# Why does Isotropy of L imply L(v^2)?

1. Mar 17, 2004

### john8197

I'm reading the first edition of Mechanics by Landau et al, published in 1960. Just before equation 3.1 on page 5 it says exactly this:

"Since space is isotropic, the Lagrangian must also be independent of the direction of v, and is therefore a function only of it's magnitude, i.e. of v(bold)^2 = v(italic)^2:

L = L(v(italic)^2) (3.1)"

This seems very cryptic to me since the magnitude is sqrt(v(bold)^2) =
v(italic).
Could someone fill in the missing details for me please?

Funky

2. Mar 26, 2004

### salsero

3. Mar 26, 2004

### Stingray

Clearly the lagrangian has to depend on the velocity somehow. Then you notice that the Lagrangian is a scalar, yet the velocity is a vector. There is only one way of converting a vector into a scalar that will be the same in all coordinate systems (which it has to be by definition): take its magnitude.

If you want to be nitpicky, then you could say v->a|v|+b, where a and b are constants, but generalization doesn't matter (you should check this).

4. Mar 27, 2004

### Kalimaa23

Why not? I've been using it for a couple of weeks now, and it's one of the nicest books on Mechanics I've ever read. It's brief, but to the point, and appeals the physical intuition without blurring the mathematical side of the story.