# I Exterior Schwarzchild Solution/Gravitational Time Dilation

1. Jun 30, 2016

### space-time

I've been looking in depth into the exterior Schwarzchild solution of the Einstein field equations. I decided to play with the line element by creating some example scenarios for myself and calculating the space-time interval between two events in these scenarios. Here is the line element:

ds2 = [1- (2GM/rc2)]c2dt2 - dr2/[1- (2GM/rc2)] - r2(dθ2 + sin2(θ)d∅2)

Now, here is the first example problem that I created:

Let's say that there is a perfectly spherical moon in a vaccum. The moon has mass M = 100 kg, and it has radius r = 6.58765432 * 10-34 Ls (Ls stands for light seconds). Now, we have two massless observers (I made them massless just for simplicity of the problem). We will call these observers Mario and Luigi. Now let's say that Mario stands on top of this tiny moon, so his distance from the moon's center of mass is r = 6.58765432 * 10-34 Ls. His angular coordinates with respect to the center of the moon are θ = 0, and ∅ = π/2.

Now, Luigi floats somewhere off in space. His distance from the moon's center of mass is r = 4 Ls. His angular coordinates with respect to the moon's center are the exact same as Mario's.

If Luigi's watch reads that 20 seconds have passed, then how much time has passed from Mario's perspective?

Now here is my answering process:

Note: I used units where c = 1 Ls/s, G = 2.47037037 * 10-36 Ls3/ (Kg * s2)

Using this line element:
ds2 = [1- (2GM/rc2)]c2dt2 - dr2/[1- (2GM/rc2)] - r2(dθ2 + sin2(θ)d∅2)

I just plugged in r = 6.58765432 * 10-34 (Mario's distance from the moon's center) , dt =20 (the amount of time Luigi counted) and dr = the difference between Luigi's r coordinate and Mario's r coordinate, which ≈ 4. θ, dθ,and d∅ should all be 0 since Mario and Luigi are at the exact same angular coordinates with respect to the center of the moon.

The final result comes out to be ds2 = dτ2 ≈ 36
The square root of 36 is 6, so this means that Mario should see a time of about 6 seconds when Luigi counts 20 seconds.

a) Is this result correct? I just want to make sure I have an accurate understanding of things here.
b) I noticed that this result differs from what I'd get using the gravitational time dilation formula. If I were to manipulate the formula t = τ /sqrt(1 - (2GM/rc2)) where t is the time that an external observer sees while τ is the proper time that the person in the gravitational field (Mario in this case) would see, then I could plug in t = 20 seconds and arrive at the result τ = 10 seconds (meaning Mario would see 10 seconds). However, upon plugging dt = 20 seconds into the line element, I noticed that 10 seconds is what I would get only if dr, dθ, and d∅ were set to 0. Would I be correct in surmising that the gravitational time dilation formula assumes that dr, dθ, and d∅ are all 0?

If a) is not correct, then please tell me where I went wrong.

I actually have some more questions pertaining to space-like intervals (I made sure the above example had a time-like interval), but this thread is getting kind of long and I am getting tired of typing, so I will just make a part 2 to this thread at a later time.

2. Jun 30, 2016

### Staff: Mentor

It's a lot more convenient to use units in which $G = 1$ as well as $c = 1$. These units are referred to as "geometric units". In these units mass will have the same units as length, so all masses will be in light-seconds. Your value of $G$ here is just the conversion factor from ordinary mass units (kilograms) to geometric units--in other words, your moon in these units will have a mass of $2.47 \times 10^{-34}$ light-seconds.

The reason this is helpful is that it makes it easy to see how close the actual radius of your massive body is to its Schwarschild radius, which is just twice the mass in geometric units. So your moon has a Schwarzschild radius of $4.94 \times 10^{-34}$. Note that this isn't much smaller than the actual radius of the moon. See further comments below.

No. You have the right value for $r$, but $dr$ is zero for the computation you are trying to do. What you are trying to compute is the "length" (elapsed proper time) along Mario's worldline (the curve describing his history in spacetime) between two values of coordinate time $t$ that differ by 20 (the time Luigi counted--see further comments on this below). $dr$ being nonzero would mean that Mario changed his radius during this time; but he didn't, he stayed on the moon's surface.

