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Lorentz transformations derivation

  1. Apr 7, 2014 #1
    I'm currently going through my courses notes for relativity. We looked at Einsteins two postulates and then said that time must therefore dilate due to constant speed of light. That I understand, however I'm still confused about the Lorentz's transformations. My notes start with a basic form of the transformations and then goes on to calculate the constants. However the basic forms they setup I don't understand where they come from. It first sets up x'=γ(x-ut). Why does distance HAVE to change? I understand why either one of distance or time has to change to compensate for constant speed of light being constant, but why both? Furthermore why is it of that form? My notes just say that these transformations can be calculated solely from Einsteins two postulates, so how does it get this form. It also says that time must therefore be t'=a*x+b*t. Where did this form come from, and why does it depend upon both x and t? If I go really far away from someone, will we have difference in time?

    I'm just confused as to where these basic forms come from, its implied straight from Einsteins two postulates but doesn't state explicitly why. I've been looking at this all day using alot of different text books. I keep flashing between understanding it and not but I ultimately have no idea how the above two forms could be found so simply it doesn't have to be stated. If you think you might help then please do, thanks :)
  2. jcsd
  3. Apr 7, 2014 #2


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    First, a quick digression that will save you endless grief down the line: it's not distance in that transformation, the ##x## that appears there is the spatial ##x## coordinate. A distance would be the difference between two ##x## coordinates (using the same frame of course - otherwise that difference would be as meaningless as subtracting a latitude from a longitude).

    OK, with that digression out of the way... The goal here is to find a transformation from the unprimed ##x## and ##t## coordinates to the primed ##x## and ##t## coordinates that keeps the speed of light equal to ##c## in both.

    We don't actually have to derive anything to do this; we could just guess what the transform might be, try it, see if it has that property... If it doesn't, we try another guess until we finally score. That would be a less than optimal approach :smile:

    So instead we make an educated guess about what these transforms might look like with various arbitrary constants in them, then go through the algebra to find the values of these constants. There are a number of good reasons (for example, they have to keep working if you move the origin) to expect that the form of these transforms will be of the form that Einstein started with: ##x'=A(x-vt)## and ##t'=B(t-vx)## but it's still an educated guess. If it hadn't worked, we would start over with something else.... But it did.

    A few more observations:
    1) It turns out that Einstein started with an unnecessarily general educated guess; ##A## and ##B## are the same. But nothing awful happened, we just worked through the algebra ended with the same value which we now call ##\gamma##
    2) If this idea of an educated guess bothers you... Check the wikipedia entry for "ansatz". :smile:
    Last edited: Apr 7, 2014
  4. Apr 7, 2014 #3


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    Additionally, one can use a more geometric and concrete approach to this problem by defining the interval (metric) ##ds^2=c^2dt^2-dx^2-dy^2-dz^2##, noticing that that interval must be conserved just like lengths in 3 space, and arrived at the conclusion that the Lorentz boosts are the rotations in the time-space axes.

