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Deriving the Lorentz Transformation

  1. Jul 20, 2014 #1
    Hello everyone. I am attempting to teach myself about special relativity, and have learned a derivation for the Lorentz transformation. Before I go any farther I want to clear up the parts I don't understand right now and check that I am correct about the parts I do understand. Here I will derive the Lorentz transformation and ask a couple of questions as I go. I would greatly appreciate it if someone can help increase my understanding of these concepts.

    There will be two observers, S and S' that will start out with their x axis' lined up and the clocks they are carrying synched at t=0 and t'=0. S' will begin moving with speed v away from S who will remain stationary so that there is distance vt in between S and S'. S' will be moving toward a box that is moving at the speed of light c in the same direction as S'. I have made a picture of the situation. The distance between S and the box is x and the distance between S' and the box is x'. x=ct.


    Here is my first question: From what I have learned it seems that not only does x=ct but x'=ct'. Is this only true because c is the same in every reference frame or would that also be true in Newtonian physics?

    According to S': x = x' + vt'
    According to S: x' = x - vt

    plugging in x=ct into the second equation you have:

    x' = ct-vt

    Since distance is a velocity times a time and we know the time, that means the velocity that S' sees the box moving is less than c. But we know that c is the same in every reference frame. So our equations must be off by some factor.

    Next question: From derivations I have seen on the internet, they add an error factor (lets call it 'f') into the two equations, so that they become this:

    x = (x' + vt')f
    x' = (x - vt)f

    My question is: how do we know that we can use the same error factor for both equations? Wouldn't we use f for one and lets say g for the other? How can we prove that we can use the same for both? Forgive me if this is obvious, I am fairly new to these concepts.

    Anyways we multiply the two equations together. (Is there a clear way to know whether to multiply, add or subtract two equations?) After multiplying them we get:

    [itex] xx' = (x' +vt')(x-vt)f^2 [/itex]

    [itex] xx' = (xx' - x'vt + xvt' - v^2tt')f^2 [/itex]

    we know that x=ct and x'=ct' so t=x/c and t'=x'/c. plugging in we get:

    [itex] xx' = (xx' - x'v(\frac{x}{c}) + xv(\frac{x'}{c}) - v^2(\frac{x}{c})(\frac{x'}{c}))f^2 [/itex]

    [itex] xx' = (xx' - \frac{xx'v}{c} + \frac{xx'v}{c} - \frac{v^2xx'}{c^2}) f^2 [/itex]

    [itex] xx' = (xx' - \frac{v^2xx'}{c^2})f^2 [/itex]

    [itex] 1 = (1 - \frac{v^2}{c^2})f^2 [/itex]

    [itex]f^2 = \frac{1}{1-\frac{v^2}{c^2}} [/itex]

    [itex]f = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}} [/itex]

    [itex]f = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

    which is the lorentz transformation. Any feedback would be greatly appreciated. :)

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    Last edited: Jul 20, 2014
  2. jcsd
  3. Jul 20, 2014 #2


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    It's not clear to me where you learned your Lorentz transformations from. So lets check to see if your source gave them correctly:

    Do you have

    x' = \frac{1}{\sqrt{1-\left( \frac{v}{c} \right)^2}}\left( x - v t \right) \quad t' = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2 }} \left( t - \frac{v\,x}{c^2} \right)

    I noticed you didn't discuss the transformation of t to t' at all, there is an important difference in the behavior of time between relativity and pre-relativity physics,
  4. Jul 20, 2014 #3


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    Actually, the time translations are contained in the pair of equations

    [itex]x' = f(x - vt)[/itex]
    [itex]x = f(x' + v t')[/itex]

    Plugging the first equation into the second gives:

    [itex]x = f^2(x - v t) + fvt'[/itex]
    which can be rearranged to give:

    [itex]t' = f[\dfrac{(1-f^2)}{f^2 v} x + t][/itex]

    which is the same as the usual LT transform, if you substitute [itex]\dfrac{1}{f^2} = 1-\frac{v^2}{c^2}[/itex] and [itex]1-f^2 = \dfrac{-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}[/itex]
  5. Jul 20, 2014 #4
    I like the derivation of the Lorentz transformations from Einstein's 4-vector.

