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The problem is in the book Mechanics of Landau's lectures in theoretical physics

Find the ratio of the times in the same path for particles having different masses but the same potential energy. (page 22 second edition)

The problem doesn't give you much information but only the fact that the potential energy along the path remains the same. So the first (and only) thing that it is given is that $$U'=U$$ where $U$ is the potential energy.

My first attempt: Thinking in paths I tried to use the given relations $$t'/t = (l'/l)^{1-k/2}$$ $$E'/E = (l'/l)^k$$ and $$v'/v = (l'/l)^{k/2}$$ Where the above notations are respectively time, total energy and velocity. As he says any mechanical quantities at corresponding times are in a ratio which is a power of $l'/l$, I tried to find a relation for time and mass such that I know the answer of the problem is $t'/t = m'/m$ so, if we define

$$\bar{f} = \lim_{\tau \rightarrow \infty}\frac{1}{\tau}\int_{0}^{\tau}f(t)\mathrm{d}t$$ we can get that $$U' = U \Leftrightarrow \bar{U'} = \bar{U}$$

But this implies, because of the relation $\bar{U} = \frac{2E}{k+2}$ that $E' = E$ and that $\bar{T'} = \bar{T}$ so $$m'\bar{v'}^2 = m\bar{v}^2$$ such that we can put $m'/m = (v'/v)^{-2} = (l'/l)^{-k} = (t'/t)^{\frac{-k}{1-k/2}}$ so

How I get that $\frac{-k}{1-k/2} = \frac{1}{2}$? Am I doing something wrong? There is a more nice approach to the problem?

My second attempt: Thinking in paths I thought that if I could find $l$ and $l'$ explicitly, may be as a function that arises from the action, the problem became easier. Is this possible?

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    $\begingroup$ Recently I found out that there already was a similar question on interpreting this section of Landau. The author of the question presented his interpretation of Landau, and laid out some steps that were omitted in the book. In addition, a comment there mentioned a book on mechanical similarity by Sedov, which might also be a good treatise on this topic. I decided to add this comment just so that should anyone stumble upon this question in the future, they would know what other resources to refer to. $\endgroup$
    – Elliot Yu
    Commented Feb 21, 2017 at 17:58

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I believe the intention of this section is to illustrate the idea that one can multiply the Lagrangian by a constant without affecting the equation of motion, therefore scaling the spatial/temporal dimensions of the problem without having to solve the equation all anew.

Thinking along these lines, I would attempt by first scaling time by a factor $\alpha$, and derive from that the scaling factor of the kinetic energy $T$. Since we are keeping the path invariant, velocity should be scaled by $\alpha^{-1}$, and since the mass is the same and that $T$ is quadratic in $v$, the kinetic energy should be scaled by $\alpha^{-2}$. Now suppose $U$ is scaled by $\beta$, then to keep the Lagrangian invariant up to a constant factor, we need $\alpha^{-2} = \beta$, which gives the desired scaling in time.

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  • $\begingroup$ I'll do the calculations and see if what you say makes sense. So what happens is that Lagrangian times a constant can just construct another path that creates dependence in other dynamical components of the system? $\endgroup$
    – user78217
    Commented Feb 11, 2017 at 20:31
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    $\begingroup$ @RafaelWagner 1) Because you want the new system to have the same length scale as the previous one, as indicated by the requirement "the same path" in both problems. 2) There is no connection, because the spatial coordinates are scaled by 1, and are not affected by the choice of $\alpha$. $\endgroup$
    – Elliot Yu
    Commented Feb 14, 2017 at 18:16
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    $\begingroup$ Also, I think we should move this conversation to the chat, since comments are not for extended discussions. I am still trying to figure how to create one, or we could also just wait until the website creates one for us when this comment section gets too long. $\endgroup$
    – Elliot Yu
    Commented Feb 14, 2017 at 18:21
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    $\begingroup$ I agree that these discussions can be useful. I was just trying to articulate a PhysicsSE policy. $\endgroup$
    – Elliot Yu
    Commented Feb 14, 2017 at 18:29
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    $\begingroup$ Finally, to drive my point home, I just want to point out that here the $\rightarrow$s should be understood as a procedure of creating a new system similar to the original one. The new system should have spatial/temporal dimesnions and other parameters like mass proportional to the original one. If the potential is a homogeneous function in time/space or other parameters, then the Lagrangian will be the original one times a constant factor. The result of that is we don't have to solve the new system, but just use the solution to the old system with all the appropriate scaling applied. $\endgroup$
    – Elliot Yu
    Commented Feb 14, 2017 at 18:31

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