Gravitational time dilation from two or four bodies

[shalayka]

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 ($M = 1\cdot10^{25}, r = 1, t = 1$), I've got it down to:

$$\tau = t \cdot \sqrt{1 - \frac{2GM}{rc^2}} \approx 0.992546\;\;(eq.1)$$

For two bodies equidistant from the test particle (each contains half the total mass), I've tried the following:

$$\tau = t \cdot \sqrt{1 - \frac{2G(M/2)}{rc^2}} \cdot \sqrt{1 - \frac{2G(M/2)}{rc^2}} = t \cdot \left(1 - \frac{2G(M/2)}{rc^2} \right) \approx 0.992574\;\;(eq.2)$$

For four:

$$\tau = t \cdot \left(1 - \frac{2G(M/4)}{rc^2} \right)^2 \approx 0.992588\;\;(eq.3)$$

For six:

$$\tau = t \cdot \left(1 - \frac{2G(M/6)}{rc^2} \right)^3 \approx 0.992592\;\;(eq.4)$$

Is this correct, or should $\tau$ be constant regardless if the mass is split into 1, 2, 4 or 6 equidistant bodies? Ex:

$$\tau = t \cdot \sqrt{1 - \frac{2G({\rm Total\;Mass})}{({\rm Average\;Distance})c^2}} \approx 0.992546\;\;(eq.5)$$

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 $\frac{d\tau}{dr} = 0$ (no acceleration), and I was curious to know if $\tau = \rm constant$ if the total mass for the points and shell and ring are equal. If that is the case, then $(eq.5)$ is correct.

 

[MeJennifer]

I think in relation to what you do (multiplying gravitational time dilations) the term first-order approximation is a misnomer here. As general relativity is a non-linear theory this approach does not make any sense.

 

[shalayka]

I think in relation to what you do (multiplying gravitational time dilations) the term first-order approximation is a misnomer here. As general relativity is a non-linear theory this approach does not make any sense.

Do you know of an approach that makes sense for calculating the gravitational time dilation experienced by a test particle at the centre of a shell (of radius r)?

My common sense says that the gravitational time dilation would be identical to the situation where the shell was changed into a single point particle, and the test particle is at a distance of r. I'm just trying to find a way to show this mathematically I suppose.

Either way, thank you for confirming that the multiplication method does not work. That's what I was suspecting. It doesn't seem to make much sense that the calculation would be related to the number of individual bodies.

 

[Jorrie]

Is this correct, or should $\tau$ be constant regardless if the mass is split into 1, 2, 4 or 6 equidistant bodies? Ex:

$$\tau = t \cdot \sqrt{1 - \frac{2G({\rm Total\;Mass})}{({\rm Average\;Distance})c^2}} \approx 0.992546\;\;(eq.5)$$

I think your eq. 5 is not far off, except that you should rather sum the specific contributions inside the square root, e.g.

$$d\tau = dt \sqrt{1 - \frac{2G}{c^2}\left(\frac{m_1 }{r_1}+\frac{m_2 }{r_2} + ...\right)}$$

I've used this before, based upon its compatibility with the Schwarzschild metric, where gravitational and velocity time dilations can be summed like this inside the square root.

 

[MeJennifer]

I've used this before, based upon its compatibility with the Schwarzschild metric, where gravitational and velocity time dilations can be summed like this inside the square root.

Any references to the literature?

 

[Jorrie]

I think your eq. 5 is not far off, except that you should rather sum the specific contributions inside the square root, e.g.

$$d\tau = dt \sqrt{1 - \frac{2G}{c^2}\left(\frac{m_1 }{r_1}+\frac{m_2 }{r_2} + ...\right)}$$

I've used this before, based upon its compatibility with the Schwarzschild metric, where gravitational and velocity time dilations can be summed like this inside the square root.

Any references to the literature?

No, I've taken a "leap of faith" and I would like here the criticisms based on the math. The Schwarzschild metric can be written (geometric units):


$$d\tau^2 = \left(1-2m/r - \frac{dr^2}{(1-2m/r)dt^2} - \frac{r^2 d\psi^2}{dt^2}\right) dt^2$$

The first two terms represent the gravitational time dilation, while the last two terms represent radial and tangential (or transverse) velocity time dilations respectively. So, the "leap of faith" doesn't look totally unjustified...

 

[MeJennifer]

Then what do you mean by: "based upon its compatibility with the Schwarzschild metric"? How do you reason you can use the Schwarzschild metric to model multiple spatially dispersed point masses?

 

[Jorrie]

Then what do you mean by: "based upon its compatibility with the Schwarzschild metric"? How do you reason you can use the Schwarzschild metric to model multiple spatially dispersed point masses?

I did say it was a "leap of faith", which I only weakly motivated by the way the Schwarzschild metric's internal structure works.

I know that the general 2-body problem is not yet solved in GR. I was hoping that someone knows or could point to a solution for this highly restricted scenario - the gravitational time dilation (redshift factor dtau/dt as viewed from asymptotically flat space) due to two black holes (hypothetically) permanently at rest relative to each other. :-)