Benefits of time dilation / length contraction pairing?

[neopolitan]

neopolitan:

Why ? The muon is at rest in it's own frame.

The muon is stored in a ring, it is tracing out a circular trajectory. There is no single rest frame for the stored muon over 2.2 milliseconds.

 

[JesseM]

Jesse,

You have an equation, right, $$t' = \gamma t$$. One t is primed, one t is not primed. The equation is discussing the effects of motion on time intervals with the underlying assumption that one value applies to a clock in one frame and one value applies another clock in another frame and so long as $$\gamma$$ is not equal to 1, the clock, and therefore the frames, are not at rest relative to each other. Look at the equation. It is taking the perspective of one clock in it's rest frame.

I don't know what it means to "take the perspective" of a frame when your equation expresses a relation between two frames. t represents the time-interval between two events on the worldline of a clock as measured in the coordinates of the clock's rest frame, and t' represents the time-interval between the same two events as measured in the coordinates of a frame where the clock is moving at speed v (which we can imagine as the rest frame of some external 'observer' although this is not really necessary). The equation relates the time intervals of the two different frames--why do you think it's "taking the perspective" of one of them?

 

[JesseM]

The muon is stored in a ring, it is tracing out a circular trajectory. There is no single rest frame for the stored muon over 2.2 milliseconds.

In that case you can't really use the time dilation equation without additional justification, since it's normally meant to tell you the time dilation of an inertial clock as measured in a clock where it's moving--as it happens the math still works for the muon since the speed is constant, but the full equation for time elapsed on an accelerating clock with speed v(t) in an inertial frame would be $$\int \sqrt{1 - v(t)^2/c^2} \, dt$$.

 

[neopolitan]

No, it doesn't. You are asking how many ticks will occur between two specific events "in the primed frame" (your words), so all that matters is the difference in time-coordinate between these events in the primed frame, the fact that you have defined "yourself" to be at rest in the unprimed frame is irrelevant to the problem as you've stated it. If you instead wanted to know how many ticks occur between these events in your rest frame, then the issue of which frame you were at rest in would be relevant, but that isn't what you asked.

Jesse,

Let me repeat the question again. I will highlight something else for your attention.

According to me, how many ticks will I calculate to occur in the primed frame between two events if:
1. $$\Delta t$$ is the number of ticks in the unprimed frame between those two events,
2. a clock in the primed frame is has a speed of v relative to me,
3. I am at rest with the clock in the unprimed frame, and
4. I remember to take into account the inertial motion of the other frame (thus adding or deleting some ticks as required)?

More ticks or less ticks than the unprimed frame?

I know the time interval between the two events in the unprimed frame. I know the speed, relative to me, of a clock in the primed frame (a clock which is at rest in the primed frame, if you prefer). I want to know how many ticks occur in the primed frame between the two events, taking into account the motion of the clock relative to those events. I think that is clear above.

What you could have justifiably called me on is that I said "speed" when I should have said "velocity" (direction and speed) and that I assumed but did not specify that I knew all about the events (not only when but also where they take place, in the unprimed frame). Mea culpa.

If I specify that I knw the velocity and that I know all about the events (time and location, in the unprimed frame), does that make things easier?

cheers,

neopolitan

 

[neopolitan]

In that case you can't really use the time dilation equation without additional justification, since it's normally meant to tell you the time dilation of an inertial clock as measured in a clock where it's moving--as it happens the math still works for the muon since the speed is constant, but the full equation for time elapsed on an accelerating clock with speed v(t) in an inertial frame would be $$\int \sqrt{1 - v(t)^2/c^2} \, dt$$.

True, blindingly irrelevant given the immediate context, but true.

 

[neopolitan]

I don't know what it means to "take the perspective" of a frame when your equation expresses a relation between two frames. t represents the time-interval between two events on the worldline of a clock as measured in the coordinates of the clock's rest frame, and t' represents the time-interval between the same two events as measured in the coordinates of a frame where the clock is moving at speed v (which we can imagine as the rest frame of some external 'observer' although this is not really necessary). The equation relates the time intervals of the two different frames--why do you think it's "taking the perspective" of one of them?

Context Jesse. Context. Do you grasp the concept?

You have an equation, right, $$t' = \gamma t$$. One t is primed, one t is not primed. The equation is discussing the effects of motion on time intervals with the underlying assumption that one value applies to a clock in one frame and one value applies another clock in another frame and so long as $$\gamma$$ is not equal to 1, the clock, and therefore the frames, are not at rest relative to each other. Look at the equation. It is taking the perspective of one clock in it's rest frame. I know you can swap the perspective over, from clock A's perspective to clock B's perspective, if you like - but still t will be the time interval for the clock whose perspective we are examining, in other words the frame in which the clock whose perspective we are examining is at rest. In terms of the equation, we could call the unprimed frame "the rest frame". But it is context.

One clock in one frame. Another clock in another frame.

You are comparing them. But you can't compare them from outside because they are equivalent, neither has precedence, neither is privileged.

What you can say is that, according to one, the other runs slow. And it doesn't matter which clock you pick as "one", they both run slow according to the other.

When I say "according to one" above, I mean the same as "from the perspective of one". They are meant to be equivalent statements. If you don't like one variant, replace it in your mind with the other.

cheers,

neopolitan

 

[JesseM]

Jesse,

Let me repeat the question again. I will highlight something else for your attention.

OK, I see that because of the way you described it, we only know the clock at rest in the primed frame has speed v in the unprimed frame by virtue of the fact that it has speed v relative to you, and you are at rest in the unprimed frame. But as I said, you could easily cut out the middleman and remove the observer from the statement of the problem, shortening the statement of the problem to the two assumptions I gave in post #28, the physical content would be exactly the same. More importantly, the "according to me" phrase at the beginning is superfluous; as long as your assumptions can be used to infer that the primed frame is moving at v relative to the unprimed frame, then the answer to the question "how many ticks will I calculate to occur in the primed frame between two events ... more ticks or less ticks than the unprimed frame?" will be the same regardless of who is doing the calculating, it will only depend on which events you pick in the unprimed frame. I thought your "according to me" might suggest a misunderstanding about that fact, that somehow the answer could be observer-dependent, which is why I said the observer was irrelevant.

I know the time interval between the two events in the unprimed frame. I know the speed, relative to me, of a clock in the primed frame (a clock which is at rest in the primed frame, if you prefer).

Are the two events supposed to be events on the worldline of that clock (so they occur at the same position in the primed frame), or two events on the worldline of an object at rest in the unprimed frame, or are they completely arbitrary?

