Benefits of time dilation / length contraction pairing?

[neopolitan]

Jesse,

Let me repeat the question. I will highlight something for you, hopefully it will answer the question you had before:

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?

cheers,

neopolitan

 

[JesseM]

Jesse,

Let me repeat the question. I will highlight something for you, hopefully it will answer the question you had before:

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.

 

[Mentz114]

neopolitan:

It seems a little odd, under these circumstances, to call t_{in. the. ring} a "rest frame" ( or the rest frame of one clock).

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

 

[neopolitan]

I've discussed this with you in the past, but when you talk about "the" rest frame in a given problem, you aren't really making much sense; you seem to have gotten in your head that when analyzing a particular problem we are supposed to pick one frame to label as "the rest frame" (or 'the observer's frame', which may be related to your insertion of an irrelevant observer above), but no such convention exists. The phrase "rest frame" is generally used only in the context of talking about some specific object; for example, if clock A is at rest in the unprimed frame and clock B is at rest in the primed frame, then the unprimed frame is "the rest frame of clock A" and the primed frame is "the rest frame of clock B".

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 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.

Take it out of context (ie don't have an equation, just compare the frames) and yes, you can't justify referring to either frame as "the rest frame". It is writing the equation that tags one of the frames as "the rest frame" (in context). I've not talked about the frames at all without referring to a specific equation.

cheers,

neopolitan

 

[JesseM]

However, you do give an example of what use time dilation has for which I thank you.

The lifetime at rest is 2.2 microseconds. The "contracted time" for the muons in magnetic ring was ... 2.2 microseconds, yes? If there was a clock stored in that magnetic ring at a gamma of 29.3 it would tick off 2.2 microseconds while clocks not stored in that magnetic ring would tick off 64.4 microseconds.

Yes, if by "stored in that magnetic ring" you mean "traveling along with a muon in the ring which has been accelerated to a relativistic velocity relative to the lab"

\Delta t_{at. rest. in. the. laboratory}=\gamma . t_{in. the. ring}

It seems a little odd, under these circumstances, to call t_{in. the. ring} a "rest frame" ( or the rest frame of one clock).

Why do you find it odd? Doesn't t_{in. the. ring} refer to the time interval in the coordinate system where the clock's position-coordinate is constant over time? That is all that "the rest frame of an object" means", the frame in which it has constant position coordinate (and is therefore at rest in that frame). You seem to be confused about the meaning of the term "rest frame", although I don't really understand what you think it means.

 

[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[/color]. Another time interval in another frame[/color]. 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[/color]. Another time interval in another frame[/color]. 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&#039; = (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&#039; = (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&#039; 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&#039; by \Delta t&#039; = (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&#039; / \Delta t&#039; 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.