Gravitational time dilation on a non-spherical object

[Dr Chaos]

A radius is required for this relativistic formula:
t=\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}
However, I notice that some objects are not spherical. Because of this, I am unsure of how to calculate gravitational time dilation for different shapes.

Could somebody help me with this and provide examples for different shapes (like a brick or a heavy stop sign)?

 

[FunkyDwarf]

You would probably need to generate the energy momentum tensor for your object, plug through the field equations to try get a metric and go from there.

 

[Passionflower]

A radius is required for this relativistic formula:
t=\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}
However, I notice that some objects are not spherical. Because of this, I am unsure of how to calculate gravitational time dilation for different shapes.

Could somebody help me with this and provide examples for different shapes (like a brick or a heavy stop sign)?

I am not sure what you are trying to do here. Calculating the gravitational time dilation for a body is very difficult and depends on more things than just the spatial dimensions of the body.

 

[Dr Chaos]

I am not sure what you are trying to do here. Calculating the gravitational time dilation for a body is very difficult and depends on more things than just the spatial dimensions of the body.

What I am looking for is the gravitational time dilation on the surface of a stationary body. Assume the body has uniform density. What other information is required?

 

[Jonathan Scott]

What I am looking for is the gravitational time dilation on the surface of a stationary body. Assume the body has uniform density. What other information is required?

The fractional time dilation is the the same as the Newtonian gravitational potential (potential energy per unit mass) divided by c².

Provided that the body in question is not as dense as a neutron star, you can just calculate the Newtonian potential by integrating the density divided by the distance from each point of the body in the usual way. For a non-spherical body, the potential normally varies with location on the surface. For a spherical body, it is the usual -Gm/rc².

 

[Passionflower]

The fractional time dilation is the the same as the Newtonian gravitational potential (potential energy per unit mass) divided by c².

Actually the second part of your posting denies this.

I would rather state:

In general relativity the fractional time dilation is not the same as the Newtonian gravitational potential (potential energy per unit mass) divided by c². However in the weak field limit it is true because here general relativity becomes equivalent with Newtonian gravity.

I think it is very interesting to study the Weyl form to see the "function" of the Newtonian potential in general relativity. (function clearly between quotes).

 

[D H]

However in the weak field limit it is true because here general relativity becomes equivalent with Newtonian gravity.

Given your assumption in post #4 ("Assume the body has uniform density"), the weak field limit most definitely applies. The only objects that can have a uniform density must necessarily be small and have negligible mass. A brick, for example. Even planets don't have uniform density, let alone stars or neutron stars. So, for an object of uniform density but of a non-spherical shape, calculating the Newtonian gravitational potential on the surface of the object will work just fine.

 

[George Jones]

Given your assumption in post #4 ("Assume the body has uniform density"), the weak field limit most definitely applies. The only objects that can have a uniform density must necessarily be small and have negligible mass.

There exists a spherically symmetric, constant-density exact solution to Einstein's equation in general relativity that does not have to have negligible mass, and for which the weak-field limit does not necessarily apply.

 

[D H]

You're apparently talking about singularities, George. A black hole will have a fairly simple geometric shape. Is there a brick-shaped solution to Einstein's field equations for black holes?

 

[George Jones]

You're apparently talking about singularities, George. A black hole will have a fairly simple geometric shape. Is there a brick-shaped solution to Einstein's field equations for black holes?

No, I wasn't talking about black holes or singularities or bricks. Schwarzschild black holes are vacuum solutions to Einstein's equation, and have zero (local) density everywhere, even inside event horizons. I gave an explicit counter-example in general relativity (but not the real world) to the general statements

constant density ==> weak limit applies,

Given your assumption in post #4 ("Assume the body has uniform density"), the weak field limit most definitely applies. The only objects that can have a uniform density must necessarily be small and have negligible mass.

General relativity allows spherical material objects which have non-zero constant-density, and for which the weak field limit does not apply.

Certainly, the weak field limit applies to a real brick.

A brick, for example. Even planets don't have uniform density, let alone stars or neutron stars. So, for an object of uniform density but of a non-spherical shape, calculating the Newtonian gravitational potential on the surface of the object will work just fine.

