Derivation of relativistic motion equations from action

[lmcelroy]

Anyone willing to derive the motion equations of special relativity from the SR action:

S = -m₀c²∫_t1^t21/γ dt
where:
m₀ = rest mass
γ = Lorentz factor =1/√(1-v²/c²)
v is the velocity as a function of time.

The full action contains terms of the vector potential and scalar potential but assume that these are negligible. I think that is all the information required. Thanks in advance

 

[elfmotat]

I've decided to use units where c=1 for simplicity. Also note that $v^2=\vec{v}\cdot \vec{v}$, where $\vec{v}$ is the three-velocity. $\vec{x}$ is the three-position (x,y,z), and $\vec{p}$ is the relativistic three-momentum.Varying the action gives:

$$\delta S= - \int \delta (m \sqrt{1-v^2}) dt=0$$

Simplifying:

$$-\int \delta (m \sqrt{1-\vec{v}\cdot \vec{v}}) dt= - \int \frac{m \vec{v}\cdot \delta \vec{v}}{\sqrt{1-v^2}}dt$$

If you integrate by parts, the integrated term will be dotted with δx, so it goes to zero when evaluated at the boundaries. What you're left with is:

$$\int \frac{d}{dt} \left ( \frac{m \vec{v}}{\sqrt{1-v^2}} \right ) \cdot \delta \vec{x}~dt=0$$

The integrand, as you can see, is the time derivative of the relativistic momentum. For δS to be zero identically, this must be zero. So we're done:

$$\frac{d}{dt} \left ( \frac{m \vec{v}}{\sqrt{1-v^2}} \right )= \frac{d\vec{p}}{dt}=0$$This tells us that objects will move inertially. Had you chosen to include the vector and scalar potentials, you would have ended up with:

$$\frac{d}{dt} \left ( \frac{m \vec{v}}{\sqrt{1-v^2}} \right )= q(\vec{E}+\vec{v}\times \vec{B})$$

Where:

$\vec{E}=-(\nabla \phi +\frac{\partial \vec{A}}{\partial t})$
$\vec{B}= \nabla \times \vec{A}$

with $\phi$ and $\vec{A}$ as the scalar and vector potential respectively.

 

[elfmotat]

Also, an easier way to go about things might be the following:

We're supposed to extremize S = -m∫L(x,v,t)dt = -m∫(1/γ) dt. We can just plug this Lagrangian into the Euler-Lagrange equation. A trick to make this easier is to realize that if a Lagrangian L satisfies the Euler-Lagrange equation, then in general any function of the Lagrangian f(L) also satisfies the Euler-Lagrange equation as long as ∂/∂λ (∂f/∂L)=0, where λ is your parameter (I can't think of an example where this wouldn't be the case, but I included it just to be safe). The reason is simple:

If:

∂L/∂q = ∂/∂λ (∂L/∂(∂q/∂λ))

, then by chain rule we get that:

∂f/∂q = (∂f/∂L)(∂L/∂q) = (∂f/∂L)(∂/∂λ)(∂L/∂(∂q/∂λ)) = ∂/∂λ (∂f/∂(∂q/∂λ))So, instead of using L(x,v,t) = 1/γ, we can use f(L) = L² = γ^-2 = (1-v²):$$\frac{\partial }{\partial \vec{x}}[1-\vec{v}\cdot \vec{v}]=0$$

and

$$\frac{\partial }{\partial t} \frac{\partial }{\partial \vec{v}}[1-\vec{v}\cdot \vec{v}]=\frac{\partial }{\partial t}[-2\vec{v}]=-2\frac{\partial \vec{v}}{\partial t}$$

So we find that:

$$\frac{\partial \vec{v}}{\partial t}=0$$.

This is equivalent to what I did in my last post. They both say that objects will move inertially (not accelerate).
One more thing to point out is that dt/γ = dτ, where τ is proper time. This means that the action S = -m∫dτ = -mΔτ. So extremizing the action is equivalent to extremizing the proper time elapsed. Some call this the principle of extremal proper time, and it generalizes nicely to general relativity. Geodesics can be found in this way, by extremizing ∫g_μν(dx^μ/dλ)(dx^ν/dλ)dλ.

 

[lmcelroy]

^Thanks for that. An easy to follow derivation and explanation even for someone new to the calculus of variations, as myself.