    We start with this..


    Then we set the spatial component to zero to put us in a proper time frame..


    Now we have [itex]ds^2=c^2(dt')^2=c^2dt^2-dx^2-dy^2-dz^2[/itex].

    So, in this instance we can call dt' (dtau), because it is a clock in its own inertial reference frame, yielding..


    Here is where the fun comes in.. Factor out the dt^2 and c^2 from the rhs of the previous equation, which gives you..

    [itex]c^2(dtau)^2=c^2(1 - \frac{1}{c^2}\frac{dx^2}{dt^2}-\frac{1}{c^2}\frac{dy^2}{dt^2}-\frac{1}{c^2}\frac{dz^2}{dt^2})dt^2=c^2(1 - \frac{v^2}{c^2})dt^2[/itex]

    Where the velocity vector is just the dx,dy,dz.

    So now you just cancel the C^2 from each side and you have your time dilation formula pulled strait from the 4 vector.

    [itex]dtau = {\sqrt{1-\frac{v^2}{c^2}}}dt[/itex]

    Or, as it is more popularly seen,

    [itex]dt = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(dtau)[/itex]

    Where [itex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex] is known as "gamma"

    Neet huh?
  6. Jul 21, 2014 #5
    Well of course life becomes much easier when you start by postulating the answer ;) In a more logical (pedagogically, I like that word!) sequence the line element would be derived (observed, really) from invariance of the Lorentz Transform, not the other way around.

    I salute the OP's approach to learning this!
  7. Jul 21, 2014 #6
    well for the spacial equation, if i plug the value for f back into x'= (x - vt)f I get:

    [itex] x' = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}} [/itex]

    so that part checks out. As for the time equation we know that x=ct x'=ct' and t=x/c. We can plug these back into x'=(x-vt)f

    [itex] x'=(x-vt)f [/itex]

    [itex] ct' = (ct-v(\frac{x}{c}))f [/itex]

    [itex] t' = (t-\frac{vx}{c^2})f [/itex]

    Plugging f back in:

    [itex] t' = \frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}} [/itex]

    so I am pretty sure that checks out too. I'm still confused as to whether using the same f for both earlier equations requires proof that f would be the same though. Is it enough that both equations are basically the same equation rearranged?
  8. Jul 21, 2014 #7
    As far as understand it, the transformation should be symmetric and independent from orientation. This leads to the same factor in both directions. In general you can start with a linear transformation like

    [itex]t' = f_{tx} \left( v \right) \cdot x + f_{tt} \left( v \right) \cdot t[/itex]
    [itex]x' = f_{xx} \left( v \right) \cdot x + f_{xt} \left( v \right) \cdot t[/itex]

    Than you need four conditions to get the four parameters. One of them is the required symmetry.

    I have to admit that I can't explain why the transformation need to be linear. In his paper "Zur Elektrodynamik bewegter Körper" Einstein wrote it is clear that the equations must be linear due to homogeneity of space. But I'm not as smart as Einstein and can't see how the linearity results from the homogeneity.
  9. Jul 21, 2014 #8


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    The principle of relativity, that all inertial frames are equivalent, requires that the factors be the same for both frames. We have two frames, F and F'. The transformation

    [itex]x' = f(x-vt)[/itex]

    implies this:
    If an object aligned with the x-axis has length [itex]L[/itex] in its own rest frame, and the object is moving in the x-direction at speed v as measured in frame F, then the object has length [itex]L/f[/itex] as measured in frame F.

    So if that's true in frame F, it should also be true in frame F'. So the factor [itex]f[/itex] must be the same for transforming from F to F' as in the other direction.

    There's actually another issue that you glossed over when you asked whether f = g. You have two observers, A and B. If B has speed v relative to A, then A has speed v relative to B. (Remember, speed is the magnitude of the velocity, so it's always positive.) If this weren't the case, then there would be an asymmetry between the two observers.

