# Fitzgerald -Lorentz contraction

by mich
Tags: contraction, fitzgerald, lorentz
P: 38
 Quote by JesseM Maybe it would be helpful to do an actual calculation here? Suppose we have a Michelson-Morley type apparatus with two arms at right angles, each 20 light-seconds long in the rest frame of the apparatus. Now consider what will happen from the perspective of an observer who measures the apparatus to be moving at 0.6c to the right along the axis of the horizontal arm. In this frame, the vertical arm will still be 20 light-seconds long, but the horizontal arm is shrunk to 16 light-seconds due to Lorentz contraction, because $$20*\sqrt{1 - 0.6c^2/c^2} = 16$$ Now suppose at t=0 some light is released at the point where the left end of the horizontal arm meets the bottom end of the vertical arm. Since the light is moving to the right at 1c while the horizontal arm is moving to the right at 0.6c, in this frame the distance between the light beam and the right end will be decreasing at a rate of 0.4c, and since the horizontal arm is 16 light-seconds long, the time for the light to reach the right end will be 16/0.4 = 40 seconds. Then the light is reflected from the right end back at the left end, and now the left end is approaching the light at 0.6c, so the distance between the light and the left end is decreasing at a rate of 1.6c. So, the time for the light to get back to the left end will be an additional 16/1.6 = 10 seconds. So, in this frame the total time for the light to go from the left end to the right end and back is 40 + 10 = 50 seconds. Now consider the light going up and down the vertical arm. The top end of the vertical arm starts out 20 light-seconds directly above the point where the light is released, then after 25 seconds it is still 20 light-seconds above the release point on the vertical axis, while it has moved (25 seconds)*(0.6c) = 15 light-seconds away from the release point on the horizontal axis. So using the pythagorean theorem, the top end is now a distance of $$\sqrt{(20)^2 + (15)^2} = \sqrt{625} = 25$$ light-seconds away from the point where the light was released. Since the light travels 25 light-seconds in 25 seconds, this must be the time when the light catches up to the top end. Similar calculations show that 25 seconds after this, the bottom end of the vertical axis will be 25 light-seconds away from the position where the light hit the top end, so this must be the time the light returns to the bottom end. So, you can see that the total time for the light to go from the bottom to the top and back on the vertical arm is 25 + 25 = 50 seconds, exactly the same as the time for the light to go from the left to the right and back on the horizontal arm, so the two light waves will indeed meet at the same point in space and time. As I said before, it is a basic rule in relativity that if one frame says two events happen at the same position and time, this must be predicted in all frames.
While I think I understand what you did, I'm not quite certain I understand why you did what you did.
Now, as granpa said, there are two ways to look at this. We can either view the problem with the light speed being c relative to us, or make the calculations accordingly with the light going at the speed relative to the moving frame. and as granpa said, we cannot switch over one frame onto another.
Now first, as you mentioned, we start with Michelson's experiment having two paths (legs), one horizontal, the other vertical, both having the same length.

The first step, I have no problems with;
At t=0, two light pulses are sent simultaneously; one on the horizontal path, the other on the vertical path. Now, you wrote:

 Since the light is moving to the right at 1c while the horizontal arm is moving to the right at 0.6c, in this frame the distance between the light beam and the right end will be decreasing at a rate of 0.4c, and since the horizontal arm is 16 light-seconds long, the time for the light to reach the right end will be 16/0.4 = 40 seconds.
Since you wrote the speed of light will be calculated as being .4c, it must be relative to the moving frame, since it moves at the velocity of c relative to you.

So, t1 = L/ gamma * (c+v) (c - v)
This frame must be kept all along .

