i am currently toying around with the behaviour of a classical relativistic point particle a bit. For a free one we get the action

\begin{align} S =\int_\tau - m\sqrt{- \dot X_\mu \dot X^\mu}. \end{align} This action is invariant to reparametrization of the worldline parameter $\tau$, which i can see mathematically, but i dont fully grasp its physical meaning (compare e.g. http://www.damtp.cam.ac.uk/user/tong/string/one.pdf ).

I would have intuitively linked this to lorentz invarianve, but we could add potential terms like $V( X_\mu X^\mu)$ that are lorentz invariant and break the reparametrization invarianve, and we can add terms that are invariant to reparametrization but not to lorentz tranfsfomrations, like $ \dot X^1$. So these two dont seem to be linked at all, but what does the reparametrization invarianvce mean then, and when is it relevant? For example, i would like to experiment a bit with simple potentials. More concrete a relativistic theory that reduces to the harmonic oscillator in the non relativistic limit. Naively i would just write down an action like this:

\begin{align} S =\int_\tau - m\sqrt{- \dot X_\mu \dot X^\mu} - \frac{m \omega^2}{2} \vec{x}^2. \end{align}

This is obviuously not lorentz invariant, which shouldnt be a problem i guess - the non relativistic version isnt galilei invariant either. But it also would break the reparametrization invariance - is that a problem?

I hope i could formulate my confusion clearly. Thanks in advance for any help!

  • $\begingroup$ Essentially a duplicate of What is the physical meaning of the affine parameter for null geodesic?. (This isn't a null geodesic, but it discusses the meaning of the affine parameter, which is generally...none) $\endgroup$
    – ACuriousMind
    Sep 3, 2015 at 11:41
  • $\begingroup$ Could you elaborate on that? Im not really familiar with general relativity, and i struggle a bit with applying these answers onto my questions. $\endgroup$
    – Peter
    Sep 3, 2015 at 12:55

2 Answers 2


In order to construct a reparameterization-invariant action, we can proceed as follows:

First, lets switch to the Polyakov-type action

$$ S[X, e] = \int d \tau \, \frac{1}{2} \left( \frac{\dot{X}^2}{e(\tau)} - e(\tau)\, m^2 \right). $$

It can be shown that it is equivalent to the (mentioned by you) Nambu-Goto type action by solving the E/L equation for $e(\tau)$ (which is algebraic) and plugging it back into the action:

$$ e(\tau) = \frac{1}{m} \, \sqrt{-\dot{X}^2} \quad \rightarrow \quad S[X]=-m \int d\tau \sqrt{-\dot{X}^2}. $$

The physical meaning of $e(\tau)$ is straightforward: it defines an einbein: a one-dimensional analogue of a tetrad. Therefore, we could instead use a one-dimensional metric (which has only one component):

$$ g_{\tau \tau} (\tau) = e(\tau)^2. $$

Reparameterizations act on $e(\tau)$ just like normal diffeomorphisms do on tetrads: they keep the product $e(\tau)\, d\tau = ds$ invariant.

Second, lets modify the Polyakov-type action in order to include the harmonic-oscillator term:

$$ S_{h.o.}[X, e] = \int d \tau \, \frac{1}{2} \left( \frac{\dot{X}^2}{e(\tau)} - e(\tau)\, m^2 - e(\tau) \cdot k \cdot \vec{X}^2 \right). $$

Here $k$ is a coupling constant (measuring the force of the spring or whatever). This term breaks Lorentz invariance, but you kind of had to expect this was going to happen (since there is a preferred rest frame in which the loose end of the "spring" attached to the particle is stationary).

But it does not break reparametrization invariance, because it was chosen in such a way. Remember that in GR a diffeo-inv. action is always of the form

$$ S = \int d^n x \sqrt{g} L $$

where $L$ transforms as a scalar? Well this is exactly what I have done here, since

$$ \sqrt{g} = \sqrt{g_{\tau \tau}} = e(\tau). $$

Equivalently, I could say that I have to write $e(\tau) d\tau$ instead of just $d\tau$ as an invariant integration measure.

Finally, lets solve the equation of motion for $e(\tau)$ and plug it back in the action:

$$ e(\tau) = \frac{\sqrt{-\dot{X}^2}}{\sqrt{m^2 + k \vec{X}^2}} \quad \rightarrow \quad S[X] = ... $$

Well, the last calculations is a little tedious for me (although pretty simple actually) so I hope you do it all by yourself. Have fun!

  • 1
    $\begingroup$ Combined with the answer from Fra, this really answered all my questions. Thank you, that helped a lot! $\endgroup$
    – Peter
    Sep 4, 2015 at 14:33

Reparametrization invariance of the worldline simply means that the physics of a problem cannot depend upon the particular choice of parameter a physicist makes, the pramaeter $\tau$ per se has no meaning, you can still choose to parametrize the wordline $x^\mu(\tau)$ with something that has a meaning like the proper time, which is only a convenient choice.

You can say the invariance is a mathematical consequence of the fact that you don't want to introduce in the theory a preferred time frame (i.e. you don't want to fix a frame of reference from the start) and that creates a redundacy: (an additional non physical symmetry, a gauge symmetry) the freedom to fix the parameter $\tau$. If for exemple you choose proper time, you would have $\tau=\tau(x^0,v)$ where $x^0$ is your coordinate time and $v$ is the velocity of the particle relative to you. This means of the $D$ coordinates you had, only $D-1$ are indipendent. In a more sophisticated language that means the Hamiltonian of the system is of pure constraint, i suggest you to read Dirac's "Lectures on quantum mechanics" if you want to better understand lagrangian and hamiltonian mechianics in the context of special relativity.

As a concluding remark on this issue, i will say that parametrization invariance means that once you compute relations between observables quantities the dependence on the particular paramter you chose must vanish, if it wasn't so, differnt observers would see different physics just by having fixed $\tau$ in a different way.

To your practical problems:

You have written $\mathcal{L}=-m\sqrt{\dot{X}^\mu\dot{X}_\mu}$ which makes for an invariant action under reparametrization. In classical mechanics in the context of special relativity the only known four force that has all the right proprieties (Lorentz+rep, you can easily check this) is the electromagnetic one $$\tag 1\mathcal{L}_I=e \dot{X}_\mu A^\mu$$ where $A^\mu$ is the usual E.M. four potential. You can think of this as: you can treat (without going insane) in a fully lorentz covariant way only the foundamental interactions of nature. If you want to study the potential $V=\frac{1}{2}kx^2$ which has a preferred frame, or others non relativistic potentials, you will have to fix from the start the frame of reference and $\tau$ and use and had oc lagrangian which will not have any obvious lorentz symmetry, for example: $$\tag 2\mathcal{L}_{adhoc}=-mc^2\sqrt{1-\beta^2}-V(x)$$ with this you can solve most practical problems. This doesn't mean that the theory with $\mathcal{L}_{adhoc}$ isn't consistent, it just means that it's valid for one particular choice of the frame of reference. If you want to know the soultion in another frame of reference, or reparametrize it, you will have to start back from scratch and define a new lagrangian for the new inertial frame.

  • $\begingroup$ glad i could help $\endgroup$
    – Fra
    Sep 4, 2015 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.