Mathematical proof for the Lagrangian function

  • #1
Silver2007
4
1
TL;DR Summary
In Landau's mechanics book, I saw them argue that due to the homogeneity of time and space, the isotropy of space leads to the Lagrangian function depending only on v^2. But i want a mathematical proof for the Lagrangian function independent of position q, time t and velocity vector.
In Landau's mechanics book, I saw them argue that due to the homogeneity of time and space, the isotropy of space leads to the Lagrangian function depending only on v^2. But i want a mathematical proof for the Lagrangian function independent of position q, time t and velocity vector.
Homogeneity of space:
We have ## \mathcal{L}(\vec{r}, \vec{\dot{r}}, t) ##, and ## \vec{r} \to \vec{r}' = \vec{r} + \vec{a} ##, because homogeneity of space so equations of motion should be the same. Therefore, the Lagrangian function differs only by a total derivative with respect to time ## \Omega(\vec{r}, t) ##:
$$
\mathcal{L}' = \mathcal{L} + \frac{d}{dt}\Omega(\vec{r}, t)
$$
We assume that ## a \ll 1 ##, and we get:
$$
\vec{a} \cdot \frac{\partial \mathcal{L}}{\partial \vec{r}} = \frac{d\Omega}{dt}
$$
At this point, I argue that if the Lagrangian function depends on position, it will be impossible to find an function ##\Omega(\vec{r}, t)## that satisfies the above equation, so the Lagrangian function cannot depend on position. Is my argument above reasonable and coherent?

And I need help with similar mathematical proofs for temporal homogeneity and spatial isotropy. Thanks
 
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  • #2
Silver2007 said:
And I need help with similar mathematical proofs for temporal homogeneity and spatial isotropy. Thanks
How do you suppose these could be proved mathematically? Space and time are related to the physical universe. You can only assume they are homogeneous and isotropic - or prove this from more fundamental physical assumptions. Mathematics would not be compromised if space were non-homogeneous. In fact, you always have to be prepared for evidence that, say, space is not homogeneous.
 
  • #3
PeroK said:
How do you suppose these could be proved mathematically? Space and time are related to the physical universe. You can only assume they are homogeneous and isotropic - or prove this from more fundamental physical assumptions. Mathematics would not be compromised if space were non-homogeneous. In fact, you always have to be prepared for evidence that, say, space is not homogeneous.
Sure, but OP is asking under the assumption that the laws of physics are say homogeneous with respect to time and space, then how do you show that this constrains the Lagrangian to be only a function of velocity squared.
 
  • #4
jbergman said:
Sure, but OP is asking under the assumption that the laws of physics are say homogeneous with respect to time and space, then how do you show that this constrains the Lagrangian to be only a function of velocity squared.
If you read my post, you'll see I answered his second question. Which was:

Silver2007 said:
And I need help with similar mathematical proofs for temporal homogeneity and spatial isotropy. Thanks
 
  • #5
Silver2007 said:
TL;DR Summary: In Landau's mechanics book, I saw them argue that due to the homogeneity of time and space, the isotropy of space leads to the Lagrangian function depending only on v^2. But i want a mathematical proof for the Lagrangian function independent of position q, time t and velocity vector.

In Landau's mechanics book, I saw them argue that due to the homogeneity of time and space, the isotropy of space leads to the Lagrangian function depending only on v^2. But i want a mathematical proof for the Lagrangian function independent of position q, time t and velocity vector.
Homogeneity of space:
We have ## \mathcal{L}(\vec{r}, \vec{\dot{r}}, t) ##, and ## \vec{r} \to \vec{r}' = \vec{r} + \vec{a} ##, because homogeneity of space so equations of motion should be the same. Therefore, the Lagrangian function differs only by a total derivative with respect to time ## \Omega(\vec{r}, t) ##:
$$
\mathcal{L}' = \mathcal{L} + \frac{d}{dt}\Omega(\vec{r}, t)
$$
We assume that ## a \ll 1 ##, and we get:
$$
\vec{a} \cdot \frac{\partial \mathcal{L}}{\partial \vec{r}} = \frac{d\Omega}{dt}
$$
At this point, I argue that if the Lagrangian function depends on position, it will be impossible to find an function ##\Omega(\vec{r}, t)## that satisfies the above equation, so the Lagrangian function cannot depend on position. Is my argument above reasonable and coherent?

And I need help with similar mathematical proofs for temporal homogeneity and spatial isotropy. Thanks
What about something like##\Omega(\vec{r}, t)=\vec{r} ^2##?

I suggest proving it first without the total derivative term. To prove there isn't a total derivative difference is harder and I am not sure it is true. I believe you would have to at least use multiple symmetries to show it.
 
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