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quasar987

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[tex]g=dt\otimes dt - (t^2+a^2)^2dy\otimes dy[/tex]

where a is a positive constant.

In previous subquestions, I have calculated the Christoffel symbols for the metric-compatible connexion and now I am asked to find the null geodesic (for a photon) starting at t=0, y=pi/a and which is directed towards y=0.

The geodesic equations write

[tex]\frac{d^2t}{dp^2}+\Gamma_{yy}^t\left(\frac{dy}{dp}\right)^2=0[/tex]

[tex]\frac{d^2y}{dp^2}+2\Gamma_{yt}^y\frac{dy}{dp}\frac{dt}{dp}^2=0[/tex]

I didn't know what a null geodesic was so I looked it up wiki and it says it's a geodesic whose tangent vector whose norm is 0.

But this is crazy because the lenght of a curve is [itex]L=\int_{\gamma}||\dot{\gamma}(p)||dp[/itex], so a null tangeant vector means null lenght, means it's just a point. :grumpy: :grumpy: :grumpy: