# Thread: Special relativity - Am I getting it right now?

1. Originally Posted by Bob Angstrom
Electrons can flow from a negative terminal to a positively charged rod so the flow is not zero
but they can't reach A to flow off, so the flow is zero. This isn't just about electrons; it's about any signal being impossible - even in principle - to reach back along the length of the moving rod.

Originally Posted by Bob Angstrom
The question is, Is there an instant in time when the circuit is physically complete? Whether the electrons have time to move or not is irrelevant. It was a bad example but I couldn't think of a better one.
Actually, I think it was a really good example. It was a really interesting extra twist on the old pole-barn paradox. The trouble here is that word "instant", since it implies that there's a universal clock ticking away somewhere that can give a time reading at both A and B. Unfortunately not, so all we're left with is "the clocks on the rod say yes, the clocks in the circuit say no".

Originally Posted by Bob Angstrom
There is no B sending a message to A. Here is the classical version of the Pole and Barn from Feynman's lectures so we are on the same page about the Pole and Barn. http://math.ucr.edu/home/baez/physic...barn_pole.html
Great link. The events at the doors are not necessarily instant - the doors can take some time (eg. if the barn is 1 light minute long, and the pole is travelling ridiculously close to c, the doors could take almost 30 seconds to cycle). OK, I had my observer sitting at B, since that's where the electrons would have been waiting to flow onto the rod, but sitting in the middle (or on top of) will have a similar effect. The link says "you close both doors simultaneously, with your switch." - ie. you send a signal to A and B. Regardless, in the barn frame, the doors at A and B operate simultaneously, after the rear of the pole has left A but before the pole has reached B. The pole, however, will see the door at B close and then open before it gets there (and, in fact, before the rear of the pole has even reached A), then after the rear has passed A it will see that door close then open.

This happens since the barn must synchronize A and B by sending signals, but those signals take differing times for a frame in motion. So are you saying you disagree with this pole-barn scenario, and want to work through it?

Originally Posted by Bob Angstrom
If it ain't philosophy, it ain't science. Science without philosophy is magic.
We can use relativity to frame what we actually observe but, when it is used to frame what we don't observe but think we are observing, that is where I draw the line.
I would say that science without philosophy is simply mathematics matching what we observe. Length contraction is the name we give the effect that we observe - does it "really" happen? Who can say - it certainly forms a nice, symmetrical picture that is consistent with a constant speed of light. It's also related to time - after all, objects retain their SpaceTime separation, so is "rotated in the time dimension" an adequate answer? It's also related to General Relativity's curvature of SpaceTime - but is SpaceTime really curved? I don't have the faintest idea - Feynman put out an alternative interpretation of objects on a hot-plate that would also match the curvature equations.

And all this is before Quantum Mechanics, where we can't even begin to conceive how the particles do what they do, but all we can do is follow where the observations and equations lead us.

Ultimately, all we have are the equations, and all we can ask of science is that the equations match observations. Beyond that, we're on our own.

2. Originally Posted by Grimble
Really? I am simply showing the derivation of length contraction and time dilation from the same diagram (Fig 2, part 2) from my OP.
The derivation of TD was accepted with no problems and I was showing that that of LC follows naturally from the first. Simple geometry/mechanics.
Oh, well okay, but there's a problem with that. There is no length contraction observed when the rod is turned perpendicular to the line of motion, so you shouldn't have measured any in that case. It is only when the rod (or at least part of the rod when angled) is turned into the line of motion that length contraction is observed, when the rod is lying along the x axis. In other words, the observers travelling with the rod will measure d for the proper length of the rod, and observers that see the rod moving at v while lying perpendicular to the line of motion will also measure d' = d, so measuring no length contraction of the rod at all when it lies along the y-z plane.

[d' / (c - v) + d' / (c + v)] sqrt(1 - (v/c)^2) = 2d/c

[2d' / (1 - v²/c²) ] √(1 - v²/c²) = 2d/c

d' / √(1 - v²/c²) = d/c

γ d' = d
Okay, good. You have now derived 2/3 of SR. The last part is of course relativity of simultaneity, which is a difference in readings observed between two moving clocks with some distance between them along the direction of motion, along the x axis. Let's use Bob's scenario for this. In the frame of the two metal targets, each target has its own clock and they are separated by a distance d along the x axis. Observers in the the target frame say that the two clocks are synchronized, that each always reads the same time as the other. The target frame says that the rod travels at v with a length of d, the same as the distance between the targets, with the rod lying along the x axis also. So at some point in time, both ends of the rod will coincide simultaneously with each of the targets, so the ends of the rod will reach each of the targets at the same time. That will be at the same time according to the readings of each of the two target clocks.

Now let's look at what a frame that sees the rod stationary and the targets moving will observe of the two target clocks. First of all, we know from the last exercise that the target frame measures the length of the rod contracted by sqrt(1 - (v/c)^2) when it lies along the x axis, the direction of motion, so the rod frame will measure the proper length of the rod to be d / sqrt(1 - (v/c)^2). We also know that the rod frame will now measure the proper distance between the targets to be contracted, so measures the distance between the targets to be sqrt(1 - (v/c)^2) d. So from the perspective of the rod frame, the rod is stationary and the two targets are moving at v toward the rod, but the distance between the two targets is measured to be smaller then the rod, so there is no way the rod frame can measure the targets coinciding with the ends of the rods simultaneously. Rather, the rod frame will see the second target coincide with one end of the rod, then some time later the first target coincide with the other end. But all frames must agree that when each target coincides with its perspective end of the rod, the reading upon either of the target clocks will be the same.

Okay, so in order for this to occur, for the readings upon the clocks to read the same when they coincide with the ends of the rods at different times according to the rod frame, then the two clocks must read different times according to the rod frame. Let's say that the second target clock reads some time tl greater than the first clock according to the rod frame. Let's also say that when the second clock coincides with one end of the rod, the reading upon that clock is T2 = 0. So when the second clock coincides with the end of the rod, the rod frame must agree it reads T2 = 0, so since the rod frame says that the first clock reads tl less than the second clock, the reading upon the first clock at this time in the rod frame is T1 = - tl. We also know that the target frame observes the ends of the rods coinciding with the targets simultaneously, so at the same time according to both target clocks, whereby T1 = T2 = 0 when they coincide.

ros001.PNG

We can see in the diagram that the first target clock reads T1 = - tl and will now take a time of t = [d / sqrt(1 - (v/c)^2) - sqrt(1 - (v/c)^) d] / v to reach the other end of the rod according to the rod frame. The first target clock will time dilate during this time according to the rod frame, so a time will pass upon the first clock of sqrt(1 - (v/c)^2) t. We also know that when the first target clock reaches the end of the rod, its reading will be T1 = 0, the same as the second target clock read when it coincided with the other end of the rod, so we have

T1 = (- tl) + sqrt(1 - (v/c)^2) t = 0

- tl + sqrt(1 - (v/c)^2) [d / sqrt(1 - (v/c)^2) - sqrt(1 - (v/c)^2) d] / v = 0

Solve for tl to find the difference in readings of the target clocks that the rod frame observes, in terms of the proper distance d the target frame measures between its own two clocks, even though the target frame says the clocks read the same.

