# Question on the Lagrangian

1. Apr 11, 2015

### acegikmoqsuwy

Hi, I have a very basic question about the Lagrangian that I can't seem to understand: why is it dependent on both the position function and the time derivative? I know that it is the difference between the kinetic and potential energy, but why? Is there a derivation of this, is it a definition, or somehow based on experimental evidence?

In fact, when I try to derive the Euler-Lagrangian equations (based on the assumption that the Lagrangian is a function of both position and velocity), I get $\displaystyle A=\int_{t_1}^{t_2} \sum\limits_{i}\left(\frac{\partial \mathcal L} {\partial x_i}-\frac d{dt} \frac{\partial \mathcal L} {\partial \dot x_i}\right) dt$ and using that kinetic energy is dependent on velocity, and potential energy is dependent on position, I seem to arrive at the conclusion that $\mathcal L (t,x,\dot x)=U-T$.

Where have I gone wrong? And why doesn't the Lagrangian depend on $\ddot x$ or only $x$ ?

2. Apr 11, 2015

### refind

I am not 100% sure I understand your question but the dependence on the time derivative is there because kinetic energy is a function of it, while potential energy is a function of the position.

If you want a good derivation of Hamilton's principle you can take a look at Goldstein's Classical Mechanics. Essentially, a stationary curve of the action satisfies Newton's laws of motion.

3. Apr 13, 2015

### sufuan

see the link: http://en.wikiversity.org/wiki/Advanced_Classical_Mechanics/Continuum_Mechanics

4. Apr 14, 2015

### mac_alleb

No derivation or experiment, pure theorie supposition....

5. Apr 15, 2015

### samalkhaiat

One one hand, you know that $E = T + U$. On the other hand, you should know that the energy is related to the Lagrangian by
$$E = v \cdot \frac{\partial L}{\partial v} - L .$$
So, you have the following identity
$$L = v \cdot \frac{\partial L}{\partial v} - T - U .$$
So, if you write $L = \alpha T + \beta U$ and substitute it in the above identity, you find $\alpha = 1$ and $\beta = - 1$. Or, if you apply Euler theorem to the Lagrangian, you find $v \cdot \frac{\partial L}{\partial v} = 2 T$.

If no body told you yet, you will soon learn that the form of the Lagrangian is determined by symmetry principles. In classical mechanics, this is the principle of (Galileo) relativity, which states that the equations of motion must have the same form in all inertial frames of reference. This means that the functional form of the Lagrangian must be invariant up to total time-derivative of some function of the coordinates and time, $$L^{'} ( x ) = L ( x ) + \frac{d F}{d t} .$$
As an exercise for you, consider a free particle with velocity $v$ and the transformation $v^{'} = v + u$, where $u$ is the infinitesimal relative velocity between two inertial frames. Show that $$\frac{\partial L}{\partial v^{2}} = \mbox{constant} .$$

6. Apr 15, 2015

### Staff: Mentor

The reason is quantum and comes from Feynmans sum over histories approach.

You start out with <x'|x> then you insert a ton of ∫|xi><xi|dxi = 1 in the middle to get ∫...∫<x|x1><x1|......|xn><xn|x> dx1.....dxn. Now <xi|xi+1> = ci e^iSi so rearranging you get ∫.....∫c1....cn e^ i∑Si.

Focus in on ∑Si. Define Li = Si/Δti, Δti is the time between the xi along the jagged path they trace out. ∑ Si = ∑Li Δti. As Δti goes to zero the reasonable physical assumption is made that Li is well behaved and goes over to a continuum so you get ∫L dt.

Each <xi|xi+1> depends on xi, t and Δxi, so Si depends on xi, t, and Δxi, hence so does ∑Li Δti. But for a path Δxi depends on the velocity vi = Δxi/Δti so its very reasonable to assume when it goes to the continuum L is a function of x, t, and the velocity v. So the Lagranian depends on position, time and velocity.

If you haven't studied QM don't worry about it - just bear it in mind for when you do. Its quite a striking result really.

Thanks
Bill

Last edited: Apr 15, 2015
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