Not Understanding Time Dilation - mk 2.

[cmb]

Further to the other current thread, I have a different 'don't understand' issue for time dilation. I've thought it over and I might have a slight grasp of 'the answer', but it is still a bit murky for me:

Two space ships of the future pass each other in opposite directions, each traveling at 0.5c relative to some arbitrary point. Each one observes the other to be a spaceship going 0.7c away from them [if my presumption that relativistic speeds add up in quadrature?]. Both therefore see the clock of the other ship running slower than their own.

How can both on-board clocks be running slower than each other?

 

[CDCraig123]

Ill take a crazy stab at it this. As this sounds like what if i go back and time and kill grandpa what will happen. As i see it Time Dilation is only relevant to the speed at which you are passing through space time. Not relevant to any other body. Crazy theory of mine, as you move through space-time, space-time applies a force on you the faster you go the greater the force, and the greater this force the faster you will proceed through time. So on your two ships there clocks will be ticking according to there speed through space time, and not according to the speed difference between them. I wish NASA would put a clock into space and give it zero velocity and see what the clock did but this would be pretty hard to do considering we are moving at 580 km/s.

 

[cmb]

As far as I have read, time dilation is an actual issue for ultra-accurate GPS [I think] satellite clocks, and ps corrections need to be fed in.

This is what actually prompted me to think of this in the first place; why does the GPS clock run slower than ground-based ones, when as far as the GPS clock is concerned it is the ground-based clock that's moving!? In this case, I suspect any such effects have more to do with gravity dilation than motional dilation, but I'd value any knowledgeable comments on this.

 

[CDCraig123]

Well if you could find out what the clock says on the voyager spaces crafts said, relative to speed and gravity on the spacecraft and clocks on Earth I bet you would have your answer anyone got a friend at NASA?

 

[HallsofIvy]

Further to the other current thread, I have a different 'don't understand' issue for time dilation. I've thought it over and I might have a slight grasp of 'the answer', but it is still a bit murky for me:

Two space ships of the future pass each other in opposite directions, each traveling at 0.5c relative to some arbitrary point. Each one observes the other to be a spaceship going 0.7c away from them [if my presumption that relativistic speeds add up in quadrature?].

Well, actually at $(0.5c+ 0.5c)/(1+ (0.5)^2)= 0.8c$

Both therefore see the clock of the other ship running slower than their own.

How can both on-board clocks be running slower than each other?

They don't- none of this is "absolute". Each person sees the other person's clock running slower. A third person, at rest relative to your "arbitrary point", would see the two clocks running at the same speed.

 

[HallsofIvy]

Well if you could find out what the clock says on the voyager spaces crafts said, relative to speed and gravity on the spacecraft and clocks on Earth I bet you would have your answer anyone got a friend at NASA?

Clocks on satellites and even fast moving jet planes do run slower than on the earth. That has already been verified.

 

[cmb]

Well, actually at $(0.5c+ 0.5c)/(1+ (0.5)^2)= 0.8c$

OK, so that's how the add up. Thanks,

They don't- none of this is "absolute". Each person sees the other person's clock running slower. A third person, at rest relative to your "arbitrary point", would see the two clocks running at the same speed.

So, er, do the GPS clocks on the satellites see the clocks running slower on earth, or faster? Maybe I read this wrong, but why this, then:

For GPS satellites, GR predicts that the atomic clocks at GPS orbital altitudes will tick faster by about 45,900 ns/day because they are in a weaker gravitational field than atomic clocks on Earth's surface. Special Relativity (SR) predicts that atomic clocks moving at GPS orbital speeds will tick slower by about 7,200 ns/day than stationary ground clocks. Rather than have clocks with such large rate differences, the satellite clocks are reset in rate before launch to compensate for these predicted effects.

http://www.metaresearch.org/cosmology/gps-relativity.asp

The gravity bit I can see, gravity is what it is at anyone point. But the velocity is relative, and if the GPS is running 7,200 ns/day to a ground-based clock watching it because it is moving relative to it, then does the GPS clock 'see' the ground-based clock running faster or slower?

