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Can anyone explain this?

  1. Sep 24, 2005 #1
    Is it generally accepted that the primes associated with the Lorentz Transformations signify something different to the primes associated with the relativistic effect equations?

    Imagine you have Inez in an "inertial frame" with a ruler of length x -when measured in a "stationary frame". She points the ruler at Stan, who is considered to be stationary, so that is it "longitudinal". Stan wants to work out how long the ruler is according to Inez, in terms of his frame of reference. Stan considers two "events" Inez' end of the ruler (0,t) and the end she's pointing at him (x,t), therefore:

    Δx'= x' - 0 = (x-vt)/sqrt(1-v^2/c^2) - (0-vt)/sqrt(1-v^2/c^2)


    Δx' = x' = x/sqrt(1-v^2/c^2)

    This is upside down ... why?

    Note that when you do the same thing with the temporal Lorentz equations, you don't run into the same problem.

    Imagine that Stan now considers two other events, the first tick of Inez' clock (x,0) and the second tick of the same clock (x,t), so:

    Δt'= t' - 0 = (t-vx/c^2)/sqrt(1-v^2/c^2) - (0-vx/c^2)/sqrt(1-v^2/c^2)


    Δt'= t' = t/sqrt(1-v^2/c^2)

    This is the correct equation for "time dilation". What's going on?

    PS - my suspicion is that our concept of time dilation is screwy, or rather that time dilation happens on part of the stationary observer in relation to the "inertial" observer and that the primes are not used consistently, but I am willing to listen and learn if anyone has a better explanation.
    Last edited: Sep 24, 2005
  2. jcsd
  3. Sep 24, 2005 #2


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    The length of an object is the difference in its position coordinate when both ends of the object have the same time coordinate. This is where the method you propose above fails, you get burned by the relativity of simultaneity. Whe Inez whats to compute the length in the I frame, she insists that both ends of the ruler have the same time coordinate in the I frame. You have specified two events which have the same time in the S frame and different coordinates in the I frame. Thus you have not computed the length of the object in the I frame.

    To work out Lorentz contraction properly, it would help to draw a space-time diagram. Unfortunately, I can't do that easily here, so I will make do with a verbal description and encourage you to draw your own diagram.

    The key is that to solve the problem, you have to transform lines, not points. A line is just an infinite set of points. To avoid confusion, it helps to parameterize the lines in terms of an affine parameter lambda1 (for line1), and lambda2 (for line2).

    i.e you write

    ts = lambda1, xs = v*lambda1

    and transform this into the I frame, and then you wrte

    ts = lambda2, xs = v*lambda2+L

    and transform that into the I frame.
  4. Sep 24, 2005 #3
    Thanks pervect,

    however, why does what I have done work for time and not for space? (or vice versa as the case more likely is) and what about the primes?
  5. Sep 24, 2005 #4


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    Whose coordinate system are you using in this sentence? If it's Stan's, shouldn't the x-coordinates be different since Inez is moving? And why is the t-coordinate primed but the x-coordinate unprimed? Usually primed vs. unprimed is used to indicate two separate coordinate systems...
  6. Sep 24, 2005 #5
    Thanks JesseM,

    Thought I'd fixed that before I posted the message. Have done so now.

    Note that Stan initially doesn't care about simultaneity in the first instance (ie whether the two ends of Inez' ruler share the same instant of time) nor whether Inez thinks her clock stands still or moves with her (it's actually fixed to Stan's frame because to make my point I need to eliminate the x component in the Lorentz Transformation, so Δx = 0). In the first instance he cares about lengths only in the second instance he cares about duration only. He just wants to see if the Lorentz Transformations can be used to give him the equations for spatial contraction and time dilation.

    [Please also note that here and in the first question it is Stan who is doing the pondering, not Inez.]

    The point is that Inez' ruler contracts in relation to Stan's frame of reference and the equations show us that something odd happens - or something funny happens with the primes - and the same manipulation of the spatial Lorentz Transforms to get the spatial contraction equation applied to the temporal Lorentz Transform results not in "time dilation" but what could be called "temporal contraction".

