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We have two inertial coordinate systems, $K'$ and $K$. $K$ is moving with infinitesimal velocity ${\epsilon}$ relative to $K'$. Using Galilean relativity we can transform this into $v'=v+{\epsilon}$. Here Lagrangian is only a function of speed $L=L(|v|^2)$. $L'$ may differ from $L$ only by a total derivative of function of coordinates and time, i.e. $L'=L+\frac{d}{dt}f(q,t)$

$"$We have $L'=L(|v|^2+2{v}·{\epsilon}+{\epsilon}^2)$. [1. ] Expanding this expression in powers of ${\epsilon}$ and neglecting terms above the first order, we obtain $L(|v'|^2)=L(|v|^2)+\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ [2.] The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $v$ $"$

I understand nothing in the quoted part. How do we get [2.]? Why does $\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ being a linear function of velocity imply that it's a total time derivative?

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So you started with $L' = L(|v|^2 + 2v\bullet \epsilon + \epsilon^2) $

If you treat $ 2v\bullet \epsilon + \epsilon^2 $ as a variation of $|v|^2$, you may use the taylor's theorem treating $|v|^2$ as a variable, and you get the following expression:

$ L(|v'|^2) = L(|v|^2) + \frac{\partial L}{\partial |v|^2} (2v \bullet\epsilon + \epsilon^2)$ + higher order derivatives.

There is still one more term in this equation that is above first order, namely $\frac{\partial L}{\partial |v|^2} \epsilon^2$

After you get rid of that term you are left with $\frac{\partial L}{\partial |v|^2} (2v\bullet \epsilon)$ which agrees with the textbook.

Now to the second point: why does this last term have to be linear?

So take a look at this equation that you wrote: $L' = L + \frac{d}{dt} f(q,t) $

You are trying to answer the question: Can we find an $f(q,t)$ such that $\frac{d}{dt} f(q,t) = \frac{\partial L}{\partial |v|^2} (2v\bullet \epsilon)$

Since f is a function only of $q$ and $t$, the left hand side should only involve terms that are products of functions of $q,t$ and first derivative of $q$ (which is $v$).

Take an example: $f(q,t) = q^2 + \frac{q^3}{q_0} + \frac{q^2}{t_0}t$(completely random function)

Then $\frac{df}{dt} = \frac{\partial f}{\partial q} \frac{dq}{dt} + \frac{\partial f}{\partial t} = v(2q + \frac{3q^2}{q_0}+ \frac{2q}{t_0}t) + \frac{q^2}{t_0}$

As you can see, the total derivative can only involve linear terms in $v$. So if the right hand side is not linear in $v$, this equation cannot possibly hold. Notice that the statement "only if" doesn't imply "if". So linearity is a necessary but not sufficient condition. My intuition is that sufficient condition should be a case-by-case discussion. But maybe I am wrong on that. Maybe someone else can clarify that.

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  • $\begingroup$ I looked this question up because I understand the derivation but I don't understand where the assumptions come from. I.e., in 1. Why can we drop the higher order terms? In 2 why does the term have to be a total derivative? $\endgroup$ – William Oliver May 29 '16 at 18:01

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