Special Relativity Clocks

JM

George and DaleSpam,
Are we done here?

The discussion here leads to the suggestion that the 'slow clock' idea, with τ<t, is only one possible result for the moving clocks. Choices of x other than x=vt lead to different relations, such as τ>t for x=0. Synchronization means that the result applies to all the moving clocks, including the one at the origin of the moving frame.

Do you all accept this idea, or do we have some more to talk about?

JM

DaleSpam

No, I don't accept the idea.

If you have a clock which is moving in an arbitrary fashion (including, but not limited to, x=vt) you use the following formula to calculate the time displayed on the clock:
[tex]\tau = \int \sqrt{1-v(t)^2/c^2} dt[/tex]
http://en.wikipedia.org/wiki/Proper_...ial_relativity

The integrand is always less than or equal to 1, so you never get [itex]d\tau>dt[/itex] where t is the time coordinate in an inertial frame and v is the clock's velocity in that frame.

ghwellsjr

In the nomenclature of the transform that Einstein developed in section 3 of his 1905 paper, he uses τ (tau), not t or t', as the time on the moving clock. [NOTE: in his version of the LT, he uses β, beta, as the Lorentz factor instead γ, gamma, which is in common usage today. We now use β to mean v/c. Also, we commonly use t' to refer to the transformed time. Just don't get confused by this difference in nomenclature.]

In any case, I explained what Einstein means in the part of my quote from post #28 that you left out: t is the time on a clock that was at rest in the stationary frame prior to t=0 and then at t=0 it instantly accelerates to velocity v and so becomes at rest in the frame moving at v where the transformed time is represented by τ. For any given time t in the stationary frame, you can calculate the time τ on a clock at the spatial origin of the moving frame using the simple formula τ=t√(1-v²/c²).
It is justified because he takes a clock from the stationary frame located at the origin (t=0 and x=0) and puts it at the origin of the moving frame (τ=0 and ζ=0) and the origin of the moving frame is defined as x=vt according to the stationary frame. This is simply what using the Lorentz Transformation is all about. Since we know where the origin of the moving frame is at any time in the stationary frame, we also know where the moving clock is since it is at rest at the origin of the moving frame.
The whole point of Einstein's derivation is to eliminate x from the equation but if you want, you can include it and say for any given x we can calculate both t and τ. Or you could say that for any given t we can calculate x and τ. We are of course assuming that v is constant and that we only care about t≥0, τ≥0 and x≥0.

After having developed the relationship between the time on a moving clock relative to the times on the stationary coordinate clocks, we extrapolate to the more general case of delta times so that we don't have to be restricted to the origin of a specific frame or even a specific speed and we can determine the instantaneous tick rate of an accelerating and moving clock compared to the coordinate time.

But because √(1-v²/c²) can only be a number less than 1, we know that a moving clock will tick slower than the stationary coordinate clocks and that's why we say "slow clocks is a universal truth in Special Relativity".
You're welcome and I apologize for taking so long to respond to your questions--I just don't recall seeing your post until now.

DaleSpam

I hope I didn't cause any confusion. I was using τ as proper time. I don't know what usage JM intended.

ghwellsjr

I was showing Einstein's derivation of Proper Time in post # 28 so it shouldn't have been confusing.

JM

George and DaleSpam,
I have read your last replies. I dont understand clearly the reasoning of 1905 where he starts a clock from rest to develop the standard formula for 'slow clock'. Let me state in detail my reasoning.
1. Stay within SR, ie all moving clocks are stationary in the frame moving toward +x at speed v.
2.The transform equation relating the time τ of the moving frame to the coordinates x and t of the stationary frame is τ = ( t - vx/c[SUP]2[SUP])/√(1-v[SUP]2[SUP]/c²)
3. x and t are independent variables, ie they can take on any values both + and -.
4.For example let x=0.5ct (v/c = 0.8) in the equation to find τ =t.
5. Since the moving clocks are synchronized this result applies to all the moving clocks, including the one at the origin of the moving frame.
6. Therefore the moving clocks are not always slow.
Comments?Is there a relation between the above and the section of 1905 on 'slow clocks'?
JM

