# Confusion about the gravitational time dilation formula

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1. Jan 17, 2016

### Whitehole

1. The problem statement, all variables and given/known data
I'm reading the book Relativity, Gravitation, Cosmology by Ta-Pei Cheng. I'm in the part where he derived the gravitational time dilation formula for static gravitational field,

τ1=[1+(Φ12)/c22.

This implies that clocks at a higher gravitational potential will run faster.

He stated that, for two positions r1 = r and r2 = ∞, with r2 being the reference point Φ(∞) = 0, while τ(r) is the local proper time, the clock at r = ∞ gives the coordinate time t ≡ τ(∞). By the equation above this then implies,

dτ=[1+(Φ(r)/c2]dt

My first question is why set r2 = ∞ as Φ=0, it is more natural in mechanics to set the position r1 = r (say, near earth) at Φ=0. I know this is just the same but my point is other books use the other way such that
My second question is he gave an example of the case of a black hole and uses the above formula to expound. He said that, "for an observer, with the time t measured by clocks located far from the gravitational source (taken to be the coordinate time), the velocity of the light appears to this observer to slow down. A dramatic example is offered by the case of black holes. There, as a manifestation of an infinite gravitational time dilation, it would take an infinite amount of coordinate time for a light signal to leave a black hole. Thus, to an outside observer, no light can escape from a black hole, even though the corresponding proper time duration is perfectly finite."

2. Relevant equations
τ1=[1+(Φ12)/c22
dτ=[1+(Φ(r)/c2]dt

3. The attempt at a solution
For the first question, I think it will just be a matter of sign difference but I'm not sure if there are other things involved.
For the second question, I understand that an observer far from the black hole sees that his clock ticks faster since he is in a higher gravitational potential but he sees the clock of someone near the black hole as ticking very slowly, also someone near the black hole will see his own clock to run slow. My problem was when I tried to correspond this concept to the formula, I'M CONFUSED. If it requires an infinite coordinate time for light to escape a black hole and the gravitational potential Φ(r) is also infinite, doesn't that give an infinite dτ? Any help?

2. Jan 18, 2016

### Simon Bridge

It is actually the convention in physics to put the zero of potential an infinite distance away. The books which do it otherwise (for gravitational time dilation) are probably doing you a disservice.

It's just more general: there is no special reason to pick "near the Earth" as one of the points, the equation is supposed to work everywhere.
Mathematically, the zero potential location is arbitrary - so you always pick your zero to be someplace makes the maths easy. This makes the maths easy in the general case. In some specific cases it is easier to use another place so we do.

No - the person near the black hole will see his own clock run at the same speed it always did.

Look at the case where things are not quite infinity.

What is your education level? Have you covered the concept of limits in calculus?

IMO. The author's description in that passage is poor.
I'm having trouble untangling it.

3. Jan 18, 2016

### shinobi20

Ain't this just the gravitational time dilation and the inside is the Schwarzchild metric?

4. Jan 18, 2016

### Whitehole

I have a B.S. in Physics. For the case of massive and compact objects, the observer near it would see his own clock ticking normally, but would see clocks of observers far from it as ticking faster. On the other hand observers far from it would see their own clocks as ticking normally, but would see the clock of the observer near it as ticking slowly. Is this correct? For a black hole, an observer near it would see light escaping the black hole with finite time but observers far from it would see time slowed down near the black hole, but since black holes have infinite gravitational pull, from the formula it would take an infinite amount of time for the observers far from it to see light escape it. Is this also correct?

5. Jan 18, 2016

### Simon Bridge

OK so you have special relativity and the calculus of limits ... excellent.

Pretty much - both observers stationary in the frame of the object... which means they locally experience acceleration.

In relativity you need to be pedantic ... you don't observe light: light is what you observe things with.'
You need to describe things in terms of events - maybe Bob fires a light pulse at Alice, which Alice detects.
You can work out the time interval between those events from the reference frames of Bob and Alice ... then consider that Bob is close to the event horizon of a black hole and Alice is far away.
This is a one-way trip for the light pulse, which can be confusing ... so maybe Alice sends a pulse right back. Now Alice and Bob can each measure their own time.
In fact...

How about the situation where your standard light clock you know and love from special relativity is aligned vertically in gravity and consider what happens with different orientations in the field but no lateral motion. 1st the light source and detector is at the bottom and next the light source and detector is at the top.