    A point that people are sometimes confused about is the fact that in the two equations:

    [itex]x' = f (x - vt)[/itex]
    [itex]x = f (x + vt)[/itex]

    there seems to be an asymmetry between the two equations, because [itex]v[/itex] appears with a negative sign in the first equation, and with a positive sign in the second equation. That's an artifact of the particular choice of the direction of the x-axis in the two frames.
  10. Jul 21, 2014 #9
    So if I am understanding you correctly it is because of the principle of relativity that the error factor is the same? In other words since A is an inertial reference frame and B is moving with uniform velocity and direction relative to it it is also an inertial frame. So it would make no difference whether A was moving with -v relative to B with B stationary, or if B was moving with v relative to A with A stationary. The magnitude of the velocity between them is still the same?
  11. Dec 27, 2014 #10
    I am reviving this topic because there is something related I am still confused about. The principle of relativity states that the laws of physics are the same in every reference frame. In pre-relativity physics this is easy to show by differentiating the galileaon-newtonian transformations

    [itex] x' = x-vt [/itex]

    [itex] \frac{dx'}{dt} = \frac{dx}{dt} - \frac{d(vt)}{dt} [/itex]

    the time derivative of position is velocity. lets call the velocities of the reference frames u and u'

    [itex] u' = u - v [/itex]

    we differentiate again to get acceleration

    [itex] \frac{du'}{dt} = \frac{du}{dt} - \frac{dv}{dt} [/itex]

    v is constant so its derivative is 0 so

    [itex] \frac{du'}{dt} = \frac{du}{dt} [/itex]


    [itex] a' = a [/itex]

    since a given mass will also be the same for both reference frames, then it follows that F=ma will hold for both reference frames and the principle of relativity holds.
    however when I try to do this for the lorentz transformations:

    [itex] x' = γ(x-vt) [/itex]

    [itex] x' = γx-γvt [/itex]

    since v and c are constant , γ is constant

    [itex] \frac{dx'}{dt} = γ\frac{dx}{dt} - γ\frac{d(vt)}{dt} [/itex]

    [itex] u' = γu - γv [/itex]

    [itex] \frac{du'}{dt} = γ\frac{du}{dt} - γ\frac{dv}{dt} [/itex]

    [itex] a' = γa [/itex]

    Since the accelerations are not the same, doesn't that also mean that F=ma is not the same for each reference frame and that the principle of relativity does not hold? I am confused about this if anyone can clear this up it would be much appreciated :)
  12. Dec 27, 2014 #11


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    No, it means that ##\vec{F} = m \vec{a}## is not a correct relativistic law of physics; it is only a non-relativistic approximation. The correct relativistic law is ##F = dp / d\tau##, where ##F## is the four-force, ##p## is 4-momentum, and ##\tau## is proper time. For an object with constant rest mass, we have ##p = m u##, where ##u## is the 4-velocity, and ##F = m a##, where ##a## is the 4-acceleration; but this is still not the same as the ordinary 3-vector form of ##\vec{F} = m \vec{a}##. To derive the latter, you need to use the non-relativistic approximation in the instantaneous rest frame of the object; so, unlike the relativistic laws above, the 3-vector law is frame-dependent.
  13. Dec 27, 2014 #12
    tau being t and not t' in this case right?
  14. Dec 27, 2014 #13


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    No. ##\tau## is the object's proper time; ##t## and ##t'## are coordinate times in two different frames. Since the object is accelerating, its proper time is not equivalent to coordinate time in any inertial frame (since it is not at rest in any inertial frame for more than a single instant).
  15. Dec 27, 2014 #14


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    Neither, unless you choose your coordinate system such that one of them is equal to ##\tau## along the world line of the object being accelerated, and this can only be true for a single moment.
  16. Dec 27, 2014 #15
    You forgot the transformation of time. The correct result is

    [itex]a' = a \cdot \left( {\frac{{\sqrt {1 - u^2 } }}{{1 - u \cdot v}}} \right)^3[/itex]

