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Why is length contracted, and not ELONGATED?

  1. Nov 8, 2012 #1
    Silly question, I assume, but I can't quite figure it out. I've just started reading up on Special Theory of Relativity, and only have knowledge about Lorentz Transformations.

    Anyway, supposing a rod is at rest in the frame S', which is moving with velocity v with respect to frame S. Then the length of the rod in frame S' is the true length L(0), and is given by x2' - x1', where x2' and x1' are the end points of the rod in frame S'.

    Now we have to find the length of the rod in frame S, which is given by x2 - x1, where x2 and x1 are the end points of the rod in frame S.
    x2 is related to x2' by x2 = γ(x2' - vt)
    and x1 is related to x1' by x1 = γ(x1' - vt)

    Then length in frame S = x2 - x1 = γ(x2' - x1')
    Or length L = γL(0)

    Hence length to an observer in frame S is elongated.

    May I know where I went wrong?
     
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  3. Nov 8, 2012 #2

    ghwellsjr

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    If you would have also calculated the times of the two events in frame S you would see that they are at different times meaning that the length that you are calculating for the rod is while it is moving. What you need to do is find two events that are at the same time in frame S (but different times in frame S') that correspond to the end points of the rod. Do you think you can figure that out?
     
  4. Nov 8, 2012 #3
    No sorry, I don't get it. I thought I had taken care of that part by putting t1 = t2 = t in the Lorentz transformation for x2 and x1?

    EDIT : Are you saying that the measurement of x1 and x2 is not simultaneous, while measurement of x1' and x2' is ?

    What is wrong with this method? All the texts that I have consulted find out x2' and x1' in terms of x2 and x1, and then subtract THOSE ; but to find the length of the rod in frame S, shouldn't we find out x2 - x1 ?
     
  5. Nov 8, 2012 #4

    ghwellsjr

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    You said the rod is at rest in frame S', correct? If that is true, then it doesn't matter what time coordinates you use to specify the ends of the rod in frame S'--you'll still get the same length, which is called the Proper Length, not the true length, of the rod. So now if you change the time coordinate for one end of the rod in frame S' and transform the events for both ends of the rod, you'll get any length you want but for different time coordinates for the two ends. What you need to do is pick a pair of time coordinates for the two ends of the rod in frame S' such that after you transform the events into frame S, the events are at the same time and that will give you the correct contracted length for the moving rod.

    It is probably easiest to select one end of the rod to be at the origin of frame S' when you do this.
     
  6. Nov 8, 2012 #5

    ghwellsjr

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    You're transforming from frame S' to frame S which means you have set t1' = t2' = t.
    The way you did it, x1 and x2 are not simultaneous (but they should have been), while x1' and x2' were simultaneous (but they should not have been).
    In whichever frame the rod is not moving, you can allow the events to be non-simultaneous so that the events for the frame in which the rod is moving will be simultaneous.
     
  7. Nov 8, 2012 #6
    I think I understand where I went wrong. While transforming x2' and x1' to x2 and x1, I neglected the fact that the measurements of x2 and x1 would not end up being simultaneous, while x2' and x1' would be. If I had transformed x2 and x1 to x2' and x1' instead, I would have gotten simultaneous measurements for x2 and x1, while x2' and x1' would be not simultaneous - which doesn't matter because the rod is at rest in frame S'.

    Is that correct?
     
  8. Nov 8, 2012 #7

    ghwellsjr

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    Well, technically that is correct, but that only works if you already know the contracted length of the rod in frame S. If the idea is that you're supposed to figure it out from its rest length in frame S' and its speed in frame S, then you have to transform from S' to S and not the other way around.

    Of course, since you probably already know the contracted length, you can use it to determine two events in S' that are the correct length and at different times such that when you transform them back to S, they will be simultaneous. But it seems like you ought to figure it out from S' without knowing the contracted length in S.
     
    Last edited: Nov 8, 2012
  9. Nov 8, 2012 #8
    I'm afraid I don't quite understand how I would find the contracted length (assuming I don't already know it) from the rest length and the speed in frame S? If I transform from S' to S, then I end up with length elongation and not contraction...
     
  10. Nov 8, 2012 #9

    ghwellsjr

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    Well one way is to do it by trial and error. But as I said before, you ought to set one end of the rod at the origin of both frames to minimize the computation. Then start with the length of the rod as the x' coordinate for the other event and the t' coordinate can be zero. Transform just the t' coordinate to t and see if it is zero which it obviously won't be. Then increase it and repeat. See if t gets closer to zero. If it does, then repeat until you get zero. If it doesn't, then you have to decrease t' until t comes out zero. When you finally get a value of t' that produces t=0, then you can use that value of t' to calculate x which will be the correct contracted length of the rod.