 

[cmb]

Clocks on satellites and even fast moving jet planes do run slower than on the earth. That has already been verified.

Still struggling here. The clock comes back to Earth and we find it has lost a few seconds. But if you traveled with the clock, wouldn't the clock on Earth (that has been moving relative to you) be the one that's lost the time?

 

[Dale]

Are you considering gravitational effects, or just kinematic effects? Both play a role, and near the Earth the gravitational effects often dominate and they are often in opposite directions. However, I don't want to go into GR if you are struggling to understand SR.

 

[cmb]

Are you considering gravitational effects, or just kinematic effects? Both play a role, and near the Earth the gravitational effects often dominate and they are often in opposite directions. However, I don't want to go into GR if you are struggling to understand SR.

Please see my previous comment. I'm happy with the gravitational issue - according to the article I link to this constitutes 45,900 ns/day. Disregard that, I see that bit.

So now I am talking about the 7,200 ns/day kinematic effect. Which clock runs slower by 7,200 ns/day, the GPS with a guy sitting next to a ground based one, or the ground-based clock as observed by an astronaut floating along with the GPS?

 

[Dale]

So now I am talking about the 7,200 ns/day kinematic effect. Which clock runs slower by 7,200 ns/day, the GPS with a guy sitting next to a ground based one, or the ground-based clock as observed by an astronaut floating along with the GPS?

In which frame? Neglecting the gravitational effects (which can only be done for a very short time), in the ground frame the satellite clock is running slow and in the satellite frame the ground clock is running slow. The satellite frame is not used for GPS navigation in any way, but you certainly could set it up and determine the speed of the ground clocks in that frame.

 

[cmb]

In which frame?

I undertand the question, but there again there is something about that question that makes me hesitate. One clock must be sped up relative to the other so that they stay in sync. I'm finding it tough to fathom that you can slow one OR other, whichever, and still end up with a 'consistent' time frame where both match up.

 

[Dale]

I undertand the question, but there again there is something about that question that makes me hesitate. One clock must be sped up relative to the other so that they stay in sync. I'm finding it tough to fathom that you can slow one OR other, whichever, and still end up with a 'consistent' time frame where both match up.

Since we are not interested in the gravitational aspect let's not talk about ground and satellite clocks, but just clocks A and B moving inertially in space.

Suppose that A had not just one clock but a whole system of clocks, all synchronized with each other in A's frame. And suppose that B had not just one clock but a whole system of clocks, all synchronized with each other in B's frame.

Now each one says: "My clocks are properly synchronized and running normally, but his clocks are not synchronized and they are running slowly".

Each one also says: "If he wants to make his clock always match the clock that is momentarily next to his then he will have to run his clock faster to counteract the fact that it is running slowly".

Finally, each one says: "If I want to make my clock always match the clock that is momentarily next to mine then I will have to run my clock faster to counteract the fact that his clocks are not synchronized".

 

[Makep]

taking into consideration the directions of the ships relative to each other, we can say that each will be traveling towards the past of the other. this means that:

1. As the ships approach each other, time relative to each other will dilate
2. At the point of passing, time relative to each other will be equal and
3. After passing, time relative to each other will contract

these are always true for relativity. the effect will be the same for two ships with different speeds traveling in the same direction. only this time, the faster ship will be traveling towards the future of the slower ship and the slower laging in the past of the faster ship.

 

[Dale]

Makep, almost none of that is correct.

 

[cmb]

I was with you up to the line;

each one says: "If I want to make my clock always match the clock that is momentarily next to mine then I will have to run my clock faster to counteract the fact that his clocks are not synchronized".

Sorry, I've read it 5 times slowly, and I still don't understand what you are saying here.

 

[pervect]

Have you ever heard of space-time diagrams, cmb? I think they will answer your question.

Here is a space-time diagram for the case where one observer is stationary. The stationary observer concludes that the moving observer is slow, because he uses the green lines to compare his clock to the moving observer.

The moving observer concludes the stationary observer's clock is slow, because she uses the red lines to compare clocks, not the green ones.

This is strange, but not paradoxical. We haven't gotten yet into why the moving obserer uses the green lines, but once you accept that they do, I hope you can see how it resolves the apparent paradox.