PAllen

For one, you are confusing coordinate time and proper time. Your equation in (1) relates coordinate times of separated clocks. More precisely, it relates: if observer A synchronizes distant clocks using Einstein synchronization, how will observer B (moving relative to A) describe the results if they also use Einstein synchronization between clockes. Proper time (tau) is a completely different animal. It is only defined along the history of a single clock. As shown in Einstein's 1905 paper, every observer perceives every clock (moving or not), to go either the same rate as theirs (if not moving relative to said observer), or slower than said observer's clock. Note, especially, that if A interprets B's clock as slow, then B interprets A's clock as slow.

ghwellsjr

Based on your post #26, I can see that you know how Einstein got from this:



to the first part of this:



and he did it by replacing x with vt but remember, there is more to the Lorentz Transformation than just the formula for τ. There are also the formulas for the spatial co-ordinates and if we plug x=vt into x'=γ(x-vt) we get:

x'=γ(vt-vt)=0

So this tells us that it is not all the clocks in the moving frame that the time co-ordinate applies to but only the one at the spatial origin of the moving frame which is where the moving clock that we are considering is located.

Now, concerning 3, x and t are not independent of each other in this situation, they are related by x=vt, as you pointed out in your post #26.

Concerning 4, I can't tell what you are doing, can you provide more detailed steps?

Points 5 and 6 were covered in my earlier comments.

DaleSpam

OK, so x=0.5ct is the worldline of a clock which is moving at 0.5c in the +x direction in the stationary frame. Boosting by 0.8c gives you a clock which is moving at 0.5c in the -x direction in the moving frame. So yes, you have correctly determined that a clock which is moving at .5c in the +x direction in the stationary frame is slowed by the same amount as a clock which is moving at .5c in the -x direction in the moving frame.

morrobay

I just posted a numerical proper time - velocity vs acceleration - problem
in the Homework Intro Physics section (page one) see update
With the evaluation of the above integral and it was not a good answer since the
recorded proper time of the (constant) accelerating clock ( with respect to earth clock )
was greater than earth clock ? So I have questions on that integral.
Once again , discussing proper time is confusing , so numerical problems might help

DaleSpam

I checked your math, and it seems all right, but your answer was wrong. I don't know if you accidentally plugged it in to the integrator wrong or if the integrator had some numerical problems.

yuiop

Slightly odd way of expressing things, but otherwise OK. Let us try and define things a little more clearly. We have two reference frames S and S' in the standard configuration, which have a relative speed v in the x direction. Clocks at rest in S' have have a velocity of +v in the +x direction as measured in S, and clocks at rest in S appear to be moving in the -x' direction as measured in S'. Time measured by clocks at rest in S' are denoted by primed variables such as t'. The v mentioned in the standard Lorentz transforms is always the relative velocity of the the two reference frame as measured in S.
You are using the symbol tau which normally stands for proper time, but the symbol on the left of that equation is actually a coordinate time as measured in the in frame S'. The equation is beter expressed as:

t' = ( t - vx)/√(1-v²)

where I am using units such that c=1 to make things more manageable.
Seems O.K.
Finding tau = t does not make much sense except in the case there is no relative motion. We can however find what the value of t' is when t=0, x = 0.5 and v=0.8 when the clock at the origin of S is next to the clock at the origin of S'.

t' = ( t - vx)/√(1-v²)

t' = ( 0 - 0.8*0.5)/√(1-0.8²)

t' = ( 0 - 0.4)/0.6 = -0.66666 seconds.

No it does not. When t=0, x = 0 and v=0.8 when the clock at the origin of S is next to the clock at the origin of S':

t' = ( t - vx)/√(1-v²)

t' = ( 0 - 0.8*0)/√(1-0.8²)

t' = ( 0 - 0)/0.6 = 0 seconds.

There is a difference of -0.6666 seconds between the times of the clocks at rest in S' when the clocks at rest in S are all reading 0 according to the observers at rest in S.
I don't think anyone knows how you arrived at this conclusion and you have not shown any algebraic or numerical examples of a situation where the clocks at rest in S' are not slower than clocks at rest in S. I think you are not clear in your mind about the differences between coordinate times that label events, elapsed times that measure the time interval between different events and the differences between proper times and coordinate times.