    However it was the primes that I primarily asked about and it is in them that I remain interested:

    What is happening with the primes? if they are consistent can someone show how?
  7. Sep 24, 2005 #6


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    There are two coordiante systems - I called them S and I for clarity. You can call them primed and unprimed if you like, but it's less clear than calling them S and I in my opinion. There is no significance to the names used, you could call the coordiante system "red" and "blue" if you wanted to (as well as primed and unprimed, or my choice of S and I) - as long as you were consistent about the usage.

    If you work out the math for time dilation in detail, and draw the diagrams, you'll hopefully see why you happened to get the right answer the first time around using your shortcut technique. Basically you picked the right synchronization convention.
  8. Sep 24, 2005 #7


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    But then he's not really figuring out how long the ruler looks in Inez's reference frame. The standard Lorentz contraction formula is based on the assumption that each frame measures length by looking at the distance between the two ends of the object at a single moment in time, in that frame. This is obvious intuitively--if the back end of a car is right next to me at 5 PM, and the front end is 60 miles away from me at 6 PM, I wouldn't conclude the car is 60 miles long, would I?
    If the clock is fixed in Stan's frame, then it won't be ticking at the correct rate in Inez's frame--she will measure the clock to be ticking slow. So it doesn't really make sense to call it "her clock" in this case. The time dilation formula assumes you're talking about clocks moving along with each observer, and that therefore are measured to tick at the correct rate by that observer.
    The Lorentz transformation can be used to give the equations for spatial contraction and time dilation, if you have the standard understanding of what those equations are supposed to apply to (see my comments above for why your assumptions don't seem to match this standard understanding). For example, Stan can observe that Inez's clock reads 0 at [tex]x=0,\,t=0[/tex] in his frame, and that her clock reads [tex](1/\sqrt{1 - v^2/c^2})(t_0 - (v^2 t_0 ) / c^2 ) = t_0 \sqrt{1 - v^2/c^2}[/tex] at [tex]x=vt_0 , \, t=t_0[/tex] in his frame. You can see that this is just what the time dilation formula would tell you, and the reading on Inez's clock at [tex]t_0[/tex] can be found by plugging into the Lorentz transformation equation [tex]t' = (1/\sqrt{1 - v^2/c^2})(t - vx/c^2)[/tex]..

    As for length contraction, you have to pick two events involving the front and back of Inez's ruler that are not simultaneous in Stan's frame, but which are simultaneous in Inez's frame. For example, say that the back of her ruler passed by [tex]x=0[/tex] at time [tex]t=0[/tex] in Stan's frame, while the front of her ruler passed by [tex]x=(1/\sqrt{1 - v^2/c^2})L[/tex] at time [tex]t=(1/\sqrt{1 - v^2/c^2})vL/c^2[/tex] in his frame. If you plug these two sets of coordinates into the Lorentz transformation equation [tex]t' = (1/\sqrt{1 - v^2/c^2})(t - vx/c^2)[/tex], you can see that both these events happened at time [tex]t'=0[/tex] in Inez's frame. And if you plug them into the Lorentz transformation equation [tex]x' = (1/\sqrt{1 - v^2/c^2})(x - vt)[/tex] you can see that the first event happened at position [tex]x'=0[/tex] in Inez's frame, while the second happened at position [tex]x' = (1/(1 -v^2/c^2 ))(L - v^2 L/c^2) = L[/tex] in her frame--since these events are simultaneous, this means the ruler has length L in her frame. Meanwhile, in Stan's frame, the front of the ruler will have travelled vt in time t, so if it was at position [tex]x=(1/\sqrt{1 - v^2/c^2})L[/tex] at time [tex]t=(1/\sqrt{1 - v^2/c^2})vL/c^2[/tex], then at earlier time [tex]t=0[/tex] it must have been at position [tex]x=(1/\sqrt{1 - v^2/c^2})L - v*(1/\sqrt{1 - v^2/c^2})vL/c^2 = (1/\sqrt{1 - v^2/c^2})(L - v^2 L /c^2) = \sqrt{1 - v^2/c^2}L[/tex]. At this same time, the ruler's back was at position [tex]x=0[/tex] in his frame, so the ruler must have length [tex]\sqrt{1 - v^2/c^2}L[/tex] in his frame, just as is predicted by the Lorentz contraction formula.
    Last edited: Sep 24, 2005
  9. Sep 25, 2005 #8
    Thanks, JesseM and pervect

    JesseM's answer confirms what I thought, there's an inconsistency.