We can get into the why of it in another post, or perhaps you have some other questions or issues...

 

[CDCraig123]

Clocks on satellites and even fast moving jet planes do run slower than on the earth. That has already been verified.

Considering I don't no what role gravity plays in time dilation, I thought of a man made object with a clock that would be under the smallest amount of gravity influences. Hence the voyager space crafts.

 

[cmb]

Have you ever heard of space-time diagrams, cmb? I think they will answer your question.

Sorry, I don't understand that diagram. (What are the dots meant to represent, a clock tick? Why are they at different gap lengths on the lines, what is the significance of the angle chosen for the red line? I'd have thought the way to draw it would be dots of equal separation along each path, then the 'observation' line is one that intersects the other path at 90deg, no?)

1) OK, let's deal with the 'real world' example above. So a satellite is launched with clock correction factors of 45,900 ns/day retardation, to counter the gravitational field effect, and 7,200 ns/day advance to account for the kinematic effects. It is launched into space and after 10 years an observer on the ground is still receiving time-stamped messages (inclusive of time-of-flight correction) from it that match the ground-based clocks, because the correct correction factor was fed in at launch. This appears to be what actually happens today, in real life. So, firstly, is my understanding of any of that incorrect?

2) But now for the thought experiment; a satellite is launched tomorrow with an astronaut on board with a life-support capsule sufficient for a 10 year mission, and he stays there for 10 years. He has no audio connection with the ground, he is a space-hermit! Each and every day, as far as he is concerned, he checks the signal from the ground based clock (inclusive of tof correction). He notes that it is losing 14,400ns/day relative to his clock, because the ground based one is running 7,200 ns/day slower, plus his clock has already been set to run 7,200 ns/day faster as well.

3) At the end of his 10 year mission he comes back to Earth and says to his flight director "over the 10 years, I observed the ground based clock gradually fall behind mine by 14,400ns/day, and the signal I got from it before I left orbit was 52 milliseconds behind mine" and the flight director says "that's funny, your clock matched ours for all of the 10 years". They compare clocks and find ...? What do they find? Did the astronaut's clock correct itself during the descent back to earth, or are the clocks at different times?

I presume paragraph 2 is where there must be an error, but I cannot see it? You might argue paragraph 3 but, obviously, we could do the thought experiment for a million years and end up with an hour's difference before he came back down to ground. The act of coming back down surely can't have a 'variable' effect on the astronaut's clock, according to how long he's been up, can it!?

So, my question boils down to: When the astronaut and flight director compared clocks, what did they find?

 

[Dale]

I was with you up to the line;

each one says: "If I want to make my clock always match the clock that is momentarily next to mine then I will have to run my clock faster to counteract the fact that his clocks are not synchronized".

Sorry, I've read it 5 times slowly, and I still don't understand what you are saying here.

Here is a diagram that may help:

Suppose that I am the clock at rest at x=2 (black vertical line). There are a bunch of clocks (white nearly vertical lines) passing by me at .6 c. These clocks are synchronized in their rest frame, but in my frame they are not correctly synchronized. In fact, at t=0 I look and I see that the t'=2 clock reads -1.5, the t'=0 clock reads 0, and the t'=-2 clock reads 1.5. So the closest clock is set behind, the next clock is set OK, and the clock after that is set ahead. If I want to adjust my clock so that it always reads the same as the clock passing me then when the x' clock passes me, my clock reads .66 and his reads -.66, so I set mine back to match his. But then, because I had to set mine back to match the first one, by the time the next clock (x'=0) reaches me my clock reads 2. and this next one reads 2.66. So I have to run my clock faster in order to catch up. The clock after that was set ahead, so I also have to run my clock faster in order to catch up with that one. I have to run my clock fast, not because his clocks are fast, but because they are not synchronized.

 

[Dale]

2) But now for the thought experiment; a satellite is launched tomorrow with an astronaut on board with a life-support capsule sufficient for a 10 year mission, and he stays there for 10 years. He has no audio connection with the ground, he is a space-hermit! Each and every day, as far as he is concerned, he checks the signal from the ground based clock (inclusive of tof correction). He notes that it is losing 14,400ns/day relative to his clock, because the ground based one is running 7,200 ns/day slower, plus his clock has already been set to run 7,200 ns/day faster as well.