 

[TCS]

No, I wasn't talking about black holes or singularities or bricks. Schwarzschild black holes are vacuum solutions to Einstein's equation, and have zero (local) density everywhere, even inside event horizons. I gave an explicit counter-example in general relativity (but not the real world) to the general statements

constant density ==> weak limit applies,



General relativity allows spherical material objects which have non-zero constant-density (constant stress energy tensor), and for which the weak field limit does not apply.

Certainly, the weak field limit applies to a real brick.

I think that you could model each shell of an atom as a constant density spehereically symmetric solution to Einstein's equation for an atom. although, I guess that electrons would create density variations in the shell and not all atoms spherical.

 

[Dr Chaos]

There seems to be some discrepancy in regard to how this is calculated; could somebody please provide an example for me?

 

[TCS]

A radius is required for this relativistic formula:
t=\frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}
However, I notice that some objects are not spherical. Because of this, I am unsure of how to calculate gravitational time dilation for different shapes.

Could somebody help me with this and provide examples for different shapes (like a brick or a heavy stop sign)?

You can do it numericly. Create a mesh, assume some linear approximations within each cell of the mesh, solve within the cell, and match the conditions at the boundaries of the cells. If your cells are small enough, you should iterate to a solution.


The book at the following link provides a discusion of the common numerical methods.

http://books.google.com/books?id=Uk...Bw#v=onepage&q=runge kutta relativity&f=false

This paper is better. the other link isn't the full book.

http://www.tesisenxarxa.net/TESIS_UIB/AVAILABLE/TDX-0923109-130054//tdda1de1.pdf

 

[Jonathan Scott]

There seems to be some discrepancy in regard to how this is calculated; could somebody please provide an example for me?

As I said before, there's no point in using anything other than a Newtonian approximation for any ordinary object. The Newtonian potential at a reference point due to an object is the volume integral over the object of -G times the mass density divided by the distance from the reference point, but I haven't got the patience to set it in LaTeX right now.

In my earlier post, I intended to be exact, but my terminology may have been non-standard. I meant that "effective Newtonian potential" in a static field (that is, the potential energy per unit mass) divided by c² is the same thing as the fractional difference in time rate compared with infinity. However, the calculation of that potential only reduces to the usual Newtonian expression of sum(-Gm/r) in the weak approximation.

 

[Dr Chaos]

As I said before, there's no point in using anything other than a Newtonian approximation for any ordinary object. The Newtonian potential at a reference point due to an object is the volume integral over the object of -G times the mass density divided by the distance from the reference point, but I haven't got the patience to set it in LaTeX right now.

In my earlier post, I intended to be exact, but my terminology may have been non-standard. I meant that "effective Newtonian potential" in a static field (that is, the potential energy per unit mass) divided by c² is the same thing as the fractional difference in time rate compared with infinity. However, the calculation of that potential only reduces to the usual Newtonian expression of sum(-Gm/r) in the weak approximation.

I'm sorry but I still do not understand. Please forgive my limited physics knowledge but I would still like an example.

 

[TCS]

I'm sorry but I still do not understand. Please forgive my limited physics knowledge but I would still like an example.

Why don't you draw the biggest sphere that fits inside, calculate the dialtion for that sphere, then inscribe spheres in each the remaining sections and calculate and correct, and repeat.

 

[Dr Chaos]

Why don't you draw the biggest sphere that fits inside, calculate the dialtion for that sphere, then inscribe spheres in each the remaining sections and calculate and correct, and repeat.

That would probably work but it seems very tedious - I would have to keep drawing spheres until I decide that they are small enough, and then calculate each individual sphere's mass and then their time dilation and then multiply it all together. Is there an easier way?

 

[Jonathan Scott]

I'm sorry but I still do not understand. Please forgive my limited physics knowledge but I would still like an example.

This is just Newtonian gravity and integration.

If the point where the potential is being measured is the origin, then the potential is given by the following integral over the object:

phi = triple integral of (- G rho/sqrt(x^2+y^2+z^2)) dx dy dz

where rho is the mass per unit volume.

The time dilation relative to infinity is then

(1 + phi/c^2)

I tried entering this using LaTeX, but I couldn't get it to work at all today.

