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I have a question, more related to a mathematical aspect of physics, which seems I am not understanding very well.

So, by applying Galilean transformation between two reference frames, which move at the speed $\epsilon$ relative to each other, the Lagrangians of a free particle looked from these two systems differ by

$$\Delta L=\frac{\partial L}{\partial (v^2)}2\vec{v}\cdot\vec{\epsilon}, \qquad \vec{v}=\frac{d \vec{x}}{d t},\qquad v:=|\vec{v}|. $$

On the other hand, it has to be

$$\Delta L=\frac{d F}{d t}.$$

Now in many texts, I see the argument that this is true only if $\frac{\partial L}{\partial (v^2)}$ is independent of $v$. It might be due to the lack of math skills, but this is not obvious for me.

Example. On contrary, let's assume that $$\frac{\partial L}{\partial (v^2)}=a v$$ and 1D motion, then we have the condition

$$\frac{d F}{d t}=2\epsilon a\frac{d x}{d t}|\frac{d x}{d t}|$$

while

$$\frac{d F}{d t}=\frac{\partial F}{\partial t}+\sum_i\frac{\partial F}{\partial u_i}\frac{d u_i}{d t}$$

Where $u_i$ are all possible time dependent functions, on which $F$ is dependent.

Could anyone help me, and explain why can't we express any of the partial derivatives in $\frac{d F}{d t}$ via $\frac{d x}{d t}|\frac{d x}{d t}|$?

References:

  1. Landau & Lifshitz, Mechanics, $\S$3.
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That's a very good question. I have explicitly done the analysis suggested by OP in Section 2 of my Phys.SE answer here. The conclusion is, unsurprisingly, that imposing that all Galilean transformations should be quasi-symmetries implies that the Lagrangian $L(v^2)$ must be an affine function of $v^2$.

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  • $\begingroup$ Thanks Qmechanic, your reply on the topic you linked was so far giving me the most satisfied answer. Also, in the approach I took in the beginning, to show that the partial derivatives of $\frac{dF}{dt}$ can be expressed through $\frac{\partial x}{\partial t}\frac{\partial x}{\partial t}$, what I get in the end is that the function $F$ is expressed as an integral, which depend on the path taken by the particle. This is in the clear contradiction with the fact that variation of the action $S$, must be the same for all observers along the unique path. $\endgroup$ – 600nebo Nov 27 '18 at 8:54

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