Fitzgerald -Lorentz contraction

[mich]

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

 

[JesseM]

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).

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!

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.

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?

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.

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.

 

[mich]

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

 

[JesseM]

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.

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.

 

[mich]

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.

Thanks again for your time

Andre