Looking to understand time dilation

[Grimble]

Are observer A and the clock co-moving?

Yes the observer is permanently adjacent to the clock, he may even be holding it.

grimble said:
"So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical."

Isn't he saying that the person and clock are moving together at some indeterminant speed?

One could certainly say that, but would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

 

[ghwellsjr]

...would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

In Special Relativity, things move or are stationary in a defined frame of reference. So you could say that this observer and his clock are stationary in a frame at the co-ordinates of x=1000 m, y=-2345 m, z-0 m at time 345 s and he is facing in the x direction with y to his right and z above his head OR you could define a different scenario in which he is moving in a frame with the same conditions as before but that he is traveling at vx=345 m/s, vy=0 m/s and vz=-4235 m/s, but most people just like to say he is stationary in an inertial frame or he's traveling a .6c in the x direction or not even specify a direction. We usually know what they mean.

But even if there is nothing else in the universe, you can still define him to be stationary or have any kind of motion and/or acceleration you desire (short of c) within a frame.

 

[Grimble]

In Special Relativity, things move or are stationary in a defined frame of reference. So you could say that this observer and his clock are stationary in a frame at the co-ordinates of x=1000 m, y=-2345 m, z-0 m at time 345 s and he is facing in the x direction with y to his right and z above his head OR you could define a different scenario in which he is moving in a frame with the same conditions as before but that he is traveling at vx=345 m/s, vy=0 m/s and vz=-4235 m/s, but most people just like to say he is stationary in an inertial frame or he's traveling a .6c in the x direction or not even specify a direction. We usually know what they mean.

But even if there is nothing else in the universe, you can still define him to be stationary or have any kind of motion and/or acceleration you desire (short of c) within a frame.

OK, then let us define him as stationary. Then can I say:
The FoR, that he is permanently at the origin of, is an Inertial Frame of Reference.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical.?

 

[Dale]

Yes the observer is permanently adjacent to the clock, he may even be holding it.

Then the proper time measured by the clock is indeed equal to the coordinate time in A's frame.

 

[phyti]

Yes the observer is permanently adjacent to the clock, he may even be holding it.



One could certainly say that, but would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

I don't accept Newtons definitions about motion and rest. Rest is not a state of motion, but a relation about the difference in motion of two objects. For me, when two objects have the same velocity (vector), one is at rest relative to the other, i.e., rest is a special case of motion but not a 'lack of motion' as Newton defines it.
Since as you say, there is no outside reference object (excepting dark matter), there is no motion of the frame, which by the previous definition eliminates 'rest'.
I agree with your 2nd statement in red.

But...if we look deeper, there is motion at the molecular and atomic levels, therefore the inertial condition is only an approximation, although a practical one.

Now if you could just get light-clock 101...

 

[Grimble]

Let us now add a second clock, 'held' by observer B, at rest at the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

This is necessary as they are both keeping proper time and as that is a requirement of Einstein's first postulate.

 

[Dale]

Let us now add a second clock, 'held' by observer B, at restat the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

Yes. Assuming that B is also inertial then the proper time measured by the clock held by B will be equal to the coordinate time in B's frame.

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

How does the independent observer judge the duration of their units of time? This is a non-standard phrase so you need to specify what you mean by this.

 

[Grimble]

Yes. Assuming that B is also inertial then the proper time measured by the clock held by B will be equal to the coordinate time in B's frame.

How does the independent observer judge the duration of their units of time? This is a non-standard phrase so you need to specify what you mean by this.

Well you are right of course.

So let us specify an independent inertial observer, who can measure the time passing in A and B's own frames and that, by applying the LT transforms, he can calculate how time passes in each of those frames according to an observer in that frame.

Secondly, as each clock is keeping proper time each tick of those clocks must be an event in space-time. And as JesseM pointed out

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event.

So if each clock's one second ticks are events in space-time they must be ticking at identical rates.

 

[Dale]

Well you are right of course.

So let us specify an independent inertial observer, who can measure the time passing in A and B's own frames and that, by applying the LT transforms, he can calculate how time passes in each of those frames according to an observer in that frame.