I want to know how many ticks occur in the primed frame between the two events, taking into account the motion of the clock relative to those events. I think that is clear above.

As I said in post #28, in general for arbitrary events the answer would be $$\Delta t' = \gamma (\Delta t - v \Delta x /c^2)$$. If you specify that the events occur at the same position in one of the two frames, then an answer in terms of the time dilation equation can be given.

 

[JesseM]

Context Jesse. Context. Do you grasp the concept?

Yes, I understand the concept of "context", you just aren't making your notion of the context particularly clear.

You have an equation, right, . One t is primed, one t is not primed. The equation is discussing the effects of motion on time intervals with the underlying assumption that one value applies to a clock in one frame and one value applies another clock in another frame and so long as is not equal to 1, the clock, and therefore the frames, are not at rest relative to each other. Look at the equation. It is taking the perspective of one clock in it's rest frame. I know you can swap the perspective over, from clock A's perspective to clock B's perspective, if you like - but still t will be the time interval for the clock whose perspective we are examining, in other words the frame in which the clock whose perspective we are examining is at rest. In terms of the equation, we could call the unprimed frame "the rest frame". But it is context.

One clock in one frame. Another clock in another frame.

You are comparing them. But you can't compare them from outside because they are equivalent, neither has precedence, neither is privileged.

First of all, I'm not really comparing the measurements of two physical clocks, I'm comparing the time intervals between a pair of events as defined in two different frames (if you're dealing with a pair of events that don't happen at the same location in one frame, you need two different clocks which are at rest and 'synchronized' in that frame to measure the time interval between them). But that's easy enough to fix, we can just rewrite your two sentences above as A time interval in one frame. Another time interval in another frame. But with this modification I can't make sense of your subsequent statement "you can't compare them from outside because they are equivalent, neither has precedence, neither is privileged." Of course, neither is privileged physically, but they're just two numbers, why can't I compare them? If in one frame the time-interval between events A and B is 5 seconds, and in a second frame the time-interval between the same events is 10 seconds, then clearly I can say "the time-interval between these events is twice as large in the second frame as it is in the first frame". Have I somehow "privileged" one of these time-intervals in making this statement? If so, which one?

What you can say is that, according to one, the other runs slow. And it doesn't matter which clock you pick as "one", they both run slow according to the other.

OK, so your argument does depend critically on the notion that you want to compare the rate that two physical clocks are ticking rather than comparing the time-intervals between a specific pair of events? If so, then in this case I agree that we can't say which is ticking faster or slower without first picking a frame. But the time dilation equation as it's normally written is about time-intervals, not instantaneous rates of ticking. Maybe a failure to realize this could be the source of much of your confusion? If we have an unprimed clock moving at velocity v relative to the primed clock, then the normal time dilation equation $$\Delta t' = \Delta t * \gamma$$ can be written in words like this:

"time interval between two events on worldline of unprimed clock as measured in primed frame = time interval between two events on worldline of unprimed clock as measured in unprimed frame (which is of the same as the time interval between those events as measured by the unprimed clock itself, since it's at rest in the unprimed frame) * gamma"

On the other hand, if we wanted to talk about clock rates in the primed frame, then we could use $$d\tau' /dt'$$ to represent the rate the primed clock is ticking relative to the primed frame's coordinate time (this would just be equal to 1 of course), and $$d\tau / dt'$$ to represent the rate the unprimed clock is ticking relative to the primed frame's coordinate time (this would be less than 1), in which case the equation would be:

$$d\tau' / dt' = (d\tau / dt' )/ \gamma$$

So in this equation, we are dividing by gamma rather than multiplying by it as in the standard time dilation equation. And in this equation we are clearly looking at things "from the perspective" of a particular frame, namely the primed frame where the unprimed clock is moving. This equation could be written in words like this:

"Rate that primed clock is ticking in primed frame = rate that unprimed clock is ticking in primed frame divided by gamma"

Does this distinction between clock rates and time intervals help at all?

 

[JesseM]

True, blindingly irrelevant given the immediate context, but true.

The relevance was to your statement "It seems a little odd, under these circumstances, to call $$t_{in. the. ring}$$ a "rest frame" ( or the rest frame of one clock)."--I didn't understand that you were specifically saying it was odd because the muon was moving non-inertially, I thought you considered it odd because the muon wasn't at rest in the lab observer's frame or something. If your argument there specifically depended on the notion that we were accelerating the muon rather than talking about a muon moving at high velocity in a straight line (like the muons from cosmic ray showers), then the simple answer is that we don't call $$t_{in. the. ring}$$ the time in the "rest frame", my earlier statement about the unprimed referring to time in the clock's rest frame was meant to refer specifically to the usual time dilation equation, which is defined to relate time-intervals measured on one inertial clock compared with the time-intervals in an inertial frame where the clock is moving. You can come up with an identical-looking equation that applies to the special case of a clock that's moving non-inertially as viewed from the perspective of an inertial frame where its speed is constant, but this is not really "the time dilation equation", and in this equation $$t_{in. the. ring}$$ would have to refer to the proper time on the non-inertial clock between two events on its worldline, compared with the coordinate time interval between those same two events in the inertial frame where the clock's speed is constant.

 

[neopolitan]

Yes, I understand the concept of "context", you just aren't making your notion of the context particularly clear.

First of all, I'm not really comparing the measurements of two physical clocks, I'm comparing the time intervals between a pair of events as defined in two different frames (if you're dealing with a pair of events that don't happen at the same location in one frame, you need two different clocks which are at rest and 'synchronized' in that frame to measure the time interval between them). But that's easy enough to fix, we can just rewrite your two sentences above as A time interval in one frame. Another time interval in another frame. But with this modification I can't make sense of your subsequent statement "you can't compare them from outside because they are equivalent, neither has precedence, neither is privileged." Of course, neither is privileged physically, but they're just two numbers, why can't I compare them? If in one frame the time-interval between events A and B is 5 seconds, and in a second frame the time-interval between the same events is 10 seconds, then clearly I can say "the time-interval between these events is twice as large in the second frame as it is in the first frame". Have I somehow "privileged" one of these time-intervals in making this statement? If so, which one?