However, for the vertical path, you wrote:

 ...after 25 seconds it is still 20 light-seconds above the release point on the vertical axis, while it has moved (25 seconds)*(0.6c) = 15 light-seconds away from the release point on the horizontal axis.
This is where you lost me; notice that you are identifying the "diagonal" path. This path is the one that is observed by you, and is not relative to the moving frame.The light in this path is moving at "c" as you rightly identified as travelling 25 light seconds in 25 seconds.

 while it has moved (25 seconds)*(0.6c) = 15 light-seconds away from the release point on the horizontal axis.
On the moving frame, the light is certainly not moving away from the release point but remains in a straight vertical path.
In other words, the pythagorean triangle is reversed. The hypotheneuse is c, and the horizontal leg is v, leaving the vertical path as being the velocity of light relative to the moving frame....which is what we want. This leg or path alone is needed is this case....
Again, I'm not claiming to be right, but am only describing how I personally perceive the problem.
Now, since the path is not contracted, it must be L, while the light's velocity would be
c^2 = v^2 + C^2( C being the speed of light relative to the moving frame).

Therefore, C = SQRT (c^2 - v^2); SQRT (c^2-v^2) being gamma. The time t2 would therefore be 2L/ SQRT (c^2 - v^2), or 2L/ gamma

t1 = 2L / gamma * (c+v) (c-v)

t2 = 2L/ gamma

Therefore : t1/t2 = (c+v) (c-v)

Andre
P: 8,470
 Quote by mich While I think I understand what you did, I'm not quite certain I understand why you did what you did. Now, as granpa said, there are two ways to look at this. We can either view the problem with the light speed being c relative to us, or make the calculations accordingly with the light going at the speed relative to the moving frame. and as granpa said, we cannot switch over one frame onto another.
What post of granpa's are you talking about? It's certainly not true that you can't switch from one frame to another, the whole point of relativity is that any physical situation can be analyzed in any inertial frame using the same laws of physics, and all frames will agree on physical predictions about local events.
 Quote by mich Since you wrote the speed of light will be calculated as being .4c, it must be relative to the moving frame, since it moves at the velocity of c relative to you.
The speed of light is c in every inertial frame in relativity. What I was referring to was the rate that the distance between the end of apparatus and the light ray would be shrinking--this is sometimes called "closing speed" (the speed that one object is closing in on another one), it's different from the speed of a single object in a given frame. The speed of any single photon is always c in every inertial frame, but the closing speed between the photon and some other object can be less than c.
 Quote by mich So, t1 = L/ gamma * (c+v) (c - v)
Not quite, t1 = (L/gamma*(c+v)) + (L/gamma*(c-v))
 Quote by mich On the moving frame, the light is certainly not moving away from the release point but remains in a straight vertical path.
Where did you get that idea? As I've told you twice, all frames must agree in their predictions about local events. If their is a photodetector directly above the release point which is at rest in the frame of the apparatus, and in the frame of the apparatus the light goes straight up and hits the photodetector, you can't have other frames predicting that the light still goes straight up and therefore misses the photodetector which is moving in this frame! If it hits the photodector in the rest frame of the apparatus, it must do so in all frames. In general if you have some device which shoots a light beam straight up in its rest frame, it must shoot a light beam diagonally in other frames; in principle you should be able to prove this using classical electromagnetism, considering something like an oscillating charge at the bottom of a tube.
 Quote by mich In other words, the pythagorean triangle is reversed. The hypotheneuse is c, and the horizontal leg is v, leaving the vertical path as being the velocity of light relative to the moving frame....which is what we want.
The velocity of light is always c along its path, in every inertial frame. If you don't understand this you've missed one of the most basic points about relativity.

On the other hand, in a Lorentz aether theory we can say that light only travels at c in the rest frame of the aether, while in any frame moving at v relative to the aether, it travels at c+v in one direction and c-v in another. Is this what you're trying to talk about? If so I can reanalyze the same problem from this perspective, under the assumption that objects shrink by a factor of 1/gamma when they are moving relative to the aether--it'll still work out that all frames agree that the two light rays return to the intersection of the horizontal and vertical arms at the same moment.
 P: 2,258 using c+v isnt a matter of looking at it from the frames point of view. its purely a mathematical convenience. it has nothing to do with relativity. we could be calculating the speed of cow manure hurled from one end of a moving truck to the other. or rather the time it takes to make the trip. we could use its actual speed and the actual distance it moves but why bother? just use the length of the truck and the speed relative to the truck. we are obviously looking at it from the moving observers point of view since its all rather trivial from the stationary point of view.
P: 38
 Quote by JesseM What post of granpa's are you talking about? It's certainly not true that you can't switch from one frame to another, the whole point of relativity is that any physical situation can be analyzed in any inertial frame using the same laws of physics, and all frames will agree on physical predictions about local events.
The two frames I meant was how the observer calculated the velocity of light; it's either c relative to him or it's a different speed relative to the moving frame, as seen by the observer (closing speed). One cannot switch from one perception to the other when calculation is done on both legs. One has to stick with either one.