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Originally Posted by pzkpfw
No, other observers in the same inertial frame will still agree that the strikes were simultaneous (if they were). Those other observers may not themselves be hit by those flashes simultaneously (as will, say, an observer exactly in the middle of the strikes), but by simple distance and speed of light calculation will still "know" that the strikes occurred at the same time.

(All those observers in the same inertial frame will agree on distances and times.)

This is different to the issue of what an observer moving with respect to the first observer will know to have happened.
I get the point that you and RobA are making but I am unclear about semantic distinction if there is one between a simultaneity that can be directly observed or observed to be casually connected and a simultaneity that is calculated. A calculated simultaneity is what I would consider to be a simultaneity of relativity as opposed to an observed simultaneity which I refer to as "simultaneous." And if two inertial frames have observations in common they can each calculate what may be simultaneous in the other frame or simultaneous in both so a calculated simultaneity is not unique to a single reference frame.

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Originally Posted by RobA
The events at the doors are not necessarily instant - the doors can take some time (eg. if the barn is 1 light minute long, and the pole is travelling ridiculously close to c, the doors could take almost 30 seconds to cycle).
(snip)
This happens since the barn must synchronize A and B by sending signals, but those signals take differing times for a frame in motion. So are you saying you disagree with this pole-barn scenario, and want to work through it?
I agree with your conclusions but there are no signal times to be considered. The signals to the doors are instant as part of the “props.” If one wants to consider signal time, the observer on top of the barn would have to anticipate the event by the considering the distances involved. This is superfluous to the point Feynman is trying to illustrate so Feynman makes the doors able to close and open instantly for the sake of discussion and is done with it. The prior setup of the experiment and any communication between doors A and B is over thinking the problem.

5. Originally Posted by grav
Oh, well okay, but there's a problem with that. There is no length contraction observed when the rod is turned perpendicular to the line of motion, so you shouldn't have measured any in that case. It is only when the rod (or at least part of the rod when angled) is turned into the line of motion that length contraction is observed, when the rod is lying along the x axis. In other words, the observers travelling with the rod will measure d for the proper length of the rod, and observers that see the rod moving at v while lying perpendicular to the line of motion will also measure d' = d, so measuring no length contraction of the rod at all when it lies along the y-z plane.
I see your problem, so I am including another diagram:

In this diagram I have added, in green, the distance travelled by the Moving clock, as measured by the stationary observer, at the time when the light hits the mirror. This is 1.25 seconds as measured by the stationary observer, meaning that the mirror will have travelled 0.75 light seconds at v = 0.6c in that 1.25 seconds. Compared with the 0.6 light seconds in the 1 second that has passed for that same stationary observer in his own frame.
So 0.75 seconds in the moving frame (in the direction of travel) has contracted to 0.6seconds in the stationary frame. Length Contraction.
Okay, good. You have now derived 2/3 of SR.
I have solved the equation you asked me to solve, does that mean that I have understood what it represents? - I don't know.
That is why I began this thread to question how well I, as a visual thinker, see things. Is the way I see them valid?

So showing me simultaneity is begging the question. Do I understand it? As I said this thread is for me to express it and you to say if I have it right; explaining it again, achieves nothing.

So let's leave that until I know I have part 2, length contraction, right?

6. Gentlemen! Thank you for your time and your consideration. May I, however, say that while the current discussion is fascinating, it is distracting from my purpose for this thread, i.e. the question of whether my understanding and my visualisation of it, is correct now.

May I suggest that these questions about poles and barns and rods and contacts continues in another tread? Indeed I will open one myself, and pose a question for you at the same time.

Last edited by Grimble; 2012-Jun-15 at 09:39 AM.

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Originally Posted by grav
Now let's look at what a frame that sees the rod stationary and the targets moving will observe of the two target clocks. First of all, we know from the last exercise that the target frame measures the length of the rod contracted by sqrt(1 - (v/c)^2) when it lies along the x axis, the direction of motion, so the rod frame will measure the proper length of the rod to be d / sqrt(1 - (v/c)^2). We also know that the rod frame will now measure the proper distance between the targets to be contracted, so measures the distance between the targets to be sqrt(1 - (v/c)^2) d. So from the perspective of the rod frame, the rod is stationary and the two targets are moving at v toward the rod, but the distance between the two targets is measured to be smaller then the rod, so there is no way the rod frame can measure the targets coinciding with the ends of the rods simultaneously. Rather, the rod frame will see the second target coincide with one end of the rod, then some time later the first target coincide with the other end. But all frames must agree that when each target coincides with its perspective end of the rod, the reading upon either of the target clocks will be the same.
I have questions about the highlighted parts of the explanation. Shouldn't the rod see its proper length as simply d or was this a reference to the target frame? It is my understanding that "proper" measurements are invariant and they only apply to the immediate inertial reference frame. If the rod sees its own length as d, then that is a proper "proper" The highlighted "proper" refers to a contracted proper which makes it a relativistic distance as opposed to a proper distance.

8. Originally Posted by Bob Angstrom
I get the point that you and RobA are making but I am unclear about semantic distinction if there is one between a simultaneity that can be directly observed or observed to be casually connected and a simultaneity that is calculated. A calculated simultaneity is what I would consider to be a simultaneity of relativity as opposed to an observed simultaneity which I refer to as "simultaneous." And if two inertial frames have observations in common they can each calculate what may be simultaneous in the other frame or simultaneous in both so a calculated simultaneity is not unique to a single reference frame.
Two different frames may always calculate what may have been simultaneous in each others frames, but if they are in relative motion events which are simultaneous (whether calculated or directly experienced simultaneously) for one can't be simultaneous for the other.

In the famous train-embankment thought experiment, the observers are positioned in the exact middle of what they each consider the sources of the flashes of light, but that's really just to make it "easy" to understand what is simultaneous (or not) for each of them.

Say there are two observers on the embankment, one in the exact middle of the sources (according to them) of the two flashes, and one who is positioned one third of the distance between the two sources. If those two observers are at rest with each other, they can agree on clocks, and both can agree on what the clocks at the sources of the flashes would say. If we assume the flashes were simultaneous in the frame of the embankment, the observer in the exact middle will see the flashes at the same time. The observer at one third will see one flash in half the time it takes for the other flash to arrive; but both will know that the flashes occured at the same time - in their frame.

