B Time Dilation: Does Observer See Your Time Speed Up?

[AdvaitDhingra]

If time slows down for an observer traveling at some speed relative to your proper time, shouldn't the traveling observer also see your time slow down relative to his proper time? Or does the observer see your time speed up relative to his proper time.

Also, is dilation exactly the same in general relativity?

 

[PeroK]

If time slows down for an observer traveling at some speed relative to your proper time, shouldn't the traveling observer also see your time slow down relative to his proper time? Or does the observer see your time speed up relative to his proper time.

Your question reveals a fundamental misunderstanding of time dilation.

Velocity-based time dilation is symmetric: two observers in uniform inertial motion measure each other's clock to be running slow, according to the frame of reference in which they are at rest.

 

[Ibix]

If time slows down for an observer traveling at some speed relative to your proper time, shouldn't the traveling observer also see your time slow down relative to his proper time?

Yes, because both people can regard themselves as "at rest in an inertial frame". Note that what you actually see is that a clock moving away slows down and a clock moving towards you speeds up because the Doppler effect due to changing distance dominates the time dilation effect. But if you correct for that, both observers will calculate that the other's clock runs slow in either direction. Also note that if one observer turns round and comes back they were not always at rest in an inertial frame and the results are not symmetric.

Also, is dilation exactly the same in general relativity?

No. Special relativistic results rely on spacetime being flat, and thus they do not usually generalise to curved spacetimes. Sometimes you can get away with it, but some experience is needed to know when you can do it.

 

[AdvaitDhingra]

Your question reveals a fundamental misunderstanding of time dilation.

Velocity-based time dilation is symmetric: two observers in uniform inertial motion measure each other's clock to be running slow, according to the frame of reference in which they are at rest.

Ah ok. thanks

 

[AdvaitDhingra]

Yes, because both people can regard themselves as "at rest in an inertial frame". Note that what you actually see is that a clock moving away slows down and a clock moving towards you speeds up because the Doppler effect due to changing distance dominates the time dilation effect. But if you correct for that, both observers will calculate that the other's clock runs slow in either direction. Also note that if one observer turns round and comes back they were not always at rest in an inertial frame and the results are not symmetric.No. Special relativistic results rely on spacetime being flat, and thus they do not usually generalise to curved spacetimes. Sometimes you can get away with it, but some experience is needed to know when you can do it.

Ok thanks for clearing that up.

 

[Myslius]

Let's say a spaceship moves away from you at the 0.8 speed of light. Time slows down by the factor of
1 /sqrt(1 - 0.8^2)=5/3. Let's say you have a very powerful telescope and are able to see the clock on the spaceship very far away. After 1 year what time would you see on the spaceship? 3/5 year? Not really, the clock actually shows sqrt((1 + 0.8)/(1 - 0.8) = 1/3 year, this is because of relativistic doppler effect and the fact that it takes some time for light to travel back at you. If after 1 year ship turns away and travels towards you with velocity of 0.8c, for some time (not 1 year) through the telescope you would see the clock running 3 times faster than your own clock. The combination of both seeing time running slower and seeing time running faster gives 5/3 ratio when the ship travels back to where it started. When you say a word "see" you have to include doppler effect. Classical doppler effect * relativistic time dilation = relativistic doppler effect.
Escape velocity on the surface of the Earth is 11186 m/s, time on Earth ticks slower in the gravity well for an observer far far away by 1 /sqrt(1 - (11186 m/s/c)^2) = 1.0000000007 times slower.

Let's say you're are in flat spacetime and someone is moving 11186 m/s away from you:
1 /sqrt(1 - (11186 m/s/c)^2) = 1.0000000007 times slower

The ratio is the same.

Both observers see proper time slowing down, and both observers see time speeding up when traveling towards each other when looking thru telescope.

 

[Ibix]

The ratio is the same.

...assuming you are in a stationary spacetime where time dilation can be defined so casually. And being aware that the Doppler effect is different in the two cases. And noting that the gravitational time dilation formula applies between clocks at rest at different heights, not clocks free-falling from infinity at those heights.

The underlying mechanism for the two phenomena is also quite different. In one case the two clocks don't agree on simultaneity and time dilation arises from the "angle" between the simultaneity planes. In the other they measure different times between simultaneity planes that they do agree on and time dilation arises from the curvature of spacetime.

Don't read too much into a similarity between two formulae. And do note that the time dilation formula in Schwarzschild spacetime is usually specified in terms of the mass rather than the escape velocity.

 

[Ibix]

Both observers see proper time slowing down,

This, on the other hand, is meaningless.

 

[jbriggs444]

Both observers see proper time slowing down

This, on the other hand, is meaningless.

If I squint just right I can read that [overly] pithy sentence to mean:

Each observer observes that elapsed proper time for the remote observer is less than elapsed coordinate time for the remote observer, measured using the local observer's inertial rest frame.

 

[Ibix]

If I squint just right I can read that [overly] pithy sentence to mean:

Each observer observes that elapsed proper time for the remote observer is less than elapsed coordinate time for the remote observer, measured using the local observer's inertial rest frame.

Maybe. Generalising your comment slightly, to be clearly written, the sentence needs to state whose proper time is being talked about and as compared to what time standard by what methodology.

 

[AdvaitDhingra]

Let's say a spaceship moves away from you at the 0.8 speed of light. Time slows down by the factor of
1 /sqrt(1 - 0.8^2)=5/3. Let's say you have a very powerful telescope and are able to see the clock on the spaceship very far away. After 1 year what time would you see on the spaceship? 3/5 year? Not really, the clock actually shows sqrt((1 + 0.8)/(1 - 0.8) = 1/3 year, this is because of relativistic doppler effect and the fact that it takes some time for light to travel back at you. If after 1 year ship turns away and travels towards you with velocity of 0.8c, for some time (not 1 year) through the telescope you would see the clock running 3 times faster than your own clock. The combination of both seeing time running slower and seeing time running faster gives 5/3 ratio when the ship travels back to where it started. When you say a word "see" you have to include doppler effect. Classical doppler effect * relativistic time dilation = relativistic doppler effect.
Escape velocity on the surface of the Earth is 11186 m/s, time on Earth ticks slower in the gravity well for an observer far far away by 1 /sqrt(1 - (11186 m/s/c)^2) = 1.0000000007 times slower.

Let's say you're are in flat spacetime and someone is moving 11186 m/s away from you:
1 /sqrt(1 - (11186 m/s/c)^2) = 1.0000000007 times slower

The ratio is the same.

Both observers see proper time slowing down, and both observers see time speeding up when traveling towards each other when looking thru telescope.

Ahh ok thanks for clearing that up