The Reality of Relativity

1. yogi

Einstein asserts in his 1905 paper that if two separated clocks at rest in the same frame are synchronized, and one is put in motion at a constant velocity v, it will be out of sync upon arrival at the second clock. No turn around is involved - so since the amount of time difference depends upon the velocity and the length of travel - we can presume that during the travel time the clock that is put in motion runs slower.

Now we take two clocks A and B at the origin of the xy axis and a third clock C located at some distance L along the positive x axis. A, B and C are all initially sychronized while at rest in the same frame - then A and B are accelerated to a uniform velocity v in the direction of C (along the x axis). When A and B reach their cruise velocity (no longer accelerating), it is presumed that their reference frame is equally as good as that of C which has not moved.

As A and B pass C, B is accelerated in the -X direction until B reaches a velocity v relative to A. Therefore, from C's inertial perspective, clock A will be running slower as measured by clock C. From A's perspective, B will be running slower as measured by the rate of clock A. So we have A running slower than C and B running slow than A. But since B is no longer in motion with respect to C, B and C should be running at the same rate.

2. JesseM

8,489
It's running slower in the frame where they were synchronized to begin with. But Einstein would say you are free to analyze the whole situation from the perspective of any frame, including one where they were not synchronized at the moment the clock was accelerated, and in which it ran faster as it approached the other clock. Do you disagree that any frame's analysis of this situation would be equally valid, according to Einstein or any other physicist?
So you're saying B would now be at rest in C's frame, right?
Well, this is only confusing if you don't say which frame you're talking about when you say one clock is running slower than another. In the inertial frame where C is at rest at the end, A is running slower than B and C; in the inertial frame where A is at rest at the end, B and C are running slower.

3. yogi

Hi Jesse - I do not agree with the last statement of yours - something we have discussed previously - When the A clock is compared to the C clock, it will read less - that is the fundamental significance of time in SR as opposed to Lorentz's development wherein Lorentz regarded the time factor in the transforms as more or less illusory ... in SR time dilation is not just an observational (apparent) phenomena - clocks that are put in motion really do lose time - they run slower - that is why we must compensate every orbiting GPS clock that is put in motion relative to the earth centered reference frame - I cannot find a single reference from Einstein that says that the C clock will be running slower than A ...or that it would be observed to run slower. Clock A will read less than C at the end of the experiment. Einstein's theory recogonized time dilation as real - and that means it is not a reciprocal situation - even though you will find many references by others that each clock appears to be running slow when viewed from the other frame - there is no experimental verification that time dilation is reciprocal - if it is reciprocal it can't be actual (two clocks cannot each be running slower than the other) - and Einstein's concluding statement in Part 4 of his paper that "a clock at the equator will be found to run slower than a clock at the pole" is meaningless if it is only observational (apparent). (Actually because of the earths oblateness they run at the same speed, but that is incidental).

4. Hurkyl

15,998
Staff Emeritus
Then, simply put, you are not discussing Special Relativity. You are discussing something else that has a universal notion of time.

Lorentz's formulation assumed that there was a universal notion of time, and had to explain why clocks don't measure it.

Einstein's formulation says that time is what clocks measure, nothing more, nothing less.

5. JesseM

8,489
Fair enough, but if you don't think Einstein would have agreed with it, then it's obvious you're misinterpreting him, and you have no basis for doing so in anything he wrote.

6. yogi

Hurkyl - not so - We are looking at two clocks, A and C during the first part of the experiment - when A reaches C, both clocks will have logged a certain number of minutes - the owner of the C clock looks at the A clock as it passes by his C clock and observes the number of minutes on the A clock and compares them to the reading on his C clock - Einstein and every experiment that that has been conducted to measure particle lifetimes, satellite clocks or airplane clock dilation confirms that the A clock will read less - why do you think this involves universal time? What is measured is the relative rate of time passage in the two frames.

7. yogi

Jesse - Is it fair to conclude you believe time dilation to be reciprocal in the one way travel scenereo Einstein described in Part 4 of his Electrodynamics paper of 1905? If after A and C are set to the same reading in the same frame, then A travels to C in the manner above described, and is stopped at C, and both clocks are read when there is no motion between them - will there be a difference in their readings?