OK, so your argument does depend critically on the notion that you want to compare the rate that two physical clocks are ticking rather than comparing the time-intervals between a specific pair of events? If so, then in this case I agree that we can't say which is ticking faster or slower without first picking a frame. But the time dilation equation as it's normally written is about time-intervals, not instantaneous rates of ticking. Maybe a failure to realize this could be the source of much of your confusion? If we have an unprimed clock moving at velocity v relative to the primed clock, then the normal time dilation equation $$\Delta t' = \Delta t * \gamma$$ can be written in words like this:

"time interval between two events on worldline of unprimed clock as measured in primed frame = time interval between two events on worldline of unprimed clock as measured in unprimed frame (which is of the same as the time interval between those events as measured by the unprimed clock itself, since it's at rest in the unprimed frame) * gamma"

On the other hand, if we wanted to talk about clock rates in the primed frame, then we could use $$d\tau' /dt'$$ to represent the rate the primed clock is ticking relative to the primed frame's coordinate time (this would just be equal to 1 of course), and $$d\tau / dt'$$ to represent the rate the unprimed clock is ticking relative to the primed frame's coordinate time (this would be less than 1), in which case the equation would be:

$$d\tau' / dt' = (d\tau / dt' )/ \gamma$$

So in this equation, we are dividing by gamma rather than multiplying by it as in the standard time dilation equation. And in this equation we are clearly looking at things "from the perspective" of a particular frame, namely the primed frame where the unprimed clock is moving. This equation could be written in words like this:

"Rate that primed clock is ticking in primed frame = rate that unprimed clock is ticking in primed frame divided by gamma"

Does this distinction between clock rates and time intervals help at all?

Ok, your words do sort of encapsulate the issue.

Getting back to the original concern about confusion that the pairing of length contraction and time dilation:

if you pick L and $$\Delta t$$, such that L/$$\Delta t$$= c,

and you had a length (with rest length L) and a clock (with a rest tick-tick time of $$\Delta t$$) in a frame in motion, the equations for working out what the speed of light is in that frame in motion using the measurements in that frame in motion are not:

$$\Delta t' = \Delta t * \gamma$$
$$L' = L / \gamma$$

but something closer to (but not quite):

$$d\tau' / dt' = (d\tau / dt' )/ \gamma$$
$$L' = L / \gamma$$

cheers,

neopolitan

 

[JesseM]

if you pick L and $$\Delta t$$, such that L/$$\Delta t$$= c,

and you had a length (with rest length L) and a clock (with a rest tick-tick time of $$\Delta t$$) in a frame in motion, the equations for working out what the speed of light is in that frame in motion using the measurements in that frame in motion are not:

$$\Delta t' = \Delta t * \gamma$$
$$L' = L / \gamma$$

You can figure out what the speed of light in the second frame is using these equations along with the equation for the relativity of simultaneity, as I showed in post #22 (and if you're talking about two-way speed of light you don't have to worry about simultaneity, as I showed in #20).

but something closer to (but not quite):

$$d\tau' / dt' = (d\tau / dt' )/ \gamma$$
$$L' = L / \gamma$$

I don't really see how the second pair of equations would be more helpful in calculating the speed of light in the primed frame than the first pair was, you'd still have to worry about simultaneity along with the fact that the object with length L' is moving in this frame. Also, if you know the $$\Delta x$$ and $$\Delta t$$ between the event of the light being sent and the event of it being received, the mathematically simplest thing to do is to calculate:

$$\Delta x' = \gamma (\Delta x - v \Delta t)$$
$$\Delta t' = \gamma (\Delta t - v \Delta x /c^2)$$

And in this case it will be true that $$\frac{\Delta x'}{\Delta t'} = c$$.

 

[neopolitan]

I don't really see how the second pair of equations would be more helpful in calculating the speed of light in the primed frame than the first pair was ...<snip>

Because I know that in the frame in motion relative to me lengths contract and clocks slow down. I know that relative to that frame in motion relative to me, I am the one in motion, with contracted lengths and slow clocks.

Using my contracted lengths and slow clocks (relative to the frame which is in motion relative to me), I should be able to calculate that the speed of light to be invariant. Similarly, I expect that anyone in motion relative to me will calculate, using their contracted lengths and slow clocks (relative to me), that the speed of light is invariant.

I'm not really trying to work out what photons are doing relative to me (which is what your work in posts #20 and #22 are about). I'm trying to work out that the relativistically affected lengths and time, in another frame, still give you an invariant speed of light.

Why? Because that is where people tend to get confused, doing just that using lc and td - and these are the people who care enough to try to work it out, the world is full of people who really don't care. (See https://www.physicsforums.com/showthread.php?t=289509" for recent examples.)

cheers,

neopolitan

 

[whybother]

"Can you think of any benefits?"
"Benefits? No, he'll die."
"I get that, but have there ever been any fundraisers?"

 

[JesseM]

Because I know that in the frame in motion relative to me lengths contract and clocks slow down. I know that relative to that frame in motion relative to me, I am the one in motion, with contracted lengths and slow clocks.

And you don't think the time dilation equation tells you that? A slow clock should naturally take longer to tick out a certain number of seconds, no?

I'm not really trying to work out what photons are doing relative to me (which is what your work in posts #20 and #22 are about). I'm trying to work out that the relativistically affected lengths and time, in another frame, still give you an invariant speed of light.

Don't really understand the distinction you're making here. My posts only made use of photons indirectly--I was picking two different clock readings which both lay on the path of the photon in the unprimed frame where the rulers and clocks were at rest, and then from then on I didn't say anything about photons, I just figured out the distance and time between these same two clock readings in the coordinates of the primed frame, based on the idea that the clocks were slowed down (and out-of-sync in the case of one-way speed) and the rulers were shrunk, and they were all moving at speed v in this frame. How would you use "relativistically affected lengths and times" to show "an invariant speed of light" if you weren't even allowed to select events on the worldlines of clocks in the primed frame such that the distance between the clocks in that frame divided by the difference in the clocks' readings at those events should be c?

Why? Because that is where people tend to get confused, doing just that using lc and td

Sorry, what do lc and td refer to?

- and these are the people who care enough to try to work it out, the world is full of people who really don't care. (See https://www.physicsforums.com/showthread.php?t=289509" for recent examples.)

Those are long threads, can you give some specific quotes which you think are examples of the type of confusion you're trying to describe? (I still don't understand quite what the confusion is that you're talking about--is it just the confusion between clock rates and time intervals, or something else?)

 

[neopolitan]

Sorry, what do lc and td refer to?

Those are long threads, can you give some specific quotes which you think are examples of the type of confusion you're trying to describe? (I still don't understand quite what the confusion is that you're talking about--is it just the confusion between clock rates and time intervals, or something else?)

lc = length contraction (or Lorentz contraction)
td = time dilation

The threads are examples. One of them you posted to. I'm not going to pick statements from either and post here out of context.

