Special Relativity Clocks

ghwellsjr

All the moving clocks read the same value for those events where τ is the same. But in general two events in the first FoR that have the same value of t will not have the same value of τ in the second FoR. Of course we can change one of those events in the second FoR so that it has the same value of τ as the other event, but now you will have two events that have the same clock reading in the second FoR but different clock readings in the first FoR.
Thanks for the added explanation. I now see what you are doing. You start with a clock moving at 0.5c along the x-axis in the first FoR and you want to see what happens in a second FoR moving at 0.8c.

So let's take as an example the time at 10 seconds. Since the clock is moving at 0.5c, that means its location along the x-axis will be vt or (0.5c)(10) or 5c seconds (or 5 light-seconds). Note that none of this has anything to do with the time on the moving clock. OK, now let's plug these values into the LT. First τ:

τ=(t-vx/c²)/√(1-v²/c²)=(10-0.8c*5c/c²)/√(1-0.8²)=(10-4)/√(1-.64)=6/√(.36)=6/0.6=10

Now ξ:

ξ=(x-vt)/√(1-v²/c²)=(5-0.8*10)/√(1-0.8²)=(5-8)/√(1-.64)=-3/√(.36)=-3/0.6=-5

So this is telling us the location of the moving clock in the second frame and when it arrived at that location. Notice that it is moving at a velocity of -0.5c in this second frame because ξ/τ = -5/10=-0.5. But it is not telling us the time on the moving clock.

We could use Einstein's formula in the first FoR and determine that the time on the moving clock is equal to:

t√(1-0.5²)=10√(1-0.25)=10√(0.75)=10(0.866)=8.66 seconds

Or we could use his formula in the second FoR and calculate the same thing:

τ√(1-(-0.5²))=10√(1-0.25)=10√(0.75)=10(0.866)=8.66 seconds

And as we can see in both cases, 8.66 seconds is less than 10 seconds.

JM

yuiop-Please check back, I gave x=0.5t, with c=1. Entering this in the transform leads to t' = t.

JM

Dale- Thats my point. x in the LT does not describe the position of any of the moving clocks. The position of the moving clocks is given in the formulation of the problem. See section 3 of 1905 " to the origin of one of the systems let a constant velocity v be imparted...and ..communicated...to the clocks." No symbol is given to indicate the position of the moving clocks. In my view x in the LT is an independent variable, perhaps indicating the position of some event such as a light flash.
With this in mind, I am questioning the use in section 4 of the variable x to indicate the position of the single moving clock.
JM

DaleSpam

The variable x is simply a coordinate. There is nothing wrong with specifying that x=0.5 ct is the x-coordinate of some clock in a given frame. The only problem is that it contradicts your assertion that the clock is at rest in a frame moving at 0.8 c wrt the first. In that frame you would have x'=-0.5 ct' representing the coordinate of the clock. This represents a clock moving at -0.5 c, not at rest, and explains why you get equal time dilation.

JM

Dale- When I refer to the 1905 model I mean that presented in section 3. Section 4, which you refer to, provides no theoretical basis for eg the use of x to indicate the position of one of the moving clocks, or the use of the LT, which refers to a single pair of frames, to a series of frames linked together and changing direction.
The purpose of this thread is to find out if anyone can provide the theory that supports the idea that moving clocks always run slow. So far I haven't seen it.
I hope you see from my added descriptions thay my analysis doesn't contradict.
JM

JM

George- First, all the clocks of the stationary frame are stationary, none move. All the clocks of the moving frame are at rest in that frame and move with the speed v. Thats all the clocks there are. So if you allow that τ can be greater than t then: all the moving clocks are synched so all read τ, including the one at the origin, which is the clock described above, and so the moving clock is not running slow.
Are you adding the condition that the x value chosen must result in the moving coordinate being 0? If so then under these conditions the moving clock 'always' runs slow.
But what about the other conditions where the moving clocks (including the one at the origin) are not slow? Suppose that I am the observer stationed at the moving origin to record the time on my clock. From the above it seems that I would record a range of values, some greater and some smaller than the stationary clocks,depending on the x values chosen by the stationary observer. How would I separate out the slow ones as being valid, and the fast ones as being not valid? Wouldn't I deny that my clock was always slow?
Jm

JM

[QUOTE=PAllen;3866528]PAllen- Can you give a specific example in terms of Einsteins stationary and moving frames?
JM

ghwellsjr

I'm not adding that condition--Einstein is (from section 4 if his 1905 paper):
Note the he is talking about "one of the clocks" and as far as I can tell, he meant that it was at rest in the stationary system for negative times and at rest in the moving system for positive times but nothing changes if he instead meant that this clock could have been at rest in the stationary system and it would have behaved like any of the other clocks at rest in the stationary system. But the important thing to note is that he is talking about just one clock, not all the clocks.
Good, I'm glad you see that.
You can pick any one clock at rest any where and at any time in any frame and compare its rate of ticking to all the clocks in any other frame moving with respect to the first frame. That one clock will tick at a slower rate in the first frame than all the clocks in the second frame. I invite you to try the Lorentz Transform to see that this is true.

