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Lorentz Transformation

  1. Nov 14, 2005 #1
    I'm reading Einstein's "Relativity, the Special and General Theory" And it just talked about the Lorentz transformations, but I don't quite understand what it's used for. I understand how the time dialation and length contraction equations are derived from it but I don't understand how to use the Transformation itself. Could anybody enlighten me?
     
  2. jcsd
  3. Nov 14, 2005 #2

    JesseM

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    For any reference frame, you can come up with a coordinate system for that frame, which is used to label the position and time of any event in spacetime--three spatial axes labeled x,y,z and a time axis labeled t. An object which is at rest in that frame will have constant x,y,z coordinates as time passes, which means the origin of the spatial axes is at rest in that frame. This means that the spatial origins of different frames' coordinate systems will be moving relative to one another. Let's say we have two observers, Alice and Bob, moving apart at velocity v. Call Alice's coordinates x,y,z,t and call Bob's coordinates x',y',z',t'. If you arrange the spatial axes so that they're all parallel--Alice's x-axis is parallel to Bob's x'-axis, and so on--and you also arrange them so that Alice sees the origin of Bob's coordinate system moving along her x axis at velocity v, with the two origins coinciding at t=t'=0, then the Lorentz transform would look like:

    [tex]x' = \gamma (x - vt)[/tex]
    [tex]y' = y[/tex]
    [tex]z' = z[/tex]
    [tex]t' = \gamma (t - vx/c^2)[/tex]
    with [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex]

    If you know the coordinates of a particular event in Alice's x,y,z,t coordinate system, you can plug them into this to figure out what the coordinates of the same event will be in Bob's x',y',z',t' system. And if you know an event's coordinates in Bob's system and want to know what they are in Alice's system, you use:

    [tex]x = \gamma (x' + vt')[/tex]
    [tex]y = y'[/tex]
    [tex]z = z'[/tex]
    [tex]t = \gamma (t' + vx'/c^2)[/tex]
    with [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex]

    Deriving time dilation from the Lorentz transform is pretty simple--just pick two events in Alice's system separated by a time interval of t, then figure out what the time interval between the two event's would be in Bob's system. Lorentz contraction is a little more subtle, because the different coordinate systems disagree about simultaneity (two events that happen at the same time-coordinate in one happen at different time-coordinates in the other), and "length" is about looking at the distance between the front and back of an object at a single moment. So you could do it by picking a ruler at rest in Alice's system and of length L, then picking two events at the front and back end of the ruler which happen at different times in Alice's system but the same time in Bob's system, then seeing what the distance between these two events is in Bob's system.
     
    Last edited: Nov 14, 2005
  4. Nov 14, 2005 #3

    jtbell

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    [added] Bah, see what I got for going off to fix a pot of coffee in the middle of writing this... JesseM got in ahead of me! :-)

    In relativity, the basic "thing" that we work with is the "event" which you can think of a point in spacetime, or a very short occurrence like a light bulb going on or off, or a clock located at a specific point registering a specific time. In a particular reference frame an event has a set of coordinates x, y, z, t. If we're talking about two different inertial reference frames S and S', then the same event has two different sets of coordinates x, y, z, t and x', y', z', t' in the two reference frames. The Lorentz transformation lets us calculate one set of coordinates if we know the other set and the relative velocity of the two frames.

    For example, suppose we have a galactic-class pole vaulter who can run at a speed of 0.8c, that is, 0.24 meters per nanosecond (the speed of light being 0.3 m/ns). She is carrying a pole that is 5.00 m in length when measured at rest, and is running in the +x direction with the pole parallel to the axis. In her reference frame (call it frame S'), the pole is stationary, and its two ends are located at [itex]x^\prime_{rear} = 0[/itex] and [itex]x^\prime_{front} = 5.00[/itex]. She sets her watch so that t' = 0 when the rear end of the pole passes a marker at x = 0 on the ground.