 The speed of light is c in every inertial frame in relativity. What I was referring to was the rate that the distance between the end of apparatus and the light ray would be shrinking--this is sometimes called "closing speed" (the speed that one object is closing in on another one), it's different from the speed of a single object in a given frame. The speed of any single photon is always c in every inertial frame, but the closing speed between the photon and some other object can be less than c.
I agree.

 Not quite, t1 = (L/gamma*(c+v)) + (L/gamma*(c-v))
Sorry...forgot my brakets. t1 = L/ gamma * [(c+v) (c - v)]

 Where did you get that idea? As I've told you twice, all frames must agree in their predictions about local events.
What about when two light signals are sent by a source located at the centre of the frame of refrence. The local events are the time when the light signals will hit the walls. An second observer, on a different frame, will view the events differently than the observer within the stationnary frame, relative to the source. Now, you might say, this is not one local event but two different events since the light pulse are separated. However for the M&M experiment, the source of light pointing towards the horizontal path could be located on one side of the frame while the other source, pointing towards the vertical path might be located on the opposite end of the frame, so I personally don't see any difference.

 If their is a photodetector directly above the release point which is at rest in the frame of the apparatus, and in the frame of the apparatus the light goes straight up and hits the photodetector, you can't have other frames predicting that the light still goes straight up and therefore misses the photodetector which is moving in this frame! If it hits the photodector in the rest frame of the apparatus, it must do so in all frames. In general if you have some device which shoots a light beam straight up in its rest frame, it must shoot a light beam diagonally in other frames; in principle you should be able to prove this using classical electromagnetism, considering something like an oscillating charge at the bottom of a tube.
By release point, I know that you did not mean the source itself, but the coordinate where the source was located when it released the photon.
Therefore, at t =25 second, neither the detector nor the source is located there anymore....the frame at that particular coordinate no longer exists. Because of this, you cannot speak of "closing speed" anymore. You are now refering to the light speed, relative to the observer....which is c.

 The velocity of light is always c along its path, in every inertial frame. If you don't understand this you've missed one of the most basic points about relativity.
When light is directly measured, Relativity promises us that it will always be c; I believe I know this, yes.

 On the other hand, in a Lorentz aether theory we can say that light only travels at c in the rest frame of the aether, while in any frame moving at v relative to the aether, it travels at c+v in one direction and c-v in another. Is this what you're trying to talk about? If so I can reanalyze the same problem from this perspective, under the assumption that objects shrink by a factor of 1/gamma when they are moving relative to the aether.
No we're stricly speaking of Relativity, and not LET. I hope that you understand that I'm not claiming anything to be false or true, but am just trying to identify the problems I have.

Andre
P: 38
 Quote by granpa using c+v isnt a matter of looking at it from the frames point of view. its purely a mathematical convenience. it has nothing to do with relativity. we could be calculating the speed of cow manure hurled from one end of a moving truck to the other. or rather the time it takes to make the trip. we could use its actual speed and the actual distance it moves but why bother? just use the length of the truck and the speed relative to the truck. we are obviously looking at it from the moving observers point of view since its all rather trivial from the stationary point of view.

I agree with everything you wrote granpa.