For example, let's pick the clock at x=321 and t=654 and transform it to a frame moving at 0.6c. The co-ordinates in the second frame are x'=-89.25 and t'=576.75. Now we increment the time on the first clock to t=655 and now x'=-90.00 and t'=578.00. Note that the t' has advanced by 1.25 while t has advanced by 1. And note also that it's a different clock that we are comparing the time to (x' has changed from -89.25 to -90.00).

Note that we are actually working the problem backwards. If we treat the second frame as the "stationary" frame and the first clock as moving in it, then the first clock is ticking at a slower rate than the co-ordinate time of the second frame. Use any other example and the same thing will hold true.

DaleSpam

Relativity is more than one section of one paper. This is an absurd restriction. The Lorentz transform is a transform between different inertial coordinate systems. You can use it to analyze clocks following any timelike worldline, and you can use as many frames as you like. That is firmly established in the theory, regardless of if it was specifically included in one section of one specific paper.

This kind of extreme censorship is not acceptable.

I did, with the formula on proper time that I posted. It applies for all inertial frames in flat spacetime, as you have been discussing.

I missed it.

JM

George- I don't understand this. Doesn't the synchronization procedure guarantee that all clocks at rest with each other must read the same value of time?
I tried the idea of following the path of a single clock and using the x transform, but the result is the same, the slow clock formula applies only for the case of x = v t, but there are other relations between x and t for which the moving clock is not slow. See my example in post 42. If moving clocks are always slow then these other values of x and t must be set aside and no event be allowed to occur there. And if events are allowed everywhere in the stationary frame there will be some events where t'-moving is not less than t-stationary.
JM

yuiop

Hi JM, I completely misread that you were specifying x as variable dependent on t but DaleSpam eventually straightened me out . So yes, when x = 0.5 ct and the relative speed of frame S and S' is 0.8c then t' = t. In this case t' is the coordinate time measured in S' by 2 clocks at rest in S' and t is the coordinate time measured in S by 2 clocks at rest in S. Neither t' or t is a proper time interval measured by a single clock. By specifying x = 05 ct you are saying the events are equivalent to the end points of a particle moving at 0.5 c relative to S and this particle would not be at rest in either S or S'. If you measured the proper time between the two events using a single clock moving inertially and present at both events, then the proper time would be 0.6t. This proper time is less than the coordinate time measured in S or S' or any other reference frame with relative motion.

You have touched on the subject of whether x and t are independent or not several times and I think this is part of where the confusion lies. x and t can be completely independent and just label the coordinates of events, or you can if you wish, make them dependent as you have done. For example let us say we have a particle at coordinates (x,t) = (10,0) and one second later it is at coordinates (x,t) = (10.5,1). You can see that in this case that Δx = 0.5 cΔt but x≠0.5ct and is actually x=10+0.5ct.

Another source of confusion is the the statement "a moving clock always reads less time than a stationary clock" applies to a single moving clock and not to calculations obtained from multiple clocks.

Here is another example. Let us say that Δx=0, v=0.8 in the equation at the top, then we get Δt' = 1.666 Δt and conclude that the time measured by the frame in which the clock is moving (S') is greater than the time measured in the frame in which the clock is at rest (S).

OK, now if you allow Δx≠0 in the equation at the top, then we could have an extreme example where the relative velocity of the two frames is 0.8c and Δx = 0.8 and calculate that Δt' = 0.6 Δt and possibly mistakenly conclude that the time measured by the frame in which the clock is moving (S') is less then the time measured in the frame in which the clock is at rest (S). The mistake here is that by specifying Δx = 0.8 is no longer at rest in S but is now at rest in S'. When neither Δx or Δx' are zero, there is no clear definition of which frame is the frame in which the clock is moving and in which frame the clock is at rest in.

ghwellsjr

It's not enough that they are at rest with each other--they also must be at rest in the frame in which they were synchronized and they must remain at rest in that frame forever.
Remember, Einstein's goal in his paper:
He's not concerned about the actual time displayed on the clock but how its rate of ticking compares to the rate of ticking of the clocks in the stationary system. You are looking at the actual times on the clocks. What you need to do is what I showed you in my previous post which is to compare two events in both frames where the the clock is stationary in the moving frame and moving in the stationary frame.