8. JesseM

8,489
Of course there'll be a difference, but each frame's analysis will make the same prediction what the two clocks will read at the moment they meet, yet each frame will disagree on the ratio between the rates the two clocks were ticking. There are certainly frames where A is ticking faster than C--but in these frames, A and C were not synchronized to begin with (you always have to remember that different frames define simultaneity differently), instead A was well behind C when it first accelerated in C's direction, so even with a faster rate of ticking it was still behind C when they met. Einstein, with all mainstream physicists, would say that any inertial frame's analysis of the problem is equally valid.

Last edited: Jan 9, 2006
9. Hurkyl

15,998
Staff Emeritus
That's certainly not what you're saying in your original post.

"Therefore, from C's inertial perspective, clock A will be running slower as measured by clock C."

You're making a comparison between clock A and clock C, and specifying relative to what that comparison is made. Or maybe you're making a comparison between clock A and the coordinate time in C's reference frame. Either way, that's fine. In particular, you are not making any sort of comparison between two reference frames. (I don't think such a notion can even make sense!)

Then, you turn around and say:

"So we have A running slower than C and B running slow than A."

You've suddenly dropped any reference to inertial frames, and are saying absolutely that A is running slower than C and B is running slower than A.

As you've said, "A running slower than C" is true from C's inertial perspective. "B is running slower than A" is true from A's perspective.

These two statements are about different "inertial perspetives" -- you simply cannot combine them to conclude that B is running slower than C.

10. RandallB

Of course Einstein recognized both clocks would observe the other clock as running slower and that time dilation was real! That what the relative observation means, and why he used the name relativity. What references made you think otherwise??

Have you not read of his main points that none of the clocks never actually run slow. They all see light cover a one foot distance in the same one nano-sec of time within their own reference frame. But that they both “observe” the time on any individual clock in the other frame a running slow. The key thing to see is that this means the “apparent” local time in that other reference frame will seem to be running FAST! Now be clear here, the time doesn’t run fast, just the times observed on the “slow” clocks passing by in the moving reference frame are only observed once each locally going by, as running fast. To determine the rate and time on the individual clocks you need to collect data from the other station locations in your own reference about an individual traveling clock. To see this you need to focus on his points about simultaneity to understand it.

Do yourself a favor and work though a simultaneity problem.
Using 0.8c or 0.866c for the speed of a train, pick a pair of stations, clocks synced of course, have them work out a synchronized bright light flash from both stations that all can see. One and only one station will receive these signals together at the same time, half way between the sending stations right. Beginning of the train “car #0” has passed two of these stations and just reached the last station at the time of the signal. That time zero for that car and station. Likewise this car has arranged to send its own flash of light for all to see and one of other cars following it is picked to also send a signal at the same time BASED ON THE TRAINS SYNCED CLOCKS i.e. train time zero.

Now obviously this has to be prearranged, planning that for the stationmasters and car conductors is simple based only on their own clocks and observations of what the time should be.
Also obvious is the only train car that receives both train light signals at the same time will be half way between the two cars sending them.

Now the information to calculate, all station and train numbers a based on distance from car and station “0”, Time passes one unit for light moving one station or one train car length.

1) Based on train time what station AND WHAT TIME IS IT AT THAT STATION are all three cars at a) train time zero, b) at the time the middle car receives both signals from the other cars, and c) when the sending cars receive the signal from the other car.

2) Based on station time which train car is passing AND THE TIME SEEN IN THAT CAR AS IT PASSES for all three stations at station times a) zero when the station flashes are sent, b) the middle station receives both signals from the other stations, and c) when each sending station receives the signal from the other station.

3) Now working calculations in station time and using the stations identified in step #1) as being local to the sending of the train signals which station should receive both train signals simultaneously and when, what train car is passing and it’s time when it does.

4) Also working calculations in train time for the cars identified as local to the signals send by sending stations in step #2). Which train car should receive both station signals simultaneously, which station is it passing and the time in that station it when it does.

Every car and station will always have a station or car passing by it from which to pick up a number and time in that other reference.

It’s a lot to work out, BUT when you do:
Do step #3 results agree with step #1?
Do step #4 results agree with step #2?
IF not, before your claim relativity is broken you need to carefully check your math.

Rigorously going though this classical analysis should convince you that at least SR is real.
When done correctly I don’t see how a flaw can be found in it to argue against it, but if you want to build a convincing example to argue against SR you will have to do so in this kind of detail.
It’s all straight forward classical linear time and distance dilation calculations, just a lot of number crunching.