    The original definition of force is F=dp/dt and this remains the same in every frame of reference both with Galilei and Lorentz transformation.
  17. Dec 27, 2014 #16
    If its possible can you show me how to get that or where I went wrong in my calculations? '

    also if anyone can answer: how is tau defined mathematically?
  18. Dec 28, 2014 #17
    You started with the Lorentz transformation of x:

    [itex]x' = \frac{{x - u \cdot t}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}[/itex]

    but that is only half the truth. The time needs to be transformed too:

    [itex]t' = \frac{{t - \frac{{u \cdot x}}{{c^2 }}}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}[/itex]

    The transformation of velocity is

    [itex]v' = \frac{{dx'}}{{dt'}} = \frac{{\frac{{dx'}}{{dt}}}}{{\frac{{dt'}}{{dt}}}}[/itex]


    [itex]\frac{{dx'}}{{dt}} = \frac{{v - u}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}[/itex]


    [itex]\frac{{dt'}}{{dt}} = \frac{{1 - \frac{{u \cdot v}}{{c^2 }}}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}[/itex]

    This results in the well known equation

    [itex]v' = \frac{{v - u}}{{1 - \frac{{u \cdot v}}{{c^2 }}}}[/itex]

    The transformation of acceleration works the same way:

    [itex]a' = \frac{{dv'}}{{dt'}} = \frac{{\frac{{dv'}}{{dt}}}}{{\frac{{dt'}}{{dt}}}}[/itex]


    [itex]\frac{{dv'}}{{dt}} = a \cdot \frac{{1 - \frac{{u^2 }}{{c^2 }}}}{{\left( {1 - \frac{{u \cdot v}}{{c^2 }}} \right)^2 }}[/itex]

    this results in

    [itex]a' = a \cdot \frac{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} ^3 }}{{\left( {1 - \frac{{u \cdot v}}{{c^2 }}} \right)^3 }}[/itex]

    and with c=1 in my simplified version above.
  19. Dec 28, 2014 #18


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    In units where c=1.
  20. Dec 28, 2014 #19


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    You have already got an answer but I wonder how you are learning SR. If you have a text book it must cover the definition of proper time. Even online there are complete introductory texts. Or are you trying to piece things together for yourself?
  21. Dec 28, 2014 #20
    Ok I see where I went wrong. Thank you all. I am learning SR on my own, but I do have my textbook for my intro physics classes that has a chapter on SR (giancoli) although it does not go into very much detail. It seems to just gloss over alot of stuff. I also have tried out some online lectures from yale and stanford but these assume more knowledge than I have in some cases, they are probably discussing things from a reading I haven't read or are for more advanced students.
  22. Dec 28, 2014 #21


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    Imho you'd be better off getting a book aimed at learning SR from scratch. There are plenty out there.
  23. Dec 28, 2014 #22
    any good not that expensive/free ones? I still gotta buy all my books for next semester =\
  24. Dec 28, 2014 #23


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  25. Dec 28, 2014 #24

    Philip Wood

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    Forgive me if this has already been covered, but the expression for [itex]\gamma[/itex] emerges very simply from your proposed transforms,
    [tex]x' = \gamma \ (x-vt) \ \ \ \ and \ \ \ \ x = \gamma \ (x'+vt')[/tex]
    Simply insert the special case of light emitted from the origin of S and S' at t = t' = 0, for which [itex]x = ct[/itex] and [itex]x' = ct'[/itex].
    [tex]ct' = \gamma \ (ct-vt) \ \ \ \ and \ \ \ \ ct = \gamma \ (ct'+vt')[/tex]
    Clearly multiplying the equations will enable one to get rid of the common factor tt', leaving [itex]c^2= \gamma^2 \ (c^2-v^2)[/itex].
    Rindler gives this symmetry method in his Introduction to SR. Resnick gives a painstaking justification of the linearity of the equations in his Introduction to SR.
  26. Dec 28, 2014 #25


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    Certainly it is no more nor less logical to start with the transforms and derive the line element than it is to start from the line element and derive the transform. Personally, pedagogically I prefer the line element -> transform approach since it prepares the groundwork quite well for GR.
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