So the independent observer calculates the ratio of (proper time A)/(coordinate time A) in A's frame and (proper time B)/(coordinate time B) in B's frame? If that is your intention then what is the purpose of the independent observer? Since he is transforming into each frame he doesn't add anything to the question just makes the problem more confusing. And we have already specified that both of those ratios are 1.

 

[Grimble]

So the independent observer calculates the ratio of (proper time A)/(coordinate time A) in A's frame and (proper time B)/(coordinate time B) in B's frame? If that is your intention then what is the purpose of the independent observer? Since he is transforming into each frame he doesn't add anything to the question just makes the problem more confusing. And we have already specified that both of those ratios are 1.

Yes, you are quite right of course that is all unnecessary!

 

[ghwellsjr]

Let us now add a second clock, 'held' by observer B, at rest at the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

This is necessary as they are both keeping proper time and as that is a requirement of Einstein's first postulate.

Why do you want to introduce a duplicate of your first scenario with no realtionship between them? If you want to introduce as second observer/clock, then you should define their position/motion/whatever in your first frame OR you could define your second frame in relation to your first frame and then define the second observer/clock in that second frame and then you could transform the observer/clock from the second frame to see how they appear in the first frame. Eventually you need to "put" all observers/clocks into a single frame, that is, if you want to discuss this in the context of Special Relativity.

 

[Grimble]

Why do you want to introduce a duplicate of your first scenario with no realtionship between them? If you want to introduce as second observer/clock, then you should define their position/motion/whatever in your first frame OR you could define your second frame in relation to your first frame and then define the second observer/clock in that second frame and then you could transform the observer/clock from the second frame to see how they appear in the first frame. Eventually you need to "put" all observers/clocks into a single frame, that is, if you want to discuss this in the context of Special Relativity.

Because now we have established that as two individual inertial FoR A and B are, as far as can be determined in identical situations. The one second 'ticks' of one clock keeping proper time, and therefore being Time-space Events, have to be equal in duration and frequency to the one second ticks of the the second clock also keeping proper time, and its 'ticks' therefore, also being Time-space Events.
There is nothing to say, so far concerning the simultaneity of the said cloaks' 'ticks'.

Let us therefore have these two clock's, and as they are at rest at the origin of their respective frames, the frames themselves moving at 0.6c relative to one another.

 

[ghwellsjr]

Because now we have established that as two individual inertial FoR A and B are, as far as can be determined in identical situations. The one second 'ticks' of one clock keeping proper time, and therefore being Time-space Events, have to be equal in duration and frequency to the one second ticks of the the second clock also keeping proper time, and its 'ticks' therefore, also being Time-space Events.

They are identical simply and only because you have defined them to be identical. That's why I asked why you wanted to make a duplicate of your first scenario with no relationship to your original one. I think your understanding of Special Relativity would be advanced more directly if you simply follow Einstein's example from your referenced book in which he looks at an object first at rest in one frame and then from another frame in motion relative to the first one. But if you want to take this indirect approach, then let's see where you go with it.

There is nothing to say, so far concerning the simultaneity of the said cloaks' 'ticks'.

Let us therefore have these two clock's, and as they are at rest at the origin of their respective frames, the frames themselves moving at 0.6c relative to one another.

Are you implying that now there is something to say concerning the simultaneity of the said clocks' 'ticks'?

 

[Grimble]

They are identical simply and only because you have defined them to be identical. That's why I asked why you wanted to make a duplicate of your first scenario with no relationship to your original one. I think your understanding of Special Relativity would be advanced more directly if you simply follow Einstein's example from your referenced book in which he looks at an object first at rest in one frame and then from another frame in motion relative to the first one. But if you want to take this indirect approach, then let's see where you go with it.

Are you implying that now there is something to say concerning the simultaneity of the said clocks' 'ticks'?

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

 

[Dale]

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

Actually, relative to observer C the other two observers will be going at 0.33333333 c.

