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I Lorentz transformation in 2 dimensions

  1. Apr 3, 2017 #1
    Hi folks,

    This is the Lorentz transformation in 1D, x axis:

    2metgzd.jpg

    I want to get the second term of the time t equation, I mean vx/c2, in two dimensions, I mean for a point in the XY plane.

    I know this term arises because if we want to syncronize a point B with the origin what we do is sending a light signal from the middle point, for the frame at rest the origin is traveling to the light ray whereas point B is traveling away so the light will contact sooner the origin than the point B, that is why the clocks are not sincronized, according to the external observer.

    If you have a point on the X axis you get the term v x/c2, for a point on the Y axis obviously because of the symmetry light will get in both frames of reference at point B and at the origin O at the same time, syncronization will be the same. But what happens with a point on the XY plane?

    I made a picture:

    2s964br.jpg

    The middle point is A, the point we want to syncronize is B. For an external observer this systems travels to the right with speed v, and an observer will think light will get a lot sooner to point O at C, that is moving to the right, than with point B at D, because B is traveling away of the light ray. If I calculate these distances I have:

    OC = v t1
    BD= v t2
    AC2 = AO2+ OC2 - 2 (OC) (BD) cos (oc-bd)
    AD2= AB2+ BD2 - 2 (AB) (BD) cos (ab - bd)

    THen I use simply: c= space/time, to get the time:

    t1= AC/c ( in this case c is the speed of light and AC is a distance from A to C)

    t2= AD/c

    And when I calculate:

    t2 - t1 using this expressions I don't get the result I see in the books. For the the distances have square roots so I cannot simplify these expressions.



    According to wikipedia these are the Lorentz transformations:

    3005xjs.jpg

    What I am doing wrong?

    Thanks!
     
  2. jcsd
  3. Apr 3, 2017 #2

    Ibix

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    What you have right at the beginning is the full 4d Lorentz transform for the case where the relative velocity between frames is in the x direction. It is valid for all x, y, z and t.

    The equation you reproduced at the bottom is the same thing for a general velocity, in the direction given by the unit vector ##\vec n##. If you let that vector be parallel to the x axis (i.e. ##\vec r=(x,y,z)^T##, ##\vec {r}'=(x',y',z')^T##, ##\vec n=(1,0,0)^T##) you will recover the first form.

    I think what you are doing wrong is not realising that what you actually wanted is what you started with. Unless I misunderstand what you're trying to do.
     
  4. Apr 3, 2017 #3
    An image will help I think:
    View attachment 118897


    Imagine that at t=t'=0 the origin of two coordinate systems F and F' are at the same point x=x'=0.
    Points x, y and xy are points moving at the same speed as F'.

    1.- IF we want to syncronize the clock at the origin of F' with the clock at point x' that is moving with F', as seen from F' there is no mystery at all since the light to syncronize them is emitted from the middle point of that reference system, so the light gets to the origin and to point x' at the same time.

    But as seen from F, the origin of F' is traveling towards the light and point x' is traveling away from the light, so light will get to the origin of F' sooner than to point x'. To calculate this term, we simply need the distance traveled as seen from F:

    distance from middle point to point x = L/2 + vt2

    so:

    t2= (L/2 + vt2 )/ c

    distance from middle point to origin F' = L/2 - vt1


    t1 =(L/2 + vt1)/c

    From here, using simply c=space/time we have:

    t2-t1= (L v γ2 )/c2 , as seen from F, if you use time dilation and space contraction you get the lorentz term.

    If you want to syncronize a clock on the y axis, I called it point y', you send a light signal from the middle point, obviously in this case because of the symmetry as seen from F the light rays travel the same distances to both the origin F' and the point y', no matter how fast F' is moving, so they are syncronized equally in both reference systems.

    But, what happens if the point is neither on the x axis, nor on the y axis?

    We have the point xy', and I tried to calculate the distance light travels from the middle point to the origin of F' and to point XY' but I get a time syncronization quite weird. Shouldn't I get the right result?

    DO you understand my point now?
     
  5. Apr 3, 2017 #4

    Ibix

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    So, to synchronise two clocks you are adding a source at the midpoint of the two. It emits a pulse, and you are saying that the time that the clocks receive the pulse is what the clocks will call zero. The frame where the clocks are at rest is F', and where they are moving is F. Right?

    The square root terms are the same when I try it, so drop out when I take the difference between the reception events and I get ##\Delta t=\gamma^2vX/c^2##. If you don't get that, post your working (LaTeX will help a lot in laying it out - see the link below the reply box for a guide if you don't know it) and we'll see where you've gone wrong.
     
  6. Apr 3, 2017 #5
    Regarding the first part yes, that's it.
    But it is F' who is moving to the right at speed v.
    I add another picture to make that clear:

    2zptp36.jpg

    I will put the calculation I did again, with all the steps.

    Again we are talking now about this image, point B is not on the X axis, is not on the Y axis, it is at a random point of the XY plane:

    2s964br-jpg.118496.png

    OC = v t1
    BD= v t2
    AC2 = AO2+ OC2 - 2 (OC) (BD) cos (oc-bd)
    AD2= AB2+ BD2 - 2 (AB) (BD) cos (ab - bd)
    Now c= space/time, so time = space / c :
    (If we call L=OB, the distances AO and AB are L/2)
    t2 = (1/c) * √(L/2)2 + (vt2)2 -2 (L/2) (vt2) cos ( AB - BD)

    t1 = (1/c) * √(L/2)2 + (vt1)2 -2 (L/2) (vt1) cos ( AO - OC)

    And now I have to rest t2 - t1:
    t2 - t1 =[ (1/c) * √(L/2)2 + (vt2)2 -2 (L/2) (vt2) cos ( AB - BD) ] - [ (1/c) * √(L/2)2 + (vt1)2 -2 (L/2) (vt1) cos ( AO - OC) ]

    And this is where I cannot see how to remove the square roots.
     
  7. Apr 4, 2017 #6

    Ibix

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    Yes, but you've got some clocks at rest in F' which are what you are trying to sync. If they're at rest in F then the problem is trivial.
    ##AD^2=AB^2+BD^2-2(AB)(BD)\cos (ABD)##
    c is not space divided by time. It is the distance travelled by light in some time interval divided by that time interval. Why not say ##AD=ct_2## and ##AC=ct_1##, which is what you mean?
    As written, this is wrong. You've written ##t_2=(1/c)\sqrt {L^2/4}+(vt_2)^2-2 (L/2)(vt_2)\cos (ABD)##, which is wrong. You mean ##t_2=(1/c)\sqrt {L^2/4+(vt_2)^2-2 (L/2)(vt_2)\cos (ABD)}## (Edit: or##t_2=(1/c)(L^2/4+(vt_2)^2-2 (L/2)(vt_2)\cos (ABD))^{1/2}## if you're on a tiny phone screen). That's why I strongly suggest you learn and use LaTeX. It's not difficult - quote my post to see how I did it.

    Anyway. This is just a quadratic expression for ##t_2##. Substitute the appropriate value for the cosine and solve it. Likewise the equivalent expression for ##t_1##, and then take the difference. Simply taking the differences of the expressions you have doesn't help you because you've got a t on both sides and you don't know what the cosine terms are.
     
    Last edited: Apr 4, 2017
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