# Gravitational time dilation

1. Jul 15, 2011

### Zman

Einstein introduced the concept of gravitational time dilation.
I get the impression that it was determined to be a consequence of gravitational red-shift.
Can anybody tell me the logical reason why gravitational time dilation follows on from gravitational red-shift?
Or was the existence of gravitational time dilation determined in some other way?

2. Jul 15, 2011

### TrickyDicky

It's the same thing, just two ways of calling the same phenomenon. One ways focus on frequency shift and the other on time rate differences at different locations, so if light frequency decreases from the point of view of an observer at rest looking at another object at rest that is closer to a gravitational source, in a conserved energy scenario, to the observer at rest that sees the light from the other redshifted it also appears that the other's clock ticks slower (time dilates).

3. Jul 15, 2011

### Zman

But I can derive gravitational red shift mathematically without any reference to clock rates.

And yes, the frequency of the light differs for observers either at different potentials in a gravitational field or;

In Einstein’s accelerating elevator where there is no external gravitational field, an equivalent field is generated by the acceleration according to the equivalence principle.
In both cases, why should a shift in the frequency of light be explained away by clocks running at different rates.

I know that it can be, but what is the logical leap that needs to be made?

When an object is going away from an observer at a constant velocity, there is a red shift from the object. But we don’t then assume that the objects clock is running faster than the observers. In fact we know for other reasons (SR) that it is running more slowly.

4. Jul 15, 2011

### harrylin

In GRT, red shift is *defined* wrt to a reference clock rate ("time"); all definitions are operational definitions, based on physical measurements of "time" and "distance".

Einstein argued as follows in 1916, under the header "Behaviour of Rods and Clocks [...]":

"Thus the clocks goes more slowly if set up in neighbourhood of ponderable masses. From this it follows that the spectral lines of light reaching us from the surface of large stars must appear displaced towards the red end of the spectrum."

- p.198 of the English translation of
http://www.alberteinstein.info/gallery/gtext3.html [Broken]

Not in both cases.

1. In the case of acceleration, there is no gravitational field and so SR should be valid.

Applying SR, ideal identical clocks in (almost) the same state of motion must be running at (almost) the same rate. However, an observer next to one clock will observe a Doppler effect of the signal coming from the other clock. This is due to his change of velocity between the time that the signal is emitted and the time that the signal is received.

2. According to the EP, one should observe the same phenomena for the case of no acceleration and, instead, an equivalent stationary situation with a gravitational field (with g=a).

Thus now the same frequency shift should be observed as with acceleration.

Einstein used the wave model of light (=conservation of cycles and a constant time delay for the light signal to reach the observer).
Based on that, the observed frequency shift for this case must be due to a difference in clock rates (which in turn is due to a difference in gravitational potential).

Does that help?

That is a very different scenario and to good approximation you can neglect that effect for small accelerations starting from v=0; a classical Doppler calculation suffices.

Cheers,
Harald

Last edited by a moderator: May 5, 2017
5. Jul 15, 2011

### yuiop

Actually red shift is an indication of the frequency of the source slowing down rather than speeding up. When an source with a constant frequency in the rest frame of the source is moving relative to another observer then the source appears to blue shifted when it is coming towards that observer and red shifted when the source is going away from the observer. All the while if the source is moving with constant velocity relative to the observer the time dilation factor is the constant, so the Doppler shift (blue or red) is not the cause of time dilation. Rather, time dilation is one cause of Doppler shift.

Also, to be picky, we do not know that the tick rate of a clock with relative motion is running more slowly. If we have two clocks with relative motion then observers at rest with each clock thinks the other clock is ticking more slowly so there is no definitive way to be certain which clock is actually ticking slower than the other if they maintain constant inertial motion. The only becomes clear which clock was ticking slower than the other if one turns around (becoming non inertial) and they come alongside each other again. Which clock actually slows down depends on which clock turns around. Real (non reciprocal) differences in clock rates only manifest themselves when acceleration is involved.

Last edited: Jul 15, 2011
6. Jul 15, 2011

### nitsuj

Could you expand on that some without equations? (Im a simpleton)

I have no idea of a real (and fake?) difference in clock rates.

7. Jul 15, 2011

### Zman

Thanks harrylin

What does 'conservation of cycles' refer to in this case?
I've never heard of the term but I suspect its what I'm after.