    Once you do this, you will quickly see what value of t' to use to get t=0, especially if the rest length of the rod is 1.
     
  11. Nov 8, 2012 #10
    Okay, I understood your method. But how then is the standard derivation given in most texts correct? I still don't quite understand why to find the length of the rod in frame S we do not find out x2 and x1, but instead find out x2' and x1'.

    If we consider time dilation in the same scenario for two events, then we will find out t2 and t1 which are in the frame S, and then subtract them to get the time interval for frame S. Should we not operate on coordinates of the S frame for length contraction as well?
     
  12. Nov 8, 2012 #11

    ghwellsjr

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    Let me work it out for you and see if this make sense:

    One end of the rod is at the origin of the rest frame. The other end of the rod is at x' so the rest length (or the Proper Length) of the rod is x'.

    Assuming that we are using units such the c=1, the time coordinate in the moving frame is t = γ(t'-vx'). We want to make this evaluate to zero. Obviously, we want to set t'=vx'.

    Now we calculate the x coordinate using x = γ(x'-vt') where we substitute vx' for t' and we get x = γ(x'-vvx'). We do a little manipulation to get x = γx'(1-v2). We remember that γ = 1/√(1-v2) and so γ2 = 1/(1-v2) or (1-v2) = 1/γ2. Substituting this into our previous equation we get x = γx'/γ2 = x'/γ, showing that the moving rod is contracted by the reciprocal of the Lorentz factor.
     
  13. Nov 8, 2012 #12
    Yeah I understood your method earlier itself. What I'm having trouble understanding is the proof/derivation given in most texts (which is the same as this : http://en.wikipedia.org/wiki/Length_contraction#Derivation )

    What I don't get is why we are transforming x1 and x2 to x1' and x2'. Did my time dilation example/question make sense? In that to find the interval we find out the difference between t2 and t1 in the SAME frame as that of which we need the interval ; why shouldn't we find out the difference between x2 and x1 in the SAME frame as that of which we need the length?
     
  14. Nov 8, 2012 #13

    ghwellsjr

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    I don't know what the standard derivation in most texts looks like. Can you link to an online reference?

    We start with the length of the rod in its rest frame S'. In the previous post, I put one end of the rod at the origin instead of calling it x1. I put the other end at x' instead of calling it x2'. This is just to reduce the computation.
    We want to make sure that t2 is equal to t1 so they subtract to zero. We make this happen by selecting values of t1' and t2' such that when we transform the two events from frame S' to frame S, they come out equal. The time dilation will calculate automatically if we use the Lorentz transformation correctly.

    As you can see in my previous post, the computation is complicated enough when I set one end of the rod at the origin. Trying to do this more generally with the two ends of the rod located anywhere will be even more complex and I leave that exercise to you. If you understand what I did in my previous post, you should be able to extrapolate that to the more general case.
     
  15. Nov 8, 2012 #14

    ghwellsjr

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    This is the first time I have ever looked at that derivation and it is quite interesting. It doesn't really transform x1 and x2 to x1' and x2', it merely is manipulating the equations to derive the end result. It might have been a little more helpful if they had included a few more intermediate steps but if you mentally do the substitutions, you can see how the math works out.
    I like my derivation (which isn't original to me) better than the wikipedia one because it addresses the issue that in the rod's rest frame, we are free to select any time coordinates for the end point events of the rod which allows us to pick some that result in the event in the moving frame being at the same time. So in my derivation, we are forcing the time difference between the two events to be zero in the same frame in which we are calculating the position difference which is the contracted length.

    Again, I want to make it clear that the wikipedia derivation is not actually transforming from one frame to another, it is mathematically manipulating the equations in a clever (and not so obvious) way to produce the desired result.
     
  16. Nov 8, 2012 #15
    To be honest, I like your way more too. It's much more logical.

    The wikipaedia one (and the one I've found in Kleppner, Berkeley and a few national texts) is, like you said, mathematical manipulation which doesn't make sense to me in a logical manner. Still, if you can manage to extract a logical reason from that method, please do so, because I will have to be writing this method in my examinations.

    However, how can you say that they are not transforming from one frame to another?
     