Also, it's not really true that the value of $dt$ is 20. What is true is that you are integrating a function along a curve of constant $r$ between two values of $t$ that differ by 20. So the right answer will be an integral.

Finally, the answer we want is $\tau$, but the line element, when integrated, would give the square of this, so we really want its square root. So the right answer is given by (using geometric units as I described above):

$$\tau = \int_{0}^{20} \sqrt{1 - \frac{2M}{r}} dt = 20 \sqrt{1 - \frac{2M}{r}}$$

(Note that, since $r$ is constant, not a function of $t$, we can just pull that factor out of the integral.) We have $2M = 4.94 \times 10^{-34}$ and $r = 6.58765432 \times 10^{-34}$, so the square root factor becomes $\sqrt{1 - 0.75} = 0.5$. So the final answer is $20 \times 0.5 = 10$; Mario should see 10 seconds elapse while Luigi sees 20 seconds elapse.

Yes, that's because the gravitational time dilation formula is the same as the one I just derived above. Hopefully the above explains why that formula is correct.

One other note: strictly speaking, we should calculate the time dilation formula for Luigi as well, since he is at a finite radial coordinate, so the 20 seconds he sees elapse is not exactly the same as 20 seconds of coordinate time, which is what we are supposed to use in the integral above. However, if you compute the factor $\sqrt{1 - 2M / r}$ for Luigi, you will see that it is so small (of order $10^{-34}$) that we can neglect it, as I did in the computation above.

3. Jun 30, 2016

### space-time

Thank you for this helpful info!

So from what I understand, dr, dθ and d∅ would only be non-zero if Mario ever changed his position? Is that what you are saying?

Also, you said that dt wouldn't exactly be 20. If that is the case, then what exactly would I plug into dt when using the line element for problems like this? Furthermore, if τ = ∫t0t1 sqrt(gμνdxμdxν) dt , then what information would just ds2 itself give you (without integrating)?(By the way, that t1 in the integral above is just the upper bound. I'm saying this because it looks a little off to the side).

4. Jun 30, 2016

### Staff: Mentor

That's not what I said. What I said is that you have to do an integral to get the answer, and $dt$ is a differential that gets put under the integral sign. Look again at the integral I wrote in post #2 (which appears in what you quoted). Notice how $dt$ appears in it. Notice also the limits of integration--that's where the 20 comes in.

In a more complicated situation in which Mario was moving, to do the integral the same way, we would have to know how Mario's other coordinates varied with coordinate time. For example, if Mario were moving radially, we would need to know $dr/dt$, the rate of change of his $r$ coordinate with respect to his $t$ coordinate. Or we would have to have some other way of relating the different coordinate differentials so that we end up with an integral over one variable.

5. Jun 30, 2016

### Staff: Mentor

This is not correct. If you are writing the integral in this form, it is just

$$\tau = \int \sqrt{g_{\mu \nu} dx^\mu dx^\nu}$$

and you can't put limits of integration on until you have an expression in just one integration variable, as I said in my previous post. As it stands, the expression $g_{\mu \nu} dx^\mu dx^\nu$ is just the line element $ds^2$; for example, the RHS of the expression for $ds^2$ that you gave in the OP of this thread. But that means there are multiple variables--multiple $dx^\mu$'s--and you have to find functions for some of them in terms of others, like the $dr/dt$ I referred to in my last post.

6. Jun 30, 2016

### MeJennifer

Depends on what you mean by floating in space.
When r stays constant I would not call that floating.

Obviously unless he uses some kind of a rocket motor he will not stay with a fixed r during the 20 second clock measurement. Alternatively his r value could remain constant if he were in a stable circular orbit but such an orbit is only possible at one single r value in a Schwarzschild solution.

The problem becomes more interesting if Luigi would start to radially free fall from a stationary position and would measure 20 seconds on his watch.

7. Jun 30, 2016

### Staff: Mentor

This can't be right as you state it; a stable circular orbit is possible at any value of $r$ greater than 3 times the Schwarzschild radius.