As to confusion about clock rates and time intervals, yes. Not quite the words I would use, but pretty much yes. I do struggle for terminology which you would understand and which is still what I mean. Perhaps "displayed time" and "time interval" - displayed time is what is on the display of your clock, time interval is the period between ticks.

I know you didn't want an observer, but I need one. Because I need someone to look at the clock face, take a measurement and calculate the speed of light. Then, if that observer is in motion (in a primed frame), and looks at the clock face (in the primed frame), takes a measurement (in the primed frame) and calculates the speed of light again (using a primed time value and a primed length), that speed of light won't have changed.

All I am saying is that to do that, that observer (in the primed frame) won't be using what is to an unprimed frame (ie one notionally at rest) a dilated time interval and a contracted length, but rather a contracted displayed time value (t/gamma) and a contracted length (but in the frame in which those measurements are actually made, neither the displayed value nor the length will be contracted, they will be normal).

cheers,

neopolitan

 

[JesseM]

The threads are examples. One of them you posted to. I'm not going to pick statements from either and post here out of context.

Can you give me a post # from one of those threads where you think someone is getting confused by the time interval/rate of ticking distinction, at least? If that is indeed the confusion you're talking about?

All I am saying is that to do that, that observer (in the primed frame) won't be using what is to an unprimed frame (ie one notionally at rest) a dilated time interval and a contracted length, but rather a contracted displayed time value (t/gamma) and a contracted length (but in the frame in which those measurements are actually made, neither the displayed value nor the length will be contracted, they will be normal).

Why do you think that, if you haven't even done the calculation yourself?

I realize in retrospect that I did refer to the photon in my post #22 a bit more than just calculating the clock readings in the unprimed frame. But let me give an altered calculation which doesn't. Say in the unprimed frame we have two clocks at either end of a rod at rest in this frame with length L, the clock on the left reads some time T when the photon leaves it, and the clock on the right reads T + L/c when the photon reaches it. Now I'll show that the distance/time between these two events must be c in the primed frame too, using only the two clock readings and the length of the rod and its velocity, along with the time dilation, length contraction and relativity of simultaneity equations (no further reference to a photon). If the rod is moving to the right with velocity v in the primed frame, then using the relativity of simultaneity equation, we know that at the "same moment" that the left clock reads T ('same moment' according to the primed frame's definition of simultaneity), the right clock must read T - vL/c^2. So, by the time the right clock reads T + L/c, it has ticked forward by (T + L/c) - (T - vL/c^2) = L/c + vL/c^2 = cL/c^2 + vL/c^2 = (c+v)*L/c^2. So, this must be the time interval in the unprimed frame between the event of the right clock reading (T - vL/c^2) and the event of the right clock reading (T + L/c), so we can use the time dilation equation with $$\Delta t = (c+v)*L/c^2$$ to conclude that the time interval between these events is $$\Delta t' = (c+v)*L/c^2 \sqrt{1 - v^2/c^2}$$ in the primed frame. And since the event of the left clock reading T is simultaneous with the event of the right clock reading (T - vL/c^2) in the primed frame, this must also be the time interval between the event of the left clock reading T and the event of the right clock reading (T + L/c), the same two events we were considering in the unprimed frame.

Now we just have to find the spatial distance between these two events in the primed frame. Well, using the length contraction equation we know that the right clock was initially a distance of $$L * \sqrt{1 - v^2/c^2}$$ from the left clock at the moment the left clock read T. We also know that the time between these events in the primed frame was $$\Delta t' = (c+v)*L/c^2 \sqrt{1 - v^2/c^2}$$, and the right clock was moving at velocity v to the right the whole time, so by the time of the second event (the right clock reading T + L/c) the right clock will have moved an additional distance of v times that time interval, or $$v*(c+v)*L/c^2 \sqrt{1 - v^2/c^2}$$. So, adding that additional distance to the distance of $$L * \sqrt{1 - v^2/c^2}$$ that the right clock was from the left clock initially when the left clock read T, the total distance $$\Delta x'$$ between these two events in the primed frame must be $$[L * \sqrt{1 - v^2/c^2}] + [v*(c+v)*L/c^2 \sqrt{1 - v^2/c^2}]$$, or $$[L*c^2*(1 - v^2/c^2) + v*(c+v)*L]/[c^2 \sqrt{1 - v^2/c^2}]$$. So, dividing this $$\Delta x'$$ by $$\Delta t' = (c+v)*L/c^2 \sqrt{1 - v^2/c^2}$$ which we found earlier gives:
$$[L*c^2*(1 - v^2/c^2) + v*(c+v)*L]/[(c+v)*L]$$
or
$$[L*(c+v)*(c-v) + v*(c+v)*L]/[(c+v)*L]$$
or
(c-v) + v = c. So, that completes the demonstration that $$\Delta x' / \Delta t'$$ in the primed frame for the event of the left clock reading T and the event of the right clock reading T + L/c must also be equal to c. And as I said initially, nowhere did I have to talk about photons except at the very beginning when finding that these two clock-readings would both lie on the path of a photon moving at c in the unprimed frame.

Can you show how you would demonstrate that distance/time in the primed frame between these same two events would be c, using your alternate equation for the instantaneous rate that clocks at rest in the unprimed frame will be seen to be ticking in the primed frame, as opposed to the time dilation equation I used? If you're claiming that the demonstration would somehow be a lot simpler using this equation, then you need to show how this simpler demonstration would actually work in order to convince me of this; I don't believe it would actually make things any simpler.

 

[neopolitan]

Here you go Jesse.

I have an experimental apparatus. Using that apparatus myself (no other frame, just me), I can measure a distance, L and the measured time it takes a photon to travel that distance, t (the specifics of the measurement is immaterial). I specifically made this apparatus so that L / t = c.

Then I give it to my buddy, I put him on a carriage with a velocity of v. According to his measurements, the distance I measured to be L would be L and the measured time it takes a photon to travel that distance would be t.

However, I know that the ruler he took with him has shrunk, relative to me. So really what he is measuring to be L, is really a shorter version of L, which I call L'.

Then, because I know that I set up my apparatus so that c = L/t, and the speed of light is invariant, I can therefore work out that t' = L' /c.

If I work out that L' = L / gamma, then I must work out that t' = t /gamma.

If I want to work out something different, specifically how much time has passed on my clock while an observed t has passed on my buddy's clock, I have to use a different equation to get:

T' = t * gamma

cheers,

neopolitan

 

[JesseM]

Here you go Jesse.

I have an experimental apparatus. Using that apparatus myself (no other frame, just me), I can measure a distance, L and the measured time it takes a photon to travel that distance, t (the specifics of the measurement is immaterial). I specifically made this apparatus so that L / t = c.