 

[Dr Chaos]

I tried entering this using LaTeX, but I couldn't get it to work at all today.

Could you please post the code? I can always use a LaTeX sandbox to view it properly.

\phi=\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}} <== is this it?

And what is the full time dilation formula?

 

[espen180]

And what is the full time dilation formula?

There is none.

The way the gravitational time dilation is obtained is to solve Einstein's Equation (google it) and obtain a metric, from which it can be obtained. If I were to give a "general formula", it would have to be \tau=\frac{\sqrt{g_{tt}}}{c}t where \tau is proper time (measured locally) and t is coordinate time (measured from infinity).

 

[Jonathan Scott]

Could you please post the code? I can always use a LaTeX sandbox to view it properly.

\phi=\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}} <== is this it?

And what is the full time dilation formula?

That's right apart from missing dx dy dz (or dV where V is volume). I hope this is right:

\phi=\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}} \, dx \, dy \, dz

The relative time rate of a clock at that location relative to a clock a long way away is then given by the other formula I tried to post:

(1 + \phi/c^2)

Note that \phi is negative, so a clock deeper in the potential runs slower. The simple case of a spherical object this reduces to the usual expression:

(1 - Gm/rc^2)

(I see LaTeX is working again, although I have to refresh to get the right stuff in preview).

 

[Jonathan Scott]

There is none.

The way the gravitational time dilation is obtained is to solve Einstein's Equation (google it) and obtain a metric, from which it can be obtained. If I were to give a "general formula", it would have to be \tau=\frac{\sqrt{g_{tt}}}{c}t where \tau is proper time (measured locally) and t is coordinate time (measured from infinity).

Yes, of course that's the general solution, but provided you're not close to something like a neutron star you can get an accurate enough result from the Newtonian approximation:

\frac{\sqrt{g_{tt}}}{c} \approx 1 + \phi/c^2

where \phi is the Newtonian potential, typically \sum_i -Gm_i/r_i.

 

[espen180]

Yes, of course that's the general solution, but provided you're not close to something like a neutron star you can get an accurate enough result from the Newtonian approximation:

\frac{\sqrt{g_{tt}}}{c} \approx 1 + \phi/c^2

where \phi is the Newtonian potential, typically \sum_i -Gm_i/r_i.

Yes, I understand that, but wouldn't it be \tau\approx t\sqrt{1+\frac{\phi}{c^2}} ? Or maybe I'm mistaken.

 

[starthaus]

Yes, I understand that, but wouldn't it be \tau\approx t\sqrt{1+\frac{\phi}{c^2}} ? Or maybe I'm mistaken.

You are off by afactor of 2, the connection between proper and coordinate time is really d\tau=dt\sqrt{1-\frac{r_s}{r}}, while the time dilation between two arbitrary observers is more complicated than your formula. Note that \frac{r_s}{r}=\frac{2\Phi}{c^2}.See https://www.physicsforums.com/blog.php?b=1954 for the case of a spherically symmetric case.
Having said the above, I see absolutely no reason why the formulas derived for spherically symmetric bodies would apply for a cube. The right thing to do is to derive the metric for the cube and to derive time dilation from the metric, the way one derives this for a spherically symmetric body (see post 2 by FunkyDwarf). I am not sure that g_{tt}=1+\frac{2\Phi}{c^2} for a cube.

 

[espen180]

You are off by afactor of 2, the connection between proper and coordinate time is really d\tau=dt\sqrt{1-\frac{r_s}{r}}, while the time dilation between two arbitrary observers is more complicated than your formula. Note that \frac{r_s}{r}=\frac{2\Phi}{c^2}.See https://www.physicsforums.com/blog.php?b=1954 for the case of a spherically symmetric case.
Having said the above, I see absolutely no reason why the formulas derived for spherically symmetric bodies would apply for a cube. The right thing to do is to derive the metric for the cube and to derive time dilation from the metric, the way one derives this for a spherically symmetric body (see post 2 by FunkyDwarf). I am not sure that g_{tt}=1+\frac{2\Phi}{c^2} for a cube.

We are discussing weak field approximations. No need to derive any metric.

