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Linearity of LT

  1. Jun 3, 2012 #1
    I often read sentences like, "if space is homogeneous, then the Lorentz transformation must be a linear transformation." What exactly does it mean to say that space is homogeneous, and how does it imply that the Lorentz transformations are linear?
     
  2. jcsd
  3. Jun 5, 2012 #2
    Hi dEdt,

    With regard to the Maxwell equations, I think the homogeneity of space (or EM medium to be a little more consistent with Maxwell theory) means that electric permittivity and magnetic permeability values [itex]\epsilon[/itex] and [itex]\mu[/itex] are:

    1. Scalars (as opposed to vectors or tensors)
    2. Constants (as opposed to variables)

    The effect of #1 and #2 is that an EM wave will propagate in all directions with equal speed.

    I don't believe that implies that all transformations between positions of particles and positions of wavefronts, when traveling at different relative velocities, must be linear. But intuitively, it does indicate that a linear transformation might be possible. However, the assumption must be made that linear relationships exist between:

    1. Particles (particle to particle)
    2. Wavefronts (wave to wave)
    3. Particles and wavefronts (wave to particle)

    Also the linear transformations for each of those must be mutually linear (linear across all) to be a completely linear theory.

    The Lorentz transformation is one means of accomplishing #2. I don't think either #1 or #3 has been proven using the LT. The Galilean transformation accomplishes #1.
     
    Last edited: Jun 5, 2012
  4. Jun 5, 2012 #3
    Hi Phil,

    As I understand it, the Lorentz transformations can be derived independently of Maxwell's equations -- you just need the principle of relativity and the light postulate. Is there a way to formulate "homogeneity of space" without recourse to E&M?
     
  5. Jun 5, 2012 #4

    Matterwave

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    Homogeneity of space means that there are no preferred points in space. Every point in space is just like any other.

    One uses Homogeneity of space (and isotropy of space and homogeneity of time and constancy of the speed of light) to prove that the interval ds^2=dt^2-dr^2 is invariant (i.e. ds=ds'). It arises as the implication that the transformation between ds and ds' cannot involve the spatial coordinates since space is homogeneous (the other restrictions will go on to prove that the transformation between ds and ds' is simply the identity).

    Once one proves that ds=ds' using the postulates of relativity, one can obtain the Lorentz transformations as simply those rotations in space-time which preserve the length of ds. Rotations are of course linear transformations and so the Lorentz transformations are linear basically by definition.
     
  6. Jun 6, 2012 #5
    That's interesting. But how would you go from the invariance of ds to the Lorentz Transformations?
     
  7. Jun 6, 2012 #6
    The homogeneity of spacetime gives you the freedom to choose the origin. The isotropy of spacetime ("every direction is equivalent") gives you the ability to choose an arbitrary basis. Put together, these allow us to set up a Minkowski spacetime as a vector space.

    Consider a 1+1 spacetime where [tex]e_t \cdot e_t = -1[/tex] and [tex]e_x \cdot e_x = 1[/tex]. At one's discretion, one can choose a new basis for this space. Let's say that,

    [tex]e_x' = a e_x + b e_t[/tex]

    Where a, b are constants. We'll leave normalization for later. It's important to note that the process of choosing a new basis is always a linear operation.

    Now, what is the transformed timelike vector, [tex]e_t'[/tex]? There are several ways you can compute it; I prefer a geometric algebra approach. The homogeneity of spacetime tells us that there is a bivector (a "directed plane") that is constant everywhere. We denote this bivector [tex]e_{tx} \equiv e_t \wedge e_x[/tex]. (See wikipedia on wedge products if you like; a quick explanation is to call them the generalized analogue to a cross product.)

    To find the new timelike vector, we just take this plane and eliminate the part in the [tex]e_x'[/tex] direction. This is accomplished with the "geometric product":

    [tex]e_t' = e_{tx} e_x' = e_{tx} (a e_x + b e_t) = a e_t + b e_t[/tex]

    I must emphasize that there are several ways to find the other basis vector; I choose geometric algebra because the invariance of the unit bivector is a physical thing you can understand--the area the two vectors subtend isn't changing no matter how you choose the basis.

    Anyway, you can see that since this uses the same a, b as before, we have the symmetry inherent to the Lorentz transformation.

    Now, let's look at the normalization. In the above, I've assumed the coefficients were normalized. Let's instead consider if the coefficients aren't. If we want the timelike basis vector to be proportional to [tex]f e_t + g e_x[/tex], then the normalized basis vector is

    [tex]e_t' = \frac{1}{f^2 - g^2} (f e_t + g e_x)[/tex]

    Let [tex]f=1, g=\beta[/tex], and you get the usual form of the Lorentz transformations.
     
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