You cannot ignore GR over the course of 10 years. You can only pretend that the astronaut and the ground observer are equivalent inertial frames in flat spacetime for a very short time. The main problem is that in flat spacetime they would continue to get further away from each other, whereas on average the distance is not changing in this scenario.

For a beginner I would not recommend this example or this thought experiment. There are too many GR pitfalls. Stick with a zero gravity example instead.

 

[cmb]

You cannot ignore GR over the course of 10 years.

I didn't.

1) ... a satellite is launched with clock correction factors of 45,900 ns/day retardation, to counter the gravitational field effect, and 7,200 ns/day advance to account for the kinematic effects. ...
...This appears to be what actually happens today, in real life.

 

[Dale]

Yes, I saw that, but I was talking about this part:

2) But now for the thought experiment; a satellite is launched tomorrow with an astronaut on board with a life-support capsule sufficient for a 10 year mission, and he stays there for 10 years. He has no audio connection with the ground, he is a space-hermit! Each and every day, as far as he is concerned, he checks the signal from the ground based clock (inclusive of tof correction). He notes that it is losing 14,400ns/day relative to his clock, because the ground based one is running 7,200 ns/day slower, plus his clock has already been set to run 7,200 ns/day faster as well.

That part only happens if we have two inertial frames in flat spacetime, which is not the case here. We would have to use some GR math to answer this question.

 

[cmb]

Yes, I saw that, but I was talking about this part:That part only happens if we have two inertial frames in flat spacetime, which is not the case here. We would have to use some GR math to answer this question.

So, are you thinking this is likely to all sum up to zero net effect on the synchronisation because of these effects, between the astronaut observing his fast-running clock compared to the go-slow clock on the ground?

 

[pervect]

Sorry, I don't understand that diagram. (What are the dots meant to represent, a clock tick? Why are they at different gap lengths on the lines, what is the significance of the angle chosen for the red line? I'd have thought the way to draw it would be dots of equal separation along each path, then the 'observation' line is one that intersects the other path at 90deg, no?)

Yes, the dots represent a clock tick. The detailed reasoning for the spacing of the dots is not necessarily required to understand why the twin pardox isn't a pardox, but it's interesting, I think.

Relativity tells us that $\Delta t^2 - \Delta x^2$ is constant for all observers (using units where c=1, and light moves at a 45 degree angle on the graph).

The spacing of the dots is 4 graph units for the stationary observer making $\Delta t$ =4. Since $\Delta x$ is zero, the interval is $\Delta t$^2 =16 in this case, For the moving observer, the spacing is $\Delta x$ = 3 and $\Delta t$ =5, and 5^2 - 3^2 = 25-9 = 16. So the dots are spaced at equal 4 unit intervals. The dots which occur at $\Delta t$ =4 have an identical spacing of "proper time", computed by the formula above, said formula giving the results which an actual clock would measure, as the dots which have $\Delta x$ = 3 and $\Delta t$ =5.

The angle of the red-line can be derived by another space-time diagram, I'll do that in another post since you're interested. Again, while it is interesting to know why the lines are drawn the particular way they are, the exact reasoning isn't needed to be known to understand why the twin paradox isn't an actual paradox.

1) OK, let's deal with the 'real world' example above. So a satellite is launched with clock correction factors of 45,900 ns/day retardation, to counter the gravitational field effect
and 7,200 ns/day advance to account for the kinematic effects. It is launched into space and after 10 years an observer on the ground is still receiving time-stamped messages (inclusive of time-of-flight correction) from it that match the ground-based clocks, because the correct correction factor was fed in at launch. This appears to be what actually happens today, in real life. So, firstly, is my understanding of any of that incorrect?

I haven't gone through the numbers you quote in detail, but it sounds basically correct. However, you are moving into the grounds of general relativity when you talk about gravitational effects, it would be best to postpone that until you understand special relativity fully.

