Solving Geodesics with Metric $$ds^2$$

In summary, the Christoffel symbols are related to the Ricci tensors. The Ricci tensors are related to the second derivative of the area.
  • #1
edoofir
6
1
I have the following question to solve:Use the metric:
$$ds^2 = -dt^2 +dx^2 +2a^2(t)dxdy + dy^2 +dz^2$$

Test bodies are arranged in a circle on the metric at rest at $$t=0$$.
The circle define as $$x^2 +y^2 \leq R^2$$

The bodies start to move on geodesic when we have $$a(0)=0$$

a. we have to calculate the second derivative of the area of the circle $$S = \int{\sqrt{g^(2)}dxdy}$$ respected to time and express your answer using the Ricci tensor.

b. calculate the second derivative respected to time of the ratio of the diagonals $D_1$ and $D_2$ and express it using Weyl tensor.

The Christoffel symbols:
$$ \Gamma^{t}_{xy} = \Gamma^{t}_{yx} = a(t)a'(t)$$
$$\Gamma^{x}_{tx} = \Gamma^{x}_{xt} = \frac{a^3(t)a'(t)}{a^4(t)-1} $$
$$\Gamma^{y}_{ty} = \Gamma^{y}_{yt} = \frac{a^3(t)a'(t)}{a^4(t)-1} $$
$$\Gamma^{x}_{ty} = \Gamma^{x}_{yt} = \frac{a(t)a'(t)}{1-a^4(t)} $$
$$\Gamma^{y}_{tx} = \Gamma^{y}_{xt} = \frac{a(t)a'(t)}{1-a^4(t)} $$

the Ricci tensors:

$$R_{tt} = \frac{2a^2(2a'^2 -aa''(a^4-1))}{(a^4-1)^2}$$
$$R_{xx} = R_{yy} = \frac{2a^2a'^2}{a^4-1}$$
$$R_{xy}=R_{yx} = a'^2 +aa''$$where $$a = a(t), a' = \frac{da}{dt}, a'' = \frac{d^2a}{dt}$$

and i calculated the second derivative of the area of the circle:
$$ \frac{ds^2}{dt^2} = \int{[\frac{d^2}{dt^2}\sqrt{1-a^4}dxdy + 2\frac{d}{dt}\sqrt{1-a^4}(\frac{dx}{dt}dy + dx\frac{dy}{dt})+\sqrt{1-a^4}(\frac{d^2x}{dt^2}dy+\frac{dx}{dt}\frac{dy}{dt}+dx\frac{d^2y}{dt^2})]} $$
I am not sure what should i do next. any suggestions?
1680466875788.png
 
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  • #2
You seem to have some sign errors in ##\Gamma^x_{ty}## and ##\Gamma^y_{tx}##.

I'm not sure how you got that expression for ##d^2S/dt^2##. What's your expression for ##S##? And what is the Ricci tensor and the geodesics followed by the test particles?
 
  • #3
thank you for your reply.
I have calculated the Christoffel symbols using Mathmatica. are you sure there is a sign error? isn't it that: $$\Gamma^{y}_{ty} = \Gamma^{y}_{yt}$$?

If it's not the case i will try to calculate agin but it according to Mathmatica it should be the signs (else i don't understand what you ment).
Second, the expression for $S$ is: $$S = \int{\sqrt{g^(2)}dxdy}$$ as i mentiond.
 
  • #4
Yes, the signs were just an overall sign that you lost - just a typo it seems, fine.

How are you getting from that expression for ##S## to the second derivative of area, was what I wanted to know. I was hoping you would show a few steps in the derivation.
 
  • #5
i attach here the calculation in the photos.

This is how i understood to take the second derivative respected to time. I understood that dx and dy are also being effected.

I also show there the Geodesic equations i have calculated using the Christoffel symbols.
can you please explain what overall sign i lost in the Christoffel symbols?
 

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  • #6
I'll double check my Christoffel symbols.

The problem with your approach to the derivative of the area is that I think the limits of integration are time dependant, so you need to carry out the integral before computing the derivative, I think.
 
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  • #7
Rechecked my Christoffel symbols and you are correct - apologies. I missed that you'd taken the minus sign into the denominator.
 
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  • #8
Thank you very much for responding. i am not sure i understood what you are saying about the derivative of the area. I carry the integral all the way.
the integration of the integral should be:
for y: $$ -\sqrt{R^2-x^2}<y<\sqrt{R^2-x^2}$$
and for x: $$-R<x<R$$

That's how i understad it so farm, so what do you mean by saying they are time dependant?
 
  • #9
edoofir said:
i am not sure i understood what you are saying about the derivative of the area.
The point is that ##R## depends on ##t##, so I don't think ##\frac{d^2}{dt^2}\int\sqrt{g^{(2)}}dxdy## is the same as ##\int\frac{d^2}{dt^2}\sqrt{g^{(2)}}dxdy##.

In any case, do you have an expression for ##R(t)##?
 
  • #10
I understood it is not the same, this is why I used the derivation of a product rule and got the expression i got. (I attached a photo of how i got it).

Yes, i have expression for the Ricci tensors (if that's what you mean):

$$R_{tt} = \frac{2a^2(2a'^2 -aa''(a^4-1))}{(a^4-1)^2}$$
$$R_{xx} = R_{yy} = \frac{2a^2a'^2}{a^4-1}$$
$$R_{xy}=R_{yx} = a'^2 +aa''$$where $$a = a(t), a' = \frac{da}{dt}, a'' = \frac{d^2a}{dt}$$
 
  • #11
I'm struggling to read your screenshots on my phone. Deriving the geodesic equations myself, I now think that the particles do not change coordinates, so their positions always satisfy ##x^2+y^2=R^2## where ##R## is an arbitrary constant, nothing to do with any tensor.

I think you are supposed to be calculating the second derivative of the area enclosed by the ring of test particles, which is the region defined as ##x^2+y^2\leq R^2##. I don't understand why you don't do this integral directly - neither ##a## nor ##R## is a function of ##x## or ##y## so it looks to me to be trivial. Then you can simply differentiate; I assume this will lead you to something that looks like the Ricci scalar.
 

1. What is a geodesic?

A geodesic is the shortest path between two points on a curved surface. It is equivalent to a straight line on a flat surface.

2. How is a geodesic related to metric $ds^2$?

The metric $ds^2$ is a mathematical expression that describes the distance between two points on a curved surface. By using this metric, we can calculate the shortest path (geodesic) between two points on the surface.

3. What is the process for solving geodesics with metric $ds^2$?

The process involves using the metric $ds^2$ to calculate the geodesic equation, which is a set of differential equations. These equations can then be solved to find the geodesic path between two points on the surface.

4. What are some applications of solving geodesics with metric $ds^2$?

One application is in navigation, where geodesics can be used to determine the shortest path between two locations on a curved surface, such as the Earth. It is also used in physics and engineering to study the behavior of particles and objects moving on curved surfaces.

5. Are there any limitations to solving geodesics with metric $ds^2$?

One limitation is that the metric $ds^2$ may not accurately describe the curvature of highly complex surfaces. In these cases, alternative methods may need to be used to solve for geodesics. Additionally, the calculations involved in solving geodesics with metric $ds^2$ can be complex and time-consuming, making it difficult to apply in real-time situations.

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