# Deriving special relativity free particle Lagrangian using infinitesimal boost?

At the very beginning of Landau and Lifshitz Mechanics they derive the form of the Lagrangian for a free particle in Newtonian mechanics.

I want to see how to do the analogous derivation in special relativity.

For Newtonian mechanics they say that, first, the Lagrangian $$L$$ can only be a function of $$v^2$$ and not $$\vec{x}$$ or $$\vec{v}$$ since there are no special locations or directions. Then they look at the Lagrangian in another inertial frame moving at relative velocity $$\vec{\epsilon}$$ and say that it must differ from the original by a total time derivative in order to keep the equations of motion the same.

The square of the velocity in the new frame is $${v'}^2 = v^2 + 2 \vec{v}\cdot\vec{\epsilon} + \mathcal{O}(\epsilon^2),$$ so the Lagrangian in this frame is $$L({v'}^2)=L(v^2)+\frac{\partial L}{\partial v^2} 2 \vec{v}\cdot\vec{\epsilon}.$$ For the last term to be a total time derivative it has to be a linear function of the velocity, i.e. $$\partial L/\partial v^2$$ is constant. And then you get $$L \propto v^2$$.

$$---$$

I tried doing the same thing for special relativity, the only change being the different formula for the velocity in the moving inertial frame. I find $$\frac{{v'}^2}{c^2} = 1 - \left(1-\frac{v^2}{c^2}\right)\left(1-\frac{\epsilon^2}{c^2}\right)\left(1-\frac{v_x \epsilon}{c^2}\right)^{-2},$$ for a boost by velocity $$\epsilon$$ along the $$x$$-axis (can get this quickly writing the invariant interval $$c^2 dt^2 - dx^2 = c^2 dt'^2 - dx'^2$$ and using the Lorentz transformation for the time coordinate to get $$dt'/dt$$).

For small $$\epsilon$$ I get $${v'}^2 = v^2 - 2 \left(1-\frac{v^2}{c^2}\right) v_x \epsilon,$$ so $$L({v'}^2)=L(v^2)-2 \left(1-\frac{v^2}{c^2}\right) v_x \epsilon \frac{\partial L}{\partial v^2} .$$

I think the argument that the last term be a total time derivative again means that the last term must be linear in the velocity. Therefore, $$\frac{\partial L}{\partial v^2} \propto \left(1-\frac{v^2}{c^2}\right)^{-1}.$$

But integrating this gives $$L \propto \log\left(1-\frac{v^2}{c^2}\right),$$

not the right answer, which is $$L\propto \sqrt{1-\frac{v^2}{c^2}}$$.

Where am I going wrong?

Here is one line of reasoning:

1. One new feature of Lorentz transformations as opposed to Galilean transformations is that time $$t$$ transforms as well. Therefore we should consider invariance properties of the Lagrangian one-form $$\mathbb{L}~=~L \mathrm{d}t~=~ L \dot{t}\mathrm{d}\lambda \tag{1}$$ rather than just the Lagrangian $$L$$ itself as OP does. Here $$\lambda$$ denotes a world-line (WL) parameter, and a dot denotes differentiation wrt. $$\lambda$$.

2. Next, Lorentz invariance suggests that the Lagrangian one-form should be $$\mathbb{L}~=~ f(\dot{x}^2)\mathrm{d}\lambda, \qquad \dot{x}^2~:=~\eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}, \qquad x^0~\equiv~ct, \tag{2}$$ for some function $$f$$. (This is a 4D relativistic analogue of Landau & Lifshitz's argument above eq. (3.1) claiming that the non-relativistic Lagrangian should be a function of 3D speed $$v\equiv |{\bf v}|$$ due to 3D rotational invariance.)

3. Conversely, eq. (2) is as far as we can get from Lorentz symmetry as eq. (2) is already manifestly Lorentz invariant. Note that Lorentz symmetry is implemented as strict symmetry of the action (as opposed to the Galilean quasisymmetry, cf. e.g. this and this Phys.SE posts).

4. Finally, WL reparametrization invariance implies that the function $$f~\propto ~\sqrt{\cdot}\tag{3}$$ is proportional to a square root. This is the right answer: The Lagrangian is $$L~\propto ~\sqrt{1-v^2/c^2}\tag{4}$$ for a massive relativistic point particle.

• Thanks! I didn't think about transforming the $dt'$. It looks like you can get the right Lagrangian using Landau & Lifshitz procedure now. I'll write it up as a separate answer.
– Alex
Commented Jan 31, 2019 at 15:54

@Qmechanic's first point was the key. The moving observer will write down the action integral as $$L(v'^2)dt'$$, not as $$L(v'^2) dt$$ as I was implicitly assuming.

Therefore, the Landau & Lifshitz argument is that in order to get the same equations of motion in the two inertial frames we have to have

$$L(v'^2) dt' = L(v^2)dt + \frac{df}{dt}dt,$$ for some function $$f(x,t)$$.

Relating $$dt'$$ back to $$dt$$ you get $$dt' = \gamma \left(dt - \frac{\epsilon dx}{c^2} \right) = \gamma \left(1 - \frac{\epsilon v_x}{c^2} \right)dt \approx \left(1 - \frac{\epsilon v_x}{c^2} \right)dt,$$

to first order in $$\epsilon$$ (the velocity of the moving frame).

Then using the expression for $$L(v'^2)$$ in terms of $$v$$ that I had before:

\begin{align} L(v'^2)dt' &= \left[ L(v^2)−2\left(1−\frac{v^2}{c^2}\right) v_x \epsilon \frac{\partial L}{\partial v^2}\right] \left(1 - \frac{\epsilon v_x}{c^2} \right)dt \\ &= L(v^2)dt - \left[2\left(1−\frac{v^2}{c^2}\right)\frac{\partial L}{\partial v^2} + \frac{L(v^2)}{c^2}\right] v_x \epsilon dt \end{align} to first order in $$\epsilon$$.

In order for the the second term to be a total time derivative the expression in brackets must be a constant. This gives a differential equation for $$L$$:

$$2\left(1−\frac{v^2}{c^2}\right)\frac{\partial L}{\partial v^2} + \frac{L(v^2)}{c^2} = A,$$ for some constant $$A$$. It can be solved by multiplying through by the integrating factor $$(1-v^2/c^2)^{-3/2}$$.

\begin{align} 2\left(1−\frac{v^2}{c^2}\right)^{-1/2}\frac{\partial L}{\partial v^2} + \left(1−\frac{v^2}{c^2}\right)^{-3/2}\frac{L(v^2)}{c^2} &= A\left(1−\frac{v^2}{c^2}\right)^{-3/2}\\ \frac{\partial}{\partial v^2} \left[ 2\left(1−\frac{v^2}{c^2}\right)^{-1/2} L \right]&= \frac{\partial}{\partial v^2} \left[ 2A c^2 \left(1−\frac{v^2}{c^2}\right)^{-1/2} \right] \\ 2\left(1−\frac{v^2}{c^2}\right)^{-1/2} L &= 2A c^2 \left(1−\frac{v^2}{c^2}\right)^{-1/2} + B, \end{align} using some integration constant $$B$$.

Finally, solving for $$L$$ gives the free particle Lagrangian:

$$L(v^2) = \mathrm{const} +\mathrm{const}\sqrt{1-\frac{v^2}{c^2}}.$$

Works great!