# I Speed of time

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1. Feb 9, 2017

### Jens K Munk

Dear Forum,

The title "Speed of time" is the best I can come up with. We know that time isn't constant, rather it depends on the speed at which we move as well as the gravity we are subjected to (see https://en.wikipedia.org/wiki/Gravitational_time_dilation for a reference to the latter). As a consequence of the latter, and as stated on the Wikipiedia page, Earth is stated as being 2.5 years younger at its core (due to higher gravity in there).

Thoughts on this:
1. Isn't gravity zero at Earth's core (disregarding the gravity of the Sun)? I would argue that at the core, you have an equal amount of Earth at each side of you, so they cancel out.

2. Instead of "gravitational time dilation", do we experience "density-dependent time dilation"? Higher density (such as that in Earth's core) slows down time. This would explain why clocks in space record time going faster than at Earth's surface. It may also explain why extreme density such as that in black holes causes light to reach zero speed (relative to observer's perspective).

Let's stop here. Please comment. Thanks.

2. Feb 9, 2017

### Orodruin

Staff Emeritus
Gravitational time dilation depends on the gravitational potential, not the gravitational field.

No. Also, there is no such thing as light reaching zero speed and the Schwarzschild black hole is a vacuum solution to the Einstein field equations, i.e., the density is zero at the event horizon.

Also note that you should not be using the A-level tag unless you have knowledge in the subject equivalent to that of a graduate student or higher. It is clear from your post that you do not. The level system is in place for you to tell us your level of expertise so that the answers can be aimed at that level of understanding. <Moderator's note: level changed>

Last edited by a moderator: Feb 9, 2017
3. Feb 23, 2017

### Albrecht

I do not understand the comment to the time dilation and the speed of light in a gravitational field.
The Schwarzschild metrics say about the proper time in a gravitational field
d(tau) = (1 - r0/r)1/2dt where r0is the Schwarzschild radius.
This means that the flow of time stops completely at r = r0 where is the event horizon. Correspondingly the speed of light reaches zero at that range.

4. Feb 23, 2017

### Orodruin

Staff Emeritus
No it doesn't. It is just a matter of the coordinate t not being a good coordinate at the Schwarzschild radius. The "speed of light" that you are referring to is not the invariant speed of light - it is a coordinate velocity.

5. Feb 24, 2017

### Albrecht

So, what means: d(tau) = 0 ?
Doesn't it mean that the local, proper time stops at that point? Whatever the general meaning of t might be? - I have listened to a talk of nobel price winner Gerard t'Hoft about cosmology. And he just stated this. Was he wrong?

With respect to the speed of light I of course meant the coordiante velocity. The one which was for instance measured in the Shapiro experiment. That should go to zero. The invariant speed of light is the one measured by a local observer. It is easy to see that this observer will always measure the nominal speed of light as his tools (clocks and rulers) change accordingly in the gravitational field. - This is at least my understanding.

6. Feb 24, 2017

### Orodruin

Staff Emeritus
You do not generally get $d\tau = 0$ at $r = r_S$ - you only get this for null world lines. It is not any stranger than having a general null world line. The only locally peculiar thing is that you have chosen to use coordinates such that the coordinate lines for the $t$ coordinate are null world lines.

7. Feb 25, 2017

### Albrecht

So I understand that all points on the event horizon are world null lines. Which consequences?

An observer outside the black hole and outside the event horizon will see any motion stopping at the event horizon. For him (and so for us) there is c = 0 and as well any other motion and also - as a general conclusion - dτ = 0. From which the question follows how a black hole can collect material.

To say it again: We are the observers looking from outside and we have the expectation that material moves into the black hole. But how?

8. Feb 25, 2017

### Orodruin

Staff Emeritus
There is no local Lorentz frame where the event horizon is stationary. No observer can be stationary at the event horizon.

This is an empty statement. The observer is not located at the event horizon and does not measure the coordinate speed of light there. The symbol $c$ is the invariant speed of light in vacuum, not the coordinate speed of light.

9. Feb 27, 2017

### Albrecht

I think that it is just the other way around. If an observer would be located at the event horizon then for him anything would look normal. But seen from our view, so from our Lorentz frame, any speed at the event horizon goes to zero. That is what also Gerard t'Hoft has said in his talk which I have mentioned earlier.
If we transport a clock into a gravitational field and take it back later then this clock will have a delay compared to our clocks. That is according to General Relativity and also the result of measurements. If we would transport a clock close to the event horizon and would later be able to take it back that this clock would have a considerable delay. This is in my understanding a clear result of the Schwarzschild metric. What else could happen?

10. Feb 27, 2017

### Orodruin

Staff Emeritus
You do realise that this is a GR equivalent of referring to an observer travelling at the speed of light in SR? It simply does not make sense. The most probable is that t'Hooft was making some sort of popularisation or that you misunderstood him (or you took the popularisation too far). Without seeing the talk, there is no way we can tell.

11. Feb 27, 2017

### Albrecht

It is a clear result of the Schwarzschild metric that
dτ = dt*sqrt(1-rs/r) and c = c0*sqrt(1-rs/r) (for tangential motion)

It has in fact a singularity for r = rs.

One can of course avoid the singularity by introducing a coordinate system which adapts gradually with the approximation to the horizon so as to avoid a singularity (like Eddington-Finkelstein coordinates). But if we discuss the development of the universe we use our coordinates, for instance to say that its age since the Big Bang is 13 billion years. Such an adapted coordinate system will tell us something different. But what would be the use of that for our understanding?

12. Feb 27, 2017

### Orodruin

Staff Emeritus
No it won't. The physical predictions of GR are coordinate independent. This is the entire point.

13. Feb 27, 2017