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Question about length contraction (geometric way)

  1. Sep 8, 2010 #1
    I'm doing some independent study with a professor at my university. He has taken one GR class and he said that it was not a good experience. So we are in a sense learning together as we pursue this.

    We are using the book "A First Course in General Relativity" by Schultz. We are doing problem 17 in chapter 1. (It is the classic "pole in the barn" problem.)

    The pole is 20 m and is moving at .8c. What is the length it has for a stationary person..? (I left out the details of the "setting" but this the gist of it.)

    Now he and I both know how to do it using the standard old formula: L=L(o) sqrt(1-v^2/c^2)

    But we were trying to get to the same answer using geometrical means. Even in the book Schultz tries to emphasize getting in this "geometrical way of thinking" but when he got to the length contraction part, he used the formula above instead.

    I've attached an image that I made in Photoshop to demonstrate how we did it the geometric way. The tan(phi)= v came from the book. Also of course c=1 which would reduce the above formula to: L=L(o) sqrt(1-v^2)

    But the problem we run into is that when we use the original equation we get 12. But when we do it the geometrical way we get 12.5. Now as far as "theoreticals" go, doing it two different way should give you almost the exact number. (Not off by .5)

    I was wondering if any of you can point out what we are not taking into account or doing something fundamentally wrong.

    Please, if you need more information just ask.

    Thank you.

    Attached Files:

  2. jcsd
  3. Sep 8, 2010 #2


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    I made a space-time diagram for the barn-pole scenario. I've attached the diagram seen from the pole frame and the barn frame. I think I can read off the lengths of the barn in the pole frame and the length of the pole in the barn frame. The relative velocity is 0.563.
    This gives gamma = 0.83 and the ratios of rest length to boosted length of 0.81( pole) and 0.83 (barn). Reading values off a small diagram is tricky but it's near enough.

    Attached Files:

    Last edited: Sep 8, 2010
  4. Sep 8, 2010 #3


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    This formula assumes that L0 is the rest length, whereas in your diagram you seem to have it backwards, the horizontal black bar appears to be moving in the x, t frame which would indicate that its length in the x, t frame is contracted from what it is in the bar's rest frame, but you label the length of the bar in the x, t frame as L0. So I think you should relabel the horizontal bar as L, then the rest length L0 would have to be along a line of simultaneity in the bar's own rest frame (since 'length' in any frame is determined by simultaneous measurements of the front and back of the object in that frame), i.e. draw an angled black bar along the x' axis, with the ends corresponding to the places that the x' axis intercepts with the axes you have labeled t' and t'L. But note that the geometric length of this angled black bar in the diagram is not exactly equal to the bar's rest length, because aside from the fact that the x' axis of the other frame is rotated relative to the x axis of the first, you also have to consider that the increments along the x' axis also get stretched by a certain factor (try drawing dashes at x'=1, x'=2, x'=3 etc. on your diagram using the Lorentz transformation to see this). To figure out what this factor is, consider what happens if you plot x'=1, t'=0 in the x,t diagram:

    x = gamma * (x' + vt') = gamma*1
    t = gamma * (t' + vx'/c^2) = gamma*v*1/c^2

    And we're using units where c=1, so the point x'=1, t'=0 would have coordinates x=gamma, t=gamma*v in the x,t diagram. So, the geometric length of the segment from the origin to this point will be given by the pythagorean theorem sqrt(gamma^2 + gamma^2*v^2) = gamma*sqrt(1 + v^2), even though in the x',t' frame the distance between this point and the origin is only 1! So, the actual length in the x',t' frame of an angled line segment that is parallel to the x' axis and has geometric length of G in your diagram would only be G/gamma*sqrt(1 + v^2).
  5. Sep 8, 2010 #4


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    Incidentally, this page has a good spacetime diagram for length contraction:


    Again note the key point that "length" in either frame is defined in terms of simultaneous measurements in that frame--AB defines the length in the frame where A and B are simultaneous, and AC defines the length in the frame where A and C are simultaneous. But keep in mind again that the geometric length of the segment AC is not the same as the distance between A and C in the frame where they're simultaneous, for that you have divide the geometric length in the diagram by gamma*sqrt(1 + v^2).
  6. Sep 8, 2010 #5
    @Mentz and Jesse

    I think I see where you guys are coming from. But the diagram that I posted may be misleading. The L(0) that touches the t' and t'L is just there to show the magnitude of L(0). Not that it shows any movement. The L' that touches x' and x'L axis is the one that from the x-t frame of reference shows movement in those axis.

    I guess what I'm misunderstanding is that wouldn't the proper length be equal to the length(L') when t=t'=0 ...? That is why the label of L(0) is placed at the bottom one. (Where the x,x' and t,t' axis meet.) Because basically they are in the same reference frame at time=0=t=t' they both would agree with the same Length.