(Simultaenity isn't about when we see events. If your camera flash goes off at the same time as you see a flash from a lightening strike on a distant hill; you consider that you saw the flashes simultaneously, but you wouldn't (normally) consider the lightening strike itself to have occured at the same time as you took the picture.)

You might argue then that the train observer should be equivalent to an embankment observer who isn't standing in the exact middle of the strikes, and who could calculate that the flashes were generated simultaneously in their frame too - but that would ignore that in their frame they do consider (in the usual set-up) that they were in the middle of the two strikes, so should see them at the same time. (Or translate this to any other position on the train, e.g. that one third position).

Code:
```A, B = strikes according to N, observer on train.
C, D = strikes according to M, O observers on embankment

====[A     N     B]=====
------------------------
C     M O   D

Later (train moved to the right):

======[A     N     B]===
------------------------
C     M O   D```
N might be next to O when the light from the leading strike hits N, but this light according to them came from B, whereas for O it came from D. So while O might see the light from D before the light from C, he or she will still calculate that C and D occured at the same time (if they did) just as M would if the flashes hit him or her at the same time; but - N can't get that same result in their frame. (Remember N can validly consider themselves as at rest, sitting still exactly in the middle (in this set up) of A and B).

9. Originally Posted by pzkpfw
Two different frames may always calculate what may have been simultaneous in each others frames, but if they are in relative motion events which are simultaneous (whether calculated or directly experienced simultaneously) for one can't be simultaneous for the other.

In the famous train-embankment thought experiment, the observers are positioned in the exact middle of what they each consider the sources of the flashes of light, but that's really just to make it "easy" to understand what is simultaneous (or not) for each of them.

Say there are two observers on the embankment, one in the exact middle of the sources (according to them) of the two flashes, and one who is positioned one third of the distance between the two sources. If those two observers are at rest with each other, they can agree on clocks, and both can agree on what the clocks at the sources of the flashes would say. If we assume the flashes were simultaneous in the frame of the embankment, the observer in the exact middle will see the flashes at the same time. The observer at one third will see one flash in half the time it takes for the other flash to arrive; but both will know that the flashes occured at the same time - in their frame.

(Simultaenity isn't about when we see events. If your camera flash goes off at the same time as you see a flash from a lightening strike on a distant hill; you consider that you saw the flashes simultaneously, but you wouldn't (normally) consider the lightening strike itself to have occured at the same time as you took the picture.)

You might argue then that the train observer should be equivalent to an embankment observer who isn't standing in the exact middle of the strikes, and who could calculate that the flashes were generated simultaneously in their frame too - but that would ignore that in their frame they do consider (in the usual set-up) that they were in the middle of the two strikes, so should see them at the same time. (Or translate this to any other position on the train, e.g. that one third position).

Code:
```A, B = strikes according to N, observer on train.
C, D = strikes according to M, O observers on embankment

====[A     N     B]=====
------------------------
C     M O   D

Later (train moved to the right):

======[A     N     B]===
------------------------
C     M O   D```
N might be next to O when the light from the leading strike hits N, but this light according to them came from B, whereas for O it came from D. So while O might see the light from D before the light from C, he or she will still calculate that C and D occured at the same time (if they did) just as M would if the flashes hit him or her at the same time; but - N can't get that same result in their frame. (Remember N can validly consider themselves as at rest, sitting still exactly in the middle (in this set up) of A and B).
Are you saying then that being in the middle between A and B, M will see the two lights at the same time (each traveling the same distance in the same time)?
Because the test for simultaneity is not that the lights hit somewhere at the same time but they have to hit someone in the centre between the strikes of lightening and that is at M.
N was there but has moved, along with his whole frame, toward B, and so is no longer at M and therefore the light from the two strikes arriving simultaneously do not denote simultaneity of the lightning strikes themselves?

10. Originally Posted by Grimble
I see your problem, so I am including another diagram:

In this diagram I have added, in green, the distance travelled by the Moving clock, as measured by the stationary observer, at the time when the light hits the mirror. This is 1.25 seconds as measured by the stationary observer, meaning that the mirror will have travelled 0.75 light seconds at v = 0.6c in that 1.25 seconds. Compared with the 0.6 light seconds in the 1 second that has passed for that same stationary observer in his own frame.
So 0.75 seconds in the moving frame (in the direction of travel) has contracted to 0.6seconds in the stationary frame. Length Contraction.
Well, not quite, no. I'm still not sure about what you've got there, I wish Grant was still around because he is good with Minkowski diagrams, but from your description, you're saying that 1 second passes for the light to travel to the mirror and that is 1.25 seconds according to the frame that sees the mirrors moving. The moving frame, however, will only measure 1.25 seconds for the one way trip when the mirrors are positioned perpendicularly to the line of travel, but in that case, there is no length contraction. If positioned at any other angle to the line of travel, there will be length contraction, but then there will also be relativity of simultaneity. For instance, as portrayed in the image for length contraction I gave, the mirror frame will measure 1 second for the light to travel each direction between the mirrors, but the moving frame will measure a time of d' / (c - v) = (.8) d / (c - .6 c) = 2 seconds for the light to travel in the same direction as the mirrors are moving, not 1.25 seconds, and a time of d' / (c + v) = (.8) d / (c + .6 c) = .5 seconds for the light to travel between the mirrors in the opposite direction. However, the time to travel away from a clock at one mirror, bounce off of the second mirror, and then travel back to the same single original clock, is 2 seconds + .5 seconds = 2.5 seconds for the two way trip. With a time dilation of .8, this means that (.8) (2.5 seconds) = 2 seconds has passed within the mirror frame. This is the time that has passed upon the single mirror frame clock.

I have solved the equation you asked me to solve, does that mean that I have understood what it represents? - I don't know.
That is why I began this thread to question how well I, as a visual thinker, see things. Is the way I see them valid?

So showing me simultaneity is begging the question. Do I understand it? As I said this thread is for me to express it and you to say if I have it right; explaining it again, achieves nothing.

So let's leave that until I know I have part 2, length contraction, right?
Okay. I couldn't be sure, but I was hoping it was clear enough. Do you understand the visual representations shown in the length contraction image for what is occurring? Do you understand the meaning of all of the physical dimensions given within that representation? Do you see how equalities can be taken directly from those dimensions to form equations to be solved?