2) But now for the thought experiment; a satellite is launched tomorrow with an astronaut on board with a life-support capsule sufficient for a 10 year mission, and he stays there for 10 years. He has no audio connection with the ground, he is a space-hermit! Each and every day, as far as he is concerned, he checks the signal from the ground based clock (inclusive of tof correction). He notes that it is losing 14,400ns/day relative to his clock, because the ground based one is running 7,200 ns/day slower, plus his clock has already been set to run 7,200 ns/day faster as well.

I'm getting confused by the space-traveler using a tweaked clock. If he's a hermit, and not communicating with the ground, I'd give him a standard clock. I'm not positive I understand how you intended his clock to be tweaked, (or why you'd even want to tweak it in the first place, unless he communicates with the ground constantly).

3) At the end of his 10 year mission he comes back to Earth and says to his flight director "over the 10 years, I observed the ground based clock gradually fall behind mine by 14,400ns/day, and the signal I got from it before I left orbit was 52 milliseconds behind mine" and the flight director says "that's funny, your clock matched ours for all of the 10 years". They compare clocks and find ...? What do they find? Did the astronaut's clock correct itself during the descent back to earth, or are the clocks at different times?

I presume paragraph 2 is where there must be an error, but I cannot see it? You might argue paragraph 3 but, obviously, we could do the thought experiment for a million years and end up with an hour's difference before he came back down to ground. The act of coming back down surely can't have a 'variable' effect on the astronaut's clock, according to how long he's been up, can it!?

So, my question boils down to: When the astronaut and flight director compared clocks, what did they find?

I can answer that for a non-tweaked clock, which represents proper time, easily enough, and hopefully if you know how you intended your clocks to be tweaked (though I still don't see why you'd want to), you can perhaps figure out the answer to your question.

Perhaps you are assigning some sort of special philosophical significance to the clock-tweaking? As far as I'm concerned what is of interest is what an untweaked clock would measure, as this would represent the actual passage of time for an astronaut. If I'm reading Neil Ashby right, tweaking the clocks isn't even done directly nowadays, they let the clock keep it's own proper time, and send a polynomial back and let the calculator at the receiver due the necessary corrections.

If we use http://relativity.livingreviews.org/Articles/lrr-2003-1/ as a source, and we assume a geostationary orbit, then the orbiting clock is too fast, by a factor of 4.4647*10^-10. A year is approximately 3.1557 10^7 seconds, so if we take a 10 year period, the astronaut's clock will gain .14 seconds over those ten years. Nothing special happens when the astronaut lands, he's just .14 seconds older than his twin on Earth - I'm not sure why this is confusing you, or what's confusing about it.

 

[pervect]

OK, onto the part where we show why the lines have the slope they do. First we have to talk about clock synchronization. There is a specified way of synchronizing clocks, called the Einstein convention, that's fairly easy to understand,that we need to follow to define the notion of "simultaneity" in relativity.

You take the worldlines of three equally spaced observers. The middle observer is moving along with the other two, and has an equal distance from both. If the middle observer emits a light pulse, the arrival time where it intersects the worldlines of the other two observers is the same, by definition,because it's traveling equal distances to both observers, and because light has a constant velocity relative to all observers.

But first, a quick review. Before we draw a complex diagram with the worldline of three obserers, let's review the diagrams for one observer.

In space-time diagrams, time points upward on the diagram, and the horizontal axis represents position. The diagrams I'll draw will all be scaled so that light is represented by a 45 degree line on the diagram, i.e. so that c=1.

Quick quiz questions for understanding:

A) Is the worldline below that of a stationary or moving observer?

B) The same as above: is the worldline below that of a stationary or moving observer?

Now, we get to the more complex diagrams. They'll give away the answers to the above questions, but hopefully you already knew the answer anyway.

We need to draw the worldlines of three equally spaced stationary observers, and we want the middle one to emit two light rays. We draw it out, and we get this:

[/quote]

The red line connects two points that occur "at the same time" in the stationary frame. It's what we expect in the stationary frame, events that occur at the same time have the same t coordinate.

Now we do the same thing in the moving frame. We get this:

We can see that the concept of "at the same time" is _different_ for the moving observer than it is for the stationary one.

