Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz invariant

  1. Jul 6, 2005 #1
    What makes an equation lorentz-covariant? When is an equation lorentz-invariant?
  2. jcsd
  3. Jul 6, 2005 #2
    I'm not sure but I believe that an equation is Lorentz invariant if it has the same form in all inertial coordinate systems and that Lorentz covariant equation is one which is Lorentz invariant and that all the constants in the equation are Lorentz scalars.

  4. Jul 6, 2005 #3


    User Avatar
    Science Advisor

    Here's one version of the Lorentz transformation equations, which covers the case where the primed frame is moving at velocity v along the x-axis of the unprimed frame:

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


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

    So what Lorentz-invariance means is that if you take some equation for a law of physics written in terms of x,y,z,t coordinates and use the Lorentz transform to make substitutions and rewrite this equation in terms of x',y',z',t' coordinates, the equation will end up looking the same as if you had just replaced x with x', y with y', z with z' and t with t'--the equation should have exactly the same form in both coordinate systems.

    Here's an example, involving "Galilei-invariance" rather than Lorentz-invariance because the math is simpler. The Galilei transform for transforming between different frames in Newtonian mechanics looks like this:

    [tex]x' = x - vt[/tex]
    [tex]y' = y[/tex]
    [tex]z' = z[/tex]
    [tex]t' = t[/tex]


    [tex]x = x' + vt'[/tex]
    [tex]y = y'[/tex]
    [tex]z = z'[/tex]
    [tex]t = t'[/tex]

    To say a certain physical equation is "Galilei-invariant" just means the form of the equation is unchanged if you make these substitutions. For example, suppose at time t you have a mass [tex]m_1[/tex] at position [tex](x_1 , y_1 , z_1)[/tex] and another mass [tex]m_2[/tex] at position [tex](x_2 , y_2 , z_2 )[/tex] in your reference frame. Then the Newtonian equation for the gravitational force between them would be:

    [tex]F = \frac{G m_1 m_2}{(x_1 - x_2 )^2 + (y_1 - y_2 )^2 + (z_1 - z_2 )^2} [/tex]

    Now, suppose we want to transform into a new coordinate system moving at velocity v along the x-axis of the first one. In this coordinate system, at time t' the mass [tex]m_1[/tex] has coordinates [tex](x'_1 , y'_1 , z'_1)[/tex] and the mass [tex]m_2[/tex] has coordinates [tex](x'_2 , y'_2 , z'_2 )[/tex]. Using the Galilei transformation, we can figure how the force would look in this new coordinate system, by substituting in [tex]x_1 = x'_1 + v t'[/tex], [tex]x_2 = x'_2 + v t'[/tex], [tex]y_1 = y'_1[/tex], [tex]y_2 = y'_2[/tex], and so forth. With these substitutions, the above equation becomes:

    [tex]F = \frac{G m_1 m_2 }{(x'_1 + vt' - (x'_2 + vt'))^2 + (y'_1 - y'_2 )^2 + (z'_1 - z'_2 )^2}[/tex]

    and you can see that this simplifies to:

    [tex]F = \frac{G m_1 m_2 }{(x'_1 - x'_2 )^2 + (y'_1 - y'_2 )^2 + (z'_1 - z'_2 )^2}[/tex]

    In other words, the equation has exactly the same form in both coordinate systems. This is what it means to be "Galilei invariant". More generally, if you have any physical equation which computes some quantity (say, force) as a function of various space and time coordinates, like [tex]f(x,y,z,t)[/tex] [of course it may have more than one of each coordinate, like the [tex]x_1[/tex] and [tex]x_2[/tex] above, and it may be a function of additional variables as well, like [tex]m_1[/tex] and [tex]m_2[/tex] above] then for this equation to be "Galilei invariant", it must satisfy:
    [tex]f(x'+vt',y',z',t') = f(x',y',z',t') [/tex]

    So in the same way, an equation that's Lorentz-invariant should satisfy:

    [tex]f( \gamma (x' + vt' ), y' , z', \gamma (t' + vx' /c^2 ) ) = f(x' ,y' ,z' , t')[/tex]
    Last edited: Jul 6, 2005
  5. Jul 6, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    A function "f" is Lorentz invariant,if its spinor rank is 0, i.e. is a Lorentz/[itex] SL(2,\mathbb{C}) [/itex] scalar. Lorentz covariance is achived when putting in the same equation quantites which transform under [itex] O(1,3) [/itex] after a finite dimensional representations.

  6. Jul 7, 2005 #5
    Thanks for answering. I found this at Wikipedia:

    Lorentz covariance is a term in physics for the property of space time, that in two different frames of reference, located at the same event in spacetime but moving relative to each other, all non-gravitational laws must make the same predictions for identical experiments. A physical quantity is said to be Lorentz covariant if it transforms under a given representation of the Lorentz group. Quantities which remain the same under Lorentz transformations are said to be Lorentz invariant (i.e. they transform under the trivial representation).

    According to the representation theory of the Lorentz group, Lorentz covariant quantities are built out of scalars, four-vectors, four-tensors, and spinors.

    The space-time interval is a Lorentz-invariant quantity, as is the Minkowski norm of any four-vector.

    Equations which are true in any inertial reference frame are also said to be Lorentz covariant (some use the term invariant here). Lorentz covariant equations can always be written in terms Lorentz covariant quantities. According to the principle of relativity all fundamental equations of physics must be Lorentz covariant.

    Note: this usage of the term covariant should not be confused with the related concept of a covariant vector. On manifolds, the words covariant and contravariant refer to how objects transform under general coordinate transformations. Confusingly, both covariant and contravariant four-vectors can be Lorentz covariant quantities.

    I didn't know how great this Wikipedia site is...looks I have to spend some more time there.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Lorentz invariant
  1. Lorentz invariants (Replies: 3)

  2. Lorentz invariance (Replies: 4)