The equations shown so far only concern coordinate times that label events and says nothing about the relative rates at which clocks with relative motion run.

To obtain the elapsed time (t2-t1) in frame S, between two events when the elapsed time interval between those two events in frame S' is (t2'-t1') we use:

[itex] t2-t1 = \Delta t = ( t2' + v*x2')/\sqrt{1-v^2} - (t1' + v*x1')/\sqrt{1-v^2} [/itex]

[itex] \Delta t = (\Delta t' + v*x2' - v*x1')/\sqrt{1-v^2} [/itex]

[itex] \Delta t = (\Delta t' + v* \Delta x')/\sqrt{1-v^2} [/itex]

Since we after the proper time in the primed frame we only use a single clock at rest in that reference frame, so x2' must equal x1' so we can say:

[itex] \Delta t = \Delta t' /\sqrt{1-v^2} [/itex]

[itex] \Delta t' = \Delta t * \sqrt{1-v^2} [/itex]

[itex] \Delta \tau = \Delta t * \sqrt{1-v^2} [/itex]

Now using the above equation can you find a single instance when [itex] \Delta \tau > \Delta t * \sqrt{1-v^2} [/itex]?

Maybe what you are getting at is that the coordinate time interval as measured in S' between between two events may be longer than the coordinate interval between those two events as measured in S? For example using:

[itex] \Delta t' = \Delta t *\sqrt{1-v^2} - v* \Delta x' [/itex]

[itex] \Delta t'[/itex] can be greater than [itex] \Delta t [/itex] if [itex] \Delta x' [/itex] is negative, but this does not mean individual clocks in S' are running faster than individual clocks in S according to observers at rest in S. It is simply a result of how clocks are synchronised and the relativity of simultaneity (What appears simultaneous in one rest frame is not simultaneous in another reference frame with relative motion).

DaleSpam

I just noticed this. These two conditions are mutually contradictory.

yuiop

Please explain. I am not seeing it.

DaleSpam

If the clocks are stationary in the moving frame then their worldline is x=0.8ct, not x=0.5ct.

yuiop

Still not seeing it. X is just a coordinate. If he said the clock started at the origin at (t,x) = (0,0) and specified a time duration of delta T = 1 then yes I would expect the clock to be at coordinates (1.0,0.8) but he did not specify a time duration. X depends on t.

Even if he specified a duration of 1 in S, the location of the moving clock is not necessarily 0.8 if the clock did not start at the origin, (which he did not specify). If he had made it clear that he meant [itex]\Delta x = 0.5[/itex] when [itex]\Delta t = 1.0[/itex] then there would be a contradiction when v=0.8, but he did not specify a time interval or a starting coordinate or that he talking about intervals (differences) rather than coordinates of individual events.

Perhaps you mean that the single statement:
is self contradictory if we interpret it to mean [itex]\Delta x = 0.5 c\Delta t \implies \Delta x / \Delta t = 0.5c \implies v/c = 0.5[/itex]?

JM

[QUOTE=DaleSpam;3854185]No, I don't accept the idea.

If you have a clock which is moving in an arbitrary fashion (including, but not limited to, x=vt) you use the following formula to calculate the time displayed on the clock:
[tex]\tau = \int \sqrt{1-v(t)^2/c^2} dt[/tex]
[\QUOTE]
Dale--I see where this formula comes from, and I think it is a valid result. The question is whether it is a 'universal' result, ie 'moving clocks always run slow, meaning t (moving)
<T(stationary)'. Lets use x,y,z,t = moving coordinates and X,Y,Z,T =stationary coordinates. Either X and T are independent coordinates and some values of X lead to t ≥ T, or X is restricted to X=vT and t is always < T. The two appear to be mutually exclusive. The 1905 paper is careful and rigorous in section 3 where the Lorentz transforms are derived and uses those transforms in Part 2 to analyze the Maxwell equations. But section 4 is vague and puzzling ( I have heard many interpretatons of what Einstein meant) and the results are not used elsewhere. So I prefer that 'slow clocks' is only one example, but not universal. I am hoping to get a clear answer from this discussion.
JM