Thanks

8. Jul 15, 2011

### PAllen

The special relativistic doppler may be thought of as classical doppler modified by time dilation (or, equivalently, invariant speed of light). Classically, you would expect infinite blueshift if you go speed c towards a light emitter (faster, and you 'break the light barrier', as it were), and infinite redshift as you go at infinite speed away from the emitter. In SR, you get infinite blue-shift as you approach the emitter asymptotically at c, but the formula is different; and you get infinite redshift as you recede from the source asymptotically toward c. Also, in SR there is transverse doppler, which doesn't exist classically.

Gravitational redshift is most simply treated as purely due to (gravitational) time dilation, but, via equivalence principle, it can be related to SR doppler shift for accelerating observers.

9. Jul 15, 2011

### pervect

Staff Emeritus
Einstein realized that if you consider a moving elevator, you'd excpect relativistic redshfit due to the doppler shift. He also relaized that the person on the elevator would need a different explanation than "doppler shift" than velocity. Hence, gravitational red-shift.

Now the relation between "doppler shift" and time dilation is pretty simple - they're basically the same, most especially in the gravitational case, less so in the doppler case. In the gravitational case, you have a constant round-trip time for light signals. So your observed red-shift is equal to the time dilation. You don't have this constant round-trip time in the case of relative motion, so you start trying to seprate out what you actually see (the redshift), to assign some moment that ignores the speed-of-light delays for when the signal was emitted, so you need to do some calculations to convert red-shift into time dilation in the doppler case. But in the gravity case, you don't need any calculation, the redshift is the same as the time dilation.

There's another important point to be made here about redshift.

Suppose you have a light beam, or better yet, a radio signal. You've got a carrier frequency, and a modulation on the radio signal. Suppose your radio signal is 1 Mhz, it's 1 million cycles per second, and it's amplutide changes at 1 Khz. Every 1000 cycles of the carrier, the ampltiude will go through a complete cycle.

Now, suppose you have a redhsift,so your 1 Mhz signal get's reduced to 900,000 hz rather than 1,000,000

Every 1000 cycles, you'll see a "peak" in the signal, so your modulation signal of 1000 hz got reduced to 900 hz.

So, when you redshift a carrier, you also redshift any modulation, by the same factor. If the carrier was showing a video signal, you'd see the video signal in slow motion, after you demodulated it of course.

So, if you have a flashing light controlled by a clock, the flashing of the light is its modulation, and the rate at which the light flashes will slow down when the light redshifts. So you conclude that the light, the clock, and everything must be "time dilated".

So, we've just seen that not only will the light be shifted, but that any signal it carries will be slowed down when you look at something in a gravity well. When you had doppler shift, you had to worry about the changing travel times, which made the red shift different from time dilation in that case. But in the gravitational case, the light travel times are constant between the "low" and "high" areas.

10. Jul 16, 2011

### harrylin

It's a continuity condition.
I don't recall how Einstein phrased it exactly, I think that he wrote (in 1911) something to the effect that the number of emitted wave crests cannot change when in transit in vacuum. If 1000 cycles per second are sent from an emitter, a distant detector should also receive these 1000 cycles, and in the same time; else wave crests could accumulate or go missing in "mid-air". :tongue2:

As a result, Einstein concluded that the frequency (as determined with a single, distant reference system) of an EM wave cannot change when in transit in vacuum.

Now, for the considered case of clocks in a gravitational field, the distance between source and detector also doesn't change.

Consequently the only remaining option to account for the predicted frequency shift, is that the reference "time rate" at the two locations differs.

However, according to GR this effect cannot be observed locally, as all resonators are affected equally. Therefore this difference in time standards affects just as much the frequency spectrum that we measure coming from a star.

Cheers,
Harald

Last edited: Jul 16, 2011
11. Jul 16, 2011

### harrylin

I'm sure that yuiop will elaborate on this, but here's a primer:

It's not possible to make a determination of the rate of a clock that is moving relative to you, about which all observers (more precisely: all independent inertial reference systems) agree. This is because, due to the changing distance, you must account for the delay time of at least one of the measurements, and different "observers" assume different delay times.

As a consequence, the "rate" of a moving clock is said to be a "relative" quantity (not exactly fake, but nevertheless its rate is "apparent" as it depends on your choice of reference system). The effect of inertial motion is even reciprocal: you measure that the other's clock runs slow, but the other measures that your clock runs slow.

That problem can be overcome by letting a clock move in a closed path, so that you can compare its average rate between departure and return with your clock which is in uniform motion. Then you don't need to account for time delay, and the result is "absolute" in the sense that all inertial "observers agree that the clock that went for a round trip accumulated less seconds than yours. In that sense one may say that this difference is "real".

