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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:
     
  22. Nov 8, 2012 #21

    ghwellsjr

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    Now that I have learned how the derivation works, especially with the added steps and using gamma, I agree. It turns out to be very simple.
     
  23. Nov 8, 2012 #22
    The math has been handled very well here, so I can't add anything more to that. However, dreamLord, it might be good to keep track of the physical picture implied by the Lorentz transformaions as I'm trying to illustrate with the sketch below. We have red and blue meter sticks moving in opposite directions along the X1 axis of the black rest frame. But, the larger implication of the Lorentz transforms in the context of Minkowski 4-dimensional space is that the red and blue meter sticks are 4-dimensional objects. So,we actually have here the 3-D cross-section views of the 4-D universe at one instant of time for each observer. And the reason for the perceived length contraction is immediately obvious.
    Pythorean_Lorentz_3.jpg
     
    Last edited: Nov 8, 2012
  24. Nov 8, 2012 #23

    ghwellsjr

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    It's not immediately obvious to me even after many minutes of trying to figure out what you are trying to show.

    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 x-axis 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.
     
    Last edited: Nov 9, 2012
  25. Nov 8, 2012 #24

    ghwellsjr

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    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 = √(1-v2) = √(1-0.82) = √(1-0.64) = √(0.36) = 0.6.

    Isn't that fun?
     
  26. Nov 9, 2012 #25
    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 space-time 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 space-time diagrams. I got in the habit of thinking of the universe as 4-dimensional, although my prof never used the term, "block universe" and never discussed the reality of the 4-dimensional objects.

    Does the little tutorial on 4-dimensional space and space-time diagrams in this thread help any?

    https://www.physicsforums.com/showthread.php?p=4139454#post4139454

    If not I can add in some basic analytic geometry on the front end.

    I'm trying to show two 4-dimensional meter sticks. Each one is slanted in the opposite direction relative to the black coordinate system (the vertical black line is the black X4 axis and the horizontal black line is the black X1 axis).

    Yes. I wanted to show red and blue moving in opposite directions with respect to the black rest frame.

    The two slanting blue lines represent the front and rear ends of the blue meter stick, displayed as a 4-dimensional object (X2 and X3 coordinates are supressed for ease of viewing). Likewise the two red slanted lines represent front and rear ends of the 4-dimensional red meter stick.

    If you follow the blue line of simultaneity, you will see the front and rear points on the blue meter stick that are simultaneous. And continuing along that blue line of simultaneity you will see the front and rear points of the red stick that are simultaneous in blues simultaneous space (an instantaneous 3-dimensional cross-section vew, cut across the 4-dimensional universe). So, just working within blue's inertial frame you can directly compare the lengths of the blue and red sticks as they appear in blue's 3-dimensional world. Then you can do the same thing for the red world, using the red simultaneous space.

    That is correct, they are moving along the black X1 axis in opposite directions at the same speed. That's why the 4-dimensional stick objects are slanted in opposite directions with the same slant angle relative to the black X4 axis. Note that as you move along the blue 4-dimensional stick in the blue X4 direction, you are increasing the distance along the black X1 axis. Or, another way of saying it, if you are at rest in the black rest frame, that means you are moving straight forward into time, i.e., moving along the black X4 axis at the speed of light--and as you move along black X4, the blue meter stick increases its distance away from you (black) along the black X1 axis. Also if you are red guy, moving along the red X4 at light speed, you also see the blue meter stick increasing its distance away from you along the red X1 axis.

    As mentioned above, the blue guy "sees" the red guy's stick contracted as compared to his own (just comparing lengths within the blue simultaneous space) and the red guy "sees" the blue guy's stick contracted as compared to his own (comparing lengths just within the red simultaneous space). And the reason I've used the symmetric Loedel space-time diagram is that you are able to compare the lengths directly, since the line lengths in the sketch have exactly the same actual distance calibrations (you can't compare distances directly between black and the moving frames without resorting to the hyperbolic calibration curves).

    Take another look at the diagram. If the blue and red 4-dimensional sticks were aligned parallel to the black X4 axis, that would mean that the blue and red sticks were at rest in the black rest system. For example, follow the blue stick along the 4th dimension, assuming the blue stick is parallel to the black X4 axis: In that case, as time passes (as you move along blue's stick along the direction of the X4 axis) you would see that the blue stick front and rear positions along black's X1 axis have not changed.

    Apparently that is the case. I'll have to give more thought as to how to do a better job of presenting the space-time diagrams. Any suggestions?
     
    Last edited: Nov 9, 2012
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