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I have this metric: $$ds^2=-dt^2+e^tdx^2$$ and I want to find the equation of motion (of x). for that i thought I have two options:

  1. using E.L. with the Lagrangian: $L=-\dot t ^2+e^t\dot x ^2 $.

  2. using the fact that for a photon $ds^2=0$ to get: $0=-dt^2+e^tdx^2$ and then: $dt=\pm e^{t/2} dx$.

The problem is that (1) gives me $x=ae^{-t}+b$ and (2) gives me $x=ae^{-t/2} +b$.

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This is just 1+1-dimensional Minkowski space written in unusual coordinates, so a third method would be just to do a change of coordinates to make it look like the usual Minkowski coordinates. –  Ben Crowell Jun 26 '13 at 21:11
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@BenCrowell It can't be. I just used Hartle's mathematica notebooks to check and I get a nonzero Riemann tensor. A look at the conformal flat form of the metric in Qmechanic's answer seems to confirm this. –  Michael Brown Jun 28 '13 at 8:31
    
@MichaelBrown: Oops, thanks for the correction. –  Ben Crowell Jun 28 '13 at 19:06

3 Answers 3

If your solution is not a null geodesic then it is wrong for a massless particle.

The reason you go astray is that Lagrangian you give in (1) is incorrect for massless particles. The general action for a particle (massive or massless) is:

$$ S = -\frac{1}{2} \int \mathrm{d}\xi\ \left( \sigma(\xi) \left(\frac{\mathrm{d}X}{\mathrm{d}\xi}\right)^2 + \frac{m^2}{\sigma(\xi)}\right), $$

where $\xi$ is an arbitrary worldline parameter and $\sigma(\xi)$ an auxiliary variable that must be eliminated by its equation of motion. Also note the notation

$$ \left(\frac{\mathrm{d}X}{\mathrm{d}\xi}\right)^2\equiv \pm g_{\mu\nu} \frac{\mathrm{d}X^\mu}{\mathrm{d}\xi}\frac{\mathrm{d}X^\nu}{\mathrm{d}\xi}, $$

modulo your metric sign convention (I haven't checked which one is right for your convention). Check that for $m\neq 0$ this action reduces to the usual action for a massive particle. For the massless case however, you get

$$ S = -\frac{1}{2} \int \mathrm{d}\xi\ \sigma(\xi) \left(\frac{\mathrm{d}X}{\mathrm{d}\xi}\right)^2. $$

The equation of motion for $\sigma$ gives the constraint

$$ \left(\frac{\mathrm{d}X}{\mathrm{d}\xi}\right)^2 = 0, $$

for a null geodesic. This is necessary and consistent for massless particles, as you know.

The equation of motion for $X^\mu$ is (EDIT: Oops, I forgot a term here. Note that $g_{\mu\nu}$ depends on $X$ so a term involving $\partial_\rho g_{\mu\nu}$ comes into the variation. Try working it out for yourself. I'll fix the following equations up later):

$$ \frac{\mathrm{d}}{\mathrm{d}\xi}\left(\sigma g_{\mu\nu}\frac{\mathrm{d}X^{\mu}}{\mathrm{d}\xi}\right)=0, $$

but you can change the parameter $\xi\to\lambda$ so that $\sigma \frac{\mathrm{d}}{\mathrm{d}\xi} = \frac{\mathrm{d}}{\mathrm{d}\lambda}$, so the equation of motion simplifies to

$$ \frac{\mathrm{d}}{\mathrm{d}\lambda}\left(g_{\mu\nu}\frac{\mathrm{d}X^{\mu}}{\mathrm{d}\lambda}\right)=0, $$

which you should be able to solve to get something satisfying the null constraint.

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Nice answer, the Lagrangian you start with looks a bit unfamiliar to me. For example where does the auxilliary field $\sigma(\xi)$ come from and why do you need this? At the first reading I thought it is a reparameterization of the world line ... If it is an auxilliary field as they appear for example in supersymmetric Lagrangians, should it then not have its own term in the Lagrangian too? –  Dilaton Jun 26 '13 at 11:25
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@Dilaton The $m^2/\sigma$ term counts as "its own term in the Lagrangian" yeah? $\sigma(\xi)$ is called the "einbein" (also denoted $e$), the one dimensional version of the vierbein. If you are familiar with strings it is the worldline analog of the string worldsheet metric, and the action I've written is related to the usual square root action as the Polyakov action is related to the Nambu-Goto action. For massive particles it is just a trick to eliminate the square root, but it has the additional advantage that the $m\to 0$ limit also works. String theory texts usually talk about this stuff. –  Michael Brown Jun 26 '13 at 13:37
    
Aah thanks Michael, now I see :-) –  Dilaton Jun 26 '13 at 13:46

There is an elegant way of doing this using symmetries.

Notice that this metric is space translation invariant, so it has a killing vector $\partial_x$. There is a corresponding conserved quantity $c_x$ along geodesics $x^\mu(\lambda)$ given by \begin{align} c_x = g_{\mu\nu}\dot x^\mu(\partial_x)^\nu = e^t\dot x \end{align} Where an overdot denotes differentiation with respect to affine parameter. On the other hand, the fact that the desired geodesic is a (null) photon geodesic along which $ds^2 = 0$ gives \begin{align} 0=-\dot t^2 + e^t\dot x^2 \end{align} This forms a set of coupled differential equations that is not actually that hard to solve. Hint: Try solving the first equation for $\dot x$, and then plugging it into the second equation.

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+1 for elegant use of symmetry! –  Alex Nelson Jun 26 '13 at 15:38
    
@AlexNelson Well thanks Mr. Nelson. –  joshphysics Jun 26 '13 at 15:44

I) Well, in 1+1 dimensions the light-cone (based at some point) is just two intersecting curves, which are precisely determined by the condition

$$\tag{1} g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}~=~0,$$

and an initial condition cf. OP's second method. However, this eq. (1) will not determine light-like geodesics in higher dimensions.

II) OP's first method, namely to vary the Lagrangian

$$\tag{2} L~:=~ g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu} $$

is in principle also correct. It is a nice exercise to show that Euler-Lagrange equations are the geodesic equation. However, it seems that OP mistakenly identifies the parameter $\lambda$ of the geodesic with the $x^0$-coordinate. These are two different things! In 1+1 dimensions, we have two coordinates $x^0$ and $x^1$. There are two Euler-Lagrange equations. The complete solution for $\lambda\mapsto x^0(\lambda)$ and $\lambda\mapsto x^1(\lambda)$ will be all geodesics: time-like, light-like and space-like.

Since we are only interested in light-like geodesics, we would also have to impose eq. (1) in the Euler-Lagrange method.

III) If one makes a coordinate transformation

$$\tag{3} u~=~\exp(-\frac{x^0}{2})\quad\text{and}\quad v~=~\frac{x^1}{2}, $$

then OP's metric becomes

$$\tag{4} \frac{4}{u^2}(-du^2+dv^2)$$

which is e.g. also considered in this Phys.SE post (up to an overall constant factor). Obviously, the light-like geodesics are of the form

$$\tag{5} v-v_0~=~ \pm (u-u_0). $$

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