Solving Friedmann's Dust-Filled Equation for Radial Geodesics

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The discussion focuses on solving Friedmann's dust-filled equation for radial geodesics, specifically the equation ds^2 = -dt^2 + a(t)^2 (dX^2 + X^2 dO^2). Participants clarify that radial geodesics imply dO^2 = 0, leading to the simplified form ds^2 = -dT^2 + a^2 dX^2. The key relationships derived include a^2 (dX/dT) = k and (dt/dT)^2 = 1 + k/a^2, with the latter being validated through the use of the metric tensor g(ab)t(a)t(b) = -1 for timelike geodesics. The discussion also touches on the conservation of quantities related to (dt/dT) in different spacetimes.

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bjaw
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I understand what geodesics are and how to calculate them from Christoffel symbols and all that. But I've just come across a question I have no idea about. I've been given the dust filled Friedmann solution:

ds^2 = -dt^2 + a(t)^2 (dX^2 + X^2 dO^2) (O=omega)

And been told to show that radial geodesics obey:

a^2 (dX/dT) = k where k constant and T is proper time.

And hence (dt/dT)^2 = 1 + k/a^2

How do I introduce proper time? And where to I get an expression for dX/dT from? I'm completely stumped...and help would be much appreciated!
 
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On a timelike worldline, dT^2 = -dt^2.

What does "radial" give you?
 
It should mean that dO^2 = 0 so that there's no angular component to the worldline. So this gives:

ds^2 = -dT^2 + a^2 dX^2

I've also realized that for the second part I can use g(ab)t(a)t(b) = -1 for timelike geodesics, which means that (dt/dT)^2 = 1 + a^2(dX/dT)^2 which gives the second part assuming that (dX/dT) = k/a^2, but I still can't think how to show this first part...
 
bjaw said:
It should mean that dO^2 = 0 so that there's no angular component to the worldline. So this gives:

ds^2 = -dT^2 + a^2 dX^2

I think you meant

ds^2 = -dt^2 + a^2 dX^2.

Using my substitution gives,

-dT^2 = - dt^2 + a^2 dX^2,

which is the same as

I've also realized that for the second part I can use g(ab)t(a)t(b) = -1

but I still can't think how to show this first part...

Even though it's for a different situation, https://www.physicsforums.com/showpost.php?p=621802&postcount=32" might help.
 
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Wonderful. I managed to obtain the required relation by extremising the action under X. Thanks so much for the fantastic help you give!

Just one more quick question about that link in your last post: can you always equate the conserved quantity relating to (dt/dT) to E/m or is that only for the Schwarzschild metric?
 
bjaw said:
Just one more quick question about that link in your last post: can you always equate the conserved quantity relating to (dt/dT) to E/m or is that only for the Schwarzschild metric?

For large r, Scwharzschild spacetime is like Minkowski spacetime, and in Minkowski spacetime, dt/dT = E/m. This led to the choice of label. I think it is common to use this label in any spacetime for which this is true. Not all spacetimes, however, have this property.
 

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