    But later on at some time t'>0,t>0, is when they would disagree with each other. Right...? Or am I looking at it the wrong way..?

    If someone can tell me how to go about solving it geometrically (Using only magnitudes and angles like how we tried to do.), that would help also. Because that is what we are aiming to master.

    Here are some calculation/logic in detail that we pursued.

    phi = angle from the t and x axis.
    tan(phi) = v , v=velocity (ie. .8c)
    Pole's proper length = 20 m
    velocity of the pole = .8c
    c=1 which implies that at the speed of light, (inverse tangent):tan^-1(1)= 45 degrees. (When both the x' and t' axis form one line with slope 1 in the x-t frame.

    Calibrating the moving frame of reference's axis (x'-t') in the stationary frame of reference (x-t):
    tan(phi) = .8
    So phi = about 38.66 degrees. (Meaning the t' axis is 38.66 degrees from the t axis and the x' axis is 38.66 degrees from the x axis.)

    The pole:
    Now since all of the "points" along the rod are moving together, we added another set of axis that the end of the pole "point" moves in. (We can forget about all the points in between since they are all "connected" from the x'-t' "point" to the x'L-t'L "point".)
    So since the two "points are touching at x=0 and x=xL, they must also be touching later on (t>0) in the x'-t' frame of reference: From some x' to x'L.

    So when drawing a straight line from those point, we can see that they are shorter in length from the x-t POV. That is not surprising since that is what it is suppose to do. Now to find that L' we use all the data we know:
    L(0)=20 m which means we can use the magnitude of L(0) to move it to conjoin with the L' (like in the picture). Since that makes a triangle we can use phi=38.66, L(0)=20, to find L'. That just means a simple trig problem which we got 12.5 m for the L'. But using the L'=(20)sqrt(1-.8^2) we get about 12 m.

    Now I know that I explained things very carefully. Take no offense to that. I'm not trying to baby anyone. I just want to show the thought process that we took to do this problem. So that way if there is any false logic we may have took, someone can point it out. Hopefully this detailed explanation helps you guys to help me.
    Last edited: Sep 8, 2010
  7. Sep 8, 2010 #6


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    On reflection, I don't think my post helps, because it's just a Lorentz transformation, and not quite what you were asking about.
  8. Sep 8, 2010 #7
    Yes, I was wondering what that was about. :)

    But I edited my last post with more details on how we approached it.
  9. Sep 8, 2010 #8


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    You'll know when you come across the barn-pole 'paradox'.:smile:

    I've attached two pics, one shows the ends of the rods in the rod frame, the other shows the ends boosted. The two points are no longer on a horizontal line. So the two measurements are no longer simultaneous. I think your original diagram has the boosted ends at the wrong positions in the boosted frame x', t'.

    I think this is what JesseM is saying, and also the diagram he posted, where AC corresponds to the line between the boosted events on my diagram.

    Attached Files:

    Last edited: Sep 8, 2010
  10. Sep 8, 2010 #9


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    And, at the risk of becoming a bore, here is the radar measurement of the rod made by sending a beam in both directions and timing the round trip. You can see in the boosted diagram that the light travel time is shorter.

    Attached Files:

  11. Sep 8, 2010 #10
    :smile: I have already come across it in my Modern Physics last year.

    And of course you are not becoming a bore. I think I see what you are talking about. The original L(0) that is shifted up in my figure in the x'-x'L ...? If that is the case, then you should know that the horizontal L(0) that forms the triangle with L' is just there for measurement sake. That is not the boosted L(0). The boosted L(0) would be the L'...

    I hope I'm making sense. It is really hard to convey what I'm talking about over a forum.

    By the way, what program do you use to make those graphs...?
  12. Sep 8, 2010 #11


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    It looks to me as if the length you've labelled L0 on the diagram is not the rest length but the transformed length. The problem with your diagram is that L0 is not there.

    I understand what you're trying to do, but I can't see how to do it without assuming the LT to transform to new axes. I'm baffled by your diagram. Which probably means nothing because that's my default state.

    [later] I think if one accepts that the worldlines of the ends of the rods are parallel slanted lines, then using the radar measurement boosted diagram, which assumes that the speed of light is independent of the motion of the source/receiver, it may be possible to use geometry to find the light travel time as a function of the angle.

    [The program I use to draw the diagrams runs under Windows and you can get a copy by looking in my blog]

    Attached Files:

    Last edited: Sep 8, 2010
  13. Sep 8, 2010 #12
    Okay, I see what we did wrong. So basically there would be no way to get length information with just pure geometry on a piece of paper... We are forced to use the derived formulas..? (L'=L*sqrt(1-v^2) , t'=...)

    Thanks for everything. I'm probably going to post more throughout this semester while we go through this book.

    Umm... Do you have a link to your blog...? I went to your profile but couldn't find a link...


    Okay I found it. I just didn't look hard enough. I will test it out. Thanks again.
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