Satellite orbiting around Earth - Spacetime Metric

  • #1
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Homework Statement



The metric near earth is ##ds^2 = -c^2 \left(1-\frac{2GM}{rc^2} \right)dt^2 + \left(1+\frac{2GM}{rc^2} \right)\left( dx^2+dy^2+dz^2\right)##.

(a) Find all non-zero christoffel symbols for this metric.
(b) Find satellite's period.
(c) Why does ##R^i_{0j0} \simeq \partial_j \Gamma^i_{00} - \partial_0 \Gamma^i_{0j} = \frac{1}{2}\partial_i \partial_j g_{00}##?
(d)Why does the separation grow in time?

2012_B5_Q1.png



Homework Equations




The Attempt at a Solution



Part(a)
Let lagrangian be ##-c^2 d\tau^2 = -c^2 \left(1-\frac{2GM}{rc^2} \right) \dot {t}^2 + \left(1+\frac{2GM}{rc^2}\right)(\dot x^2 + \dot y^2 + \dot z^2)##. The corresponding geodesic equations are
[tex] \ddot t + \frac{\left( \frac{GM}{r^2c^2} \right)}{1-\frac{2GM}{rc^2}} \dot r \dot t = 0[/tex]
[tex]\ddot r - \frac{\left( \frac{GM}{r^2c^2} \right)}{1+\frac{2GM}{rc^2}} (\dot r)^2 + \frac{\left( \frac{GM}{r^2} \right)}{1+\frac{2GM}{rc^2}} (\dot t)^2 = 0 [/tex]
The christoffel symbols are given by ##\Gamma^t_{rt} = \Gamma^t_{tr} = \frac{\left( \frac{GM}{r^2c^2} \right)}{1-\frac{2GM}{rc^2}}, \Gamma^r_{rr} = -\frac{\left( \frac{GM}{r^2c^2} \right)}{1+\frac{2GM}{rc^2}}, \Gamma^r_{tt} = \frac{\left( \frac{GM}{r^2} \right)}{1+\frac{2GM}{rc^2}} ##.

Part (b)
Given ##x=R \cos(\omega \tau)## and ##y = R \sin (\omega \tau)## and ##z=0## we have ##dx^2 + dy^2 + dz^2 = (R\omega)^2 d\tau^2##. The metric now becomes
[tex]-c^2 d\tau^2 = -c^2 \left(1-\frac{2GM}{Rc^2} \right)dt^2 + \left(1+\frac{2GM}{Rc^2} \right)\left( (R\omega)^2 d\tau^2 \right)[/tex]
This relates the time duration on earth ##dt## with proper time on satellite ##d\tau##.
[tex]dt = \sqrt{\frac{1+ \left( \frac{R\omega}{c}\right)^2 \left( 1 + \frac{2GM}{Rc^2} \right) }{1 - \frac{2GM}{Rc^2}}} d\tau [/tex]

Part(c)
Under the approximation ##\frac{GM}{rc^2} \ll 1## the christoffel symbols become ##\Gamma^t_{rt} \approx \frac{GM}{r^2c^2}, \Gamma^r_{rr} \approx -\frac{GM}{r^2c^2}, \Gamma^r_{tt} \approx \frac{GM}{r^2}##.
Not sure why ##R^i_{0j0} \simeq \partial_j \Gamma^i_{00} - \partial_0 \Gamma^i_{0j} = \frac{1}{2}\partial_i \partial_j g_{00}## holds. Is there some trick here?

Regardless, we have
[tex]\frac{1}{2}\partial_i \partial_j g_{00} = -\frac{GM}{c^2} \partial_i \partial_j \left[ (x_kx_k)^{-\frac{1}{2}}\right][/tex]
[tex] = \frac{GM}{c^2} \partial_i \left[ (x_kx_k)^{-\frac{3}{2}} x_j \right][/tex]
[tex]= \frac{GM}{c^2} \left[ \delta_{ij} (x_kx_k)^{-\frac{3}{2}}-3(x_kx_k)^{-\frac{5}{2}} x^i x^j \right][/tex]
[tex] \frac{1}{2}\partial_i \partial_j g_{00} = \frac{GM}{c^2 r^3} \left[ \delta_{ij} -3\frac{x^i x^j}{r^2} \right][/tex]

Part(d)
I suppose one is nearer to the earth and experiences a stronger field than other. Thus the one closer to earth experiences a greater acceleration, so the spatial distances between them increases.
 
Last edited:

Answers and Replies

  • #3
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bumpp on part (b) - satellite's period. The time experienced on earth is ##dt## while proper period is ##d\tau##, so is the period of the satellite observed on earth simply
[tex]T = \sqrt{\frac{1+ \left( \frac{R\omega}{c}\right)^2 \left( 1 + \frac{2GM}{Rc^2} \right) }{1 - \frac{2GM}{Rc^2}}} \left(\frac{2\pi}{\omega}\right)[/tex]

?
 
  • #4
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bump on period in (b)
 
  • #5
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solved.
 

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