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Fitzgerald Lorentz contraction 
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#55
Sep508, 05:33 PM

P: 38

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: So, t1 = L/ gamma * (c+v) (c  v) This frame must be kept all along . However, for the vertical path, you wrote: However, notice your statement: 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^2v^2) being gamma. The time t2 would therefore be 2L/ SQRT (c^2  v^2), or 2L/ gamma t1 = 2L / gamma * (c+v) (cv) t2 = 2L/ gamma Therefore : t1/t2 = (c+v) (cv) Andre 


#56
Sep508, 06:08 PM

Sci Advisor
P: 8,470

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 cv 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 aetherit'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. 


#57
Sep508, 06:09 PM

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. 


#58
Sep508, 07:03 PM

P: 38

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. Andre 


#59
Sep508, 07:07 PM

P: 38

I agree with everything you wrote granpa. Andre 


#60
Sep508, 07:52 PM

Sci Advisor
P: 8,470

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. 


#61
Sep508, 11:05 PM

P: 38

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/(cv) = 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) (cv) = 1.6 * .4 = .64 Andre 


#62
Sep508, 11:32 PM

Sci Advisor
P: 8,470




#63
Sep608, 08:31 AM

P: 38

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 


#64
Sep608, 10:31 AM

Sci Advisor
P: 8,470

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. 


#65
Sep608, 11:44 AM

P: 38

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 


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