How does special relativity account for the time on a single moving clock?

[Dale]

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

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:
$$\tau = \int \sqrt{1-v(t)^2/c^2} dt$$
http://en.wikipedia.org/wiki/Proper_time#In_special_relativity

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

 

[ghwellsjr]

Yes, slow clocks is a universal truth in Special Relativity.
So we always know the tick rate of a clock moving in a Frame of Reference by the simple formula expressed above.

George: I believe I understand your explanation. I have studied section 4 and my questions are:
Einstein refers to a clock qualified to mark the time t (my notation) when at rest relativily to the moving system and so adjusted that it marks the time t. This adjustment seems to mean that the moving clock displays the time t given by the transform, doesn't it?

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).

Then he says "Between the quantities x,t,and τ, which refer to the position of the clock,..." (his notation), x and t being the coordinates of the stationary frame and τ being the time of the moving frame. By what justification does x refer to the position of the clock?

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.

In the transforms, as they are usually viewed, x and t are independent variables allowed over the range -∞ to +∞. If slow clocks is universal then x must be permanently restricted to the values x=vt. If x is an independent variable then there is no significance to where x is 'pointing' because all clocks read the same value wherever located.

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".

Thanks for your participation.
JM

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.

 

[Dale]

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.

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

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 don't 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/c2)/√(1-v2/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]

George and DaleSpam,
I have read your last replies. I don't 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/c2)/√(1-v2/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

^{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]

George and DaleSpam,
I have read your last replies. I don't 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/c²)/√(1-v²/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

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.

 

[Dale]

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)/√(1-v2/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.</sup>

^{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]

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:
$$\tau = \int \sqrt{1-v(t)^2/c^2} dt$$
http://en.wikipedia.org/wiki/Proper_time#In_special_relativity

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

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

 

[Dale]

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

 

[yuiop]

1. Stay within SR, ie all moving clocks are stationary in the frame moving toward +x at speed v.

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.

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/c²)/√(1-v²/c²)

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.

3. x and t are independent variables, ie they can take on any values both + and -.

Seems O.K.

4.For example let x=0.5ct (v/c = 0.8) in the equation to find τ =t.

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.

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.

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.

6. Therefore the moving clocks are not always slow.

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:

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

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

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

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:

$\Delta t = \Delta t' /\sqrt{1-v^2}$

$\Delta t' = \Delta t * \sqrt{1-v^2}$

$\Delta \tau = \Delta t * \sqrt{1-v^2}$

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

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:

$\Delta t' = \Delta t *\sqrt{1-v^2} - v* \Delta x'$

$\Delta t'$ can be greater than $\Delta t$ if $\Delta x'$ 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).

 

[Dale]

1. Stay within SR, ie all moving clocks are stationary in the frame moving toward +x at speed v.
...
4.For example let x=0.5ct (v/c = 0.8) in the equation to find τ =t.

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

 

[yuiop]

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

Please explain. I am not seeing it.

 

[Dale]

Please explain. I am not seeing it.

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

 

[yuiop]

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

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 $\Delta x = 0.5$ when $\Delta t = 1.0$ 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:

4.For example let x=0.5ct (v/c = 0.8) in the equation to find τ =t.

is self contradictory if we interpret it to mean $\Delta x = 0.5 c\Delta t \implies \Delta x / \Delta t = 0.5c \implies v/c = 0.5$?

 

[JM]

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:
$$\tau = \int \sqrt{1-v(t)^2/c^2} dt$$
[\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)'. Let's 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

 

[JM]

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

Dale-- The clocks are moving at 0.8cT, but X is not an indicator of the position of the clocks, its the independent space variable of the stationary frame. There is no dedicated symbol indicating the position of the moving frame.
JM

 

[yuiop]

... some values of X lead to t ≥ T,

This is correct if we consider time intervals measured by spatially separated clocks rather than time intervals measured by a single clock.

The statement 'moving clocks always run slow' applies to single clocks and not to time intervals calculated from multiple clocks far apart from each other.

 

[yuiop]

Dale-- The clocks are moving at 0.8cT, but X is not an indicator of the position of the clocks, its the independent space variable of the stationary frame. There is no dedicated symbol indicating the position of the moving frame.
JM

That is not helpful or meaningful. Please clarify what you were trying to say in statement (4) "For example let x=0.5ct (v/c = 0.8) in the equation to find τ =t."

 

[PAllen]

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:
$$\tau = \int \sqrt{1-v(t)^2/c^2} dt$$
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)'. Let's 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

I think you are confusing some concept here. Proper time as defined in the integral is not a coordinate at all. It gives time elapsed on a single clock following some spacetime path between two specific events. Two different frames may give different labels to all the events on the clocks path, but the computed proper time will come out the same (as will the time elapsed on an actual single clock between two physically defined events).