Andre
P: 8,470
 Quote by mich The two frames I meant was how the observer calculated the velocity of light; it's either c relative to him or it's a different speed relative to the moving frame, as seen by the observer (closing speed).
If you're just talking about closing speed, saying a "different speed relative to the moving frame" is overly confusing--the closing speed of the photon and the end of the apparatus not the same as the speed of the photon in the rest frame of the apparatus, that speed is still c.
 Quote by mich One cannot switch from one perception to the other when calculation is done on both legs. One has to stick with either one.
And do you think I did? On both legs I was calculating times and distances in the frame of the observer who sees the apparatus moving at 0.6c.
Quote by mich
 Not quite, t1 = (L/gamma*(c+v)) + (L/gamma*(c-v))
Sorry...forgot my brakets. t1 = L/ gamma * [(c+v) (c - v)]
Your expression is still wrong--if you're adding the fraction L/[gamma*(c+v)] to the fraction L/[gamma*(c-v)], the two fractions have different denominators, so you can fix that by multiplying both top and bottom of the first fraction by (c-v), and multiplying both top and bottom of the second fraction by (c+v), so the two fractions become [L*(c+v)]/[gamma*(c+v)*(c-v)] and [L*(c-v)]/[gamma*(c+v)*(c-v)]. Then you just add the numerators, which would give [L*(c+v) + L*(c-v)]/[gamma*(c+v)*(c-v)] = L*c / [gamma*(c+v)*(c-v)].

 Quote by mich What about when two light signals are sent by a source located at the centre of the frame of refrence.
What do you mean by "center"? A frame of reference in SR is an infinite coordinate system assigning coordinates to every point in space and time--do you just mean the origin of the coordinate system, or the physical center of a room?
 Quote by mich The local events are the time when the light signals will hit the walls.
But times are not physical events unless you are talking about readings on a particular set of physical clocks. If there are clocks mounted on the walls which are synchronized in the frame of the walls, and a flash is set off at the midpoint between the two clocks, then all frames will make the same prediction about the reading on each clock at the moment the light from the flash first hits them--they'll all predict that the two clocks read the same time at the moment the light hits them. But this doesn't mean the two events actually happen at the same coordinate time in all frames, some other frame will say that the light actually reached one clock before the other, but in this frame the clocks were out-of-sync by just the right amount so they showed the same readings at the two different times the light hit them.
 Quote by mich Now, you might say, this is not one local event but two different events since the light pulse are separated. However for the M&M experiment, the source of light pointing towards the horizontal path could be located on one side of the frame while the other source, pointing towards the vertical path might be located on the opposite end of the frame, so I personally don't see any difference.
Like I said, a frame is an infinite coordinate system, so talking about "one side of the frame" doesn't really make sense, I'd guess you are probably just using "frame" to mean the physical MM apparatus here. Anyway, if you have one light source located on the right end of the horizontal arm and pointed towards the left end (where the horizontal arm intersects with the bottom end of the vertical arm), and another light source located on the top end of the vertical arm and pointed at the bottom end, then if the two sources are turned on in such a way that one frame predicts the light rays reach the intersection point at the same time, then it must be true that all frames predict this, since the meeting of the two rays is a localized event.
 Quote by mich By release point, I know that you did not mean the source itself, but the coordinate where the source was located when it released the photon. Therefore, at t =25 second, neither the detector nor the source is located there anymore....the frame at that particular coordinate no longer exists.
Again, are you using "frame" to refer to the physical apparatus? I agree the apparatus is no longer at that coordinate, but the observer still determines the distance the light ray traveled by looking at the distance interval between the position coordinate in his frame that the ray was released and the position coordinate where it was reflected, and likewise he determines the time by looking at the time interval between the time coordinate the ray was released and the time coordinate it was reflected, and he defines speed in his frame in terms of (distance interval)/(time interval).

