# Covariance of relativistic free particle Lagrangian under Lorentz transformation

How does the Lagrangian of a free particle: $$L = - \sqrt{1- \Big(\frac{dx}{dt}\Big)^{2}}$$ transform under the co-ordinate transformation $(q,\tau)$? $$q = \cosh(\psi)x + \sinh(\psi)t,$$ $$\tau = \sinh(\psi)x + \cosh(\psi)t.$$

I've seen this worked out in the case where the boost is expressed in terms of $\gamma$ and relative velocity but not when there are hyperbolic trig functions.

My attempt:

Writing x and t interms of $q$ and $\tau$ and then trying to write $\frac{dx}{dt}$ in terms of $\frac{dq}{d\tau}$.

However I am not able to do that! Am I missing something?

• Using the chain rule you can show that $\frac{dx}{dt}$ and $\frac{dq}{d\tau}$ are both equal to $c(\tanh{\psi}+\tanh^{-1}{\psi})$, thus your Lagrangian density is indeed invariant. – G K Sep 6 '18 at 14:09
• @FredericThomas He is talking about the Lagrangian, while you are reffering to the Action. – G K Sep 6 '18 at 14:21
• @GK Thanks ! But surely you do mean $\tanh \psi + \coth \psi$ i.e. the second term is $(\tanh \psi )^{-1}$ instead of $\tanh^{-1} \psi$, right ? – Astronomer Sep 6 '18 at 16:15
• Yes thats right! I am sorry for the confusion. – G K Sep 6 '18 at 16:20

I think while doing the calculations, I have found the answer. @GK gave an answer but I think there is a subtle mistake in that approach because both q and $\tau$ are coupled transformations of x and t. Thus the total derivative is not equal to the partial one. So this is my solution.

$q$ and $\tau$ both can be expressed a function of $(x,t)$. From the definition of a total derivative $$\frac{dq}{dt} = \Big(\frac{\partial q}{\partial x}\Big)\frac{dx}{dt} + \frac{\partial q}{\partial t}$$

$$\Rightarrow \frac{dq}{dt} = \cosh(\psi) \frac{dx}{dt} + \sinh(\psi)$$

Similarly we have $$\frac{d\tau}{dt} = \Big(\frac{\partial \tau}{\partial x}\Big)\frac{dx}{dt} + \frac{\partial \tau}{\partial t}$$

$$\Rightarrow \frac{d\tau}{dt} = \sinh(\psi) \frac{dx}{dt} + \cosh(\psi)$$

From (1) and (2) we have $$\frac{dq}{d\tau} = \dfrac{\cosh(\psi) \frac{dx}{dt} + \sinh(\psi)}{\sinh(\psi) \frac{dx}{dt} + \cosh(\psi)}$$

$$\Rightarrow 1- \Big(\frac{dq}{d\tau}\Big)^{2}= 1- \Bigg(\dfrac{\cosh(\psi) \frac{dx}{dt} + \sinh(\psi)}{\sinh(\psi) \frac{dx}{dt} + \cosh(\psi)}\Bigg)^{2}$$

Going through the algebra and simplifying using (2) we get

$$\Rightarrow 1- \Big(\frac{dq}{d\tau}\Big)^{2}= \dfrac{(\sinh^{2}(\psi) - \cosh^{2}(\psi))(\frac{dx}{dt})^{2}+(\cosh^{2}(\psi) - \sinh^{2}(\psi))}{(\frac{d\tau}{dt})^{2}}$$

Using the well know trig identity $\cosh^{2}(\psi) - \sinh^{2}(\psi) = 1$ we have,

$$\Rightarrow 1- \Big(\frac{dq}{d\tau}\Big)^{2}= \dfrac{1-(\frac{dx}{dt})^{2}}{(\frac{d\tau}{dt})^{2}}$$

We know the Lagrangian $L^{\prime} = - \sqrt{1- \Big(\frac{dq}{d\tau}\Big)^{2}}$

Thus $$L^{\prime} = (\frac{dt}{d\tau})L$$

$$\Rightarrow L^{\prime}(\dot{q}) d\tau = L(\dot{x}) dt$$

$$\Rightarrow S^{\prime} [L^{\prime}] = S[L]$$

Even though the Lagrangian doesn't stay the same, the action does !