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I have to discuss the motion of a free particle in Minkowski using the Rindler metric $$ds^2=e^{2ap}(-d\tau^2+dp^2)$$ So it has to satisfy the geodesic condition $\frac{d^2x^\mu}{d\tau^2}+\Gamma_{\rho\sigma}^\mu=0$ which gives $$\ddot{p}+2a(\dot{p}^2-\dot{\tau}^2)=0$$ and, because $\tau$ is the proper time (or is prop to it?) $$\ddot{p}+2a(\dot{p}^2-1)=0$$ $$p=e^{a(1+\sqrt{3})\tau}\quad or \quad p=e^{a(1-\sqrt{3})\tau}$$ so in Rindler coordinates a free particle moves exponentially in time (maybe with a different coefficient instead of $(1+\sqrt{3})$).Graphic of the motion in Rindler coordinates.

It's all correct?

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$\let\a=\alpha \def\dp{\dot p} \def\dq{\dot q} \def\dx{\dot x}$ No. It isn't true that $\tau$ is proper time. Incidentally, I had never seen before that form of Rindler metric. It's right, but the choice of $\tau$ for one coordinate is unfortunate. So let me use $q$ instead: $$ds^2 = e^{2ap}\,(-dq^2 + dp^2).$$

In a problem like that writing gedesic equations is seldom useful. And in any case you need two equations, as you have two unknown functions: $p(\tau)$, $q(\tau)$. Usually there are shortcuts, thanks to symmetries or at least to a general identity:

$$g_{\mu\nu}\,\dx^\mu\,\dx^\nu = -1 \tag1$$ for a timelike geodesic. (Here the dot does mean derivative wrt proper time!) Did you know eq. (1)? Or can you see why it holds true?

In our case eq. (1) becomes $$e^{2ap}\,(\dq^2 - \dp^2) = 1.$$

Other useful relations derive from symmetries. Have you heard of it? Killing vectors? Noether's theorem? At least the special case when one coordinate does not appear in $g_{\mu\nu}$? This is what happens here: coefficients of metric tensor don't depend on $q$. If $g_{\mu\nu}$ is independent of say $x^\a$ then it can be shown that $g_{\a\nu}\,\dx^\nu$ is constant along any geodesic. In our case $$e^{2ap}\,\dq = \mathrm{const.} \tag2$$

Of course you could also derive (2) via geodesic equation for $q$. It's only a longer way.

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  • $\begingroup$ Thanks, i didn't think to use conserved charges because, this was an assignment given much before knowing killing vectors, etc. (just geodesics, maybe using both the geodesic for p and $\tau$ the result would be the same). Also in the equation between (1) and (2) there is a minus on the right-hand side but i cant edit that $\endgroup$ – Pancio Jan 20 at 10:57
  • $\begingroup$ @Pancio Also in the equation between (1) and (2) there is a minus on the right-hand side but i cant edit that You're right, thank you. It depends on my being used to a (1,-1,-1,-1) convention. I've edited it. $\endgroup$ – Elio Fabri Jan 20 at 14:02
  • $\begingroup$ @Pancio Of course I suspected you had not seen eq. (1) and Noether's theorem but I wanted to tell you that there are such shortcuts. Moreover, it's true that in principle you could derive those equations from geodesic equations, but it might not be so immediate (I didn't try). Observe that a geodesic equation is second order, whereas my equations are first integrals, i.e. first order, containing only $\dot q$ and $\dot p$. $\endgroup$ – Elio Fabri Jan 20 at 14:03

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