By the way, note that the coordinates of a particular observer's frame are imagined to be grounded in the readings on a grid of rulers and synchronized clocks which are at rest with respect to the observer. So the observer could look at what marking on his ruler the bottom of the vertical arm was next to at the moment the light was released, and what the clock affixed to that marker read at that moment, and then compare with the marking on his ruler that the top of the vertical arm was next to at the moment the light was reflected, and what the clock affixed to that marker read at the moment of reflection. So he's assigning coordinates based purely on local readings that were right next to each event as they occurred.
 Quote by mich Because of this, you cannot speak of "closing speed" anymore. You are now refering to the light speed, relative to the observer....which is c.
Sure. In the observer's frame, the distance between the position the light was released and the position of the top arm when it was reflected was 25 light-seconds, and the event reflection happened 25 seconds after the event of release, so the speed of light was indeed c in his frame. With the horizontal arm I did make use of the notion of closing speed to calculate the time for the light to get from the left end to the right end and back, but this wasn't really necessary, I could have just as well said that the light was released at time t=0 and position x=0, and at that moment the right end was at position x=16 light-seconds (because the horizontal arm is 16 light-seconds long in this frame), and that 40 seconds later the light would naturally be at position x=40 l.s. because it's moving at c, and since the right end was moving at 0.6c for 40 seconds it would have covered a distance of 24 light seconds, and since it started at x=16 it would now be at x=16+24=40 too, so 40 seconds must be the time for the light to catch up with the right end. This is exactly the same answer as the one you get if you take the length of the arm in this frame, 16 light seconds, and divide by the closing speed of 0.4c. Closing speed is just a convenient shortcut for figuring out when one moving object will catch up with another moving object, nothing more.
P: 38
 Quote by JesseM If you're just talking about closing speed, saying a "different speed relative to the moving frame" is overly confusing--the closing speed of the photon and the end of the apparatus not the same as the speed of the photon in the rest frame of the apparatus, that speed is still c.
I agree

 And do you think I did?
At first, I thought you did, but I was wrong.

 On both legs I was calculating times and distances in the frame of the observer who sees the apparatus moving at 0.6c.
I agree

 Your expression is still wrong--if you're adding the fraction L/[gamma*(c+v)] to the fraction L/[gamma*(c-v)], the two fractions have different denominators, so you can fix that by multiplying both top and bottom of the first fraction by (c-v), and multiplying both top and bottom of the second fraction by (c+v), so the two fractions become [L*(c+v)]/[gamma*(c+v)*(c-v)] and [L*(c-v)]/[gamma*(c+v)*(c-v)]. Then you just add the numerators, which would give [L*(c+v) + L*(c-v)]/[gamma*(c+v)*(c-v)] = L*c / [gamma*(c+v)*(c-v)].
I was not adding those two fractions at all; I was identifying a length contraction for L, making it L/gamma. This over the velocities (c+v) (c-v) gave me t1.

 What do you mean by "center"? A frame of reference in SR is an infinite coordinate system assigning coordinates to every point in space and time--do you just mean the origin of the coordinate system, or the physical center of a room?
Physical center

 But times are not physical events unless you are talking about readings on a particular set of physical clocks. If there are clocks mounted on the walls which are synchronized in the frame of the walls, and a flash is set off at the midpoint between the two clocks, then all frames will make the same prediction about the reading on each clock at the moment the light from the flash first hits them--they'll all predict that the two clocks read the same time at the moment the light hits them. But this doesn't mean the two events actually happen at the same coordinate time in all frames, some other frame will say that the light actually reached one clock before the other, but in this frame the clocks were out-of-sync by just the right amount so they showed the same readings at the two different times the light hit them.
If an observer on a moving frame can observe an event, coming from what we can claim to be a stationnary frame, as being non simultaneous while the other claims it to be simultaneous, then this is all I'm saying about the M&M experiment observed from a moving frame.

 Like I said, a frame is an infinite coordinate system, so talking about "one side of the frame" doesn't really make sense, I'd guess you are probably just using "frame" to mean the physical MM apparatus here.
Yes, and no; along with the moving apparatus, or train, if you like, I see with it, a plane of coordinates having a xyzt axis. This is the reason why I prefer to call it a frame of reference.

 Anyway, if you have one light source located on the right end of the horizontal arm and pointed towards the left end (where the horizontal arm intersects with the bottom end of the vertical arm), and another light source located on the top end of the vertical arm and pointed at the bottom end, then if the two sources are turned on in such a way that one frame predicts the light rays reach the intersection point at the same time, then it must be true that all frames predict this, since the meeting of the two rays is a localized event....