11. Originally Posted by Bob Angstrom
I have questions about the highlighted parts of the explanation. Shouldn't the rod see its proper length as simply d or was this a reference to the target frame? It is my understanding that "proper" measurements are invariant and they only apply to the immediate inertial reference frame. If the rod sees its own length as d, then that is a proper "proper" The highlighted "proper" refers to a contracted proper which makes it a relativistic distance as opposed to a proper distance.
It appears I unintentionally changed your scenario to make the ends of the rod touch the targets simultaneously, sorry about that. In your original scenario, the proper length of the rod and proper distance between the targets are both d, and each sees the other contracted. But by making the length of the rod d as the target frame measures it when I posted, that would make the proper length of the rod d / sqrt(1 - (v/c)^2) for the scenario I gave.

12. Originally Posted by Grimble
Are you saying then that being in the middle between A and B, M will see the two lights at the same time (each traveling the same distance in the same time)?
In this case, no - because the events at C and D were simultaneous in the frame of M (and O).

Events can of course be simultaneous in the frame of N, in which case N would see the two lights at the same time (from A and B, because N happens to be in the middle) - but they would be different events than the ones described (and M won't agree they were simultaneous).

Originally Posted by Grimble
Because the test for simultaneity is not that the lights hit somewhere at the same time but they have to hit someone in the centre between the strikes of lightening ...
That's a simplified test for simultaenity, which is the point being made to Bob Angstrom. Being in the exact middle of the events and seeing them at the same time just makes it simple to calculate.

But if the observer is nearer one source than the other, so detects the events at different times (like O in my diagram, compared to M) that does not change the reality of whether the events were simultaneous or not - in their frame. M and O will both agree that C and D were simultaneous (if they were simultaneous in their frame).

If O (in this situation) detects the events at the same time, that does not mean they were simultanous events, and both M and O would agree the events were not simultaneous.

Originally Posted by Grimble
... and that is at M.
Well, for M it is. But not for N (for whom the events were at A and B, the ends of the carriage, with which they are at rest). Both M an N have their own (non preferred) frame, and both have valid observations of the World.

Originally Posted by Grimble
N was there but has moved, along with his whole frame, toward B, and so is no longer at M and therefore the light from the two strikes arriving simultaneously do not denote simultaneity of the lightning strikes themselves?
In this situation...
M considers themselves as at rest with the sources of the strikes, and in the middle of the sources (C and D).
N also considers themselves as at rest with the sources of the strikes, and in the middle of the sources (A and B).
If M sees the events at the same time, they can make a simple deduction that the events occured at the same time (in their own frame).
As N sees the events at different times, they can make a simple deduction that the events occured at different times (in their own frame).

The point of my post was that the situation for N is different to that of O.
O is at rest with C and D (and M) and can use simple distance & speed of light to calculate whether or not the events occured at the same time (in their frame) - and will agree with M.
But for N, no calculation will show that the events occured at the same time, in their frame, if they occured at the same time in the frame of M.
Remember, in their frame, N happens to be in the middle of - and at rest with - A and B; which is what they consider to be the events.
(If N were not in the middle of the train carriage, they could use speed of light and distances to N-A and N-B to calculate when the strikes occured, just as O does using O-C and O-D (but N won't get the same answer as O)).
N will be happy to agree that the events were simultaneous in the frame of M (if they were).

13. Originally Posted by grav
Well, not quite, no. I'm still not sure about what you've got there, I wish Grant was still around because he is good with Minkowski diagrams, but from your description, you're saying that 1 second passes for the light to travel to the mirror and that is 1.25 seconds according to the frame that sees the mirrors moving. The moving frame, however, will only measure 1.25 seconds for the one way trip when the mirrors are positioned perpendicularly to the line of travel, but in that case, there is no length contraction. If positioned at any other angle to the line of travel, there will be length contraction, but then there will also be relativity of simultaneity. For instance, as portrayed in the image for length contraction I gave, the mirror frame will measure 1 second for the light to travel each direction between the mirrors, but the moving frame will measure a time of d' / (c - v) = (.8) d / (c - .6 c) = 2 seconds for the light to travel in the same direction as the mirrors are moving, not 1.25 seconds, and a time of d' / (c + v) = (.8) d / (c + .6 c) = .5 seconds for the light to travel between the mirrors in the opposite direction. However, the time to travel away from a clock at one mirror, bounce off of the second mirror, and then travel back to the same single original clock, is 2 seconds + .5 seconds = 2.5 seconds for the two way trip. With a time dilation of .8, this means that (.8) (2.5 seconds) = 2 seconds has passed within the mirror frame. This is the time that has passed upon the single mirror frame clock.
And I am not sure what your problem is here. The diagram is of the standard light clock with the vertical light being rotatde by the movement. The mirrors are, therefore horizontal. The length contraction is that of the distance travelled by the mirror, not the light.
In the observer's own frame the distance the mirror has moved after one second proper time is 0.6 proper light seconds.
While that same observer, measuring the moving frame, measures that distance (after 1.25 seconds coordinate time = 1 second proper time) to be 0.75 coordinate light seconds. Giving the length contraction 0.75 -> 0.6 light seconds, transformed from coordinate to proper. (the green lines I have added.

Okay. I couldn't be sure, but I was hoping it was clear enough. Do you understand the visual representations shown in the length contraction image for what is occurring? Do you understand the meaning of all of the physical dimensions given within that representation? Do you see how equalities can be taken directly from those dimensions to form equations to be solved?
Yes.

14. Originally Posted by pzkpfw
In this situation...
M considers themselves as at rest with the sources of the strikes, and in the middle of the sources (C and D).
N also considers themselves as at rest with the sources of the strikes, and in the middle of the sources (A and B).
If M sees the events at the same time, they can make a simple deduction that the events occured at the same time (in their own frame).
As N sees the events at different times, they can make a simple deduction that the events occured at different times (in their own frame).

The point of my post was that the situation for N is different to that of O.
O is at rest with C and D (and M) and can use simple distance & speed of light to calculate whether or not the events occured at the same time (in their frame) - and will agree with M.
But for N, no calculation will show that the events occured at the same time, in their frame, if they occured at the same time in the frame of M.
Remember, in their frame, N happens to be in the middle of - and at rest with - A and B; which is what they consider to be the events.
(If N were not in the middle of the train carriage, they could use speed of light and distances to N-A and N-B to calculate when the strikes occured, just as O does using O-C and O-D (but N won't get the same answer as O)).
N will be happy to agree that the events were simultaneous in the frame of M (if they were).
Fine, I thought at first you were saying something different ...

