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

[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 &#039; = \frac{t-vx}{\sqrt{1-v^2}}

we get:

t &#039; = \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&#039; = \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&#039; + v \Delta x&#039;}{\sqrt{1-v^2}}

and note that if the clock is at rest in S', then \Delta x&#039; = 0 and we obtain:

\Delta t = \frac{\Delta t&#039; }{\sqrt{1-v^2}}

then \Delta t is always greater than \Delta t&#039; 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 &#039; = \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.

 

[JM]

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

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

 

[Dale]

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.

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]

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.

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

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]

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.

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]

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).

PAllen- Can you give a specific example in terms of Einsteins stationary and moving frames?
JM

 

[ghwellsjr]

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.

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?

I'm not adding that condition--Einstein is (from section 4 if his 1905 paper):

Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time τ when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time τ. What is the rate of this clock, when viewed from the stationary system?

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.

If so then under these conditions the moving clock 'always' runs slow.

Good, I'm glad you see that.

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

You can pick anyone 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.

 

[Dale]

Dale- When I refer to the 1905 model I mean that presented in section 3.

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.

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 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 hope you see from my added descriptions thay my analysis doesn't contradict.

I missed it.

 

[JM]

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.

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]

OK you have given that x = 0.5 and v = 0.8 and given the Lorentz transform:
t &#039; = \frac{t-vx}{\sqrt{1-v^2}}
we get ... 0.625

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

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]

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.

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?

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.

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

Remember, Einstein's goal in his paper:

What is the rate of this clock, when viewed from the stationary system?

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]

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.

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.

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 referring 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 into 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]

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 into 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

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.

 

[Dale]

Who knew there were two theories, a multi-clock one in section 3 and a single clock one in section 4?

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.

 

[Dale]

If t' = t √(1-v²/c²) means that the moving clock is slow, then t' = t means that the moving clock is not slow.

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

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&gt;dt where t is the time coordinate in an inertial frame and v is the clock's velocity in that frame.

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

 

[yuiop]

If t' = t √(1-v²/c²) means that the moving clock is slow, then t' = t means that the moving clock is not slow.

With the conditions you specified, yes, t'=t but neither t or t' measurements were obtained using a single clock. The single clock is important. In your example the single clock is moving at 0.5c relative to frame S. If we make the first measurement in S when the moving clock passes the origin so x1=0, and another measurement 1 second later, we make the second measurement when the moving clock is at x2=0.5. The time measurement in frame S requires 2 clocks (one at x1=0 and the other at x2=0.5), so it is not a proper time measurement. In frame S' measurements are made at x1'=0 and x2'=-0.5 so in frame S' the time interval has to be measured using 2 clocks. The time interval measured by the single clock moving at 0.5c relative to frame S is tau=1*sqrt(1-v^2) = 0.866 seconds which is less than time interval measured in S or S'.

The important concept is that the time interval between two events measured by a single clock that is present at both events is always less than the time interval measured in any other reference frame using two clocks. A time interval measured by a single inertial clock is called a proper time measurement.

This idea can be expressed another way. If we have two events and can find a reference frame where those two events happen in the same place, then the time measured in that
frame will always be shorter than the time measured in any other frame.

In your example we had a clock moving at x= 0.5 ct or 0.5c relative to frame S. In another reference frame moving at 0.5c relative to S (and co-moving with the moving clock), the time interval between 2 events will be less than the time interval between the same 2 events measured in any other reference frame. In this co-moving frame the "moving" clock is stationary. As you can see, moving is a relative concept and so the expression "the moving clock" is not very well defined. It is better to say the proper time (which is measured by a single clock) is always less than the coordinate time (which is measured by more than one clock).

Let me know if that clears things up for you.

 

[Dale]

Btw, JM, the t and t' in the Lorentz transforms are coordinate times in inertial frames. Not necessarily the time on any clock. The time on a clock is given by the expression I gave. That expression reduces to the coordinate time only for the case v=0. I.e. only clocks at rest measure coordinate time. Your v=.5c clock is not at rest in either frame so it does not measure coordinate time in either frame.

