Integrating Geodesic Equations: Kevin Brown

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    Geodesic Integral
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SUMMARY

Kevin Brown's book "Reflections on Relativity" integrates two geodesic equations, yielding specific results for the derivatives of time and angle with respect to the affine parameter. The equations are: \(\frac{dt}{ds}=\frac{kr}{(r-2m)}\) and \(\frac{d\phi}{ds}=\frac{h}{r^{2}}\). These results are derived through a systematic approach involving logarithmic differentiation and the manipulation of the original geodesic equations. The discussion highlights the complexity of understanding these integrations and their implications in the context of general relativity.

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  • Understanding of general relativity concepts
  • Familiarity with differential equations
  • Knowledge of geodesic motion in curved spacetime
  • Proficiency in mathematical manipulation of equations
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  • Study the derivation of geodesic equations in general relativity
  • Learn about the physical significance of the constants \(k\) and \(h\)
  • Explore applications of geodesic equations in astrophysics
  • Investigate the role of affine parameters in curved spacetime
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Physicists, mathematicians, and students of general relativity seeking to deepen their understanding of geodesic equations and their applications in theoretical physics.

exmarine
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Kevin Brown, in his excellent book "Reflections on Relativity" p. 409, "immediately" integrates 2 geodesic equations:

[itex]\frac{d^{2}t}{ds^{2}}=-\frac{2m}{r(r-2m)}\frac{dr}{ds}\frac{dt}{ds}[/itex]

[itex]\frac{d^{2}\phi}{ds^{2}}=-\frac{2}{r}\frac{dr}{ds}\frac{d\phi}{ds}[/itex]

to get:

[itex]\frac{dt}{ds}=\frac{kr}{(r-2m)}[/itex]

[itex]\frac{d\phi}{ds}=\frac{h}{r^{2}}[/itex]

Does anyone understand that? I certainly don't.
 
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exmarine said:
[itex]\frac{d^{2}\phi}{ds^{2}}=-\frac{2}{r}\frac{dr}{ds}\frac{d\phi}{ds}[/itex]

[itex]\frac{d\phi}{ds}=\frac{h}{r^{2}}[/itex]
They both go pretty much the same way. For the second one,

[tex]\begin{eqnarray*}\frac{\frac{d^2 \phi}{ds^2}}{\frac{d \phi}{ds}} &=& - \frac{2}{r}\frac{dr}{ds}\\<br /> \frac{d}{ds}(\ln(\frac{d \phi}{ds})) &=& -2 \frac{d}{ds} \ln(r)\\<br /> \ln(\frac{d \phi}{ds}) &=& -2 \ln(r) + const\\<br /> \frac{d \phi}{ds} &=& \frac{h}{r^2}\end{eqnarray*}[/tex]
 
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