15. Originally Posted by Grimble
And I am not sure what your problem is here. The diagram is of the standard light clock with the vertical light being rotatde by the movement. The mirrors are, therefore horizontal. The length contraction is that of the distance travelled by the mirror, not the light.
In the observer's own frame the distance the mirror has moved after one second proper time is 0.6 proper light seconds.
While that same observer, measuring the moving frame, measures that distance (after 1.25 seconds coordinate time = 1 second proper time) to be 0.75 coordinate light seconds. Giving the length contraction 0.75 -> 0.6 light seconds, transformed from coordinate to proper. (the green lines I have added.
Well, I'm having problems following it for a few reasons which I will go ahead and give. The green lines are vertical, so the mirrors are not moving, and they are separated by .15 light-seconds for some reason. The distance travelled in 1.25 seconds at .6 c compared to the distance travelled in 1 second at .6 c is still just comparing the times, 1.25 seconds to 1 second, the time dilation, not length contraction. A distance travelled does not length contract per say, it is just v t, but the lengths of objects and the distance between objects travelling together are generally said to contract. The light starts at the origin instead of at one of the mirrors. The time for the light to travel between the mirrors is not 1.25 seconds, but 2 seconds in the forward direction and .5 seconds in the opposite direction.

Here is a simple space-time diagram that shows the mirrors moving along x with respect to time t at .6 c, with the light emitted from one mirror, reflected by the other, and returning to the first. A Minkowski diagram would be very similar to this. Of course, in a diagram such as this, the length contraction is already given, however, as the .8 light-seconds shown between the two mirrors, rather than the 1 light-second measured in the mirror frame, so it is not much help for deriving the length contraction to begin with.

diag008.PNG

Yes.
Okay, good. That is the derivation for length contraction. It shows directly where length contraction comes from. Did you look through it thoroughly to see how it comes about? Do you understand it?
Last edited by grav; 2012-Jun-16 at 11:19 PM.

16. We do seem to be getting our wires crossed here, don't we?

You seem to be taking a completely different reading from my diagrams than that which I am intending them to shew.

So let me try and resolve that first of all with a couple of extra diagrams shewing the light reflecting from the mirror, first against the y axis to shoe reflection and secondly against the time axis as, obviously that will continue rather than reflecting.

Originally Posted by grav
Well, I'm having problems following it for a few reasons which I will go ahead and give. The green lines are vertical
Yes they will be! They are points dropped vertically to the x axis to define the x coordinates after 1 second of the clock's travel.
, so the mirrors are not moving, and they are separated by .15 light-seconds for some reason.
I am not sure how you are interpreting this. There is only one mirror. It is horizontal, at a fixed distance, of 1 Proper light second from the origin and travels horizontally across the diagram along that 1 second line (in Blue) of the ct (proper time) axis.
The distance travelled in 1.25 seconds at .6 c
0.75 coordinate light seconds
compared to the distance travelled in 1 second at .6 c
0.6 proper light seconds
is still just comparing the times,
as measured in light seconds?
1.25 seconds to 1 second, the time dilation, not length contraction. A distance travelled does not length contract per say, it is just v t, but the lengths of objects and the distance between objects travelling together are generally said to contract. The light starts at the origin instead of at one of the mirrors. The time for the light to travel between the mirrors is not 1.25 seconds,
there is but one mirror.
but 2 seconds in the forward direction and .5 seconds in the opposite direction.
It is a simple light clock with the light travelling vertically to a mirror 1 light second away and back to the source, triggering another pulse of light.
For the stationary observer, observing the light moving at 0.6c, the clock will have travelled 0.6 light seconds in the one second that the light takes to reach the mirror for an observer travelling with the mirror, or the 1.25 seconds (dilated time) the light takes, as measured by the stationary observer.
So, as measured by the stationary observer the clock will have travelled 0.6 x 1.25 (coordinate seconds), or 0.75 coordinate light seconds as measured by the stationary observer in the moving clock's frame.
The light will then take 1 proper second, 1.25 coordinate seconds to travel back to the light source at 0,1.5 (red coordinate scale) on the diagram. As shewn in the first part of the diagram.

Okay, good. That is the derivation for length contraction. It shows directly where length contraction comes from. Did you look through it thoroughly to see how it comes about? Do you understand it?
Do you?

17. Originally Posted by Grimble
Yes they will be! They are points dropped vertically to the x axis to define the x coordinates after 1 second of the clock's travel.
I am not sure how you are interpreting this. There is only one mirror. It is horizontal, at a fixed distance, of 1 Proper light second from the origin and travels horizontally across the diagram along that 1 second line (in Blue) of the ct (proper time) axis.
0.75 coordinate light seconds 0.6 proper light seconds as measured in light seconds? there is but one mirror.
It is a simple light clock with the light travelling vertically to a mirror 1 light second away and back to the source, triggering another pulse of light.
For the stationary observer, observing the light moving at 0.6c, the clock will have travelled 0.6 light seconds in the one second that the light takes to reach the mirror for an observer travelling with the mirror, or the 1.25 seconds (dilated time) the light takes, as measured by the stationary observer.
So, as measured by the stationary observer the clock will have travelled 0.6 x 1.25 (coordinate seconds), or 0.75 coordinate light seconds as measured by the stationary observer in the moving clock's frame.
The light will then take 1 proper second, 1.25 coordinate seconds to travel back to the light source at 0,1.5 (red coordinate scale) on the diagram. As shewn in the first part of the diagram.

Okay well, you seem to be referring to one stationary frame with proper and coordinate measurements with the light travelling at .6 c, and the light is still travelling up and down along the y axis instead of along x. First of all, you know that light always travels at c according to all inertial frames. Maybe that was misstated? The proper measurements are made in the frame where the mirrors do not move, so the light travels straight up and down along y according to your diagrams now, 1 light-second in 1 second. In another frame where the mirrors are observed to be moving away at .6 c, the mirrors are still 1 light-second apart along y, so there is no length contraction observed for the distance between the mirrors since length contraction is only observed along x. You can see from your diagrams, which are as viewed from the frame that sees the mirrors moving at .6 c, though, not the one stationary with the mirrors, that 1 light-second is still measured between the mirrors along y. This frame will observe that it takes 1.25 seconds for the light to travel at c between the mirrors positioned with 1 light-second between them while moving along the x axis at .6 c.

But this is still the time dilation only, 1.25 seconds as compared to 1 second, so it has nothing to do with length contraction and will not show length contraction. Consider that the two mirrors are connected by a rod. We want to know what length the frame that sees the rod moving at .6 c will measure for the length of the rod, how the length of the rod will be observed to be contracted. For this, you will now need to position the rod with the mirrors at each end along the x axis with the light travelling along the x axis only, nothing along the y axis.

Do you?
Sure, but you're the one asking.

18. This really is very basic.

There is one vertical light clock, mirror 1 light second from the source.
Time to mirror in the clock's frame, 1 second.

Clock is moving v = 0.6c.

