
#19
Nov812, 09:25 AM

PF Gold
P: 4,518

The rod is at rest in frame S' and moving at v in frame S. There are a lot of equations that we can write down but we arbitrarily pick two that relate the end points of the rod in S' to their events in S. Note that we are not concerned with the events in S' because we are ignoring the corresponding time coordinates in frame S'. This is because we realize that we can specify any time coordinates for the events in frame S' and it won't change the calculation of the length of the rod since it is at rest. So we specify the two ends of the rods in frame S' as x1' and x2' and we write: x1' = γ(x1vt1) and x2' = γ(x2vt2) Now we want t1 to be equal to t2 so we replace both of them with simply t. x1' = γ(x1vt) and x2' = γ(x2vt) Now we note that since L(0) is equal to x2'  x1' we can write: L(0) = x2'  x1' = γ(x2 vt)  γ(x1vt) L(0) = γx2  γvt  γx1 + γvt L(0) = γx2  γx1 + γvt  γvt L(0) = γx2  γx1 L(0) = γ(x2x1) But since we also know that L (the length in frame S) is equal to x2x1 we can make that substitution and get: L(0) = γL And rearranging we get the desired final (and correct) result: L = L(0)/γ Note that this derivation is very similar to yours in post #1 where you made the mistake of implicitly setting t = t1' = t2'. Instead, I explicitly set t = t1 = t2. So in effect, you were doing the calculation for the situation where the rod is at rest in frame S and you were calculating its length in frame S'. Remember, the equations are establishing a relationship between the coordinates of events in two different frames. You can use the Lorentz transformation equations to solve for the coordinates in either direction but it is more complicated going in the opposite direction and you have to remember that the sense of v is in the opposite direction. 



#20
Nov812, 04:22 PM

P: 3,178





#21
Nov812, 05:00 PM

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#22
Nov812, 07:51 PM

P: 848





#23
Nov812, 10:08 PM

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P: 4,518

You mentioned a black frame and I think you are showing axes for red and blue frames, so are we supposed to think about three frames at the same time? You said you have a red and a blue meter stick but I see what appears to be two red and two blue meter sticks. That confuses me. Plus you show all four meter sticks aligned with lines of simultaneity. The whole point of the preceding posts in this thread is that in the frame in which the rod is moving, we pick a pair of events at the two ends of the stick that are simultaneous but in the rod's rest frame these events are not simultaneous. Is your sketch supposed to be illustrating the process described in the wikipedia article or are you showing some other method to illustrate length contraction? Since you said that both meter sticks are moving in opposite directions in the black frame and since everything looks symmetrical to me, I'm assuming that they are moving at the same speed, in which case they should be length contracted by the same amount and I would think they should be aligned parallel to the black xaxis but I don't see that. So it's immediately obvious to me that a sketch like this needs some background training in order for it to communicate to other people. 



#24
Nov812, 11:45 PM

PF Gold
P: 4,518

Here's another way for an observer to measure the length of a moving rod, or for a moving observer to measure the length of a stationary rod. Let's say the rod is 1 unit in length (using units where c=1) stretching from coordinates [t,x] of [0,0] to [0,1]. The observer is moving in the x direction at 0.8c. In the rest frame of the rod, he will be at the origin at time 0. Since v = d/t and t = d/v, the time in the rod's rest frame that he will reach the other end of the rod will be t = d/v = 1/0.8 = 1.25. The event of him arriving at the far end of the rod in the rod's rest frame is [1.25,1]. Transforming this event into the observer's rest frame we get [0.75,0]. Now he can calculate how long the rod is using d = vt = (0.8)(0.75) = 0.6. This is the same answer we would get using the reciprocal of the Lorentz factor, 1/γ = √(1v^{2}) = √(10.8^{2}) = √(10.64) = √(0.36) = 0.6.
Isn't that fun? 



#25
Nov912, 10:54 AM

P: 848

I guess I did a pretty poor job of laying out the sketch without a useful explanation. You and the others have done such a good job of explaining the contraction with the Lorentz transformations, I thought I might be able to compliment that with the spacetime sketch. It's an old habit started in my first graduate course in special relativity when our prof required us to do so many of the old paradox problems for homework. He required us to first do it with just the math and then repeat it using spacetime diagrams. I got in the habit of thinking of the universe as 4dimensional, although my prof never used the term, "block universe" and never discussed the reality of the 4dimensional objects.
Does the little tutorial on 4dimensional space and spacetime diagrams in this thread help any? http://www.physicsforums.com/showthr...54#post4139454 If not I can add in some basic analytic geometry on the front end. 



