# Thread: How "real" is gravitational time dilation?

1. In a previous thread about black holes and their event horizons, the discussion turned to how the equation for time dilation.... Gravitational time dilation - Wikipedia, the free encyclopedia

.....only relates coordinate time measured by a distant observer to local proper time at the event horizon. It doesn't relate local proper time at the event horizon to local proper time at a location further away. Meaning that, even though the formula predicts that a distant observer would observe time to be standing still at the event horizon, it isn't exactly standing still. Things can still fall into the black hole during that observer's lifespan.

Now, looking for a more practical real world example, we have the global GPS system.

Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion.
Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.
The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time. This kind of accumulated error is akin to measuring my location while standing on my front porch in Columbus, Ohio one day, and then making the same measurement a week later and having my GPS receiver tell me that my porch and I are currently about 5000 meters in the air somewhere over Detroit.

GPS and Relativity

I'm not sure what formula they used to calculate the 45 microseconds per day from GR. Without knowing the satellites' altitude and etc, it would be hard to try and test it against the time dilation formula above.

What I'm wondering is: clearly the effect measured by the satellites is not just an artefact of our choice of coordinate system. If we didn't correct for the daily shift of 38 microseconds, then let the satellite stay up there for a long time, and then brought the satellite back down to Earth, we would surely find that in the course of it's round trip up there and back, the satellite had quite literally "aged" more than we have.

Wouldn't an observer who spent a long time at or near the event horizon of a black hole also age slower than we do on Earth? What formula describes that difference if the formula above doesn't?  2.

3. Originally Posted by kojax In a previous thread about black holes and their event horizons, the discussion turned to how the equation for time dilation.... Gravitational time dilation - Wikipedia, the free encyclopedia

.....only relates coordinate time measured by a distant observer to local proper time at the event horizon. It doesn't relate local proper time at the event horizon to local proper time at a location further away. Meaning that, even though the formula predicts that a distant observer would observe time to be standing still at the event horizon, it isn't exactly standing still. Things can still fall into the black hole during that observer's lifespan.

Now, looking for a more practical real world example, we have the global GPS system.

Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion.
Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.
The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time. This kind of accumulated error is akin to measuring my location while standing on my front porch in Columbus, Ohio one day, and then making the same measurement a week later and having my GPS receiver tell me that my porch and I are currently about 5000 meters in the air somewhere over Detroit.

GPS and Relativity

I'm not sure what formula they used to calculate the 45 microseconds per day from GR. Without knowing the satellites' altitude and etc, it would be hard to try and test it against the time dilation formula above.

What I'm wondering is: clearly the effect measured by the satellites is not just an artefact of our choice of coordinate system. If we didn't correct for the daily shift of 38 microseconds, then let the satellite stay up there for a long time, and then brought the satellite back down to Earth, we would surely find that in the course of it's round trip up there and back, the satellite had quite literally "aged" more than we have.

Wouldn't an observer who spent a long time at or near the event horizon of a black hole also age slower than we do on Earth? What formula describes that difference if the formula above doesn't?
http://www.kowoma.de/en/gps/orbits.htm

the formula for escape velocity is:

76d0f896f720f511285b72dfcf4d782e.png

the escape velocity of teh earth is
sqrt[2*(6.67×10^-11 newton m^2 / kg^2)*(5.9721986×10^24 kg )/6367.5 km] = 11180 m/s

the escape velocity from the altitude of the gps satellites is
sqrt[2*(6.67×10^-11 newton m^2 / kg^2)*(5.9721986×10^24 kg )/26560 km] = 5477 m/s

RELATIVITY CALCULATOR

gamma for 11.180 km/s is 1.000000000695364

gamma for 5.477 km/s is 1.0000000001668838

arbitrary precision calculator applet

there are 86400 seconds per day

1.000000000695364 / 1.0000000001668838 = 1.00000000052848

0.00000000052848 * 86400 = 0.000 045 66

thats 45 microseconds per day

http://www.astronomy.ohio-state.edu/...Unit5/gps.html  4. The formula you have listed gives a relation between coordinate time ( absence of gravitational field ) and proper time ( gravitational field caused by mass M, distance r from field source ). The observer on Earth and the satellite in orbit have a different r coordinate, so using the above formula you can calculate the difference between the two clock readings.
Is this what you meant to ask ? I'm a little confused, because the height of the satellite's orbit is always known, so I am not entirely sure of your point.
And yes, a clock near a BH's event horizon would run slower compared to a clock on Earth's surface, simply because the BH's gravitational field is much stronger.  5. Time dilation?

