# A The Lagrangian a function of 'v' only and proving v is constant

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1. Jul 23, 2017

### Ren Figueroa

Hi everyone. So I'm going through Landau/Lifshitz book on Mechanics and I read through a topic on inertial frames. So, because we are in an inertial frame, the Lagrangian ends up only being a function of the magnitude of the velocity only (v2) Now my question to you is, how does one prove that the velocity is constant like in (3.2)? When I first went through it, I thought it was obvious because it makes sense that the partial derivative of the Lagrangian with respect to velocity has to be a constant in order for the derivative with respect to time to equal zero. But a professor, who is mentoring me, brought it to my attention that this is not as obvious as it looks and once he explained to me why it wasn't obvious, I started to think so too.
Actually, when I think about it, if the Lagrangian is only a function of the magnitude of velocity, then the derivative with respect to time automatically should be zero. But, this doesn't automatically prove that velocity is constant. (At least it doesn't to me) I am curious to know anyone else's explanation/feedback on this subject. Thanks!!
(Also, I am not sure if this is a graduate level text or undergraduate but I have already read undergraduate texts by Morin and Taylor. I put the thread under "graduate level" based on what I found on other forums but if it is actually and undergrad text then I apologize!)

2. Jul 23, 2017

### Orodruin

Staff Emeritus
While the Lagrangian is a function of the speed only, its derivative with respect to the components of $\vec v$ are not. Generally, if $L = L(v^2)$, then
$$\frac{\partial L}{\partial v^i} = L'(v^2) \frac{\partial(v^2)}{dv^i} = 2v^i L'(v^2).$$

3. Jul 23, 2017

### Ren Figueroa

Hi. Thanks for the response. I understood that the derivative is a different type of function, but it doesn't convince me that v is constant. Do you have another explanation to help me through this? Or maybe some recommended literature?

4. Jul 24, 2017

### Orodruin

Staff Emeritus
Which part do you not understand? Whatever that derivative is is constant and it holds for all components. If you are worried about the $L'$ factor you can easily show that $v^2$ also is constant and therefore so is $L'(v^2)$.

Normally, the Lagrangian contains the kinetic term $mv^2/2$, making $L' = m/2$.