7

In chapter 1, of Landau-Lifshitz Mechanics' book, Landau through isotropy and homogeneity of space and homogeneity of time proves that the Lagrangian must depend of magnitude of velocity of the particle.

This seemed fine. But he didn't give a reason as to why it must depend of $|v|^2$ and not on some $|v|^n$ where $n\neq2$.

Qmechanic
  • 201,751

2 Answers2

3

Firstly, this is the simplest possible dependence. Second and more important this allows to reproduce the second Newton's law from variation of the action $$m \dot{v}=-\frac{\partial V}{\partial x }.$$ Where $V(x)$ is potential in which the particle is moving.

Edvard
  • 780
3

The choice of the dependence is largely arbitrary at that point. In that chapter they just choose lagrangian as $L(v^2)$. If they chose it as $L^\prime(|v|^n)$ in equation $(3.1)$, they'd just have to say $L^\prime(|v|^n)=\frac12 m\sqrt[n]{|v|^n}^2$ in equation $(4.1)$.

Also, this choice is simple enough because it's merely a dot product of velocity with itself, which is the simplest scalar function of a vector.

Ruslan
  • 28,862
  • 1
    Why is there a requirement for simplicity here ? –  Jan 19 '14 at 13:22
  • 1
    Lack of requirement for complexity (i.e. arbitrariness of choice) means freedom to choose simple way. – Ruslan Jan 19 '14 at 14:30
  • That isn't a good enough logical argument. I think. –  Jan 19 '14 at 14:53
  • @L-L I think Occam's razor is a good enough argument. – Ruslan Jan 19 '14 at 19:47
  • It is good enough for guessing, but once you know that the right answer has been guessed, you must back it up with a more reasonable argument. –  Jan 24 '14 at 10:36
  • Once again, if you'd chosen any other power, you wouldn't be mistaken. It's arbitrary, as I noted in my answer, because later you'd just use another formula in equation $(4.1)$. So, you don't have to guess — just use any form which fits your needs, i.e. which is isotropic. – Ruslan Jan 24 '14 at 13:02
  • Here you take the nth root because you know what the correct Lagrangian is, I think the question was precisely why wouldn't you have a fourth power term in the Lagrangian. Like what forbids it? – user655870 Dec 19 '20 at 00:39
  • @user655870 see the $\S4$ right up to $(4.1)$. The way $(4.1)$ is derived doesn't assume the value of $n$: this value follows from the fact that the complete term with the derivative of $L$ in the expression for $L'$ must be linear in $\vec v$. E.g. if you choose $n=1$, then $\frac1{|v|}\frac{\partial L}{\partial |v|}=A$ doesn't depend on $|v|$, which implies that $L(|v|)=\frac A2 |v|^2$, where $A$ is actually the mass $m$. – Ruslan Dec 19 '20 at 09:12
  • I'm not sure how this explains why a fourth power term is not allowed in the final Lagrangian. – user655870 Dec 19 '20 at 19:44
  • @user655870 if it has a fourth power term, then $L'-L$ will not be a total time derivative, and thus Galilean relativity principle breaks down. – Ruslan Dec 19 '20 at 21:01
  • OK, so I think the appropriate answer to the question here would have been that the Galilean principle is really what forbids a 4th power in the Lagrangian, not a "lack of requirement for complexity" or "Occam's razor". – user655870 Dec 19 '20 at 21:22
  • @user655870 ultimately, yes. But this answer just follows the logic of the chapter 4, which at first doesn't impose any constraints on the Lagrangian, being free to choose the general form $L(v^2)$. Only the derivation of $(4.1)$ actually fixes the structure, requiring that $L\propto v^2$. – Ruslan Dec 19 '20 at 21:32