8. Jun 30, 2016

### Staff: Mentor

This is a good point in general. However, in practical terms it won't make a difference with the numbers the OP has given. I encourage both the OP and you to calculate:

(a) What Luigi's proper acceleration has to be to hold him at a fixed $r$ of 4 light-seconds;

(b) How far Luigi would free-fall in 20 seconds by his own clock, starting at an $r$ of 4 light-seconds, if he had zero proper acceleration.

9. Jun 30, 2016

### MeJennifer

Oops, sorry I meant of to say minimum, indeed greater or equal to 3 times the Schwarzschild radius.

10. Jul 2, 2016

### space-time

Sorry to reply so late. Anyway, I think I am starting to understand it better now, but one thing still confuses me.

Let's say for instance that we consider a slight modification to the problem I presented in the OP. In this alternate scenario, Luigi still stays where he is and counts 20 seconds. Mario however, decides during this time that he wants to change his radial distance to the moon's center from his original r = 6.58765432 * 10-34 Ls to r = 3.29382716 * 10-33 Ls. He makes this radial change at a rate so that r(t) = (6.58765432 * 10-34)t + (6.58765432 * 10-34 ). This of course also means that dr/dt = 6.58765432 * 10-34 . The whole process of Mario moving takes 4 seconds of coordinate time t (time from Luigi's perspective). Under these circumstances, how much time does Mario experience when Luigi counts out t = 20 seconds? If it would make the problem easier or make the problem make more sense, how much time does Mario experience when Luigi counts t = 4 seconds (since that is the time it took Mario to move from Luigi's perspective)?

I'd greatly appreciate if you could help me understand what to do in this situation. The reason why this confuses me is because (whether or not this is what you meant in your first reply), your first reply kind of gave me the impression that you are not supposed to actually plug into the differential terms in the line element (dt, dr, dθ, d∅). Rather, you are supposed to choose one of these terms to integrate over, and express the other differential terms as functions of the differential term that you integrate over.

Also, your reply did not make it seem like Luigi's distance from the center of the moon had any influence on how fast or slow Mario's time ran relative to Luigi's. This doesn't make sense to me because an observer's clock gets slower and slower relative to external observers as they enter more intense gravitational fields. As Luigi gets further and further from the moon, the moon's gravitational effects on Luigi should get weaker and weaker. Therefore, relative to Luigi, the gravitational effects on Mario should seemingly be continually getting stronger as Luigi gets further away (which should mean that Mario's flow of time should be moving slower and slower relative to Luigi). The gravitational time dilation formula however, doesn't take into account the external observer's position (Luigi's position in this case). It only accounts for Mario's position.

11. Jul 2, 2016

### Staff: Mentor

This is the only part that requires any new calculation, so we'll do that one. See below.

That's correct. Here's how it goes:

$$\tau = \int \sqrt{ds^2} = \int \sqrt{ \left( 1 - \frac{2M}{r} \right) dt^2 - \frac{1}{1 - 2M / r} dr^2} = \int dt \sqrt{ \left( 1 - \frac{2M}{r} \right) - \frac{1}{1 - 2M / r} \left( \frac{dr}{dt} \right)^2}$$

I'll use $K$ to denote the value $6.58765432 \times 10^{-34}$, so we have $dr/dt = K$ and $r = K + Kt$. This gives:

$$\tau = \int_0^4 \sqrt{1 - \frac{2M}{K + Kt} - \frac{1}{1 - 2M / \left( K + Kt \right)} K^2} dt$$

This is a messy integral and I'll leave it to you to try to solve it if you want to. But this should be enough to illustrate the method.

Only because Luigi is so far away from the hole that his proper time is essentially the same as coordinate time. If you want to see this, try plugging Luigi's value for $r$ into the gravitational time dilation formula and see how close the resulting factor is to 1.

If Luigi were a lot closer to the hole, then how close he was would affect how slow Mario's clock ran relative to his.

Again, that's because it assumes that the time dilation is being evaluated relative to an observer who is so far away that his proper time is essentially the same as coordinate time. If that is not the case, things get more complicated.