Then I give it to my buddy, I put him on a carriage with a velocity of v. According to his measurements, the distance I measured to be L would be L and the measured time it takes a photon to travel that distance would be t.

However, I know that the ruler he took with him has shrunk, relative to me. So really what he is measuring to be L, is really a shorter version of L, which I call L'.

L' represents the length of the apparatus in your frame, yes?

Then, because I know that I set up my apparatus so that c = L/t, and the speed of light is invariant, I can therefore work out that t' = L' /c.

If I work out that L' = L / gamma, then I must work out that t' = t /gamma.

I thought you were supposed to be deriving the fact that the speed of light is invariant, but here you seem to simply assume it. Also, what does t' represent, physically? Is it the time-interval in your (primed) frame between the event of your buddy measuring the light passing the left end of his apparatus and the event of his measuring the right end of the apparatus? If so, you can't assume the distance between these events is L' in your frame even though that's the length of the apparatus in your frame, because of course his apparatus is moving in your frame.

If t' does not represent a time interval in your frame between two specific events, then you have to either specify in clear physical terms what it does represent or your argument is totally incoherent, which is what it appears to be right now. I can't see how the equation c = L'/t' would make any sense unless it's interpreted as (distance between two events on the worldline of a light beam in the primed frame)/(time between the same two events in the primed frame). That's what speed always means in physics, (change in position)/(change in time), where the "change" is between two events on the worldline of the object whose speed you're measuring. Like I said, if you have some other clear physical definition of what t' in your equation represents if not the time-interval in your frame between two specific events, by all means present it, but I suspect you're just playing with symbols without having really thought through what they are supposed to represent physically.

 

[neopolitan]

I thought you were supposed to be deriving the fact that the speed of light is invariant, but here you seem to simply assume it.

I would have started with L' = L / gamma and t' = t / gamma.

If the apparatus is such that L/t = c, then:

L' / t' = (L / gamma) / (t / gamma) = L/t = c

Are you happier with that?

cheers,

neopolitan

 

[JesseM]

I would have started with L' = L / gamma and t' = t / gamma.

If the apparatus is such that L/t = c, then:

L' / t' = (L / gamma) / (t / gamma) = L/t = c

Are you happier with that?

No, because you haven't defined the physical meaning of t' in any way that leads me to think L'/t' can be understood as the "speed" in any frame. As an analogy, I could define L'=L*pi/(the gravitational constant), and t'=t*pi/(the gravitational constant), and then it would be true mathematically that if L/t=c then L'/t'=c as well, but this would have no physical interpretation in terms of L'/t' representing distance/time in some inertial frame (and therefore would have nothing to do with the idea that the speed of light is frame-invariant), it would just be a meaningless math game.

 

[Mentz114]

The muon is stored in a ring, it is tracing out a circular trajectory. There is no single rest frame for the stored muon over 2.2 milliseconds.

Rubbish. You are are really missing something here.

 

[Dmitry67]

A few years ago, experimenters at Brookhaven National Laboratory put a beam of muons (mass = 105.658 MeV) into a magnetic ring and stored them at a gamma of 29.3. The measured lifetime in the ring was about 64.4 microseconds, a factor of 29.3 longer than their measured lifetime at rest (2.2 microseconds). The dilated lifetime gave the experimenters more time to make accurate QED measurements on the muons. During the 64.4 microseconds, the muons traveled about beta*gamma*c*tau meters, where c and tau are the speed of light, and tau is the lifetime at rest. So time dilation works.

Shouldnt muon decays be additionally slowed down because of very strong centrifugal forces affecting the muon? (GR time dilation)?

 

[neopolitan]

No, because you haven't defined the physical meaning of t' in any way that leads me to think L'/t' can be understood as the "speed" in any frame. As an analogy, I could define L'=L*pi/(the gravitational constant), and t'=t*pi/(the gravitational constant), and then it would be true mathematically that if L/t=c then L'/t'=c as well, but this would have no physical interpretation in terms of L'/t' representing distance/time in some inertial frame (and therefore would have nothing to do with the idea that the speed of light is frame-invariant), it would just be a meaningless math game.

I haven't defined either L or L' in the post you are responding to either.

I have to admit hovering my cursor over a section of text and wondering if I would have to define t' for you.

Here goes:

My buddy is measuring a length which is at an angle in spacetime due to his motion relative to me. In my frame, the extremities of the length that he is measuring are not simultaneous and not as far apart as he measures them to be. The overall effect is that his lengths are contracted, according to me. So my apparatus, at rest with me, has a greater length than his apparatus.

My buddy is also measuring a time interval, between two events which are colocated in his frame. His clock runs slow, so that while his clock measures off t, the "real" time elapsed (according to me) is a greater period. So on my clock, at rest with me, I have a displayed time which is greater than his displayed time.

Since I was calling my length and displayed time L and t respectively, that makes his smaller length L' and his lesser displayed time t'.

So, if L/t = c then L'/t'=c.

(Little reminder here, t' here is not derived from time dilation. It is the time displayed in my buddy's frame where t is the time displayed in my frame.)

------------------------------------

You may have a problem with "displayed time". I do understand that.

However, think about how we measure the average speed of a horse running on a race track. We press a button on the stopwatch when the race starts. We press it again when the horse passes the finish line. We know the length of the course (L) and the time it took the horse to run that distance (t) because we see the time displayed on the stopwatch.

Working out the average speed (which you seemed to have difficult coping with) is relatively simple. Take the length measured and divide by the displayed time.

 

[neopolitan]

Rubbish. You are are really missing something here.

So the muons in question had a single inertial rest frame? Is that what I am missing?

I'd be curious to know how they managed to keep that up for 64.4 ms while at a gamma factor of 29.3 (about 99.9% of the speed of light). Was their apparatus 20,000km long? I am pretty sure I would have noticed it when visiting New York.

 

[Mentz114]

So the muons in question had a single inertial rest frame? Is that what I am missing?

I'd be curious to know how they managed to keep that up for 64.4 ms while at a gamma factor of 29.3 (about 99.9% of the speed of light). Was their apparatus 20,000km long? I am pretty sure I would have noticed it when visiting New York.

You don't know the meaning of 'rest frame'. Forget the muons for a momemnt and suppose I was on spaceship accelerating away from you - are you saying I don't have a rest frame ?
It may not be inertial but I've still got one.

Your attempt at sarcasm is pathetic.