Within any one inertial frame in SR, this integral demonstrates that any moving clock, no matter its path, will seem to run slower than any stationary clock of that frame. The same will be true in every other inertial frame. The seeming discrepancy gets resolved by the frames differing on clock synchronization. Each thinks the other's clocks at different positions are out of synch.

 

[JM]

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.

I sense a confusion between x-the independent variable in the Lorentz Transforms, and x-indicating the position of the moving clock. Do you agree that x in the LT can assume any of a large range of values? Even some values that result in t-moving > t-stationary?
If x now indicates the position of the moving origin, isn't that different from the LT? Arent we entitled to an acknowledgment of this change, and an explanation of how the new meaning relates to the old, since the same transform is used in both?
JM

 

[yuiop]

Let's use x,y,z,t = moving coordinates and X,Y,Z,T =stationary coordinates.

Better still, use $\Delta x,\Delta y, \Delta z, \Delta t$ = moving frame measurements and $\Delta X,\Delta Y, \Delta Z, \Delta T$ = stationary frame coordinates, then if $\Delta x = \Delta y = \Delta z = 0$ then t will always be less than T and represents a single clock that is at rest in the moving frame and this is called the proper time and uses the symbol tau.

 

[JM]

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.

Again the confusion between x-variable and x-indicator of origin. Hasn't Einstein specified that all the clocks of a given frame are synch-ed using exchange of light signals? If this applies here then all the moving clocks read the same value. This doesn't add much for this case, but is important for other cases, eg Point 4

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

The point here was to show an example of a calculation using x-variable that resulted in t-moving not less than t-stationary. The procedure is to sub the values given into the LT for time.
JM

 

[JM]

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.

Thats what I'm questioning, particularly the universality of his result. My questions and the replies are noted. Your input is appreciated.
JM

 

[yuiop]

I sense a confusion between x-the independent variable in the Lorentz Transforms, and x-indicating the position of the moving clock.

There are not two definitions of x. X always indicates the position of a particular event.

Do you agree that x in the LT can assume any of a large range of values?

Yes.

Even some values that result in t-moving > t-stationary?

Yes.

If x now indicates the position of the moving origin,

x can represent the position of a clock or of the origin or of any other object or event. For example the origin is at x=0, the location of a clock might be x = -5 and the location of Fred might be x = 9 and the location where John crashed his car at t=7 is x=10. You just have to make it clear what you are measuring. For a single clock at rest in frame S' moving relative to S the value of x changes over time so that for example at t=0, x=0 and at t=1, x=0.8 and at t=2, x=1.6 and so on. In the moving frame spatial measurements are made relative to the origin of S' so if the clock is stationary in S', then at t=0, x'=0 and at t=1, x'=0 and at t=2, x'=0.

x and x' are just positions as measured in S and S' respectively. If we mean changes in position over a time interval then we should use $\Delta x$ and $\Delta x'$ respectively, or for brevity, just use x and x' and make it clear we mean spatial separations rather than spatial locations.

 

[JM]

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.

I sense a change of model here, from LT where all clocks move only in the + x direction, to world lines where clocks can move in other directions. My intention is to stay within the 1905 model, and use x as an independent variable of the stationary frame. If t-moving < t-stationary, as in section 4,indicates a slow clock , then t-moving = t-stationary, as above, indicates that moving clocks are not always slow.

I appreciate the comments of all.
JM

 

[yuiop]

The point here was to show an example of a calculation using x-variable that resulted in t-moving not less than t-stationary. The procedure is to sub the values given into the LT for time.
JM

OK you have given that x = 0.5 and v = 0.8 and given the Lorentz transform:

$$t ' = \frac{t-vx}{\sqrt{1-v^2}}$$

we get:

$$t ' = \frac{t-0.8*0.5}{\sqrt{1-0.8^2}} = \frac{t - 0.4}{0.6}$$

then for any value of t>1 we get t'>t.

However t and t' as used here are just coordinates or labels for an event and are not a comparison of clock rates where we have to compare intervals between events. If we mean intervals we should use:

$$\Delta t' = \frac{\Delta t-v \Delta x}{\sqrt{1-v^2}}$$

Now if v=0.8c and $\Delta x$ =0.5 then $\Delta t$ must be 0.5/0.8 = 0.625

so referring to the conclusion above it is obvious in this case that t' <t.

When we talk about intervals (deltas) then $\Delta t$ and $\Delta x$ are not independent of each other, if we are talking about clocks at rest in S' because they are related by v.

Furthermore, if we use the reverse transformation:

$$\Delta t = \frac{\Delta t' + v \Delta x'}{\sqrt{1-v^2}}$$

and note that if the clock is at rest in S', then $\Delta x' = 0$ and we obtain:

$$\Delta t = \frac{\Delta t' }{\sqrt{1-v^2}}$$

then $\Delta t$ is always greater than $\Delta t'$ for all values of v<1 where c=1.

I am sure most of the confusion is because you are not clear on whether you mean coordinate labels or space and time intervals.