To a stationary observer, the clock, mirror and all, travel horizontally v = 0.6c
Time is dilated, gamma = 1.25, so the light in the moving clock takes 1.25 coordinate seconds to reach the horizontal mirror the light travelling at angle beta.
As speed is the same in any frame (as c is constant) then the moving mirror will have travelled 0.6 light seconds in one second and 0.75 in 1.25 seconds.
To the stationary observer, in one second of his proper time, the clock will have travelled 0.6 light seconds; as another observer 0.6 light seconds away in the same frame would measure, therefore he will measure the clock to have moved 0.6 Proper Light Seconds.

To that same stationary observer, observing time dilation for the travelling clock, he will measure 1.25 coordinate seconds for the light to reach the mirror. In that dilated time the clock will have travelled 0.75 coordinate light seconds along the abscissa, the x axiswhere you agree length contraction will occur and we see the 0.75 coordinate light seconds of the moving clock contracted to 0.6 proper light seconds of the stationary observer. x = x'/γ

Moving time = stationary time x gamma - Time Dilation in any direction as time has no direction
stationary Length = moving length / gamma - Length contraction along the x axis the direction of travel.

19. Originally Posted by grav
Okay well, you seem to be referring to one stationary frame with proper and coordinate measurements with the light travelling at .6 c, and the light is still travelling up and down along the y axis instead of along x.
yes
First of all, you know that light always travels at c according to all inertial frames.
yes
Maybe that was misstated?
why would you think that?

The proper measurements are made in the frame where the mirrors do not move, so the light travels straight up and down along y according to your diagrams now, 1 light-second in 1 second. In another frame where the mirrors are observed to be moving away at .6 c, the mirrors are still 1 light-second apart along y, so there is no length contraction observed for the distance between the mirrors since length contraction is only observed along x. You can see from your diagrams, which are as viewed from the frame that sees the mirrors moving at .6 c, though, not the one stationary with the mirrors,
Yes, I have drawn it with the clock and mirrors moving and the observer stationary, is there a problem with that?

that 1 light-second is still measured between the mirrors along y. This frame will observe that it takes 1.25 seconds for the light to travel at c between the mirrors positioned with 1 light-second between them while moving along the x axis at .6 c.
??????
"that it takes 1.25 seconds for the light to travel at c between the mirrors positioned with 1 light-second between them"
1 light second in 1.25 seconds = c ???

Are you sure about that statement? Can you substantiate it?

But this is still the time dilation only, 1.25 seconds as compared to 1 second, so it has nothing to do with length contraction and will not show length contraction.
And I have not suggested that it will.

Consider that the two mirrors are connected by a rod. We want to know what length the frame that sees the rod moving at .6 c will measure for the length of the rod, how the length of the rod will be observed to be contracted. For this, you will now need to position the rod with the mirrors at each end along the x axis with the light travelling along the x axis only, nothing along the y axis.
I have made no claim to length contraction of the distance between the mirror and the source. (between the mirrors).

20. Originally Posted by Grimble
why would you think that?
Because you wrote "For the stationary observer, observing the light moving at .6 c,..." . Obviously you meant the mirrors (or source and mirror).

Yes, I have drawn it with the clock and mirrors moving and the observer stationary, is there a problem with that?
Your diagrams say "Light's path measured in the mirrors frame." , but it is the moving observer's frame.

?????? 1 light second in 1.25 seconds = c ???

Are you sure about that statement? Can you substantiate it?
From your own last set of diagrams, the mirror and source are d = 1 light-second apart along y and they travel along x at v = .6 c for t = 1.25 seconds while light traveling diagonally at c for t = 1.25 seconds also. So we have

d^2 + (v t)^2 = (c t)^2

(1 light-second)^2 + (.6 c)^2 (1.25 seconds)^2 = c^2 (1.25 seconds)^2

1^2 + (.6)^2 (1.25)^2 = 1.25^2

1 + .5625 = 1.5625

And I have not suggested that it will.
Your diagrams are labelled "Length Contraction".

I have made no claim to length contraction of the distance between the mirror and the source. (between the mirrors).
There is no length contraction whatsoever in the diagrams.

21. Originally Posted by Grimble
This really is very basic.

There is one vertical light clock, mirror 1 light second from the source.
Time to mirror in the clock's frame, 1 second.

Clock is moving v = 0.6c.

To a stationary observer, the clock, mirror and all, travel horizontally v = 0.6c
Time is dilated, gamma = 1.25, so the light in the moving clock takes 1.25 coordinate seconds to reach the horizontal mirror the light travelling at angle beta.
As speed is the same in any frame (as c is constant) then the moving mirror will have travelled 0.6 light seconds in one second and 0.75 in 1.25 seconds.
To the stationary observer, in one second of his proper time, the clock will have travelled 0.6 light seconds; as another observer 0.6 light seconds away in the same frame would measure, therefore he will measure the clock to have moved 0.6 Proper Light Seconds.

To that same stationary observer, observing time dilation for the travelling clock, he will measure 1.25 coordinate seconds for the light to reach the mirror. In that dilated time the clock will have travelled 0.75 coordinate light seconds along the abscissa, the x axiswhere you agree length contraction will occur and we see the 0.75 coordinate light seconds of the moving clock contracted to 0.6 proper light seconds of the stationary observer. x = x'/γ

Moving time = stationary time x gamma - Time Dilation in any direction as time has no direction
stationary Length = moving length / gamma - Length contraction along the x axis the direction of travel.
To an observer that sees a clock moving at .6 c along x, the clock will travel .6 light-seconds in 1 second while .8 seconds pass upon the clock, .75 light-seconds in 1.25 seconds while 1 second passes upon the clock, .9 light-seconds in 1.5 seconds while 1.2 seconds pass upon the clock, etc. It is just distance travelled = v * t. There is nothing special about that at all except for the reading on the clock which demonstrates time dilation, but not length contraction. Length contraction has to do with measuring the length of an object as shorter than its proper length, the proper length being the length as measured when it is not moving. So in the frame of a rod that is not moving, call it frame A, let's say the rod is 1 light-second long. Another frame sees the rod is moving at .6 c, call it frame B. So when the rod is moving as shown in your last diagrams according to frame B, positioned vertically along y as it moves along the x axis at .6 c, how long does frame B measure the rod to be? When the rod is turned perpendicularly so that it is positioned horizontally along the x axis while also moving along the x axis at .6 c, how long will frame B measure the rod to be?

22. Originally Posted by grav
Because you wrote "For the stationary observer, observing the light moving at .6 c,..." . Obviously you meant the mirrors (or source and mirror).
Yes, you are right; I should have written "Observing the clock moving ..."

Your diagrams say "Light's path measured in the mirrors frame." , but it is the moving observer's frame.
"Light's path measured in the mirrors frame." - from the observer's frame - it is a diagram of what the observer sees, everything is from her point of view.