    An observer on the ground (call it frame S) sets his watch so that it reads t = 0 when the rear end of the pole passes the x = 0 marker. We have x = x' = 0 when t = t' = 0, which is the required setup for the standard version of the Lorentz transformation.

    Now, suppose that each end of the pole has a light bulb. The runner arranges for them to flash simultaneously at t' = 0 in her reference frame. Therefore in her frame, the rear bulb flashes at [itex]x^\prime_{rear} = 0[/itex] and [itex]t^\prime_{rear} = 0[/itex], and the front bulb flashes at [itex]x^\prime_{front} = 5.00[/itex] and [itex]t^\prime_{front} = 0[/itex].

    The Lorentz transformation now tells us when and where the two bulbs flash in the ground observer's reference frame:

    [tex]x_{front} = \gamma (x^\prime_{front} + vt^\prime_{front})[/tex]

    [tex]t_{front} = \gamma (t^\prime_{front} + vx^\prime_{front}/c^2)[/tex]

    [tex]x_{rear} = \gamma (x^\prime_{rear} + vt^\prime_{rear})[/tex]

    [tex]t_{rear} = \gamma (t^\prime_{rear} + vx^\prime_{rear}/c^2)[/tex]

    where [itex]\gamma = 1 / \sqrt{1 - v^2 / c^2}[/itex].

    By the way, I'm using the inverse Lorentz transformation which goes from frame S' to frame S, rather than the normal Lorentz transformation which goes from frame S to frame S'. That's why I have + signs in my equations whereas you've probably seen - signs.

    I'll let you plug in the numbers and do the calculations. Remember that I'm using units of nanoseconds for time and meters for distance, so v = 0.24 m/ns and c = 0.3 m/ns. This is just to get rid of most of those pesky powers of ten that we would have to include if we used meters for distance.
     
    Last edited: Nov 14, 2005
  5. Nov 15, 2005 #4
    OK, so now I have some questions about plugging in the numbers. First of all, why can't we just assign t' to the person on the ground and t to the high-jumper? Thus wouldn't the numbers come out different? Because you would have all negative signs?

    And second question: if t and t' must always equal zero then why wouldn't the equations be:

    [tex]x^\prime=\gamma(x)[/tex]

    [tex]t^\prime=\gamma(0)[/tex]

    Because x and t would always be zero, thus the numerator for equation 4 would always equal zero. Or are x,x',t,t' not always zero?
     
    Last edited: Nov 16, 2005
  6. Nov 16, 2005 #5

    JesseM

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    You mean switch who gets the x,y,z,t coordinates and who gets the x',y',z',t', coordinates? Remember that in the Lorentz transform equations given above, the "v" refers to the velocity of the origin of the x'-y'-z' axes as seen in the x-y-z-t system. So if the high-jumper has a positive velocity along the x-axis of the ground-observer's coordinate system, then that means the ground-observer has a negative velocity along the x'-axis of the high-jumper's system. So, if you switch whose coordinate system is labeled as the primed one, then the new v that you use will be -1 times the v that you used when the high-jumper had the primed coordinate system.
    They don't always equal 0. The origins of the two coordinate systems are moving relative to each other, but there will be a particular place and time where they coincide, and both coordinate systems assign that single point in spacetime a time-coordinate of 0. Events at other positions and times in spacetime will not have a time-coordinate of 0 in both coordinate systems.
     
  7. Nov 16, 2005 #6

    jtbell

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    No, they can't always both equal zero. It's just that the usual formulas for the Lorentz transformation assume for simplicity that there is one event (point in spacetime) where [itex]x = x^\prime = 0[/itex] and [itex]t = t^\prime = 0[/itex].

    We can remove this assumption by using a more general version of the Lorentz transformation equations. Pick an event (any event!) as our "reference event" and let its coordinates be [itex]x_0[/itex] and [itex]t_0[/itex] in frame S, and [itex]x^\prime_0[/itex] and [itex]t^\prime_0[/itex] in S'. Then for any other event,

    [tex]x^\prime - x^\prime_0 = \gamma [(x - x_0) - v(t - t_0)][/tex]

    [tex]t^\prime - t^\prime_0 = \gamma [(t - t_0) - v(x - x_0)/c^2][/tex]
     
  8. Nov 16, 2005 #7

    jtbell

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    Ugh. That last bit at the end should read "if we used seconds for time". Apparently it's too late for me to go back and edit that posting.
     