What is the exact angle of the red line? Well, it turns out that the angle the red line makes with the horizontal x-axis is the same as the angle that the worldline of the moving observer makes with the vertical t-axis. This can be demonstrated formally with similar triangles.

The important conceptual point, though, even before we worry about the value of the angle, is that the notion of simultaneity is different, that the lines connecting simultaneous events in the moving frame are different (slanted) relative to the lines connecting simultaneous events in the stationary frame.

 

[cmb]

I think you are saying paragraph 2 is incorrect because the ground-based clock would not be going 45,900 ns/day slower from gravity effect and 7,200 ns/day slower from kinematic effects relative to the satellite's time.

OK, what is it? How do I work out each element of the timing difference?

 

[Makep]

Makep, almost none of that is correct.

Why not?

 

[Dale]

For one, the time dilation formula depends only on speed, not direction. It doesn't matter if they are going towards each other, away, or in the same direction, or tangentially.

 

[Dale]

So, are you thinking this is likely to all sum up to zero net effect on the synchronisation because of these effects, between the astronaut observing his fast-running clock compared to the go-slow clock on the ground?

OK, what is it? How do I work out each element of the timing difference?

The elapsed time on any clock in GR is given by:
$$\tau=\int \sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda$$
Where x is the path taken by the clock, g is the metric, and lambda is a parameter along the path. In GR this is done for the astronaut's path, and the ground clock's path. This quantity is invariant, meaning that it doesn't matter what strange coordinate system you express it in. So the astronaut and the ground clock will both agree on how much time will have elapsed for themselves and for the other.

 

[cmb]

OK, so what is it?

I've provided figures from someone else. I'll stick with them 'til there is a wiser head that uses that equation properly and modifies the +45,900 ns/day gravity effect/-7,200 ns/day kinematic effect.

I regret I cannot follow this because I'm asking straight questions to aid my understanding... and then I get an answer to a different question.

 

[Dale]

I regret I cannot follow this

I know. That is why I suggested multiple times that you avoid this question until you have the right background. You need to learn SR before jumping into GR. Avoid scenarios involving gravity for now.

 

[pervect]

I second Dale's the motion. If you don't understand SR first, you won't get anywhere with GR at all.

One way to avoid GR with something close to your original question is to place the experiment on a small asteroid, rather than Earth.

This avoids gravity, at the minor expense of the hovering spaceship having to do a powered orbit.

However, this is still not the simplest case to understand, the simplest and standard textbook case does not involve any accelerating clocks.

 

[cmb]

I know. That is why I suggested multiple times that you avoid this question until you have the right background. You need to learn SR before jumping into GR. Avoid scenarios involving gravity for now.

Sorry, but I am not following it because I am not getting any direct answers. I'm sure it would help if I knew more, but I want to work with some numbers first because to me that is easier to make 'real sense' out of this specific paradox than these space-time plots.

I do appreciate you trying, but the way I see it is that when we look up at a GPS satellite it is running Xns/day fast due to a lower gravity field, and Yns/day slow due to kinematic effects. If we were to look down from the satellite, we would see clocks running Xns/day slow due to a higher gravity field, and Yns/day slow due to kinematic effects.

I don't see why that is a wrong understanding, nor anything in your equation that says it is wrong, so I'm unable to progress this dialogue where there is no direct response to that essential question. The answer is yes or no, and maybe we can take it from there once there's a definite answer to whether that last paragraph is correct.

 

[Dale]

If we were to look down from the satellite, we would see clocks running Xns/day slow due to a higher gravity field, and Yns/day slow due to kinematic effects.

The problem is that this separation into gravitational and kinematical effects can only be done where the gravitational field is weak and not changing over time. In the astronaut's inertial frame the gravitational field changes over time, so you cannot decompose it like that. All you can do is evaluate the integral that I provided, which is guaranteed to agree with the previous overall calculation regardless of what coordinate system you use.

We are not trying to be evasive. I have, in fact, provided the answer, even though I knew you would not understand it. You simply need to learn SR first, you are going to be unsuccessful with the GR-first approach.