Solving Friedmann's Dust-Filled Equation for Radial Geodesics

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In summary, the conversation discusses the understanding and calculation of geodesics, specifically in the context of the dust filled Friedmann solution. It is shown that radial geodesics obey a certain equation and the concept of proper time is introduced. The necessary steps to show this are outlined and further questions about the use of conserved quantities in different metrics are raised.
  • #1
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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  • #2
On a timelike worldline, dT^2 = -dt^2.

What does "radial" give you?
 
  • #3
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...
 
  • #4
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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  • #5
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?
 
  • #6
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.
 

1. What is Friedmann's Dust-Filled Equation for Radial Geodesics?

Friedmann's Dust-Filled Equation is a mathematical equation used in cosmology to describe the evolution of the universe. It takes into account the effects of dust particles (also known as matter) on the curvature of space-time.

2. Why is it important to solve Friedmann's Dust-Filled Equation for Radial Geodesics?

By solving this equation, we are able to understand how matter and energy affect the expansion and evolution of the universe. This is crucial in gaining a deeper understanding of the origins and fate of our universe.

3. What is the process for solving Friedmann's Dust-Filled Equation for Radial Geodesics?

The process involves using mathematical techniques, such as differential equations and integration, to manipulate the equation and solve for the desired variable. This can be a complex and time-consuming process, requiring advanced mathematical knowledge.

4. What are some applications of solving Friedmann's Dust-Filled Equation for Radial Geodesics?

One major application is in the field of cosmology, where it helps us understand the dynamics of the universe and its expansion. It also has applications in astrophysics, as it can be used to study the motion of celestial bodies.

5. Are there any challenges in solving Friedmann's Dust-Filled Equation for Radial Geodesics?

Yes, there are several challenges. One is the complexity of the equation itself, which requires advanced mathematical skills to solve. Additionally, the equation assumes a homogeneous and isotropic universe, which may not accurately describe the universe on a larger scale.

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