  9. Nov 16, 2005 #8
  10. Nov 16, 2005 #9
    So if we were trying to derive x' from x values then it would still basically come out as x+vt rather than x-vt because the ground observer is traveling in a negative direction relative to the view of the high-jumper. OK, that makes sense.

    So in the case that we have here the numerator for the x formula will be x' and the numerator for the t formula will be 0, but that's not always the case? And if they are different then you have to use the more general formula? And thanks for all the feedback guys.
     
  11. Nov 16, 2005 #10

    JesseM

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    You're still using the same formula:

    [tex]x' = \gamma (x - vt)[/tex]
    [tex]t' = \gamma (t - vx/c^2)[/tex]

    But you can see that if you plug in x=0 and t=0, you'll get [tex]x' = \gamma (0) = 0[/tex] and [tex]t' = \gamma (0) = 0[/tex]. And if you wanted to find the coordinates in the primed system of an event that happens at t=0 and an arbitrary x-coordinate in the unprimed system, then the formulas would reduce to [tex]x' = \gamma (x)[/tex] and [tex]t' = \gamma (vx/c^2)[/tex]. I'm not sure I understand your question about the numerator (the numerator of what fraction in what formula?), but hopefully this helps.
     
    Last edited: Nov 16, 2005
  12. Nov 16, 2005 #11
    Yeah, that's what I meant, I was just making sure that was the case. But appearently I didn't state it very well. So here's a question, jtBell gave another more general equation:

    [tex]x^\prime-x\prime_0=\gamma[(x-x_0)-v(t-t_0)][/tex]

    [tex]t^\prime-t\prime_0=\gamma[(t-t_0)-v(x-x_0)/c^2][/tex]

    but [tex]x^\prime-x\prime_0[/tex] seems more like a distance rather than a certain point on a co-ordinate system, and the same thing with the second formula. So how does that work with finding a certain point where an event happened?
     
  13. Nov 16, 2005 #12

    JesseM

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    Well, the idea there is that instead of knowing that the event with coordinates x=0,t=0 has coordinates x'=0,t'=0 in the other coordinate system, instead you just know the coordinates of some other specified "reference event" in both coordinate systesm-the coordinates of the event in the unprimed system are [tex](x_0 ,t_0 )[/tex] and the coordinates in the primed system are [tex](x^\prime_0 , t^\prime_0)[/tex]. So once you are given these coordinates of the "reference event", you can plug them into those formulas to translate the coordinates of an arbitrary event (x,t) into the coordinates (x',t') of the same event in the primed system.
     
    Last edited: Nov 17, 2005
  14. Nov 17, 2005 #13

    jtbell

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    To give a specific numeric example based on the one that I gave earlier, suppose that the observer on the ground is standing at [itex]x_0 = 6[/itex] in his reference frame S, and that the pole-vaulter is holding her pole at its midpoint ([itex]x^\prime_0 = 2.5[/itex]). Recall that the pole is 5 m long and the origin of her frame S' is at the rear end of the pole. Further, suppose that when the pole-vaulter passes the observer on the ground, the pole-vaulter's watch reads [itex]t^\prime_0 = 4[/itex] and the observer's watch reads [itex]t_0 = 3[/itex]. Also the runner's speed is 0.8c so that [itex]\gamma = 1.667[/itex]. Then the Lorentz transformation from frame S to frame S' would be

    [tex]x^\prime - 2.5 = 1.667[(x - 6) - 0.24(t - 3)][/tex]

    [tex]t^\prime - 4 = 1.667[(t - 3) - 2.667(x - 6)][/tex]

    (Reminder: I'm using meters for distance and nanoseconds for time.)
     
    Last edited: Nov 17, 2005
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