So here's the process:

Pick two frames such that frame 1 is moving at v/c with respect to frame 2.
Pick any event in the frame 1. Call this event A1.
Change the time to any other value. Call this event B1.
Transform event A1 to event A2 in frame 2.
Transform event B1 to event B2 in frame 2.
Subtract the time co-ordinates for events A1 and B1 and call this Δt1.
Subtract the time co-ordinates for events A2 and B2 and call this Δt2.
Divide Δt1 by Δt2 and call this TD.

Verify that TD=√(1-v²/c²)

Here's an example with [t,x]:

We'll make frame 1 move at .8c with respect to frame 2.
We'll pick A1 to be [1234,5678]
We'll pick B1 to be [4321,5678]

A2 transforms to [-5514,7818]
B2 transforms to [-369,3702]

Δt1 is 1234-4321 = -3087
Δt2 is -5514-(-369) = -5145
TD is Δt1/Δt2 = -3087/(-5145) = 0.6

Verify that TD=√(1-v²/c²) = √(1-0.8²) = √(1-.64) = √(0.36) = 0.6

The only difference between this example and the process that Einstein was doing is that he picked the x co-ordinates for A1 and B1 to be 0 and he picked the time co-ordinate for B1 to also be 0. This just means that he doesn't have to do the subtraction process because the rates of the clocks now are identical to the actual times on the clocks.

So let's repeat with these conditions:

We'll make frame 1 move at .8c with respect to frame 2.
We'll pick A1 to be [1234,0]
We'll pick B1 to be [0,0]

A2 transforms to [2056.667,-1645.333]
B2 transforms to [0,0]

Δt1 is 1234-0 = 1234
Δt2 is 2056.667-0 = 2056.667
TD is Δt1/Δt2 = 1234/2056.667 = 0.6

JM

[QUOTE=yuiop;3872155]Hi JM, I completely misread that you were specifying x as variable dependent on t but DaleSpam eventually straightened me out . So yes, when x = 0.5 ct and the relative speed of frame S and S' is 0.8c then t' = t.[/QUOTE ]
This is progress that you agree with my analysis. The process I use is the same used to get the slow clock formula. We both start with the time transform equation. I insert x = 0.5 ct,and 'slow' inserts x = v t. So the resulting equations are on the same footing. If t' = t √(1-v²/c²) means that the moving clock is slow, then t' = t means that the moving clock is not slow.
I'm not sure what you mean. In section 3 each frame has many clocks, and x and t are allowed a wide range of values independently. The above calcs show that the moving clock can be slow or not. Do you mean that in this case an analysis of a single clock would give a different result? Or are you refering to section 4 where there is only one clock?

I will need more time to review the rest of what you have written. In any event I must turn my attention to other pressing matters. I will check back to see any new posts at this thread
Thanks again for your attention.
JM

JM

George, Thanks for your detailed post 80. I want to study it but I must attend to other matters. I may reply but I intend to look in to see any new posts.
I think I gained some new understanding from these discussions, but each step seems to raise new questions. Who knew there were two theories, a multi-clock one in section 3 and a single clock one in section 4?
Best wishes to you and to all who took the time to contribute.
JM

ghwellsjr

There aren't two theories. It's one continuous discussion with more development in each section. There are multi-clocks stationary in each frame. With two frames, there are two sets of multi-clocks. You can pick any single clock from either frame and compare its rate of ticking to the multi-clocks in the other frame, one at a time, whichever clock it is adjacent to. The single clock in the first frame will tick at a slower rate than the multi-clocks in the second frame.

You can then pick any single clock from the second frame and compare it to the multi-clocks in the first frame and it will tick at a slower rate than the multi-clocks in the first frame, one at a time, whichever clock it is happens to be adjacent to.

So there are multi-clocks all the time, we just focus our attention on any single clock from one frame compared to a succession of multi-clocks in the other frame.

DaleSpam

There are not multiple theories of SR. There is one theory and that theory can handle any number of clocks moving in any possible arrangement. Your failure to work a problem correctly even after being corrected doesn't cause SR to undergo fission.

DaleSpam

The correct expression for the proper time on an arbitrarily moving clock as viewed from any inertial frame is what I posted above.

dt=dτ only if v=0 and otherwise is strictly slow.