 

[JesseM]

My buddy is measuring a length which is at an angle in spacetime due to his motion relative to me. In my frame, the extremities of the length that he is measuring are not simultaneous and not as far apart as he measures them to be. The overall effect is that his lengths are contracted, according to me. So my apparatus, at rest with me, has a greater length than his apparatus.

So are you just talking about the standard notion of "length" in the standard length contraction equation, where L' represents the distance between ends of his apparatus at a single moment in your primed frame?

My buddy is also measuring a time interval, between two events which are colocated in his frame.

If he's measuring a time between events that are colocated in this frame, what does this have to do with measuring the speed of light? Why did you say that L/t=c? Is he measuring the two-way speed of light using a clock at one end of his apparatus and a mirror at the other?

His clock runs slow, so that while his clock measures off t, the "real" time elapsed (according to me) is a greater period.

The time elapsed in your frame between the same two events that were colocated in his frame? But if the time elapsed in your frame $$\Delta t'$$ is greater than the time elapsed in his frame $$\Delta t$$, that would imply you should be using the standard time dilation equation $$\Delta t' = \Delta t * \gamma$$, so what were you talking about when you said "If I work out that L' = L / gamma, then I must work out that t' = t /gamma"? It was the meaning of t' and t' in that equation that I wanted a physical definition of, of course I know the physical definition of $$\Delta t$$ and $$\Delta t'$$ in the normal time dilation equation.

Since I was calling my length and displayed time L and t respectively that makes his smaller length L' and his lesser displayed time t'.

Aarrgh, you never specified that L and t represented your frame, and throughout this thread I've been using the convention (and you've been quoting from things like the wikipedia page which use the same convention) that the rest length of the ruler whose length is shrunk in the length contraction equation, and the time in the rest frame of the clock whose time is dilated in the time dilation equation, is the unprimed frame, which is the reverse of what you're doing above. I even asked "L' represents the length of the apparatus in your frame, yes?" back in post #53 and you never corrected me (keep in mind that in your example, your buddy was the one carrying 'the apparatus' along with him, not you). And then in this very post you say above "His clock runs slow, so that while his clock measures off t ..." (unprimed), so now you seem to be contradicting yourself when you say your time is t and he has "lesser displayed time t' ".

In any case, if the length in his frame is L', is this supposed to be the length of something at rest in his frame, like the apparatus he's carrying? If so, it makes no sense for you to say "his smaller length L' " since of course the rest length of an object is greater than its contracted length in the frame of an observer (like you) who sees it in motion. If you wanted L' to represent the rest length of his apparatus in his frame, and L to represent the length of his apparatus in your frame, the equation would have to be L' = L*gamma, not L'=L/gamma like you wrote (note that if you want to use a weird convention where primed is the rest length, then it would make more sense to have the equation giving unprimed as a function of primed, i.e. L = L' / gamma, since the length contraction and time dilation equations are always written to give you time and distance in the frame where the ruler/clock are moving as a function of time and distance in the frame where they're at rest. I really wish we could just agree to use the standard convention that unprimed represents the rest frame of the ruler and clock though, it would lead to much less confusion when quoting sources like wikipedia, and it's the convention I've been using throughout the entire thread which you never objected).

Also, if t' represents the time in his frame between events that are colocated in his frame, and t represents the time in your frame between these same events, and you said yourself that his time t' would be smaller (because his clock has a 'lesser displayed time'), then there is nothing nonstandard about your equation "t' = t / gamma"--this is just a reshuffling of the standard time dilation equation, which if we write it in words so we avoid primed and unprimed confusion, would be (time interval between events on clock's worldline in frame where clock is in motion) = (time interval between events on clock's worldline in frame where clock is at rest) * gamma, so obviously dividing both sides by gamma will give (time interval between events on clock's worldline in frame where clock is at rest) = (time interval between events on clock's worldline in frame where clock is in motion) / gamma, which is what you seem to be expressing with your equation t' = t / gamma.

BTW, maybe to avoid any further possible confusion about notation, we should use notation like $$\Delta t_{buddy}$$ and $$L_{buddy}$$ to express times and lengths measured in your buddy's frame, and $$\Delta t_{neo}$$ and $$L_{neo}$$ to express times and lengths in your frame? In this case if your buddy says the length of his apparatus is $$L_{buddy}$$, then the length of his (moving) apparatus in your frame is given by:

$$L_{neo} = L_{buddy} / \gamma$$

And if your buddy says the time between two events which are colocal in his frame is $$t_{buddy}$$, then the time between those same two events in your frame is:

$$t_{neo} = t_{buddy} * \gamma$$

Which of these equations are you disagreeing with, if any?

So, if L/t = c then L'/t'=c.

Why does L/t=c? Again, if we're not talking about the distance and time between two events on the worldline of a lightbeam, that equation makes little sense.

You may have a problem with "displayed time". I do understand that.

If we're dealing with two events which are colocal in your buddy's frame and your buddy has a watch at the position of these events, then on his clock "displayed time" is of course exactly equal to the time interval between the events in his frame, and that's exactly what's in the standard time dilation equation. So are you saying the time in your frame where your buddy is moving is also "displayed time"? But if you're talking about events which are colocal in his frame, in your frame these events are at different positions, so if you use a single stopwatch you have to worry about light-speed delays.

However, think about how we measure the average speed of a horse running on a race track. We press a button on the stopwatch when the race starts. We press it again when the horse passes the finish line.

Yes, that works fine in a horce-race because the delay between the events of the horse departing/finishing and the events of our seeing these things happen is totally negligible, since the time for light to get from the horse's position to our eyes is miniscule compared to the time of the race. If we were measuring a horse traveling at close to the speed of light, though, then if I'm standing at a fixed position with a stopwatch while the horse races by me, then the time between my seeing the horse begin and end might be significantly different than the time interval between the horse actually beginning and ending in my rest frame, because of these delays for light to reach my eyes. Of course if I know the distance the horse was from me when it began and ended I can compensate for this (for example, if I see the horse cross the finish line when my watch reads 8 seconds but I know the finish line is at a distance of 3 light-seconds from me, then I can say the event 'really' occurred simultaneously with my clock reading 5 seconds in my frame), but this requires some more calculation than just looking at my watch and noting "displayed time" directly. The other option, of course, would be the one Einstein imagined where there are multiple clocks which are "synchronized" in my frame, so I could note the displayed time on the clock at the starting line when the horse starts, then note the displayed time on the clock at the finish line when the horse crosses it, and the second displayed time minus the first displayed time would be the time interval in my frame.

We know the length of the course (L) and the time it took the horse to run that distance (t) because we see the time displayed on the stopwatch.

But this seems to contradict what you said earlier about your buddy measuring the time "between two events which are colocated in his frame".