Is there any space dilation with that, or is that the argument, space dilation? Like "what"?  6. Originally Posted by equus Time dilation?

Is there any space dilation with that, or is that the argument, space dilation? Like "what"?
Time dilation is closely related to length contraction; what one observer experiences as time dilation, another observer might see as length contraction and vice versa. The ratio between the two is always constant for all observers ( provided their reference frames are inertial ones ).  7. Mostly I'm confused because the definition I have for proper time is "time as measured by a clock that travels the whole distance." Proper time must be what is going faster for the satellites then, because their onboard clock is the one that's physically ticking faster. If we were to send a satellite into orbit, leave it there for a while, and then bring it back, the onboard clock would have traveled the whole distance..... so that would be proper time we're measuring, right?

That means even when we compare proper time on Earth against proper time in space at a higher R, the two don't match. But I'm thinking it has to be a different equation than this one, then, right? Because this equation only relates proper time to coordinate time. So what equation would I use to compare proper time with proper time, or does it require something more complicated than just using a formula?  8. Originally Posted by kojax Mostly I'm confused because the definition I have for proper time is "time as measured by a clock that travels the whole distance." Proper time must be what is going faster for the satellites then, because their onboard clock is the one that's physically ticking faster. If we were to send a satellite into orbit, leave it there for a while, and then bring it back, the onboard clock would have traveled the whole distance..... so that would be proper time we're measuring, right?

That means even when we compare proper time on Earth against proper time in space at a higher R, the two don't match. But I'm thinking it has to be a different equation than this one, then, right? Because this equation only relates proper time to coordinate time. So what equation would I use to compare proper time with proper time, or does it require something more complicated than just using a formula?
I don't really understand what you mean. You can calculate the proper time for two separate observers, and then compare the two. What you get is the difference between their proper times, and that is all we are really interested in. In this way, you could for example compare the proper time of an earth-bound observer with the one of a satellite in orbit, and we will know the difference between the two.  9. Originally Posted by kojax I'm confused Because this equation only relates proper time to coordinate time.
Does it? Then I am even more confused than you are!
If you could say what is , what is , what is , what is someone may be able to help you.

In particular, you should able to tell us what you mean by "coordinate time"

PS Pls don't point me to the Wiki, I want you own words  10. Originally Posted by Markus Hanke I don't really understand what you mean. You can calculate the proper time for two separate observers, and then compare the two. What you get is the difference between their proper times, and that is all we are really interested in. In this way, you could for example compare the proper time of an earth-bound observer with the one of a satellite in orbit, and we will know the difference between the two.
How do you calculate that? Originally Posted by Guitarist  Originally Posted by kojax I'm confused Because this equation only relates proper time to coordinate time.
Does it? Then I am even more confused than you are!
If you could say what is , what is , what is , what is someone may be able to help you.

In particular, you should able to tell us what you mean by "coordinate time"

PS Pls don't point me to the Wiki, I want you own words
As I understand it, coordinate time is what another observer sees when they observe you from a different place and/or gravity well.

Proper time is what they would observe if they were at your location. I'll admit I'm still confused about the distinction, but apparently there are some situations where it matters. If you accelerate to a relativistic speed toward a distant planet, time at that location is presumed to suddenly advance a number of years depending on what speed you accelerated to and how far away that planet is, from your perspective. Clearly the local clocks at that location are not changing to create this effect. It's just something to do with relativity of simultaneity or something.

