# I Time dilation in special vs. general relativity

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1. May 26, 2016

### philipp

Hi all,

I was trying to understand the time dilation in special and general relativity and after much time of "overthinking" I am pretty much stuck now. My problem is, that what seems to me to be the same premises apparently imply opposite things.

In special relativity, for two inertial reference frames moving relative to each other with velocity v, we have the following formula for time dilation:

T' = γ ⋅ T0

where
• T' is the time measured in the moving reference frame
• T= is the proper time measured in the resting system
• γ = 1/√1 - (v2/c2) ≥ 1
• v is the relative velocity of the intertial reference frames
• c is the speed of light

We see that: T' ≥ T0

We further know that "moving clocks run slow."
So in the resting reference frame, the time runs faster, meaning more time passes in the resting frame relative to the moving one.

So a smaller T (in this case T0) ⇒ more time passing relative to the other frame

Now for the general relativity. Let's imagine a source of gravitaton, e.g. a planet, and T1 being a time interval measured close to that planet and T2 being a time interval measured further away. We have

T2 = gh/c2 ⋅ T1 + T1

where
• g is the gravitational acceleration
• h is the distance to the center of gravity (≈ hight above ground)
• c is the speed of light

We see that: T2 ≥ T1

We further know that "clocks close to gravitation run slow."
So for a place closer to gravitation the time runs slow meaning less time passes relative to a place further away.

So a smaller T (in this case T1) ⇒ less time passing

Now how can this be? How can a smaller time interval once imply more time being passed and another time imply less time being passed. Where am I wrong?

Thanks in advance

2. May 26, 2016

### Simon Bridge

You need to be careful about who is moving wrt who else.

Alice is stationary wrt two events that are separated by time T and distance 0 (they happen in the same place.
Alice is moving wrt Bob: bob measures the same events separated by time T' and distance vT' where $T'=\gamma T$
This means that Bob sees Alice's clock go slowly: she does not tick off as many seconds as he does between events.
Thus: Alice's moving clock is slow: she is the one who is moving wrt Bob.
That help?

3. May 26, 2016

### Mister T

Meaning that T' is always greater than the proper time. Proper time is a relativistic invariant, meaning all observers will agree on its value.

That's a far less precise way of saying it.

No. The moving observer will observe the resting clock to be running slow.

A more precise way to say it is that proper time is the time that elapses between two events that occur in the same location. Thus, if two events occur at the same place in the moving frame, the time that elapses between them is a proper time, and observations from the rest frame will indicate that the time elapsed between those events is greater than the proper time. On the other hand, if two events occur at the same place in the rest frame, the time that elapses between them is a proper time, and observations from the moving frame will indicate that the time elapsed between those events is greater than the proper time.

I think that in this case T1 and T2 are both proper times. All observers will agree on their values so if for one observer T2 is greater than T1, then all observers will agree.

4. May 26, 2016

### pervect

Staff Emeritus
A unifying principle that applies to both special and general relativity would be the principle of maximal aging. This is that if one considers multiple clocks taking different paths to the same destiation, the object that follows a path of natural motion, with no non-gravitational forces acting on it, would have the maximum indication of proper time.

You can google for this term, E.F.Taylor, the author of several relativity texts, has a number of papers on this topic on his website. (They're also published, so you can read them in the literature if you have access, but they'll likely be paywalled).

Unfortunately this is literally true only in the region where the paths of clocks in "natural motion" never meet each other. If the clocks can "cross paths", one needs to revise this statement to say that the path of natural motion extrermizes proper time, rather than maximizes it. Mathetmatically it's not much different, but it makes the topic harder to discuss.

An example. If you're on the Earth, and you have a clock sitting on the ground, the clock is not following "natural motion". Non-gravitational forces are acting on it - the ground is pushing up on it. If you throw a clock up in the air, let it rise for one second, and then fall back down to the ground again, this rising/falling clock will follow natural motion with no non-gravitational forces acting on it, and hence have the greatest elapsed (or proper) time.

The clock at higher altitudes ticks faster, so it accumulates more time than the clock that stays on the ground. But the velocity of throwing it upwards makes it slow down. The path that maximizes the time is the path when you throw the clock upwards and let it coast.

Note that if you throw a clock horiziontally, fast enough, eventually it'll orbit the Earth and return to it's starting point, in a charecteristic period that depends on the Earth's mass and altitude. However, this sort of path exists only if you wait long enough (at least one orbital period). It turns out to be one of the paths I cautioned about - a path that is mathematically extremal, without being a global maximum.

If you consider motions that are less than an orbital period, you don't get the "multiple paths" issue, and the one and only path that between any two points that is a path of natural motion is the path that maximizes proper time (the clock reading, the oldest "twin").

5. May 27, 2016

### Staff: Mentor

You'll often hear time dilation described that way, and it's not exactly wrong as long as it comes with enough additional explanation and qualification.... But without that additional explanation and qualification it can be misleading. You might try "a clock will run slow according to an observer who is in motion relative to it" instead - that wording emphasizes that it is not the motion of the clock, but rather the motion of the observer relative to the clock, that leads to time dilation.

6. Jun 2, 2016

### exmarine

?? Are you sure about that? From Einstein’s 1905 SRT paper, page 10 of the English translation that I have:

“From this there ensues the following peculiar consequence. If at points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (t*v^2/2c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.”

7. Jun 2, 2016

### Staff: Mentor

The rest of the paper is the "enough additional explanation and qualification" needed to make that passage not misleading.

8. Jun 2, 2016

### Jeronimus

If Bob would measure Alice's clock to be ticking slower compared to his clock, but Alice would measure Bob's clock to be ticking at a different ratio compared to her own clock than the ratio Bob experiences, then they would be able to decide who is the moving one.

But that is not the case. It would violate the equivalence principle. It would equate to being able to perform an experiment which would allow you to decide if you are the moving one or not.

Bob measures Alice's clock to be ticking slower by the same ratio Alice measures Bob's clock to be ticking slower while they are moving relative at a given constant speed.

The formula T' = γ*T can be applied by both. They can both declare themselves to be the prime frame in a sense.

Or to put it differently.

T' = T/γ
T2 = T2'/γ

T' being the time interval Alice measures on Bob's clock for a given time interval T Alice measures on her clock. Hence, a moving clock at vrel.

T2 being the time interval Bob measures on Alice's clock for a given time interval T2' Bob measures on his clock.

Only if the above is true, neither Bob nor Alice would be able to decide on who is the moving one.

Last edited: Jun 2, 2016
9. Jun 2, 2016

### Yukterez

When you want to compare special relativity to general relativity you should not regard uniform motion, but acceleration. While with uniform motion (which is relative) it is always the other guy who's clock are ticking slower than your own in your rest frame, acceleration (which is, like rotation, absolute) makes the accelerated clock tick slower. The same goes for gravity in general relativity, since gravity and aceleration are equivalent.

10. Jun 2, 2016

### Orodruin

Staff Emeritus
This is a very common misconception. Acceleration can be perfectly well handled by special relativity. What differs from SR to GR is the introduction of a curved space-time.

This is only true locally. It is not true in a global perspective.

11. Jun 2, 2016

### Yukterez

That's what I said. Acceleration and rotation make the accelerated clock tick slower in SR already, and gravity in GR also the clock near the gravitating mass.

Time dilation is also not global. Tidal forces can be neglected in this context.

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