    I believe though that Einstein did not have this formalism when he came up with special relativity, and it was Minkowski who actually did things this way.
  5. Apr 7, 2014 #4
    I look at it differently. I'm in the midst of learning this, so this post is worth what you paid for it :P
    Look at the geometry. Take two observers moving with different (constant, inertial) velocities. We can orient the 3-d space axes so that their velocity is along the x-axis. So now we only have one space-like dimension to worry about. So, lets graph them out on a plane. As usual time (t) is the vertical axis, x the horizontal.
    Here is the important part: draw a diagonal line representing the distance per unit of time that LIGHT travels.
    It is not necessary that the line be at 45°, BUT it gets really difficult to do the geometry if it isn't. So, pick the (arbitrary) units of length as 1 light-second and units of time as 1 second. (We could use meters and nano-seconds if we wanted, but choosing the speed of light to be at 45° has turned out to be *very useful*.
    Now we have the two axes and a diagonal line (45°, arbitrarily). I didn't specify if the line had a positive slope or a negative slope, and actually we need both. Light traveling in the positive x direction is the diagonal line with positve slope, light traveling in the -x direction has slope of -45°. OK? We know that some place and time is defined to be the origin of this coordinate system (0,0). And the diagonal lines (called the light cone, since with two spatial dimensions, instead of just one, it would form a cone). Ok, so far so good.
    Now lets draw the coordinate system of the other observer, the one who will be traveling at a non-zero speed relative to the origin of the (x,t) system. Give her some velocity, v. We know that relative to the original observer that after a time t1, that her position will have changed by v*t1 or more accurately v*(t1-t0), but we will assume t0 = 0. There is no reason for the two observers to agree on one point that their own coordinate systems have in common. We could pick ANY point (3,12.3) or (-π, e^√7) but lets keep it simple and agree that BOTH observers agree that at (x,t) = (0,0) it is also true that (x',t') = (0,0) ( it would just be a simple translation in time and space to make them so, if they were different). So (0,0) is an agreed point in space-time of both observers. OK? So what do we have? We have that x = vt for her in his coordinates (I forgot to mention observer 1 is him, 2 is her). In HER coordinates, she hasn't moved. x'=0. What the heck does that mean?? Well, relative to herself, she is motionless and so at all points on the line where x=vt we know that x'=0. Is this clear? Do you understand that we just determined the x' axis for her? The faster she is traveling (lets agree it is in the positive x direction) the larger the slope. If her slope was at 45°, then she would be traveling at c, which is NOT allowed, so her |slope| has to be less than 45° (of course, going forward (upwards) in time.) We know that her x' axis is a straight line in the (x,t) coordinate frame, since her distance is linearly increasing with time. So. Now we have to figure out where to draw the t' coordinate axis. What clues do we have? We actually have NO CHOICE in drawing the t' axis. It is fixed because the speed of light for HER must be EXACTLY the SAME as for HIM. At a given x' distance, t' must also be given for that diagonal light ray we already drew. The ONLY way to get this to work out is that if the x' axis is at an angle of Θ above the x' axis, then the t' axis MUST be at an angle of Θ as measured from the t axis. But is it to the left (negative slope) or to the right (positive slope)?? Intuitively, you would think the two axes are just rotated and still meet at 90°, just like the x and t axes, right? So just rotate the x' and t' axis counterclockwise around the origin and you'd expect everything to work out, right? Not so much. Your intuition is WRONG. The t' axis is rotated clockwise (to the right) and x' and t' do NOT meet at a 90° angle (in the coordinate system of HIM). They both move towards that line representing the (positive) speed of light. And they move towards it by the exact same Θ (in our units, see how handy they come in?) In HIS coordinate system, his coordinate grid is squares, all angles are 90°. But her coordinates drawn on the same graph will be rhombuses (rhombi?). Collapsed toward the speed of light line. Once you understand this you can solve most of the problems presented to you in special relativity. (as far as distance and veloticity go). It is not obvious that her coordinates drawn out relative to his space-time is not euclidean, but that is the ONLY way that one x unit and one t unit is the position of a photon (emitted from the origin) at time t AND that one x' unit and one t' unit also intersect that same line. Oh, I should also mention that the lines of constant x are vertical, right? and constant t are horizontal. This is
    obvious. Lines of constant x' are parallel to the x' axis. Lines of constant t' are parallel to the t' axis.
    Work it out for the twins speeding away from one another. Remember to always draw light using lines with a slope of +/- 45°. Not to be confusing, but if we started with HER coordinates (at 90°, like she would see them) then HIS coordinate system would be rhombohedral (but would be moving in the -x direction, unless we switched +x and -x). The two will view the other coordinates (except for the switch of + an - x) as indentically distorted.
    Doing it this way means that as velocity increases from 0 to c, the angle between the t,x axes decrease from 90° to (almost, for massive objects, exactly for massless objects) 0°. The faster v is, the more'squeezed together' x' and t' are. We have words for this (dilation, contraction).
    Last edited: Apr 7, 2014
  6. Apr 7, 2014 #5


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    You're right. Einstein did not have it and Minkowski did come up with it.

    The original poster's question ("Where did this ##x'=A(x-vt)## thing come from?") suggests that he's learning the pre-Minkowski approach that Einstein used. It's far less elegant but arguably requires less mathematical sophistication.
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