Apparently coordinate time isn't as "real" as proper time.  11. coordinate time is proper time measured by a stationary observer whose frame defines the coordinate system being used.  12. There's no such thing as a stationary observer, since there is no reference point to define stationary.  13. well duh! lol  14. Originally Posted by granpa coordinate time is proper timemeasured by a stationary observer whose frame defines the coordinate system being used.
Define all of these?  15.  16. Originally Posted by kojax How do you calculate that?
You calculate this by using a line integral of the form wherein G denotes a geodesic in space-time ( the satellite's or the earth-observer's ), and the differential dt is calculated from the given metric ( in this case most likely the Schwarzschild metric ).

As I understand it, coordinate time is what another observer sees when they observe you from a different place and/or gravity well.
Coordinate time is what a stationary observer at infinity - outside the gravitational field - observes.  17. Originally Posted by Markus Hanke As I understand it, coordinate time is what another observer sees when they observe you from a different place and/or gravity well.
Coordinate time is what a stationary observer at infinity - outside the gravitational field - observes.
True, but an observer who's just a parsec or two away would be far enough outside the gravitational field's reach that their observation is approximately the same.

It's not like we're going to go out there with a fine ruler and measure the distances/velocities ... etc down to the micron. A near approximation should be sufficient. Originally Posted by MeteorWayne There's no such thing as a stationary observer, since there is no reference point to define stationary.
True, but you can be stationary relative to the phenomenon you are attempting to observe, as opposed to being in motion relative to it. I suppose it's more precise to say "an observer in the same inertial frame as..".  18. Originally Posted by kojax True, but an observer who's just a parsec or two away would be far enough outside the gravitational field's reach that their observation is approximately the same.

It's not like we're going to go out there with a fine ruler and measure the distances/velocities ... etc down to the micron. A near approximation should be sufficient.
Yes, you are correct. Strictly speaking coordinate time does not actually exist, since all gravitational fields reach into infinity. One can, however, make a good approximation in the real world by being far enough from the source of a weak gravity field.

PS. I am getting an "Unknown Latex Error" in my previous post, and I can't see the maths expression. Not sure why that is. You can find a similar expression in the Wikipedia article for "Proper Time" though.  19. Post deleted as LaTex appears to be broken, and without that facility, it would be incomprehensible. Message left in Mod Corner  20. You didn't need to delete it, Guitarist. If I click "quote" I can recover the original text, and then cut and paste it somewhere else to see it. I just hadn't around to figuring out where to view it before you deleted it.  21. OK, sorry about that. Not everyone is as motivated as you are.

As I am on a campsite in rural France and the weather has turned poor, I am bored, so let's take a stroll....

First recall the Einstein field equations (in covariant form) where i.e. these are spacetime indices and (the is there to make the time component of the coordinate functions have the same dimension as the spatial components; physicists are naughty about this, and work in units where . Bad, bad boys!.

So what does this mess actually mean? Let's see....

First: on the RHS, represents the gravitational source - mass-energy to be precise. is an arbitrary constant

Second: on the LHS is the metric tensor field - I shall explain this in a mo

Third: on the LHS is the Ricci curvature tensor, a contracted form of the full 4-index Riemann curvature tensor, and is its trace. Let's see what this guy is....

Suppose an arbitrary closed curve on some 4-manifold. Take a vector (a pointed arrow) at the point and transport around our curve without turning or twisting it i.e. at every point on our curve these vectors are parallel. Now you can easily show for yourselves that, unless the manifold is globally "flat" i.e.Euclidean, the start and end points of our transported vector do not match. The Ricci tensor extremely roughly speaking gives the "difference" between the start and transported vector. BTW say "Ritchie" for Ricci, NOT "Ricky".

As you have just shown, this tensor is zero when our manifold is flat - recall our manifold is spacetime.

Now consider the case of the E.field equations in the absence of a gravitational source. We have that on the RHS so the LHS must be zero identcally. We have just shown that implies spacetime is flat, so that when the source is absent, the curvature is zero.

And if so, of course is its trace . So we now have that in the absence of a source spacetime is flat and as easily follows. But what does this tell us about the metric ?

Well you'll have to wait (or not) as my BBQ will not  Bookmarks
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