  17. Nov 8, 2012 #16
    May I suggest the following analysis from my web site:

    http://www.relativitysimulation.com/Documents/DeterminingTheLengthOfMovingObject.htm
     
  18. Nov 8, 2012 #17

    Fredrik

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    The fact that you're denoting the S' coordinate of the right endpoint of the rod by x2' suggests that you didn't realize that you need to consider three events, not two.

    I will call the end of the rod that has the smaller position coordinate in both coordinate systems the left end of the rod. Let A be any event on the world line of the left end of the rod. Let B be the event on the world line of the right end of the rod that's simultaneous in S with A. Let C be the event on the world line of the right end of the rod that's simultaneous in S' with A. (I recommend that you draw a spacetime diagram for the coordinate system S. Draw the world lines of both ends of the rod, and a simultaneity line of S' through A).

    Let L0 be the rest length of the rod. Since the rod is at rest in S', L0 is equal to the coordinate length in S'. So L0=x'C-x'A. We are looking for the relationship between L0 and the coordinate length of the rod in S, which will be denoted by L. By definition of L, we have L=xB-xA.

    For any event E, I will denote the coordinates of E in the coordinate systems S and S' by
    $$S(E)=\begin{pmatrix}t_E\\ x_E\end{pmatrix}$$ and $$S'(E)=\begin{pmatrix}t'_E\\ x'_E\end{pmatrix}$$ respectively. When I don't feel like LaTeXing matrices, I'll write S(E)=(tE,xE) and S'(E)=(t'E,x'E) instead. I will use units such that c=1. In these units, the relationship between S'(E) and S(E) is given by
    $$\begin{pmatrix}t'_E\\ x'_E\end{pmatrix}=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t_E\\ x_E\end{pmatrix}.$$ This is just the Lorentz transformation in matrix form.

    Now suppose that S was chosen so that the world line of the left end of the rod goes through the event at the origin. Then we can choose A to be that event. Then we have S(A)=(0,0), S'(A)=(0,0), S(B)=(0,L) and S'(C)=(0,L0). The coordinates in S of points on the world line of the right end of the rod are given by the map
    $$t\mapsto\begin{pmatrix}t\\ vt+x_B\end{pmatrix}.$$
    If we denote that map by R, we have
    $$S(C)=R(t_C)=\begin{pmatrix}t_C\\ vt_C+x_B\end{pmatrix}=\begin{pmatrix}t_C\\ vt_C+L\end{pmatrix}.$$ So ##x_C=vt_C+L##. We use this result in the last step of the following calculation.
    $$\begin{pmatrix}0\\ L_0\end{pmatrix}=\begin{pmatrix}t'_C\\ x'_C\end{pmatrix} =\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}\begin{pmatrix}t_C\\ x_C\end{pmatrix} =\gamma\begin{pmatrix}t_C -vx_C\\ -vt_C+x_C\end{pmatrix} =\gamma\begin{pmatrix}t_C -vx_C\\ L\end{pmatrix}.$$
    So ##L_0=\gamma L##, or equivalently, $$L=\frac{L_0}{\gamma}.$$
     
  19. Nov 8, 2012 #18
    The easiest way to do this is to write:

    x2' = γ (x2 -v t2)

    x1' = γ (x1 -v t1)

    so

    x2' - x1' = γ (x2 - x1) - γv (t2 - t1)

    But when reckoning the length of the moving rod from the S frame, you need to measure the locations of both ends of the rod at the same time, so

    t2 = t1,

    and hence, x2 - x1 = (x2' -x1') / γ

    When reckoned from the S frame, the length of the rod is shorter than in its own rest frame.
     
  20. Nov 8, 2012 #19

    ghwellsjr

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    If you just write out the equations for the Lorentz transformation process, are you transforming from one frame to another? Or are you merely establishing a mathematical relationship? In any case, let me restate the derivation from wikipedia with some extra steps and using gamma:

    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(x1-vt1) and x2' = γ(x2-vt2)

    Now we want t1 to be equal to t2 so we replace both of them with simply t.

    x1' = γ(x1-vt) and x2' = γ(x2-vt)

    Now we note that since L(0) is equal to x2' - x1' we can write:

    L(0) = x2' - x1' = γ(x2 -vt) - γ(x1-vt)
    L(0) = γx2 - γvt - γx1 + γvt
    L(0) = γx2 - γx1 + γvt - γvt
    L(0) = γx2 - γx1
    L(0) = γ(x2-x1)

    But since we also know that L (the length in frame S) is equal to x2-x1 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.
     
  21. Nov 8, 2012 #20
    I also learned to obtain it directly from the Lorentz transformations; maybe because I was taught that way, it looks simpler to me than your method. :tongue2:
     
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