 

[JesseM]

Shouldnt muon decays be additionally slowed down because of very strong centrifugal forces affecting the muon? (GR time dilation)?

Centrifugal forces are "fictitious" forces that are only introduced for bookkeeping purposes when you use accelerating (non-inertial) reference frames (see the wikipedia page on fictitious forces, along with the illustrated discussion here, especially the last box), accelerating observers can experience them in totally flat (uncurved) spacetime, and as long as you stick to inertial frames in flat spacetime you can analyze everything about the behavior of such accelerating observers within the context of SR using only SR time dilation to calculate their aging (though as this section of the twin paradox page says, even in flat spacetime you can have 'pseudo-gravitational' time dilation if you use a non-inertial frame in flat spacetime).

 

[Dale]

I have not been following this conversation very closely, but I do find it confusing that the http://en.wikipedia.org/wiki/Length_contraction" is given as:
L' = L/γ

whereas the http://en.wikipedia.org/wiki/Time_dilation" is given as:
Δt' = γΔt

when the 1-1 d Lorentz transform looks the same for time and space (in units where c=1):
t' = γ(t-vx)
x' = γ(x-vt)

There are a lot of things in the standard presentation of SR that I think are sub-optimal from a pedagogical perspective.

 

[JesseM]

I have not been following this conversation very closely, but I do find it confusing that the http://en.wikipedia.org/wiki/Length_contraction" is given as:
L' = L/γ

whereas the http://en.wikipedia.org/wiki/Time_dilation" is given as:
Δt' = γΔt

when the 1-1 d Lorentz transform looks the same for time and space (in units where c=1):
t' = γ(t-vx)
x' = γ(x-vt)

You can see visually why this is if you look at the diagram I drew up for neopolitan in a previous discussion, which he posted in post #5--basically the two equations are telling you somewhat different things if you think in terms of a spacetime diagram, and it is possible to come up with a "spatial analogue for the time dilation equation" which looks like the time dilation equation but with L substituted for $$\Delta t$$ (in this equation you are talking about the spatial distance in the primed frame between two events which occur at the same time but at a spatial separation of L in the unprimed frame, which is analogous to how time dilation tells you the time interval in the primed frame between two events which occur at the same position but at a temporal separation of $$\Delta t$$ in the unprimed frame...note that this is not what the standard length contraction equation tells you), and it's also possible to come up with a "temporal analogue for the length contraction equation" which looks like the length contraction equation but with $$\Delta t$$ substituted for L (this is more difficult to state in words, but it's basically the time-interval in the primed frame between two surfaces of constant t in the unprimed frame which have a temporal distance of $$\Delta t$$ in the unprimed frame, which is analogous to how length contraction tells you the spatial distance in the primed frame between two worldlines of constant position in the unprimed frame which have a spatial separation of L in the unprimed frame).

 

[Dale]

Yes, I understand, and I also know it is correct as defined. I just find it confusing, especially when I was first learning.

Because of my initial confusion I made it a general rule to never ever use either the time dilation or length contraction formula, and I have continued to stick to that rule since then. Instead I always use the Lorentz transform directly in order to avoid making a mistake in which frame is doing what.

 

[Rasalhague]

I’m new to SR and have also been struggling with this apparent asymmetry between the equations for time dilation and length contraction. I found the diagram https://www.physicsforums.com/attachment.php?attachmentid=17992&d=1237127533 very helpful, and Jesse’s explanations, e.g. #64, of the "temporal analogue to length contraction", and "spatial analogue to time dilation".

If I try putting this in my own words, could you tell me if I've got it right?

As I understand it, the time dilation formula takes as its input the time between two events in a frame where there’s no space between them, i.e. two events colocal in the unprimed frame. When you plug into it this time and the speed the primed frame is moving relative to the unprimed, the formula t’ = t * gamma tells you the time in the primed frame between the two events.

The length contraction formula takes as its input the distance between two events that are simultaneous in the unprimed frame. When you plug into it this distance and the speed the primed frame is moving relative to the unprimed frame, the formula L’ = L/gamma tells you the distance in the primed frame between one of these events *and a third event*, which third event is simultaneous in the primed frame with the first, and in the unprimed frame is colocal with the second.

More generally, I guess I’ve been trying to understand what exactly the asymetry is, and where it comes from: whether a physical difference between time and space, an accident of the kind of questions asked in textbooks, or of the way the idea is expressed, or if there’s something about our relationship to time and space that makes us want to ask a different question of each--that is, something that makes this pair of not-directly-analogous questions more useful to us than any other combination of the four possible questions illustrated in the diagram. Does that make any sense?

 

[neopolitan]

Note that I too have said that time dilation is correct as defined.

I have also said that the type of time used in time dilation is a different type of time to the one we usually use (one is the time between ticks, the other is the number of ticks displayed).

Jesse,

In post #51 you posted:

Can you show how you would demonstrate that distance/time in the primed frame between these same two events would be c, using your alternate equation for the instantaneous rate that clocks at rest in the unprimed frame will be seen to be ticking in the primed frame, as opposed to the time dilation equation I used? If you're claiming that the demonstration would somehow be a lot simpler using this equation, then you need to show how this simpler demonstration would actually work in order to convince me of this; I don't believe it would actually make things any simpler.

I accept that there has been some dancing around, which makes things confusing. But the bit I was trying to do, the bit I was focussed on was "using your alternate equation". I ignored the words following it, because you were taking something out of context from post #45:

but something closer to (but not quite):

$$d\tau' / dt' = (d\tau / dt' )/ \gamma$$
$$L' = L / \gamma$$

I specifically said "closer to (but not quite)" because it is not an instantaneous rate that I am talking about. So what you demanded of me was unreasonable.

The alternate equation I was using is this:

$$t' = t / \gamma$$

You want to know in which frame the events are colocated. Really, they don't need to be colocated in any frame.

Perhaps if I rephrase it will make it easier. I want to know the relativistic effects on my buddy's measurements of time and space. I can use that to calculate what the speed of light is in his frame (and I am most interested in that speed when light moves parallel to the direction of his motion). I want to use the same sort of dimensions that we usually use to calculate a speed (a length which can be traversed and a readout of time elapsed).

What effect does putting my buddy into motion have on his dimensions?

The answer is that his lengths contract and the readouts of time elapsed are reduced - according to me. (And my lengths and my readouts of time elapsed are reduced - according to him.)

Are we agreed on this? (I agree that the period between his ticks is greater than the period between my ticks, according to me.)

cheers,

neopolitan

 

[neopolitan]

I regret posting exactly when I did. Rasalhague asks a fine question in the post before mine.