 

[starthaus]

We are discussing weak field approximations. No need to derive any metric.

Well, at least get your formulas correct for the weak field, yours is wrong by a factor of 2.

 

[Dr Chaos]

Thanks guys

So the formulas are:

\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}}\, dx\, dy\, dz

and

t=\frac{1}{\sqrt{1+\frac{\phi}{c^2}}}

What is "d" in the first formula?

 

[Jonathan Scott]

Thanks guys

So the formulas are:

\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}}\, dx\, dy\, dz

and

t=\frac{1}{\sqrt{1+\frac{\phi}{c^2}}}

The second is wrong by a factor of 2 and a missing term for the local time, and should be something like:

\frac{\partial \tau}{\partial t} \approx \sqrt{1+\frac{2 \phi}{c^2}} \approx 1+\frac{\phi}{c^2}

where \tau is the proper (local) time and t is the coordinate time.

What is "d" in the first formula?

It's the usual Leibniz notation used in differential and integral calculus for an infinitesimal change in the quantity.

 

[espen180]

Thanks guys

So the formulas are:

\iiint\frac{-G\rho}{\sqrt{x^2+y^2+z^2}}\, dx\, dy\, dz

and

t=\frac{1}{\sqrt{1+\frac{\phi}{c^2}}}

What is "d" in the first formula?

"dx" is an infinitessimal increment of x. etc.

The triple integral sums together all the infinitessimal volume elements dxdydz=dV.

 

[Dr Chaos]

Can somebody provide an example with what we have established so far? Just use a brick so I can see how it is done. Thanks.

 

[Jonathan Scott]

Can somebody provide an example with what we have established so far? Just use a brick so I can see how it is done. Thanks.

I don't know a quick answer for this integral; it's rather messy and best handled by mathematical software nowadays.

You could for example use the Wolfram Online integrator to get each of the indefinite integrals in the triple integral and substitute the result into the next one (switching the variable names round cyclically if using the online integrator, as it always assumes dx). For a brick, you could substitute the limits as you do each integration, as the x, y and z limits are independent.

 

[Jonathan Scott]

well this works for metrics like the Reissner Nordstrom metric, but only for static metrics, from what I understand:
in the case of the Kerr metric, there is frame dragging, and there is not only dt^2 in the metric but dt.dphi too

furthermore, at the event horizon of a Kerr black hole, g_rr -> infinite often without g_tt =0
this means that space is infinitely contracted but time is not infinitely dilated, if only g_tt provides the time dilation.
in such case the factor of the time dilation would not be the factor of the length contraction.

in other words, the event horizon is not always the ergosphere

I assumed that this thread was referring to static situations (and outside any event horizon). If the object is at rest relative to the source and the coordinates themselves are static, I think the dt^2 term is all that is needed.

 

[George Jones]

I have the posts discussing "time dilation" and "length contration" for rotating black holes to the separate thread

https://www.physicsforums.com/showthread.php?t=416147

that relativity fan started.

 

[Jonathan Scott]

I've spotted a paper "Gravitational potential and energy of homogeneous rectangular parallelepiped" on the ArXiv which calculates the Newtonian potential for various shapes including a cube and a rectangular shape.

The PDF is at http://arxiv.org/pdf/astro-ph/0002496v1".

 

[Jonathan Scott]

Here is an example of a comparison of a cube and a sphere based on that paper:

Consider a cube of side 2a with density \rho.

The total volume of the cube is 8 a^3 so its mass is m = 8 a^3 \rho.

The potential at a vertex (corner) of the cube is given in the paper (equation 12) as a multiple of a^2 G \rho (that is, Gm/8a) as follows:

{12 \, \ln \left (\frac{\sqrt{3} + 1}{\sqrt{2}} \right ) - \pi } \approx 4.76

Now consider the potential due to a sphere of the same mass at the same distance from the center as the corner of the cube, that is a \sqrt{3}, in the same units:

\frac{8}{\sqrt{3}} \approx 4.62

This shows that the gravitational effect of a cube is not very different from the effect of a sphere of the same mass produced by smoothing off the corners and pushing the excess material onto the faces.

(The paper seems to quietly ignore the usual convention that the potential expression -Gm/r has a minus sign).