From your own last set of diagrams, the mirror and source are d = 1 light-second apart along y and they travel along x at v = .6 c for t = 1.25 seconds while light traveling diagonally at c for t = 1.25 seconds also. So we have

d^2 + (v t)^2 = (c t)^2

(1 light-second)^2 + (.6 c)^2 (1.25 seconds)^2 = c^2 (1.25 seconds)^2

1^2 + (.6)^2 (1.25)^2 = 1.25^2

1 + .5625 = 1.5625
Yes, exactly! The dilated time for the light to reach the mirror is 1.25 seconds in the moving clock. Time Dilation.
The mirror is 1 light second away.
After 1 second in the observer's frame, the light will have travelled 0.8 light seconds along the y axis and 0.6 light seconds along the x axis. Simple vector addition.
The light has to travel for another 0.25 seconds to reach the mirror making those measurements: 1 light second along the y axis and 0.75 seconds along the x axis.
Now Special Relativity shows us that this still happens in one second of the observer's time demonstrating time dilation, and travelling 0.75 light seconds in the dilated time is contracted to 0.6 light seconds in the proper time. Length contraction.

Your diagrams are labelled "Length Contraction".

There is no length contraction whatsoever in the diagrams.
The length contraction is exactly the that which one would measure with the twins paradox, only here it is the clock travelling rather than a twin. Or are you implying that the distance travelled by that twin is not contracted?
Last edited by tusenfem; 2012-Jun-19 at 09:34 AM. Reason: corrected faulty quote tag

23. Originally Posted by grav
To an observer that sees a clock moving at .6 c along x, the clock will travel .6 light-seconds in 1 second while .8 seconds pass upon the clock, .75 light-seconds in 1.25 seconds while 1 second passes upon the clock, .9 light-seconds in 1.5 seconds while 1.2 seconds pass upon the clock, etc. It is just distance travelled = v * t. There is nothing special about that at all except for the reading on the clock which demonstrates time dilation, but not length contraction. Length contraction has to do with measuring the length of an object as shorter than its proper length, the proper length being the length as measured when it is not moving. So in the frame of a rod that is not moving, call it frame A, let's say the rod is 1 light-second long. Another frame sees the rod is moving at .6 c, call it frame B. So when the rod is moving as shown in your last diagrams according to frame B, positioned vertically along y as it moves along the x axis at .6 c, how long does frame B measure the rod to be? When the rod is turned perpendicularly so that it is positioned horizontally along the x axis while also moving along the x axis at .6 c, how long will frame B measure the rod to be?
Once again let me tell you that it is not the length of the rod that is contracting

[b]The length contraction shown in these diagrams has nothing to do with your rod.

It has nothing to do with the distance between the the mirrors. It is in the x direction.

It is the length, the distance that the clock has travelled that is length contracted.[b]

In exactly the same way that the distance the travelling twin goes is contracted as measured by the stay at home, stationary twin.

The distance the clock travels in the time it takes for the light to reach the mirror is 0.75 light seconds in 1.25 seconds. But because of Special Relativity this distance, this length, is contracted to 0.6 light seconds for the stationary observer, for whom it takes only 1 second, their time not being dilated.

Can it be any plainer or simpler than that?

24. In the other thread, I emphasized how important it was for terminology to be used correctly. Ditto for diagrams, and I think this is a source of confusion, so let's look #76.

In the diagram, we have a clock moving to (0.8, 1.0), then returning back to the x axis. That returning back leg has ct decreasing - that's going back in time ! Another thing to bear in mind is that light would show as a 45 degree angle line, and this marks the boundary of our future light cone, so all the clock's motion will have to happen "above" this line.

So let's rework this into a valid Minkowski diagram. We have a clock moving at 0.6c to a mirror 1 light second away, then returning. That 1 light second is a measure of distance - so it's a vertical line at x=1. The clock travels to it, then reflects back - not back to the x axis, but to the starting location x=0 just a bit later in time (ie. ct > 0). So the clock's travels form a triangle with the "base" on the ct axis. At 0.6c, the round trip will take 3.3 seconds, so the clock returns at (0, 3.3).

Now, the angle of travel will be tan(alpha)=0.6, so alpha=31 degrees off the ct axis - so mark that the ct' axis. Likewise, the clock's x axis (call it x') will be 31 degrees above our x axis. To show length contraction, let's say the clock is on the left end of a pole 1 light-second long at rest. Since the pole is (of course) moving at 0.6c, it will be sitting along the x' axis.

So start by having a shot at drawing that up

25. Gaaah, sorry - I've been reading recently about Minkowski, and leapt in and read that into the diagrams without paying proper attention

26. OK, after that gaffe, let's see if I can redeem myself

To start with, reading through, I kept getting confused with moving clocks, lights, stationary observers (surely they're both stationary ), etc - so I'm going to recruit Alice and Bob again.

1) Alice's Frame
Alice is sitting in a room, shining a light to a mirror on the roof 1 light second away, and timing it until it returns. It takes 2 seconds.

2) Relative Motion
Bob's on an embankment watching Alice go past in a train travelling at 0.6c.

3) Bob's Frame
Bob agrees that the roof is 1 light second high, but disagrees with the distance the light travels. He says the light takes 1.25 seconds up and same again back.

Now, this light-bouncing scenario is the example that we typically use to derive Gamma - as indeed you did in the OP. However, in these diagrams, you're taking gamma as a given (eg. with quotes like "Time is dilated, gamma = 1.25, so the light in the moving clock takes ....". ). Fair enough - we can prove that the whole picture hangs together. However, there's a danger that you can find ratios where it doesn't really apply. That's especially true where you "already know" the answer.

That's especially true with statements like (original bolding removed):
Originally Posted by Grimble
To that same stationary observer, observing time dilation for the travelling clock, he will measure 1.25 coordinate seconds for the light to reach the mirror. In that dilated time the clock will have travelled 0.75 coordinate light seconds along the abscissa, the x axis where you agree length contraction will occur and we see the 0.75 coordinate light seconds of the moving clock contracted to 0.6 proper light seconds of the stationary observer. x = x'/γ
Let's translate this to the context of Alice and Bob. Bob sees the light take 1.25 seconds to reach the mirror. In that time, the train will have moved 0.75 light seconds along the track. 0.75 light seconds in Alice's frame contracts to 0.6 light seconds of Bob's.

See the problem? You start with a measurement in Bob's frame, translate that to a distance due to the relative motion (so is not an implicit feature of either frame), and then take that distance and contract it using the ratios that you know apply to length contraction. Especially note that the "0.75 light seconds in Alice's frame" springs out of nowhere - Alice's frame doesn't have a "0.75 light seconds", since that distance was derived from a time measurement that Bob made.