 

[ghwellsjr]

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

You haven't specified which frame you want to analyze these three observers in and it sounds like you might even be thinking in terms of a third frame where the third observer is at rest. You need to be focusing on a single frame before you start transforming between frames.

 

[Mike_Fontenot]

Einstein, in his little book "Relativity" by Crown publishers, derives the time dilation result in a very clear way. If you want to understand why any inertial observer concludes that a clock which is stationary in some other inertial frame is ticking slower than his own stationary clocks, THAT'S where I'd recommend that you spend some quality time.

Mike Fontenot

 

[ghwellsjr]

"some other inertial frame"

Other than what?

Every inertial observer concludes that every moving clock is ticking slower than his own stationary clocks, no matter what frame we use to define, describe, or analyze the entire scenario in. All observers and clocks "reside" equally and simultaneously in all frames we want to consider. There is no reason or need to introduce a separate frame for each and every observer and clock in a scenario.

 

[Grimble]

You haven't specified which frame you want to analyze these three observers in and it sounds like you might even be thinking in terms of a third frame where the third observer is at rest. You need to be focusing on a single frame before you start transforming between frames.

No, you are right I haven't specified any frame and I am not going to until I have agreement as to the situation/circumstance that I am concerned with.

I have tried before to ask a question or state where I have concerns and it just becomes an endless series of questions, suggestions and alternatives that I should use and what I am asking, trying to understand, or, in my previous attempt, trying to show what my logic tells me in order to find where I am going wrong.

So this time I am trying to build a picture, just as I see it, so we can all agree what I am talking about.

 

[Grimble]

Einstein, in his little book "Relativity" by Crown publishers, derives the time dilation result in a very clear way. If you want to understand why any inertial observer concludes that a clock which is stationary in some other inertial frame is ticking slower than his own stationary clocks, THAT'S where I'd recommend that you spend some quality time.

Mike Fontenot

Thank you Mike, but that is not the problem I understand the relationship and the derivation of time dilation and length contraction. Einstein derives them very clearly.

All I am doing is attempting to describe a relationship between certain bodies: clocks, rulers, and observers. When I can be sure that that arrangement is accepted I can raise the points I find trouble with logically.
Otherwise it all descends into endless arguments over minutiae.

 

[Grimble]

Actually, relative to observer C the other two observers will be going at 0.33333333 c.

Yes, of course, thank you

 

[Grimble]

Now as the one second 'ticks' of the clocks 'held' by A and B are events and they were synchronised at their passing C, then if they emitted a pulse of light at the start of each tick the light from those pulses would arrive at C simultaneously?

And if we allow for the time for the passage of light between frames A and B would each observe the pulses of light from the other to be synchronous with their own pulses of light.

(Each pulse of light from either source is being emitted once every proper second and as those emissions are events their timing will be agreed upon by any inertial observer.)

Am I right so far?

 

[ghwellsjr]

If by "synchronizing" the one-second ticks of the clocks at the moment they are all co-located, then, yes, C will from then on see the respective "one-second" ticks arriving from A and B simultaneously, but they will arrive at less than at one-second intervals as measured by C. In fact, you can use the Relativistic Doppler factor to determine that C will measure the time intervals at 1.414 seconds.

And, in a symmetrical way, A and B will "see" and measure C's one-second ticks coming in also at 1.414 second intervals.

But there is no sense in which we can say that any of the three clocks continue to be synchronized after the one monent when they are co-located. Only clocks that are at rest with respect to each other can be synchronized and remain synchronized.

Furthermore, since A and B are traveling apart from each other at .6c, they will "see" and measure the one-second ticks coming from each other at 2-second intervals.

And we can calculate the time-dilation factors that each observer will determine that each of the other clocks are running at. For the clock/observers involving C (C-A, A-C, B-C, C-B) where the relative speed is 1/3 c, the time dilation is 1.125 even though they "see" each other's clocks ticking at 1.414-second intervals, they "know" that the clocks are actually ticking at 1.125-second intervals. And since A and B have determined that their relative velocity is .6c, they "know" that the time dilation of the other one's clock is 1.25 even though the "see" each other's clocks ticking at 2-second intervals.