After reading over your post, it struck me that you were right. Of coarse it doesn't matter how the observer will measure the light's velocity c in his frame, or xc relative to another moving frame. His
time measurement would indeed be the same for the event in question no matter which frame of reference the observer would use (his or the moving frame's).
Yet, I still could not understand how the the ratio between t1/t2 could be 1 since we are clearly dealing with a length contraction for the horizontal distance and a non contracted distance for the vertical distance.I looked at what I wrote down and could not see anything that could help me see where my error was.
And so I went through your figures and found something questionable. However, I've been out of school for quite a while and so I certainly maybe doing a miscalculation....but this miscalculation seem to work my way.

It concerns your measured length contraction for the moving platform. You wrote down 16.
I got 20* SQRT (1-.6)
= 20 * sqrt (.4)
=20 * .632
= 12.64

So, with this contraction, we continue:

L/(c-v) = 12.64 / .4 = 31.6
L/ (c+v) = 12.64/ 1.6 = 7.9

31.6 + 7.9 = 39.5
Therefore, t1 = 39.5

t2/t1 = 25/39.5
= .6329

The ratio I had calculated was
(c+v) (c-v)
= 1.6 * .4
= .64

Andre
P: 8,470
 Quote by mich I was not adding those two fractions at all; I was identifying a length contraction for L, making it L/gamma. This over the velocities (c+v) (c-v) gave me t1.
But where did you get that equation? The two fractions I gave are correct for the time for the light to go from the left end to the right end of the horizontal arm, and the time for the light to go from right end to left end. After all, the length of the arm is L/gamma, and the closing speed from left to right is (c - v) while the closing speed from right to left is (c + v), and the time for a faster-moving object to get from one end to the other of a slower-moving object is always going to be (length of slower object)/(closing speed between faster object and the end of the slower object it's moving towards).
 Quote by mich If an observer on a moving frame can observe an event, coming from what we can claim to be a stationnary frame, as being non simultaneous while the other claims it to be simultaneous, then this is all I'm saying about the M&M experiment observed from a moving frame.
Disagreements about simultaneity can only happen for events at different locations--as I keep saying, if two events are predicted in one frame to coincide at the same point in space and time, then all other frames must predict this as well. Imagine that at the intersection of the two arms of the MM apparatus we place a small photosensitive bomb that will go off only if the photons from each arm hit it at the same moment--if different frames made different predictions about whether the bomb goes off, that would either mean we could find a preferred frame by seeing whether or not it actually does go off, or it would mean different frames were parallel universes with distinct histories!
 Quote by JesseM Like I said, a frame is an infinite coordinate system, so talking about "one side of the frame" doesn't really make sense, I'd guess you are probably just using "frame" to mean the physical MM apparatus here.
 Quote by mich Yes, and no; along with the moving apparatus, or train, if you like, I see with it, a plane of coordinates having a xyzt axis. This is the reason why I prefer to call it a frame of reference.
When you say "yes, and no" does that mean you are referring to the apparatus as a "frame", and the plane of coordinates as a "frame of reference"? Normally "frame" is just used as shorthand for "frame of reference", and only refers to the coordinate system, not to any physical apparatus (unless we're talking about the imaginary grid of rulers and clocks which are used to define the coordinate system physically). Regardless of how you've used these terms up until now, is it OK with you if from now on we just use "frame" in the standard way, to refer to the type of coordinate system you describe above?
 Quote by mich Yet, I still could not understand how the the ratio between t1/t2 could be 1 since we are clearly dealing with a length contraction for the horizontal distance and a non contracted distance for the vertical distance.
Yes, but the light is not just traveling the length of each arm, since the arms themselves are moving in this frame--to calculate the time you need to take into account that in this frame, the light has to travel on diagonals to go to the top end of the vertical apparatus and back, and the light going on the horizontal arm has to catch up with the right end which is moving away from where the light originates, and then has a much shorter distance to travel when it is reflected back from the right end and the left end is rushing forward towards it.
 Quote by mich And so I went through your figures and found something questionable. However, I've been out of school for quite a while and so I certainly maybe doing a miscalculation....but this miscalculation seem to work my way. It concerns your measured length contraction for the moving platform. You wrote down 16. I got 20* SQRT (1-.6) = 20 * sqrt (.4) =20 * .632 = 12.64
1/gamma is $$\sqrt{1 - v^2/c^2}$$, not $$\sqrt{1 - v/c}$$. So you have to take 0.6c/c and square it, giving 0.36, then take the square root of (1 - 0.36), which is the same as the square root of 0.64, and the square root of 0.64 is 0.8.
P: 38
 Quote by JesseM 1/gamma is $$\sqrt{1 - v^2/c^2}$$, not $$\sqrt{1 - v/c}$$. So you have to take 0.6c/c and square it, giving 0.36, then take the square root of (1 - 0.36), which is the same as the square root of 0.64, and the square root of 0.64 is 0.8.
Ha! You got me. I munched on all the meat and 'tatos you wrote and I'd like to thank you for your patience because it clarified many things.