#26
Nov912, 01:01 PM

P: 848

Post Script. I probably should just give it up as a fruitless effort. I had forgotten the lengthy question and answer sessions it took to get this concept across to some of my students.




#27
Nov912, 06:18 PM

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I think there's just too much information in that diagram. (I didn't try to determine if that information is correct or not). I would prefer one with just the necessary information drawn in. Something like this:
If you disregard my poor drawing skills (the slope of the x' axis isn't exactly right), the only problem with this diagram is that a person who doesn't understand that the scales on the axes are fixed by the invariant hyperbolas might interpret it as saying that ##L_0=x'_Cx'_A>x_Bx_A=L##. I would recommend section 1.7 in Schutz to anyone who's struggling with that. In particular fig. 1.11 at the bottom of page 17. To see that the inequality actually goes the other way, what you have to do is to draw a curve of the form ##t^2+x^2=\text{constant}## with the constant chosen so that the curve goes through C. The point where this curve intersects the x axis will have coordinates (0,L_{0}). This point will be to the left of B on the x axis. So ##L_0<L##. 



#28
Nov912, 08:52 PM

P: 848

Here's the picture with the hyperbolic calibration curves. Of course these are the timelike curves. I could generate the spacelike curves as well but just happened to have these handy, so am specifically illustrating the time dilation rather than the length contraction (a little off point, but I think you get the point about the need for hyperbolic calibration curves).




#29
Nov912, 09:22 PM

P: 848

Here are the calibration curves for timelike and spacelike. Red and blue inertial frames are included. I still think it's better to avoid the calibration curves by just showing two observers moving in opposite directions. Both time dilation and length contraction are directly viewed. I insisted my students understand these diagramsit just took a little more time for some of them.




#30
Nov912, 10:04 PM

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P: 4,858

Are you perhaps confusing the Enzyte Theory of Relativity with the Einstein Theory of Relativity?
 Sorry mentors, tried and tried to resist, couldn't any longer (stop it). Do what you may. 



#31
Nov1012, 03:12 AM

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#32
Nov1012, 08:23 AM

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I made a mistake in my comments about my spacetime diagram. I will have more to say about that when I've figured out how to draw the diagram in a math program.
Edit: I'm trying to relearn how to do this in Mathematica. That will definitely take a while, and I will have to do something else in about ten minutes. So I will have to return to this later. I have posted a reply to ghwellsjr below, but the diagram and my new comments about it will have to wait. 



#33
Nov1012, 08:58 AM

P: 848

This sketch describes the situation originally set up in the post. Here we can consider red to be the rest system, S, and blue to be the S' system moving with respect to S at relativistic speed. There is no need for calibration curves for this kind of symmetric diagram (Loedel spacetime diagram). It is clear that red will "see" a contracted rod, length L, as compared to the length, L', as "seen" in the blue frame S'.




#34
Nov1012, 09:35 AM

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I would say that the subject of this thread is length contraction, not some specific calculation. I didn't even look at the derivation in post #1. I just saw that the result was wrong, and that the notation looked weird to me. Then I posted a correct derivation, and later the spacetime diagram that I had described in words. The diagram is correct, but I made some comments about it that are incorrect. 



#35
Nov1012, 09:48 AM

P: 848

You may prefer measuring both lengths from the origin (but that's not the way it was presented in the initial post).




#36
Nov1012, 09:49 AM

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We've seen plenty of confusion here between terms proper distance, proper length, and rest length. I believe ghwellsjr is following one common usage where proper length (not distance, which applies between a pair of events along a specified simultaneity line) of an effectively rigid body is the same as rest length = length in a frame in which it is at rest. In such a frame, there is no requirement to measure length of a rigid object between simultaneous events, because nothing changes or moves in this frame. You can note the coordinate position of one end of the rigid body at time t1, and the position of the other end at time t2, and subtract, and still get proper length = rest length, because neither end is moving in this frame. With this understanding, only two events are needed (that are simultaneous for the frame in which the rod is moving). This same observation is also the basis of the 'cute' derivation in the wikipedia entry attributed to Born.



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