- #1
shalayka
- 126
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How would one go about calculating (as a first-order approximation) the gravitational time dilation generated by multiple point sources?
When generated by one point source ([itex]M = 1\cdot10^{25}, r = 1, t = 1[/itex]), I've got it down to:
[tex]
\tau = t \cdot \sqrt{1 - \frac{2GM}{rc^2}} \approx 0.992546\;\;(eq.1)
[/tex]
For two bodies equidistant from the test particle (each contains half the total mass), I've tried the following:
[tex]
\tau = t \cdot \sqrt{1 - \frac{2G(M/2)}{rc^2}} \cdot \sqrt{1 - \frac{2G(M/2)}{rc^2}} = t \cdot \left(1 - \frac{2G(M/2)}{rc^2} \right) \approx 0.992574\;\;(eq.2)
[/tex]
For four:
[tex]
\tau = t \cdot \left(1 - \frac{2G(M/4)}{rc^2} \right)^2 \approx 0.992588\;\;(eq.3)
[/tex]
For six:
[tex]
\tau = t \cdot \left(1 - \frac{2G(M/6)}{rc^2} \right)^3 \approx 0.992592\;\;(eq.4)
[/tex]
Is this correct, or should [itex]\tau[/itex] be constant regardless if the mass is split into 1, 2, 4 or 6 equidistant bodies? Ex:
[tex]
\tau = t \cdot \sqrt{1 - \frac{2G({\rm Total\;Mass})}{({\rm Average\;Distance})c^2}} \approx 0.992546\;\;(eq.5)
[/tex]
I've always wondered about this.
The reason I wonder is because I know that in between two points of equal mass (or inside of a thin homogeneous ring or shell) it is found that [itex]\frac{d\tau}{dr} = 0[/itex] (no acceleration), and I was curious to know if [itex]\tau = \rm constant[/itex] if the total mass for the points and shell and ring are equal. If that is the case, then [itex](eq.5)[/itex] is correct.
When generated by one point source ([itex]M = 1\cdot10^{25}, r = 1, t = 1[/itex]), I've got it down to:
[tex]
\tau = t \cdot \sqrt{1 - \frac{2GM}{rc^2}} \approx 0.992546\;\;(eq.1)
[/tex]
For two bodies equidistant from the test particle (each contains half the total mass), I've tried the following:
[tex]
\tau = t \cdot \sqrt{1 - \frac{2G(M/2)}{rc^2}} \cdot \sqrt{1 - \frac{2G(M/2)}{rc^2}} = t \cdot \left(1 - \frac{2G(M/2)}{rc^2} \right) \approx 0.992574\;\;(eq.2)
[/tex]
For four:
[tex]
\tau = t \cdot \left(1 - \frac{2G(M/4)}{rc^2} \right)^2 \approx 0.992588\;\;(eq.3)
[/tex]
For six:
[tex]
\tau = t \cdot \left(1 - \frac{2G(M/6)}{rc^2} \right)^3 \approx 0.992592\;\;(eq.4)
[/tex]
Is this correct, or should [itex]\tau[/itex] be constant regardless if the mass is split into 1, 2, 4 or 6 equidistant bodies? Ex:
[tex]
\tau = t \cdot \sqrt{1 - \frac{2G({\rm Total\;Mass})}{({\rm Average\;Distance})c^2}} \approx 0.992546\;\;(eq.5)
[/tex]
I've always wondered about this.
The reason I wonder is because I know that in between two points of equal mass (or inside of a thin homogeneous ring or shell) it is found that [itex]\frac{d\tau}{dr} = 0[/itex] (no acceleration), and I was curious to know if [itex]\tau = \rm constant[/itex] if the total mass for the points and shell and ring are equal. If that is the case, then [itex](eq.5)[/itex] is correct.
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