As I was trying to visualize the experiement, it dawned on me that
leaving the length contraction out of the equation for the horizontal leg would leave me with the galilean form only....and this cannot be for the constancy of light's speed. So I do agree that the transform is needed for the horizontal path.
However, my problem concerns strickly length contractions,not time dilation. Therefore I'd like to visualise the experiment a little differently. By the way, I appologize for all the mathematical mistakes; I've really out of practice. :)

Also, I've come to understand most of your other explanations as well; In the experiment, having the local event out of synch for one observer and in synch for the other would indeed violate the first postualte.So, now I also agree that Relativity cannot predict this to happen.

So in looking at the scenario again, but in keeping with only time dilations and leaving length contraction aside, I would say first:

O1 = observer stationnary to the experiment
O2 = observer on a moving frame

For the horizontal path.

For O1; In both cases (horizontal and vertical path) T1(t1)= 2L/c

For O2; The horizontal path would be
T2 (t1) =[(L + (vt)) + (L - (vt)] / c
= L[ (1+(vt) + (1 - (vt)]/c
= L /c [ 1-(vt)^2]

T2/(T1) = c/2L * L/c [ 1-(vt)^2] = [ 1-(vt)^2] /2

For O2; vertical path:

T2 = 2[SQRT L^2 +(vt)^2] / c
2[ SQRT (ct)^2 + (vt)^2]/c
2[ SQRT t^2 (c^2+v^2)]/c
2t/c [SQRT (1 + (v^2/c^2)]

....ok; I think I get now....L is measured by both observers as being
ct. However, due to the time dilation inlvolved, L cannot be the same for both frames.

Case closed... thanks again.

Andre
P: 8,470
 Quote by mich For O2; The horizontal path would be T2 (t1) =[(L + (vt)) + (L - (vt)] / c
I don't get it, what is "t" supposed to represent here? It's true that the distance the light travels going left to right is L + v*t1, where t1 is the time to get from the left end to the right end, and then the distance the light travels going right to left is L - v*t2, where t2 is the time to get from the right end to the left end, but t1 and t2 will be unequal.

Also, you seem to use L for the length of the uncontracted arms, why are you also using it for the length of the contracted horizontal arm here? You should really use a different symbol like L' = L*sqrt(1 - v^2/c^2)

