# Relativistic Lagrangian of a Free Particle

1. Apr 17, 2006

### emob2p

Hi,
As argued in Jackson p. 580, the quantity $$\gamma L$$ is invariant. So imagine a free particle. In the particle's frame, the particle can be treated non-relativistically since its v << c (it's zero). But non-relativistically we define the $$L = T - V$$. In the particle's frame, this is zero. So invariance would say L would be zero in all frames.

I think the way out of this is to argue that we can only define a Lagrangian up to a total time derivative of some function. In this case the function is $$-t mc^2$$. Consequently, in the particle's frame $$L = T - V - mc^2$$. Doing this we get the the standard relativistic Lagrangian for a free particle.

My Questions:
Why is the added function mc^2? This argument seems rather ad hoc. Also, in introductory physics courses why don't we just define the Lagrangian as $$L = T - V - mc^2$$. Thanks.

Last edited: Apr 17, 2006
2. Apr 17, 2006

### pmb_phy

T-V is not an invariant since neither T nor V is. If you transform to the frame in which the particle is temporarily at rest then V will not always be zero and if it is it will change imediately. If V is constant then T will have different values in different frames.
Regardless of what you have the equation of motion will remain the same. Adding a function as you suggest will not change the behavior of the system.
Because you chose to add it. Typically those additions are done in EM to allow for a gauge transformation.

See
http://www.geocities.com/physics_world/sr/relativistic_energy.htm

Pete

3. Apr 17, 2006

### emob2p

I think you missed my point. In the particle's frame T-V = 0. V = 0 because it is free and T = 0 because velocity = 0. So if we define the Lagrangian as T - V, then L = 0 in the particle's frame. This means $$\gamma L = 0$$ in the particle's frame. But $$\gamma L$$ is invariant. Thus L = 0 in all frames.

4. Apr 18, 2006

### pmb_phy

Oops. Sorry. In relativity the Lagrangian is not given by T - V. See the Lagrangian I gave you in that link above as an example. There you will see that, for a free particle, $$\gamma L = m_0$$. The invariant thus being the particle's proper mass.

Pete

Last edited: Apr 18, 2006
5. Apr 18, 2006

### Hans de Vries

The non-relativistic Lagrangian is an approximation of the relativistic one
in the following way:

$$-L\ = \ \sqrt{1-v^2/c^2}\ m_o c^2 \quad \approx \quad m_o c^2\ -\ \frac{1}{2}m_o v^2 \qquad .$$ (for v << c)

The large constant comming from the restmass is simple ignored in
classical mechanics.

The gamma factor in the Lagrangian corresponds to the time dillation of
an object moving at v. In QM words: the number of phase changes (ticks)
over the trajectory of the particle (the t' axis) is less by a factor gamma.

The non-relativistic Hamiltonian is an approximation of the relativistic one
as follow:

$$H\ =\ \frac{1}{\sqrt{1-v^2/c^2}}\ m_o c^2 \quad \approx \quad m_o c^2\ +\ \frac{1}{2}m_o v^2 \qquad .$$ (for v << c)

In QM words: the number of phase changes (ticks) over the t axis is
higher by a factor gamma.

That's why -L = V-T while H = V+T in the non-relativistic limit.

Regards, Hans

6. Apr 18, 2006

### Meir Achuz

The covariant {\cal L} for a free particle is a scalar, and the only variable it can depend on the the four-velocity. The only scalar you can get using only the four-velocity is its length^2=-m^2 (with c=1).
So {\cal L} must equal m times some constant. The constant is chosen as -1 so that the usual eqs rresult when interaction are added. (It could be any constant for a free particle.) The artifact T-V was a pre-=relativistic way of getting the right answer.

7. Apr 6, 2010

### chwie

The theory of special relativity can be constructed by to postulated, the theory should be covariant with respect to the Lorentz transform and in the limit v/c-> 0 should be equivalent to non-relativistic mechanics. L=0 because of the covariant condition, but like we can add a total derivative to the Lagrangian, we can use that liberty to satisfy that in the limit v/c->0 L should be that classical Lagrangian. Then we find that if we add -mc^2 we satisfy this condition.