This brings us to Length Contraction. Let's start with a definition, like "Length Contraction is the fact that lengths of objects and distances are shortened by gamma in the direction of motion".

So what object in Alice's frame are we going to examine to see length contraction? There isn't one! Read point (1) above - there's no length or distance mentioned (except for the 1 light second to the ceiling, and that's not contracted).

So, to start considering length contraction, Alice has a rod ...…

27. Originally Posted by RobA
OK, after that gaffe, let's see if I can redeem myself

To start with, reading through, I kept getting confused with moving clocks, lights, stationary observers (surely they're both stationary ), etc - so I'm going to recruit Alice and Bob again.

1) Alice's Frame
Alice is sitting in a room, shining a light to a mirror on the roof 1 light second away, and timing it until it returns. It takes 2 seconds.

2) Relative Motion
Bob's on an embankment watching Alice go past in a train travelling at 0.6c.

3) Bob's Frame
Bob agrees that the roof is 1 light second high, but disagrees with the distance the light travels. He says the light takes 1.25 seconds up and same again back.

Now, this light-bouncing scenario is the example that we typically use to derive Gamma - as indeed you did in the OP. However, in these diagrams, you're taking gamma as a given (eg. with quotes like "Time is dilated, gamma = 1.25, so the light in the moving clock takes ....". ). Fair enough - we can prove that the whole picture hangs together. However, there's a danger that you can find ratios where it doesn't really apply. That's especially true where you "already know" the answer.

That's especially true with statements like (original bolding removed):

Let's translate this to the context of Alice and Bob. Bob sees the light take 1.25 seconds to reach the mirror. In that time, the train will have moved 0.75 light seconds along the track. 0.75 light seconds in Alice's frame contracts to 0.6 light seconds of Bob's.

See the problem? You start with a measurement in Bob's frame, translate that to a distance due to the relative motion (so is not an implicit feature of either frame), and then take that distance and contract it using the ratios that you know apply to length contraction. Especially note that the "0.75 light seconds in Alice's frame" springs out of nowhere - Alice's frame doesn't have a "0.75 light seconds", since that distance was derived from a time measurement that Bob made.
Bob sees Alice is travelling at 0.6c.
Bob measures 1.25 seconds for the light to reach the mirror.
Bob measures that in that 1.25 seconds Alice will have travelled 0.6 x 1.25 = 0.75 light seconds. *Is that 'springing out of nowhere'?

We could say that if, when the light strikes the mirror it generates a flash, and Bob has assistants, with synchronised clocks, along the embankment, and that the one at 0.6 light seconds reported that flash after 1 second, then he would have measured both
one second and 0.6 light seconds for the light to hit the mirror in his own frame
and 1.25 seconds and 0.75 seconds in Alice's frame. *
From that Bob could calculate gamma.
This brings us to Length Contraction. Let's start with a definition, like "Length Contraction is the fact that lengths of objects and distances are shortened by gamma in the direction of motion".

So what object in Alice's frame are we going to examine to see length contraction? There isn't one! Read point (1) above - there's no length or distance mentioned (except for the 1 light second to the ceiling, and that's not contracted).

So, to start considering length contraction, Alice has a rod ...…
I certainly see what you are saying Rob, but isn't that making it all into something bigger than it is?

What is Length contraction? It is the difference in the way that two lengths are measured, the difference between a measurement in an observers own frame and the same measurement as seen in a moving frame. Two views, two measurements of the same length. The way that one length, one distance, is measured differently in the observer's own frame and another frame moving relative to him.

And that is exactly what we have here. The distance that Alice has travelled when the light hits the mirror. Measured by one observer using his own clocks and rods it is 0.6 light seconds. Measured by that same observer in the moving frame it is 0.75 light seconds.

If we do not call that Length Contraction, what do we call it? Putting a label on it or not changes nothing. The difference in the two lengths speak for them selves.

Or are you disagreeing with the figures?

28. Afterthought.

Consider Einstein's moving metre rod. It is contracted by a factor of gamma when measured in the observer's own frame.
How then do we consider a metre rod in that observer's own frame? How large a rod would measure that length (1 metre in the local frame) if it were in the moving frame? Would it not have to be expanded by a factor of gamma?

29. Moved to S&T.

30. Originally Posted by Grimble
Bob measures that in that 1.25 seconds Alice will have travelled 0.6 x 1.25 = 0.75 light seconds. Is that 'springing out of nowhere'?
But that's 0.75 light seconds in BOB'S frame. I'm talking about the next statement : "0.75 light seconds in ALICE'S frame contracts to 0.6 light seconds of Bob's." ALICE'S FRAME. Where does ALICE get 0.75 light seconds from? When does ALICE measure 1.25 seconds? You can't just take Bob's 0.75 light seconds, and then apply that to Alice (especially since the sentence is about how different the measurements are!)

Yes, we take a look at the "answer book", and find out at the end of the day that 0.75 ls of Alice does indeed get contracted to 0.6 of Bob's, but this example is about how we derive that relationship in the first place. The biggest mistake that people make with SR is mixing units between two frames - it is an easy mistake to make. To counteract that, we have to be very rigorous in keeping track of whose units we are dealing in. The 0.75 light seconds is in Bob's Frame. Do you agree, and do you see why you can't transfer it to Alice's?

Originally Posted by Grimble
We could say that if, when the light strikes the mirror it generates a flash, and Bob has assistants, with synchronised clocks, along the embankment, and that the one at 0.6 light seconds reported that flash after 1 second, then he would have measured both
one second and 0.6 light seconds for the light to hit the mirror in his own frame
and 1.25 seconds and 0.75 seconds in Alice's frame.
From that Bob could calculate gamma.
I'll call this "SCENARIO 2", compared to "SCENARIO 1" above.
With Scenario 2, you're getting on the right track (and yes, that pun was intended ), but you're mixing frames values again (it is a VERY easy mistake to make). How long does the light take to reach the mirror in Bob's frame, and how long in Alice's?

Originally Posted by Grimble
What is Length contraction? It is the difference in the way that two lengths are measured, the difference between a measurement in an observers own frame and the same measurement as seen in a moving frame. ... And that is exactly what we have here. The distance that Alice has travelled when the light hits the mirror. ... Or are you disagreeing with the figures?
Does 0.75ls in Alice's frame contract to 0.6ls in Bob's? Yes - given that we know that the length contraction formula. But to deriving that formula - deriving how long Bob thinks something in Alice's frame is - we need a length IN Alice's frame. That means we can't pick something that's related to the relative motion between them. That's what's missing in Scenario 1 - Alice does not have a measurement of length for Bob to compare against.

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