So, if the point where the light is first sent out is defined as x=0, then the position of the right end as a function of time is x(t) = L' + vt, while the position of the light ray is x(t) = ct, so this tells you the light reaches the right end when L' + vt = ct, and solving for t gives L'/(c - v). Then if we redefine x=0 to be the position of the left end at the moment the light is reflected from the right end, then the position of the left end as a function of time is x(t) = vt, and the position of the light as a function of time is x(t) = L' - ct, so the light will meet the left end when vt = L' - ct, and solving for t gives L'/(c + v). So the time for the light to go from left to right and back must be L'/(c - v) + L'/(c + v), or L'*(c + v)/[(c + v)*(c - v)] + L'*(c - v)/[(c + v)*(c - v)], which is equal to 2*L'*c/[(c + v)*(c - v)] = 2*L'*c/[c^2 - v^2] = 2*L'*c/[c^2*(1 - v^2/c^2)] = 2*L'/[c*(1 - v^2/c^2)]. And then if you substitute in L' = L * sqrt(1 - v^2/c^2), you get 2*L/[c * SQRT(1 - v^2/c^2)] for the total time for the light to go from bottom to top and back.
 Quote by mich For O2; vertical path: T2 = 2[SQRT L^2 +(vt)^2] / c 2[ SQRT (ct)^2 + (vt)^2]/c
It looks like you substituded L = ct here, but why? In this case L is supposed to be the length of the vertical arm, and t is supposed to be the time to get from the bottom to the top of the vertical arm, right? But because light is traveling along a diagonal path in this frame, it travels a greater distance than L in time t--specifically it travels a distance of SQRT[L^2 + (vt)^2]. And since in time t light always travels a distance ct, you should really have ct = SQRT[L^2 + (vt)^2], or (ct)^2 = L^2 + (vt)^2, which you can solve for t to give t^2 = L^2 /(c^2 - v^2) = L^2 / [c^2 * (1 - v^2/c^2)], which gives you t = L/[c*SQRT(1 - v^2/c^2)]. Of course that's just the time to go from the bottom to the top of the vertical arm, the time to get back to the bottom is twice that, or 2*L/[c*SQRT(1 - v^2/c^2)], which is exactly the same as the time we previously got for the light to go from left to right and back on the horizontal arm.
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 Quote by JesseM I don't get it, what is "t" supposed to represent here? It's true that the distance the light travels going left to right is L + v*t1, where t1 is the time to get from the left end to the right end, and then the distance the light travels going right to left is L - v*t2, where t2 is the time to get from the right end to the left end, but t1 and t2 will be unequal. Also, you seem to use L for the length of the uncontracted arms, why are you also using it for the length of the contracted horizontal arm here? You should really use a different symbol like L' = L*sqrt(1 - v^2/c^2) So, if the point where the light is first sent out is defined as x=0, then the position of the right end as a function of time is x(t) = L' + vt, while the position of the light ray is x(t) = ct, so this tells you the light reaches the right end when L' + vt = ct, and solving for t gives L'/(c - v). Then if we redefine x=0 to be the position of the left end at the moment the light is reflected from the right end, then the position of the left end as a function of time is x(t) = vt, and the position of the light as a function of time is x(t) = L' - ct, so the light will meet the left end when vt = L' - ct, and solving for t gives L'/(c + v). So the time for the light to go from left to right and back must be L'/(c - v) + L'/(c + v), or L'*(c + v)/[(c + v)*(c - v)] + L'*(c - v)/[(c + v)*(c - v)], which is equal to 2*L'*c/[(c + v)*(c - v)] = 2*L'*c/[c^2 - v^2] = 2*L'*c/[c^2*(1 - v^2/c^2)] = 2*L'/[c*(1 - v^2/c^2)]. And then if you substitute in L' = L * sqrt(1 - v^2/c^2), you get 2*L/[c * SQRT(1 - v^2/c^2)] for the total time for the light to go from bottom to top and back. It looks like you substituded L = ct here, but why? In this case L is supposed to be the length of the vertical arm, and t is supposed to be the time to get from the bottom to the top of the vertical arm, right? But because light is traveling along a diagonal path in this frame, it travels a greater distance than L in time t--specifically it travels a distance of SQRT[L^2 + (vt)^2]. And since in time t light always travels a distance ct, you should really have ct = SQRT[L^2 + (vt)^2], or (ct)^2 = L^2 + (vt)^2, which you can solve for t to give t^2 = L^2 /(c^2 - v^2) = L^2 / [c^2 * (1 - v^2/c^2)], which gives you t = L/[c*SQRT(1 - v^2/c^2)]. Of course that's just the time to go from the bottom to the top of the vertical arm, the time to get back to the bottom is twice that, or 2*L/[c*SQRT(1 - v^2/c^2)], which is exactly the same as the time we previously got for the light to go from left to right and back on the horizontal arm.

Jesse, when I wrote down "case closed", I had just realized how right you were, and everything was becoming clear. I was starting to understand my errors now....Case closed meaning you win, I loose.

The L=ct was not part of the equation I was trying out. It just dawned on me that both observers measure distances with ct. Therefore it made me see that clearly length contractions are needed in Relativity.

Andre

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