 

[Rock2shark]

The more I read the above discussion, the more it seems that the purpose of all our physical theories, instruments, and experiments, being to explore the actual physical world outside whatever distortions may take place due to our perceptions, appears to be overlooked. In short, the objective of physics appears to have been drowned by a preoccupation with psychological problems and issues closer to those of philosophy.

One problem that faced Einstein was the counter intuitive nature of the constancy of the speed of light. The other problem was the difficulty of finding empirical indicators of the true nature of the physical world. This could not have been more graphically illustrated than by the failure of the Michaelson Morley experiment to detect any variation in the speed of light, despite the undoubted high velocity of the Earth through the luminiferous aether, and which experiment did so by its very design which was such that not even the use of clocks was required, but which experiment instead used a comparison of wavelengths that could not have done other than pass through alternate reference frames relative to the aether.

I think that it is of immense importance to understand this, and to thereby understand that Einstein's second postulate, containing as it did the specification "... regardless of the state of rest or motion of its source ..." as in the English translation of his postulate, was based on the net empirical evidence accumulated in human experience by the time that he wrote down his second postulate. Due to the supposed fact of the existence of the aether, proven as it was thought by the fact that light traveled with ease through empty space yet at the same time possessed a wave quality among its characteristics, Einstein necessarily had also to deal with the aether. He was too wise to say that the aether does not exist, so he simply stated what appeared to be the case, and to do so with confidence if in the knowledge of the Michaelson Morley experiment, and that is, "No experiment of any sort can detect the aether."

So, the constancy of the speed of light was a distillation of Einstein’s of all known experience and also a translation of this in order to ascertain a fact of the physical world. All testing of that ‘fact’ hinges on comparing reference frames, and that comparison of reference frames involves the use not only of standard clocks, but also standard rulers, or rigid bodies.

I have always felt that it is hazardous to any understanding of Special Relativity not to understand the above. Furthermore, an understanding of the above I suspect could clarify any question, should such question exist, as to the purpose of the clocks, the relevance of their exact construction, or their physical accuracy, and that these only really matter to the experimental physicist.

To the theoretical physicist, the clocks are a kind of tool that is used, along with rigid bodies, to compare reference frames in motion relative to one another, whose exact construction does not matter, only their ability to be understood to represent identical ways of measuring time in the reference frames under scrutiny such that observers in those reference frames have identical experiences of time and space as provided by those tools.

This is not quite the case for the experimental physicist. Unfortunately however, for the experimental physicist, there are other problems in addition to the actual practical accuracy of the instruments, namely that of measuring the time and space of reference frames that cannot be occupied by independent observers and therefore have to be understood by inference rather than by direct measurement.

 

[JM]

The important concept is that the time interval between two events measured by a single clock that is present at both events is always less than the time interval measured in any other reference frame using two clocks. A time interval measured by a single inertial clock is called a proper time measurement.

Well finally! The usual statement that 'moving clocks run slow' says nothing about proper clocks, and it has taken 5 pages to get to it here. The description in section 4 is of a proper clock. So why isn't the phrase ' proper clocks run slow' used? It certainly seems to clear things up a lot. That I can accept.
So I see the following picture.The Lorentz transforms appear general, allowing many different arrangements of events in the stationary frame. The special case of x = v t makes the clock at the origin of k into a proper clock. This should not inhibit the use of other arrangements of events, such as the one with x = 0.5ct, where there may not be any proper clocks.

In your example we had a clock moving at x= 0.5 ct or 0.5c relative to frame S. In another reference frame moving at 0.5c relative to S (and co-moving with the moving clock),..

In my example I did not imply or mean that there was a clock at x=0.5ct. x is the independent variable describing the location of events, eg lightning strikes, light flashes, or trains arriving re the stationary frame. I think of the x = v t in section 4 the same way, as a series of events, it's only in a roundabout way that x is the clock position.

Let me know if that clears things up for you.

I will give it some thought, but I think that does it for now.

Thanks again, y'all.

 

[Dale]

Well finally! The usual statement that 'moving clocks run slow' says nothing about proper clocks, and it has taken 5 pages to get to it here. The description in section 4 is of a proper clock. So why isn't the phrase ' proper clocks run slow' used?

All clocks measure proper time, there isn't a subset of clocks called proper clocks.

Also, it didn't take five pages, I posted the Wikipedia link on proper time back in post 36. Did you not even bother to read it?