Note that this analysis is based on what each observer sees and measures of the other clock's ticks and what they can calculate. No frame has been assumed or used in this analysis which means it has nothing to do with Special Relativity. It is simply a description of what is actually going on for each observer. Now if you want, you can specify a frame of reference, any frame of reference, and you can further anayze the relative simultaneity of the clocks, but all the previous analysis will be true in any frame of reference.

 

[Dale]

For some concise notation, let Exy be the emission event from clock X as expressed in frame Y coordinates and Dxyz be the detection event of the pulse from clock X detected by observer Y as expressed in frame Z coordinates.

Now as the one second 'ticks' of the clocks 'held' by A and B are events and they were synchronised at their passing C, then if they emitted a pulse of light at the start of each tick the light from those pulses would arrive at C simultaneously?

Yes.

Dacc = (1.41 cn, 0,0,0)
Dbcc = (1.41 cn, 0,0,0)
Therefore the pulse from A is detected by C at the same time (in C's frame) as the pulse from B is detected by C.

And if we allow for the time for the passage of light between frames A and B would each observe the pulses of light from the other to be synchronous with their own pulses of light.

No.

Ebb = (cn,0,0,0)
Dabb = (2cn,0,0,0)
Therefore in B's frame the pulses emitted by B are twice as fast as the pulses detected by B from A.

Eaa = (cn,0,0,0)
Dbaa = (2cn,0,0,0)
Therefore in A's frame the pulses emitted by A are twice as fast as the pulses detected by A from B.

(Each pulse of light from either source is being emitted once every proper second and as those emissions are events their timing will be agreed upon by any inertial observer.)

The emissions are indeed events, but the time coordinate of an event is not frame invariant.

Eaa = (cn,0,0,0)
Eac = (1.06 cn, -0.35 cn, 0,0)
Therefore the time coordinate of the emission from A are not the same in A's frame and C's frame.

Ebb = (cn,0,0,0)
Ebc = (1.06 cn, 0.35 cn, 0,0)
Therefore the time coordinate of the emission from B are not the same in B's frame and C's frame.

 

[Grimble]

If by "synchronizing" the one-second ticks of the clocks at the moment they are all co-located, then, yes, C will from then on see the respective "one-second" ticks arriving from A and B simultaneously, but they will arrive at less than at one-second intervals as measured by C. In fact, you can use the Relativistic Doppler factor to determine that C will measure the time intervals at 1.414 seconds.

And, in a symmetrical way, A and B will "see" and measure C's one-second ticks coming in also at 1.414 second intervals.

But there is no sense in which we can say that any of the three clocks continue to be synchronized after the one monent when they are co-located. Only clocks that are at rest with respect to each other can be synchronized and remain synchronized.

Furthermore, since A and B are traveling apart from each other at .6c, they will "see" and measure the one-second ticks coming from each other at 2-second intervals.

And we can calculate the time-dilation factors that each observer will determine that each of the other clocks are running at. For the clock/observers involving C (C-A, A-C, B-C, C-B) where the relative speed is 1/3 c, the time dilation is 1.125 even though they "see" each other's clocks ticking at 1.414-second intervals, they "know" that the clocks are actually ticking at 1.125-second intervals. And since A and B have determined that their relative velocity is .6c, they "know" that the time dilation of the other one's clock is 1.25 even though the "see" each other's clocks ticking at 2-second intervals.

Note that this analysis is based on what each observer sees and measures of the other clock's ticks and what they can calculate. No frame has been assumed or used in this analysis which means it has nothing to do with Special Relativity. It is simply a description of what is actually going on for each observer. Now if you want, you can specify a frame of reference, any frame of reference, and you can further anayze the relative simultaneity of the clocks, but all the previous analysis will be true in any frame of reference.

You see here we go straight away with a whole lot of extra information.

C has no clock as seen by A or B.

I have already stated that I am not concerned with the time a signal takes. We may assume that all observers allow for the transmission times which only add another layer of complication.

