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edoofir

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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?

$$ds^2 = -dt^2 +dx^2 +2a^2(t)dxdy + dy^2 +dz^2$$

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?

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