidentalFourierTransform
Jun 2 at 14:13
$begingroup$
Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114
$endgroup$
– AccidentalFourierTransform
Jun 2 at 14:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Imagine you take the transformation you mentioned above:
$$x^i rightarrow x'^i = alpha x^i,\
t rightarrow t' = alpha t,$$
where $alpha in mathbbR$.
Then assuming the Newton's law holds in the new coordinates, it will be of the form
$$F^i = m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2 left(fracdtdt' right)^2.$$
As you can see, the derivative $dt/dt' = 1/ alpha$, and therefore the equality becomes
$$m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2frac1alpha^2 = fracmalpha fracd^2 (x^i) dt^2.$$
Therefore the Newton's law is actually not invariant under this transformation as you claimed before, the transformation actually transforms an object of mass $m/alpha$ to an object of mass $m$.
answered Jun 1 at 23:34
AkeraiAkerai
338110
$endgroup$
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
3
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Imagine you take the transformation you mentioned above:
$$x^i rightarrow x'^i = alpha x^i,\
t rightarrow t' = alpha t,$$
where $alpha in mathbbR$.
Then assuming the Newton's law holds in the new coordinates, it will be of the form
$$F^i = m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2 left(fracdtdt' right)^2.$$
As you can see, the derivative $dt/dt' = 1/ alpha$, and therefore the equality becomes
$$m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2frac1alpha^2 = fracmalpha fracd^2 (x^i) dt^2.$$
Therefore the Newton's law is actually not invariant under this transformation as you claimed before, the transformation actually transforms an object of mass $m/alpha$ to an object of mass $m$.
answered Jun 1 at 23:34
AkeraiAkerai
338110
$endgroup$
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
3
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
add a comment |
$begingroup$
Imagine you take the transformation you mentioned above:
$$x^i rightarrow x'^i = alpha x^i,\
t rightarrow t' = alpha t,$$
where $alpha in mathbbR$.
Then assuming the Newton's law holds in the new coordinates, it will be of the form
$$F^i = m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2 left(fracdtdt' right)^2.$$
As you can see, the derivative $dt/dt' = 1/ alpha$, and therefore the equality becomes
$$m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2frac1alpha^2 = fracmalpha fracd^2 (x^i) dt^2.$$
Therefore the Newton's law is actually not invariant under this transformation as you claimed before, the transformation actually transforms an object of mass $m/alpha$ to an object of mass $m$.
answered Jun 1 at 23:34
AkeraiAkerai
338110
$endgroup$
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
3
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
add a comment |
$begingroup$
Imagine you take the transformation you mentioned above:
$$x^i rightarrow x'^i = alpha x^i,\
t rightarrow t' = alpha t,$$
where $alpha in mathbbR$.
Then assuming the Newton's law holds in the new coordinates, it will be of the form
$$F^i = m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2 left(fracdtdt' right)^2.$$
As you can see, the derivative $dt/dt' = 1/ alpha$, and therefore the equality becomes
$$m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2frac1alpha^2 = fracmalpha fracd^2 (x^i) dt^2.$$
Therefore the Newton's law is actually not invariant under this transformation as you claimed before, the transformation actually transforms an object of mass $m/alpha$ to an object of mass $m$.
answered Jun 1 at 23:34
AkeraiAkerai
338110
$endgroup$
Imagine you take the transformation you mentioned above:
$$x^i rightarrow x'^i = alpha x^i,\
t rightarrow t' = alpha t,$$
where $alpha in mathbbR$.
Then assuming the Newton's law holds in the new coordinates, it will be of the form
$$F^i = m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2 left(fracdtdt' right)^2.$$
As you can see, the derivative $dt/dt' = 1/ alpha$, and therefore the equality becomes
$$m fracd^2x'^idt' ^2 = m fracd^2 (alpha x^i) dt^2frac1alpha^2 = fracmalpha fracd^2 (x^i) dt^2.$$
Therefore the Newton's law is actually not invariant under this transformation as you claimed before, the transformation actually transforms an object of mass $m/alpha$ to an object of mass $m$.
answered Jun 1 at 23:34
AkeraiAkerai
338110
edited Jun 1 at 23:48
answered Jun 1 at 23:34
AkeraiAkerai
338110
answered Jun 1 at 23:34
AkeraiAkerai
338110
answered Jun 1 at 23:34
AkeraiAkerai
338110
338110
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
3
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
add a comment |
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
3
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
$begingroup$
Thanks, I made a mistake in my calculation. Thank you so much for clearing up.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:40
3
3
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
$begingroup$
This make sense. If it is truly scale invariant, we probably won't need units on acceleration forces etc.
$endgroup$
– Shuheng Zheng
Jun 2 at 3:41
add a comment |
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$begingroup$
I agree that "scale" is a badly neglected degree of physical freedom.
$endgroup$
– Steve
Jun 1 at 23:50
$begingroup$
@Steve, it is not a degree of freedom in classical mechanics.
$endgroup$
– Akerai
Jun 2 at 0:18
$begingroup$
@Akerai, why exactly is that? Is it simply because no apparent practical means of changing the scale of physical things currently exists (and thus the theoretical possibility didn't need to be considered by the classical physicists)?
$endgroup$
– Steve
Jun 2 at 0:38
1
$begingroup$
@Steve It’s because, as Akerai states in an answer here, the laws of classical mechanic are in fact generally not scale-invariant as OP claims.
$endgroup$
– Thatpotatoisaspy
Jun 2 at 3:10
$begingroup$
Interestingly enough, the Schrödinger equation is both invariant under galilean transformations and dilations, even if it has a length scale (mass). It is even conformally invariant! cf. mathoverflow.net/a/270122/106114
$endgroup$
– AccidentalFourierTransform
Jun 2 at 14:13