Maybe one draw back; it may convince you that four dimensional Minkowski “space-time curves” are not needed in SR which would leave you at odds with many that believe space-time is a legit part of SR.
That’s OK; I’d enjoy the company, as I’ve never seen a requirement for space-time in SR when a rigorous classical calculation as above will always give the same result. {If anyone can contrive a SR problem and solution using space-time, I’ll be happy to match the solution with a classical one}
Space-time is certainly a requirement in GR, as its 4D provides a means of warping to explain gravity without connecting forces or particles. But SR works fine in 3D classical physics without space-time.

Take your time in working up your own example and be careful to double check your numbers. You will have a lot data that makes it hard to help review in a forum like this.

11. yogi

Randall - I don't follow a lot of your post and obviously you and some of the others have missed the point of mine.

Einstein began his development of the mechanical aspects of SR by assuming symmetry between two assertedly equal inertial frames - so length contraction is a reciprocal observational effect since each observer is as good as the other, and therefore the phenomena must be apparent only. Likewise, time dilation based upon measurements made between two relativly moving clocks (while they are in relative motion) is also reciprocal (the observers in relatively moving inertial frames are in symmetrical disagreement about which clock is running slow). What Einstein did in Part 4 of his paper is make a transition from apparent phenomena (length contraction) and dynamical time dilation, to assert that if two clocks are in sync initially in one frame, and one clock is accelerated to a velocity v where it travels some distance L, and is later returned to the original frame of the stationary clock, it will have logged less time than the stationary clock. Here there is a record left by the two clocks when they are compared later in the same frame - they have recorded different times

12. RandallB

NO, I think it’s you that have missed out on making your own point to yourself completely.
Of course a clock that speeds away from earth 10 units distance and returns those same 10 units back to earth at the same speed will read the same time as on a second clock that continued to 20 units distance. The only diff is the earth bound observer gets to record the traveler time and earth time when it happens locally and must wait for the radioed data from 20 units away to see that the other clock and earth time there were exactly the same there.

What your failing to account for in detail is dealing with all three or four reference frames. The returning clock is in a ‘fourth’ frame that since it changes direction we have no way of making easy valid distance measurements before and after the turn, but we can track the clock time easy enough. BUT, I see no effort on your part to detail accurately the locations and times in the other two frames namely a) the outbound frame and b) the inbound frame. ALL simple SR math, just more of it than you're doing so far.

Specifically: When the clock(s) depart earth in reference frame “a” exactly where are they (the clocks and earth together) as measured in reference frame “b”? It can only be one unique time and location there.

At the turn around, easy to define in the earth frame, but you don’t say where and when is the earth in reference frame “a” at that earth time of the turn around.
Also for both the earth and that location in “a”; where and when are they in reference frame “b”?
And finally in detail where and when does the clock that turns around step into reference frame “b” as measured in “b”.

Now during the return the clock doesn’t change location in frame “b” but we can see the clock will still allow time to pass.
But when it does return to earth the and a local clock comparison can be made. The traveling clock is only reading time at the same rate as frame “b” what is the “b” time.
Also where and where is the clock and earth as measured in frame “a”.

Finally for the turn around point a fixed distance from earth, where and when is that point in frame “a” and frame “b” at the start of the trip and also at the end of the trip.

Unless your willing to do a complete job of this you have no basis for claiming relativity is broke (at least nor SR), just that your evaluation is incomplete.
– If not mine that you don’t follow, at least your own, worked out completely.

And as I said, once you do the work, you can see where simultaneity is a big issue, totally dependent on reference frame.

13. Peterdevis

62
No, because you can't decide wich clock is in motion. Motion is always relative to someting. So for clock A clock B moves with a velocity B and for clock B clock A moves with velocity v

When you deal with accelleration, your clocks are not longer connected with an inertial frame and so you left SR and comes in the world of GR.