 

[Dale]

The clocks are moving at 0.8cT, but X is not an indicator of the position of the clocks, its the independent space variable of the stationary frame.

Then you cannot use x=0.5 ct to describe any of those clocks.

 

[Dale]

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.

Sure, but the equation for clocks that don't start at the origin is $x=0.8ct+x_0$. The problem is the velocity. With x=0.5ct you have a clock which is moving in both frames, not stationary in either one of them (and being stationary was specified in part 1).

 

[Dale]

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)'.

It is not universal. It applies for inertial frames in flat spacetime only. The universal formula is:
$$\tau=\int \sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

However the scenario you have described here uses only inertial frames in flat spacetime so the simplified version applies.

I am hoping to get a clear answer from this discussion.

I hope my answer has been clear.

 

[yuiop]

Sure, but the equation for clocks that don't start at the origin is $x=0.8ct+x_0$. The problem is the velocity. With x=0.5ct you have a clock which is moving in both frames, not stationary in either one of them (and being stationary was specified in part 1).

Ah OK. I concede your point now. Thanks. If x =0.5ct then the clock is moving at 0.5c in S and not stationary in S' which is moving at at 0.8c relative to S. I think I misread what JM intended, but I not convinced that JM is sure what he intended either. I think he needs to clear that up.

 

[Dale]

I sense a change of model here, from LT where all clocks move only in the + x direction

I don't know where you got that idea.

My intention is to stay within the 1905 model

In the 1905 model he analyzed a clock which goes in a circle. Such a clock goes in the +x and +y and -x and -y directions at some point and every combination inbetween. A restriction to clocks moving in the +x direction is not a part of the 1905 model, and indeed is incompatible with the Lorentz transform for boosts to arbitrary speeds.

If t-moving < t-stationary, as in section 4,indicates a slow clock , then t-moving = t-stationary, as above, indicates that moving clocks are not always slow.

Moving clocks are always slow. Your analysis above contradicts itself as I mentioned above.

 

[ghwellsjr]

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 ξ=0) and the origin of the moving frame is defined as x=vt according to the stationary frame.

I sense a confusion between x-the independent variable in the Lorentz Transforms, and x-indicating the position of the moving clock. Do you agree that x in the LT can assume any of a large range of values?

Yes, of course, x can be any value and the transform will work. But for every x, you also have y, z and t and you have to calculate all of them for a complete transformation.

The Lorentz Transform converts the four co-ordinates of an event defined according to one Frame of Reference into the four co-ordinates of the same event defined according to a second Frame of Reference moving at some speed v in the x direction with respect to the first FoR. Using Einstein's nomenclature from section 3 of his 1905 paper, the first FoR has co-ordinates with labels of t, x, y and z, while the second FoR has co-ordinates with labels of τ, ξ, η and ζ. You have to solve all four equations to get the co-ordinates in the second frame. You can't just solve for the time co-ordinate and ignore the spatial co-ordinates.

Even some values that result in t-moving > t-stationary?

In Einstein's nomenclature, you are asking if τ can be greater than t. Of course, there are many events in the first FoR with a t co-ordinate less than the τ co-ordinate in the second FoR. But in general that has nothing to do with a clock moving in a stationary frame. The only time you can use the Lorentz Transform to calculate the time on a clock moving in the stationary frame is when a clock at the origin of the second FoR moves at the same velocity that the second FoR is moving and this will be indicated by the spatial co-ordinates remaining zero in the second FoR while the time co-ordinate is changing.

If x now indicates the position of the moving origin, isn't that different from the LT? Arent we entitled to an acknowledgment of this change, and an explanation of how the new meaning relates to the old, since the same transform is used in both?
JM

The origin of the second FoR is moving along the x-axis of the first FoR at a velocity of v so for any time t in the first FoR, we can calculate the x co-ordinate in the FoR for the origin of the second FoR by using x=vt. This gives us the t and x co-ordinates of an event in the first FoR (y and z are always 0). Then we can plug both the x and t values into the LT and calculate the co-ordinates of the same event in the second FoR and we will find that the time co-ordinate will always be less in the second FoR and the location co-ordinates will be 0.

In fact the time co-ordinate, τ, in the second FoR will be t√(1-v²/c²), making τ always less than t.

 

[ghwellsjr]

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.

Again the confusion between x-variable and x-indicator of origin. Hasn't Einstein specified that all the clocks of a given frame are synch-ed using exchange of light signals? If this applies here then all the moving clocks read the same value.

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.

This doesn't add much for this case, but is important for other cases, eg Point 4

4.For example let x=0.5ct (v/c = 0.8) in the equation to find τ=t.

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

The point here was to show an example of a calculation using x-variable that resulted in t-moving not less than t-stationary. The procedure is to sub the values given into the LT for time.
JM

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]

OK you have given that x = 0.5 and v = 0.8 and given the Lorentz transform:

$$t ' = \frac{t-vx}{\sqrt{1-v^2}}$$

we get:

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