    A standard unit of time in Stan's frame, lets say the "time" for the second hand to make one full revolution of his watch, is longer than in Inez' frame, so the second hand of Inez' watch - on her wrist, which travels with her - makes a full revolution plus a bit. This surely means that Inez' seconds are shorter than Stan's, and that means Inez' seconds are contracted rather than dilated.

    Fiddle around with it as much as you like, call the phenomenon "time dilation" but when it comes down to it, time is contracted for Inez (relative to Stan) just as space is contracted for Inez (relative to Stan).

    This seems to imply that the prime for the "time dilation" equation is in the wrong spot. Note that this doesn't really matter because we can - as JesseM ably demonstrated - work it all out with the Lorentz Transformations, which are internally consistent as far as prime placement goes.

    I agree with pervect, the use of primes makes things a little unclear and no matter what you call your frames. it is consistency that matters. Therefore, let me suggest this pair of equations:

    [tex]t_(_a_c_c_o_r_d_i_n_g_t_o_I_n_e_z_i_n_t_e_r_m_s_o_f_S_t_a_n_s_u_n_i_t_s_)=t_(_a_c_c_o_r_d_i_n_g_t_o_I_n_e_z_i_n_t_e_r_m_s_o_f_I_n_e_z_u_n_i_t_s_).\sqrt{1 - v^2/c^2}[/tex]


    [tex]x_(_a_c_c_o_r_d_i_n_g_t_o_I_n_e_z_i_n_t_e_r_m_s_o_f_S_t_a_n_s_u_n_i_t_s_)=x_(_a_c_c_o_r_d_i_n_g_t_o_I_n_e_z_i_n_t_e_r_m_s_o_f_I_n_e_z_u_n_i_t_s_).\sqrt{1 - v^2/c^2}[/tex]

    Now since:
    1. a standard ruler in Inez' frame, according to Inez, would be indistinguishable from a standard ruler in Stan's frame according to Stan, and
    2. a standard unit of time in Inez's frame, according to Inez, would be indistinguishable from a standard unit of time in Stan's frame according to Stan;
    then we could argue that it is the "according to Inez in terms of Inez' units" components which should be unprimed. However, as JesseM showed, we are capable of flipping the primes over in our minds, so it's probably no problem to prime the "according to Inez in terms of Inez' units" components - as long as we maintain what pervect called for, ie consistency.
  10. Sep 25, 2005 #9

    Doc Al

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    I'm struggling to understand the issue you are raising. In general, [itex]\Delta x[/itex], [itex]\Delta x'[/itex], [itex]\Delta t[/itex], [itex]\Delta t'[/itex] as used in the Lorentz transformations do not represent measurements of the length of objects or the time as recorded by a single clock. As JesseM and pervect explain, they represent measurements of space and time coordinates for events; these measurements generally involve multiple clocks and multiple observers. Only under special conditions do they reduce to the equations for length contraction (the measurement of the length of an object implies that the measurements were done simultaneously) and time dilation (the measurement of the time elapsed on a single moving clock).
  11. Sep 25, 2005 #10


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    Nope, each one sees the other person's clocks run slow in their own frame. This is not a contradiction, it can be understood in terms of the fact that each defines simultaneity differently. Check out the illustrated example I provided in this thread, showing two rulers sliding alongside each other with clocks placed along them at regular intervals--you can see in this example how in each ruler's frame, it's the other ruler that's shrunk and whose clocks are ticking slow, yet there is no inconsistency because both frames always agree on what two clocks will read at the moment they pass each other.
  12. Sep 25, 2005 #11
    Simply put - he used the Lorentz transformation correctly in the second case because time dilation refers to comparing events which occur at the same place ([itex]\Delta[/itex]x' = 0) in the primed frame where the clock (the time interval in which the events are occuring) is at.

    pervect - You're too cool for school dude. Its sooo nice to be at a place where there are so many knowledgeable and inteligent people such as Reilly, robphy, Tom, yourself etc.

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