In GR you will learn that propertime (that is the time reading by the clock) depends of the path of that clock and that you can only compare two clocks if there in the same place

14. yogi

Randall - your in the wrong post - there is another post entitled relativity is broken - why do you keep insisting that I said that - I am not concerned with doing a complete numbers crunch using the traditional approach which presupposes that in some way what we observe in another frame is actually going to impact the clock - in almost every analysis I have seen, there is a transition from apparent phenomena to explain actual time dilation - and I know you can get to the desired result (a number that conforms to measured time dilation in GPS and particle disintegration). But its pseudo science - I have over 30 books on relativity and none of them is able to explain actual time dilation - Einstein did not have an answer and neither do you - the best attempts I have seen has been by the critics of SR - Curt Renshaw for example makes a plausible argument based upon energy considerations - but what i am looking for is something conceptual - Einstein made a bold prediction in his 1905 paper when he asserted that when one oif two clocks is put in motion relative to the other - the one in motion would arrive out of sync.
He dosn't not convey how or why this can result - becuse prior to that bold assertion he was dealing only with observational effects - whcih do not explain how a clock that is moving will run at a different rate than the one which did not - but it does - that is why we have to offset every GPS clock put into orbit in order for it to read the same as the clock in the earth centered reference frame

Staff: Mentor

That's your whole problem? You want to toss out a theory that works because you don't think the explanation for how/why it works is satisfactory?

Last edited: Jan 11, 2006
16. JesseM

8,489
As long as the fundamental equations of physics have the mathematical property of Lorentz-symmetry, it is guaranteed that they will work the same way in the different coordinate systems allowed by the Lorentz transformation, and thus that time dilation and other effects will be observed. If you were designing the laws of nature in a simulated universe on a computer, and the equations you programmed in as the fundamental laws governing the simulation happened to have this property of Lorentz-symmetry, then you would automatically see these relativistic effects in your simulation, even if you had not planned or expected this.

Why do the fundamental laws of physics all have the property of Lorentz-symmetry? I don't know, but I also don't know why they have spatial translation symmetry, or rotation symmetry, or any other symmetry. Do you need an "explanation" for these other symmetries?

Last edited: Jan 11, 2006
17. yogi

Russ - did I say i wanted to toss out SR - that is a strange comment. I pose an interrogatory in the hope it will lead to some interesting comments and what I get back is a sermon.

18. yogi

Jesse - That is the whole point - things are symmetrical as long as it is assumed there are no preferred reference frames and every inertial system is equal to all others - There is reciprocal behavior of each clock in such a world but things are not symmetrical when one clock is accelerated wrt to the other. We know things are asymmetric during the acceleration - but once the accelerated clock levels off to a constant velocity v wrt the stay behind clock, things can no longer be symmetrical if at the end they are both returned to the same original frame and the clocks compared - The question posed is whether there is complete symmetry between the two clocks during the uniform velocity phase - and if so, why do they read differently when brought together in the same frame at a later time (where the amount of difference depends upon the uniform velocity and the length of travel at said velocity). Or do you think they will both read the same when later compared in the same frame?

19. yogi

Peterdavis - I did not mean to ignor your post - but briefly - we do know which was put into motion and which remained stationary - read part 4 of Einstein's original paper where the experiment is described.

20. JesseM

8,489
No, you miss my point, I'm talking only about a mathematical property of the equations of the fundamental laws of physics. That a particular equation has the property of "Lorentz-symmetry" can be checked just by looking at the equation, without any notion of "reference frames" whatsoever. But as long as all the fundamental equations have this mathematical property, it is guaranteed that the laws of physics will be the same in the different coordinate systems provided by the Lorentz transformation. Again, if you were creating a simulated universe with its own laws of physics, then even if you gave no thought to how the laws would look in different reference frames, if the equations you wrote down were Lorentz-symmetric, this alone would be enough to be absolutely certain that relativistic phenomena like time dilation would appear in your simulation. For example, Maxwell's laws are Lorentz-symmetric, so if you programmed a simulation to be governed entirely by Maxwell's laws then this simulation would automatically obey the rules of relativity, despite the fact that physicists came up with Maxwell's laws well before they had even conceived of relativity.
I don't know what it means to ask whether the clocks themselves are symmetrical, it only seems meaningful to ask whether the way the clocks appear in different reference frames is symmetrical or not. And if you're talking about a reference frame where the non-accelerated clock is at rest vs. a reference frame where the clock that accelerates comes to rest after it finishes accelerating, then clearly their respective views of this specific situation are not symmetrical--one frame will see the two clocks reading the same time up until the first clock accelerates, the other frame will see the two clocks being out-of-sync up until the first clock accelerates. The only symmetry is in how the laws of physics work in each frame--in either frame, a clock which is moving at velocity v will be slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$.