# Is variation pricip for light-geodetic correct?

1. Sep 16, 2009

### archipatelin

Princip stacionary action for propagation of light is apply on thus definition of action:
$$S=\int\!\mbox{d}\tau=\frac{1}{c}\int\!\sqrt{\mbox{d}x_{\mu}\mbox{d}x^{\mu}}=\frac{1}{c}\int\!\sqrt{g_{\mu\nu}\frac{\mbox{d}x^{\nu}}{\mbox{d}\tau}\frac{\mbox{d}x^{\mu}}{\mbox{d}\tau}}\mbox{d}\tau$$
The Evolution coupled Euler-Lagrangian equations we get correct equation of geodetic.

BUT for light-like vectors are true $$\mbox{c^2d}\tau^2=\mbox{d}x_{\mu}\mbox{d}x^{\mu}=0$$.
Therefore hasn't sense $$\frac{\mbox{d}x^{\nu}}{\mbox{d}\tau}$$
and also whole integral for action $$S=\int\!\mbox{d}\tau$$.

Note: During evolution of E-L equs is dividing of Lagrangian in this case this is dividing of ZERO and this is incorrect.
For time-like or space-like (negativ argument of sqrt is possible neglect or work with comlex number, however E-L equs will still only real) vectors is this operation OK. But variation for light-like vectors isn't (for me) mathematically correct.

Exist another (math. correct) definition of action for light?

2. Sep 16, 2009

### Jonathan Scott

As far as I can see, the only problem is with the use of the parameter d \tau, which doesn't work for light speed. If you express the action in terms of some other parameter such as dt it still works for light speed, and the E-L equations come out the same as before.

[Edited several times in an attempt to correct incorrectly displaying LaTeX, but it seemed to be stuck with old images, so I gave up]

Last edited: Sep 16, 2009
3. Sep 16, 2009

### haushofer

Good point, but I believe you overlooked a factor of m; look at your dimensions. An action has dimensions of energy*time. What you can do is to introduce an auxiliary field to solve your problem! Let's put c=1.

The motion of a massive particle can be described by the action

$$S=-m \int ds, \ \ \ \ \ \ ds^2 = -g_{ab}dx^adx^b$$

So already we see the problem: m=0 means S=0 for the whole trajectory, which doesn't give us dynamics. Also, we have a square root in our action. This is quite nasty as soon as you start to quantize things. The idea now is to introduce an auxiliary field e to solve these problems. Sometimes people call this object an "einbein". Ofcourse, this e also obeys certain equations of motion, and these have to be consistent with what we already know!

So we introduce the action (after some thought ofcourse :) )

$$s =\frac{1}{2}\int [e^{-1}\dot{x}^2 - em^2]d\tau$$

where

$$\dot{x}^2 = g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}, \ \ \ \ e=e(\tau)$$

What you can do now, is to obtain the equations of motion for e, and put this back into the action. You'll see that this gives you the original action. However, our auxiliary field enables us to deal also with massless particles!

Ofcourse, the invariance of the action under reparametrizations now also involves this auxiliary field e. It's quite easy to show that choosing e=1/m gives you the well-known expression for the momentum. So I would say: give it a try, and if it doesn't work out, let us know ;)

4. Sep 16, 2009

### archipatelin

But, if I understand it well, then your def. of action has this same problems as previous.
1) you cannot use $$d\tau$$ (the proper time) for integration, because is for light-like vector null, but another affine parametr $$d\lambda$$ (as wrote Jonathan Scott), which isn't invariant in general like $$d\tau$$.

2) for light-like vector is your term (change $$d\tau \leftrightarrow d\lambda$$)
$$\dot{x}^2 = g_{ab}\frac{dx^a}{d\lambda}\frac{dx^b}{d\lambda}, \ \ \ \ e=e(\lambda)$$
well zero and therefore action is:
$$S =-\frac{1}{2}m^2\int\!e(\lambda)d\lambda$$
reparametrisation $$e(\lambda) d\lambda \rightarrow d{}\tilde{\lambda}$$ give
$$S =-\frac{1}{2}m^2\int\!d\tilde{\lambda}$$
For $$m=0$$ is $$S=0$$ for each trajectory, which doesn't give us dynamics.
I think, "einbein" is a utility (math. trick) for variation "square root" lagragian need to quantize, and therefore alone don't help with this.

Last edited: Sep 16, 2009
5. Sep 16, 2009

### Jonathan Scott

Apologies for an incomplete reply first time. Looking at my old notes, I forgot that as well as switching the proper time parameter, you also have to switch to a definition of the action in terms of the energy (which is variable). I don't have all the details, but I once started from something like the following and landed up with the usual equation of motion for GR (by treating c as a variable in isotropic coordinates) including the light speed case:

(I give up - once I get a LaTeX error I can't make it go away)

S = integral (E/c^2 (1 - v^2/c^2)) dt

Last edited: Sep 16, 2009
6. Sep 16, 2009

### archipatelin

I don't understand it. How this help us? This corrcet for special relativity only. General relativity term is:
$$g_{00}dt^2+2g_{0k}dx^kdt+\gamma_{ik}dx^kdx^i=0$$
equally, for energy of light is:
$$g_{00}E^2+2g_{0k}p^kE+\gamma_{ik}p^kp^i=0$$
and your action in general is
$$S=\int{\!E\psi(x,\dot{x})dt$$
where $$\psi(x,\dot{x})$$ is for consistend with mass particle action
"g_{\mu\nu}u^{\mu}u^{\nu}" - TeX error
but this term is for light-like vector null and more E isn't constant for variation (for light perhaps is, i don't know).

Last edited: Sep 16, 2009
7. Sep 16, 2009

### atyy

8. Sep 16, 2009

### Jonathan Scott

I should have remembered this before, as it helps a lot for the static case:

For a test particle falling freely in a static gravitational field the total energy is constant (so the rest energy varies with potential).

9. Sep 17, 2009

### haushofer

I believe that what you do is to plug in the equations of motion for a massless particle into the action, but the whole point of the action is that the variation with respect to x and e give you the equation of motion! It is something like staring at the Dirac equation, and noting that the action itself is 0 due to the equations of motion. Or to give a very simple example: the function $f(x)=x^2$ attains its minimum at x=0, and for this point the function is zero, but that doesn't mean that f(x)=0 for all x!

So, if I vary the action with respect to the einbein e I get

$$\dot{x}^2 + e^2m^2=0$$

Now, for massless particles this tells me that

$$\dot{x}^2 = g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau} = 0$$

And also, for this particular case the action appears to be zero. That's not a mystery; we see the same thing for eg the Hilbert action in the conformal case where the Ricci scalar R is zero, or the earlier mentioned Dirac equation;

$$L = \bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi \rightarrow \frac{\delta L}{\delta\bar{\psi}} = 0 \rightarrow (i\gamma^{\mu}\partial_{\mu}-m)\psi = 0$$

This means that for the extremum the action is zero, but only for this extremum!

By the way, note that for m=0 the einbein doesn't appear in the equations of motion because there is an m in front of it and the equations of motion for x just gives me the geodesic equation. For the case m not equal to zero I can use parametrization invariance, and put e=1/m.

Hope this helps :)