See also section 4, last part of:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Also gravitational time dilation is agreed upon by all, and not reciprocal: if you measure that the frequencies coming from near a heavy star are shifted to the red, an observer at that place will observe our signals as shifted to the blue.

Harald

Last edited: Jul 16, 2011
12. Jul 16, 2011

### yuiop

Well Harrylin did a pretty good job of saying most of what I intended to say (Thanks )

I should not have used the the word "real" so you are right to pick me up on that point. As Harrylin said is better to say absolute (something all observers agree on) and use apparent for not real. Let us say you hold you hand up to the Sun. You could say that your hand "really" is bigger than the sun because it subtends a larger angle than the Sun. A relativist might say you are entitled to your point of view and since the word "real" is not clearly defined physically the statement does not really mean anything, but most people would simply say don't be silly! Now let us say we 2 people (A and B) standing some distant apart. A says he is taller than B and B says he is taller than A. A relativisist would yes, both A and B are right! Logic says that you cannot have a "real" measurement where A>B and B>A are both true at the same time. Indeed a third observer standing in the middle might say A and B are the same height. Clearly measuring height using subtended angle without taking distance into account is a relative process and the measured heights are "not real". The same is true with measuring time dilation when 2 objects are moving at constant velocity relative to each other. The measured time dilation is just a relative measurement that depends on your point of view. In the first case we can find out which person "really" is the tallest by bringing them together so they are standing alongside each other and then everyone would agree which person is the tallest and we can call this an absolute (real) measurement rather than an apparent measurement. In the time dilation case we can only ascertain that there is real/absolute difference in clock rates and elapsed times if we start 2 clocks alongside each other and then bring them back together at some later time. Until they are back together any difference is in clock rate or elapsed time is just subjective or relative (depends on the observer) which in my informal way I called "not real".

As I said before, any absolute difference in clock rates involves acceleration. For example 2 clocks at different gravitational potentials have an absolute clock rate difference (all observers agree which clock is ticking faster) and they experience different proper accelerations. Another example is a clock that remains on top of a very tall tower compared to another clock that is orbiting at the same altitude. One experiences proper acceleration and the other does not and all observers would agree on the difference in elapsed time of the clocks each time they pass each other. Then there is the classic twins paradox with one twin turning around (accelerating) to return to the other. Also consider the case where two clocks are at the same great distance from a black hole and one clock is dropped. If the dropped clock starts to accelerate when it is very close to the event horizon and returns to its original location then hundreds of years could pass on the stationary clock while only minutes pass on the dropped clock. I cannot think of a case where there is an absolute difference in elapsed times where absolute acceleration is not involved. I am not "absolutely" sure whether or not we can have absolute time differences if only coordinate acceleration is involved.

13. Jul 17, 2011

### harrylin

Last edited: Jul 17, 2011
14. Jul 17, 2011

### yuiop

I will go along with that fine distinction.
However, I cannot agree here. An object at rest in a gravitational field does have absolute acceleration in the sense that it has proper acceleration that is measured by an accelerometer and all observers agree on its value. In the reference frame of a free falling inertial observer the object at rest in the gravitational field also has coordinate acceleration. Recall that acceleration does not always have to be moving. The Newtonian equation for force is force = mass X acceleration and it is possible to have a static force such as pressing your hand against a wall which is not moving.

I was wondering about such a situation and I think it might work. It would be interesting to do a more exact analysis involving both gravitational and kinetic time dilation such as comparing the proper time of a clock in an elliptical type orbit versus a clock in circular orbit when there paths cross from time to time. In this situation both clocks would unambiguously have zero proper acceleration.

15. Jul 18, 2011

### harrylin

Sorry but that is making it even more confusing - and even wrong, similar to saying that a car in motion is in rest! Also, Newton's second law refers to the acceleration due to force imbalance; it does not imply that a building will accelerate if you push against the wall. That's a huge misconception. If you still think so, please start a topic about it in the general physics forum.

In physics, acceleration is defined as the rate of change of velocity, and proper acceleration is defined wrt a free-falling reference system (compare http://en.wikipedia.org/wiki/Acceleration).

Mixing of references in one single statement is confusing at best. In these (simplified) discussions we contrast two opposite point of views for an accelerometer that is laying on the ground:

1. We choose a reference system in which the accelerometer is at rest (by definition v=0 and a=0). Consequently we ascribe an eventual bending of the accelerometer beam to the action of a gravitational field g.
or
2. We pretend that there is no gravitational field (g=0) and that the bending of the accelerometer beam is due to acceleration (a=/=0). For that we choose a free-falling frame as reference.

Harald