 

[JesseM]

Note that I too have said that time dilation is correct as defined.

I have also said that the type of time used in time dilation is a different type of time to the one we usually use (one is the time between ticks, the other is the number of ticks displayed).

This distinction seems meaningless--unless you are postulating some notion of absolute time, the only way we can talk about "time between ticks" is by looking at the "number of ticks displayed" on a clock (or pair of synchronized clocks) with some smaller interval between ticks; for example, if the time between ticks is 1 second, put another clock next to it that ticks once every millisecond, and find that the second clock elapses 1000 ticks between each tick of the first clock. The time between a clock's ticks is just like any other time interval, we measure it using the difference in reading between the end of the interval and the beginning (or more abstractly, we can define it using the difference between the time coordinates between the events of two successive ticks).

I accept that there has been some dancing around, which makes things confusing. But the bit I was trying to do, the bit I was focussed on was "using your alternate equation". I ignored the words following it, because you were taking something out of context from post #45

When I said "using your alternate equation" I really meant "using whatever alternate equation you wish to define", not specifically using the alternate equation I gave in post #45.

The alternate equation I was using is this:

$$t' = t / \gamma$$

You want to know in which frame the events are colocated. Really, they don't need to be colocated in any frame.

Fine, but regardless of what events you choose, you still haven't given any coherent definition of what t and t' mean in terms of coordinates assigned to the events, or in terms of actual physical measurements involving those events. Does t represent the coordinate time interval between two events in the unprimed frame, and t' represent the coordinate time interval between the same two events in the unprimed frame? If not, what do t and t' mean? You can't write down equations and say they have any relevance to physics if you can't even define the physical meaning of the variables in the equations!

Perhaps if I rephrase it will make it easier. I want to know the relativistic effects on my buddy's measurements of time and space. I can use that to calculate what the speed of light is in his frame (and I am most interested in that speed when light moves parallel to the direction of his motion). I want to use the same sort of dimensions that we usually use to calculate a speed (a length which can be traversed and a readout of time elapsed).

What effect does putting my buddy into motion have on his dimensions?

The answer is that his lengths contract and the readouts of time elapsed are reduced - according to me. (And my lengths and my readouts of time elapsed are reduced - according to him.)

Are we agreed on this?

Your words are too vague and would only make sense with specific types of elaborations. Does "his lengths contract ... according to me" mean that the length you assign to an object at rest in his frame is smaller than the length he assigns to that same object? In that case I would agree. Does "his lengths contract ... according to me" mean that the length you assign to an object at rest in your frame is smaller than the length he assigns to that same object? If so, the statement is wrong. Does "his lengths contract ... according to me" mean that the spatial distance between two arbitrary events as measured by you is smaller than the spatial distance between the same two events as measured by him? If so, this would be true in some cases but not in others.

And what does "readouts of time elapsed are reduced ... according to me" mean? Does it mean that if you consider a time interval from some time $$t'_0$$ to another time $$t'_1$$ in your frame, then for a clock at rest in his frame, the difference (his clock's readout at the event on the clock's worldline that occurs at time $$t'_1$$ in your frame) - (his clock's readout at the event on the clock's worldline that occurs at time $$t'_0$$ in your frame) is smaller than the difference $$t'_1 - t'_0$$ which represents the size of the time interval in your frame? In that case I would agree. Does it mean that if you consider two arbitrary events A and B (like two events on the worldline of a photon), with A happening at $$t'_0$$ and B happening at $$t'_1$$ in your frame, then for a clock at rest in his frame, the difference (his clock's readout at the event on the clock's worldline that occurs simultaneously with B in your frame) - (his clock's readout at the event on the clock's worldline that occurs simultaneously with A in your frame) is smaller than the time interval between A and B in your frame? If so, this is exactly equivalent to the first one, so I'd agree with this too. But does it mean that if you consider the same two events A and B, then for a clock at rest in his frame, the difference (his clock's readout at the event on the clock's worldline that occurs simultaneously with B in his frame) - (his clock's readout at the event on the clock's worldline that occurs simultaneously with A in his frame) is smaller than the time interval between A and B in your frame? If so this is not true in general. And if the two events are events on the path of a photon that occur on either end of a measuring-rod of length L in his frame, then it is that last difference in his clock's readouts that you divide L by to get c.

I regret posting exactly when I did. Rasalhague asks a fine question in the post before mine.

Don't worry, I intend to reply to it.

 

[Rasalhague]

The usual convention is that the unprimed t represents the time interval between two events on the worldline of a clock as measured in the clock's rest frame (so both events happen at the same spatial location in the unprimed frame, and t will be equal to the time interval as measured by the clock itself), whereas the primed t' represents the time interval between the same two events in a frame where the clock is moving (so the events happen at different locations in the primed frame). This is consistent with the convention for length contraction, where the unprimed L represents the distance between two ends of an object in the object's own rest frame, and the primed L' represents the distance between the ends of the same object in the frame where the object is in moving.

As someone new to SR, the use of primed and unprimed coordinates has been a source of some confusion for me. I think that’s partly because, to begin with, I didn’t know which details of a textbook explanation were going to turn out to be the significant ones, and which were accidental to a particular author’s presentation. At first, I got the impression that the convention was simply that primed frame = rest frame. But then I encountered examples such as: “Consider a rod at rest in frame S’ with one end at x’2 and the other end at x’2 [...]” (Tipler & Mosca: Physics for Scientists and Engineers, 5e, extended version, p. 1274). From this, I guessed that the desciding factor in whether to call a frame primed or unprimed was the direction it was moving, a primed frame being one that moves in the positive x direction with respect to the unprimed frame, hence the signs used in versions of the full Lorentz transformation such as this:

x' = gamma (x – ut)
t' = gamma (t – ux/c^2)

x = gamma (x' + ut')
t = gamma (t' + ux'/c^2)

I think that’s the traditional choice of signs, isn’t it? But I don’t know yet how rigid that convention is. The practice of using primed coordinates to represent a frame moving in the positive x direction agrees with their use in discussions of Euclidian rotation for a frame rotated in the positive (counterclockwise) direction. At least, that’s how Euclidian rotation was introduced in the first books where I met it. But in Spacetime Physics, Taylor and Wheeler do it the other way around, using primed coordinates for a negative (clockwise) rotation. I suppose, looking on the bright side, the advantage to having a variety of presentation methods is that it, eventually, you can see by comparing them which details are the physically significant ones, and which a matter of convention. But to begin with it’s a lot of information to take in.