Mechanical Similarity: Solving Equation 10.2

[msbell1]

Hi,

I have recently embarked on a quest to read through (and hopefully to understand) all 10 volumes of the Course of Theoretical Physics by Landau and Lifgarbagez. Well, I have already found myself hung up on section 10 in Volume I (Mechanics, 3rd Edition). Specifically, I am stuck on how equation 10.2 was derived. For those of you that do not have the book available (and for those that don't feel like pulling it off the shelf), here's my problem.

The authors first state that we are only interested in potentials which are homogeneous functions of the coordinates (homogeneous of degree k). They then say that we will carry out a transformation in which the coordinates are transformed by a factor A: r_a -> Ar_a, and time is transformed by a factor B: t -> Bt. They then state that velocities will be transformed by A/B, and the kinetic energy by A²/B². That makes sense to me. Then, since the potential U is homogeneous of degree k, in the Lagrangian U will have a factor of A^k in front of it. Since the Lagrangian can be multiplied by a constant without changing the equations of motion, we need for the kinetic energy T to also have a factor of A^k in front of it. For this to happen, we have

A²/B² = A^k

Solving for B gives B = A^1-k/2

So far so good.

Now the authors claim that the equations of motion permit a series of geometrically similar paths, and the times of the motion between corresponding points are in the ratio

t'/t = (l'/l)^1-k/2

where l'/l is the ratio of the linear dimensions of the two paths.

OK, how do they find this expression for t'/t? Thanks a lot for the insight!

 

[physicsworks]

Hi, msbell1!

This thing
$$\frac{t'}{t} = \frac{l'}{l}^{1-k/2}$$
comes from
$$B = A^{1-k/2}$$ where $$t' = Bt$$ and $$l' = A l$$.

P.S. Good choice of the book! ;)

 

[msbell1]

Thanks physicsworks! That helped a lot, and it makes sense now. I really appreciate your help.