 

[ghwellsjr]

You see here we go straight away with a whole lot of extra information.

C has no clock as seen by A or B.

I have already stated that I am not concerned with the time a signal takes. We may assume that all observers allow for the transmission times which only add another layer of complication.

I'm sorry if you think I'm adding "a whole lot of extra information" because it is very important information and I'm trying to help you. Please pay attention to what I'm about to say. You need to understand this because your questions belie a fundamental misunderstanding of the problem you're trying to understand.

You have repeated your caveat that you are "not concerned with the time a signal takes. We may assume that all observers allow for the transmission times which only add another layer of complication."

Here are the facts: no observer, no person, nobody, knows "the time a signal takes". We cannot "assume that all observers allow for the transmission times" since they don't know the transmission times. Nobody does. If we knew them or if we could figure them out, we wouldn't have Special Relativity. You need to understand this or the entire rest of this learning exercise will be a waste of time because if you do learn something, it will be the wrong thing.

What I described for you in my previous post was all that the observers can know and as I said, it has nothing to do with SR and nothing to do with transmission times.

As soon as you apply SR to a scenario, you are arbitrarily assigning transmission times and these times vary depending on the frame of reference you select. You may think that there is some reality to using each observer's rest frame to determine the transmission times, but there is not. SR arbitrarily defines the transmission times even within a single frame of reference. So your statement that "all observers allow for the transmission times" does not comport with reality, it's the same as saying that you (and your observers) believe in an absolute aether rest frame and that you know where it is.

Am I getting through to you?

 

[ghwellsjr]

Ebb = (cn,0,0,0)
Dabb = (2cn,0,0,0)
Therefore in B's frame the pulses emitted by B are twice as fast as the pulses detected by B from A.

Ebb = (cn,0,0,0)
Dbaa = (2cn,0,0,0)
Therefore in A's frame the pulses emitted by A are twice as fast as the pulses detected by A from B.

Dalespam, I believe your second pair of equations should start with Eaa rather than Ebb, correct?

And just for the fun of it, could you also calculate those two ratios in all three frames, please?

 

[bobc2]

Sometimes it can help to visualize the problem in the context of a 4-D spatial universe. Depicting the motion of a 4-D structure with respect to some system at rest (supressing X2 and X3), the velocity is manifest directly as the slope of the world line that constitutes the 4th dimension for that structure. The peculiar and mysterious aspect of Special Relativity is that if the X4 axis is rotated with respect to a rest system, then the X1 axis is rotated as well---rotated such that the photon 4-D world line always bisects the angle between the X1 and X4 axis (this goes for all observers). A little reflection on this circumstance leads to the realization that when a photon is observed (for any observer, no matter his velocity), the distance traveled by the photon along the observer's X4 axis (ct) is always the same as the distance traveled by the photon along the X1 axis. The ratio of dX1 to dX4 is always 1. Thus, the speed of light is the same for all observers. Each observer moves along his own X4 axis at the speed of light.

The first sketch above (upper left) depicts an observer moving at some velocity with respect to a rest system (X1, X4). The instantaneous 3-D universe he lives in is represented by the slanted X1' axis. Observers having different velocities will have X1' axes with different slants--they are living in different cross-sections of the same 4-D universe. Coordinate systems for observers having different velocities are shown in a sequence across the bottom sketches (they are all living in different 3-D cross-sections of the same 4-D universe). Above Right: An observer is changing velocity as time passes--his coordinate system, thus his cross-section view of the universe changes accordingly.

 

[Dale]

Dalespam, I believe your second pair of equations should start with Eaa rather than Ebb, correct?

And just for the fun of it, could you also calculate those two ratios in all three frames, please?

Oops, thanks for catching the typo, I will go edit it.

Which two ratios are you interested in? I certainly can calculate them in any frame quite easily.

 

[ghwellsjr]

Which two ratios are you interested in? I certainly can calculate them in any frame quite easily.

Dabx to Ebx and Dbax to Eax, where x is a, b, and c. You did two of the six possible calculations and got ratios of 2 for each of them, I'd like you to do the other four, please.