Crash course:
http://preposterousuniverse.com/grnotes/grtinypdf.pdf [Broken]

Last edited by a moderator: May 7, 2017
6. Jan 18, 2016

### Simon Bridge

The Schwarzchild metric is the geometry of space-time outside the event horizon of a non-rotating black hole.
http://www.jimhaldenwang.com/black_hole.htm

7. Jan 19, 2016

### Whitehole

Thanks for all your explanations. I think my confusion is really from the formula. Suppose observer A is nearer a spherical object with strong gravitational field than observer B, and observer A orbited it which took him say dτ=1 day, from the formula above observer B who is far far away from observer A would see that observer A took dt to orbit the object,

dτ=[1+(Φ(r)/c2]dt

dt = dτ / (1 - (GM / r)) where Φ(r) = -GM / r, for convenience I will make the value of GM=1 light year

If observer A is say 2 light years away,
dt = dτ / (1 - (GM / r)) = 1 / (1 - 1/2) = 2

If observer A is say 4 light years away,
dt = dτ / (1 - (GM / r)) = 1 / (1 - 1/4) = 1.3333

Why is it like that? Shouldn't dt be greater the farther observer A is from the object?

Last edited by a moderator: May 7, 2017
8. Jan 19, 2016

### Simon Bridge

... the closer A is to the object, the deeper in the gravity well she is, so the time dilation effect should be stronger.

Note: recheck your equation there... i.e. what happens when r <1?

Last edited: Jan 19, 2016
9. Jan 19, 2016

### Whitehole

I understand that as A is closer to the object, the time dilation will be more so she will experience shorter time as seen by B (farther observer). So dt should get smaller as A is getting closer, but in the calculation it's the other way. If I use 0.1ly, dt would be negative. How can time be negative?

10. Jan 20, 2016

### Simon Bridge

Indeed...
Consider that $r_s = 2GM/c^2$ ... if $r<r_s$ then A is inside the event horizon ;)
When you put $GM = 1$ you made the Schwarzschild radius 1 light-year .... so $r < 1$ means...

The equation should be: $d\tau^2 = (1-r_s/r)\; dt^2$ ... for a clock at rest.
That's just from the metric.

Check the definition of $\Phi$ in the book.

11. Jan 20, 2016

### Whitehole

Oh, now I understand when it is negative but regardless of the Φ, the answer still is smaller when A is farther. I'm still confused in this part. My expectation is B will observe the time A experiences as being smaller as A is nearer the object but from the calculation it turned out the other way. Is there something wrong with what I'm doing?

12. Jan 20, 2016

### Whitehole

I think I already know where I'm getting the confusion, dτ is the time experienced by A as seen by B (frame of B) right? So if I want to compute what A will experience as seen by everybody farther away (here it's B) then I should compute for dτ (not dt) in the frame of B. What I've done above is I've computed for dt in the frame of A(without me knowing) but either ways I'm still wrong because if I'm in the frame of A and accidentally trying to find the time in the frame of B I should still be finding dτ since I'm finding the proper time B is experiencing now. Sorry this is a bit confusing but do you think my analysis of my mistake is correct? Thanks Simon Bridge!

13. Jan 20, 2016

### Simon Bridge

You are starting to think along the right lines:
A is in the gravity well: $d\tau$ is the proper time - so it is the time between event's in A's frame as measured by A's clock. dt is the coordinate time of the event's in A's frame - as measured by B's clock.
So the time measured by B is $$dt = d\tau/\sqrt{1-r_s/r}$$ ...

The "proper time" is the time measured by clocks in the same frame as the events.

You can get further by being pedantic about what events are being detected.
So A1 and A2 are flashes of light at the same place but different times in A's frame ... B1 and B2 are the events that B detects the flashes from A1 and A2 respectively ... what is the time between B1 and B2 in terms of the time between A1 and A2?

14. Jan 20, 2016

### Whitehole

dτ = (1 + ΔΦ/c2) dt

Let dτB = time between B1 and B2, and dtA = time between A1 and A2

Suppose we are in the frame of A, so based on my understanding, B measures

B = ( 1 - ( GM/c2 )( 1/rB - 1/rA ) ) dtA

15. Jan 20, 2016

### Simon Bridge

I wonder that you keep missing out the square...