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- Thread starter Philosophaie
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- #1

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- #2

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Probably if you can write down your equations it will be much clearer what your problem actually is.

- #3

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Equation 1:

[tex]\frac{d}{d\tau}(\frac{2r}{r-2m}(\frac{dr}{d\tau})) + \frac{2m}{(r-2m)^{2}}(\frac{dr}{d\tau})^{2} - r(\frac{d\theta}{d\tau})^{2} - rsin^{2}\theta (\frac{d\phi}{d\tau})^{2} + \frac{mc^{2}}{r^{2}}(\frac{dt}{d\tau})^{2} = 0[/tex]

Equation 2:

[tex]\frac{d}{d\tau}(r^{2}\frac{d\theta}{d\tau}) - r^{2}sin\theta cos\theta (\frac{d\phi}{d\tau}) = 0[/tex]

Equation 3:

[tex]\frac{d}{d\tau}(r^{2}sin^{2}\theta (\frac{d\phi}{d\tau})) = 0[/tex]

Equation 4:

[tex]\frac{d}{d\tau}(\frac{r-2m}{r}(\frac{dt}{d\tau})) = 0[/tex]

where m and c are constants

[tex]\frac{d}{d\tau}(\frac{2r}{r-2m}(\frac{dr}{d\tau})) + \frac{2m}{(r-2m)^{2}}(\frac{dr}{d\tau})^{2} - r(\frac{d\theta}{d\tau})^{2} - rsin^{2}\theta (\frac{d\phi}{d\tau})^{2} + \frac{mc^{2}}{r^{2}}(\frac{dt}{d\tau})^{2} = 0[/tex]

Equation 2:

[tex]\frac{d}{d\tau}(r^{2}\frac{d\theta}{d\tau}) - r^{2}sin\theta cos\theta (\frac{d\phi}{d\tau}) = 0[/tex]

Equation 3:

[tex]\frac{d}{d\tau}(r^{2}sin^{2}\theta (\frac{d\phi}{d\tau})) = 0[/tex]

Equation 4:

[tex]\frac{d}{d\tau}(\frac{r-2m}{r}(\frac{dt}{d\tau})) = 0[/tex]

where m and c are constants

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- #4

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If I'm not mistaken, for the Schwarzchild metric, there is analytic solution.

But if you still insist on numerical solution you can try Rungge-Kutta fourth order. One of the easiest to program. But you need to rewrite your equations as system of first order DE of the form.

[tex]X'(\tau) = A(\tau) X(\tau)[/tex]

where X is a column vector and A is a matrix.

By the way why do you say your equation is 3rd order?

- #5

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Yes the equation is 2nd order not 3rd.

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- #6

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[tex]\frac{d}{d\tau}(r^{2}\frac{d\theta}{d\tau}) = 2r\frac{dr}{d\tau}\frac{d\theta}{d\tau} + r^{2}(\frac{d\theta}{d\tau})^{2}[/tex]

Is this correct?

- #7

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[tex]\frac{d}{d\tau}(r^{2}\frac{d\theta}{d\tau}) = A \cot (\theta) [/tex]

where A is the arbitrary constant from eq. 2

- #8

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Any suggestions on how to solve for r and [tex]\theta[/tex] from the first two equations if the solution from equation 3 is :

[tex]\frac{d\phi}{d\tau} = \frac{constant3}{r^{2}sin^{2}\theta}[/tex]

and equation 4 is:

[tex]\frac{dt}{d\tau} = \frac{(constant4)r}{r-2m}[/tex]

[tex]\frac{d\phi}{d\tau} = \frac{constant3}{r^{2}sin^{2}\theta}[/tex]

and equation 4 is:

[tex]\frac{dt}{d\tau} = \frac{(constant4)r}{r-2m}[/tex]

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- #9

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- #10

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- #11

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tau takes the place of s in geodesic equations and is called proper time in a "local free-fall frame replacing s by:

[tex]ds^{2} = -c^{2}d\tau^{2}[/tex]

[tex]ds^{2} = -c^{2}d\tau^{2}[/tex]

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