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Proving d'Alembertian Invariant under Lorentz Transformations

  1. May 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that (D'Alembertian)^2 is invariant under Lorentz Transformation.


    2. Relevant equations

    The book (E/M Griffiths) describes the D'Alembertian as:

    [tex]\square^2=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}[/tex]


    3. The attempt at a solution

    I don't really know what it is asking me to do here. Any guidance at where to even start would be very helpful. I am not very good at writing out proofs.
     
  2. jcsd
  3. May 2, 2009 #2
    I could be wrong, but it seems that it's asking you to show how the operator manifests itself in the Minkowskian Metric (0,0,0,1); from there, i would assume that any ''rotations'' made in the Cartesian Product always remain invariant... but it's vague, so, i am not entirely sure.

    Since the Operator maintains the four dimensions (t, x, y, z), basic rotations in such a ''spacetime'' are considered in special relativity as being invariant. For instance: Take the following Cartesian Distance System

    [tex]s^2=(\Delta x*)+(\Delta y*)+(\Delta z*)+(\Delta t*)[/tex]

    Here, the asterisks represent the ''rotations'' i spoke of to you, where the rotations are orientated from the original coordination:

    [tex]s^2=(\Delta x)+(\Delta y)+(\Delta z)+(\Delta t)[/tex]

    So in this sense, it is said to be that distance is invariant under any such coordinational changes. Moving on, the covarient expression of the operator is given as [tex]2=\nabla^{\mu}\nabla_{\mu}[/tex] which is crucial when taking into account the four functions of [tex]x^{\mu}[/tex], but they are merely functions themselves and not exactly vectors. But this is why i first started off with the Cartesian Map, which is a flat spacetime which is identified as a ''harmonic function''.

    So there is my bit, and i hope it helps in any way?
     
  4. May 2, 2009 #3

    This might also give some insight, which maybe more correct towards the answer you are looking for:

    http://en.wikipedia.org/wiki/Alternatives_to_general_relativity

    ''Nordström (1912)

    The first approach of Nordström (1912) was to retain the Minkowski metric and a constant value of but to let mass depend on the gravitational field strength . Allowing this field strength to satisfy...

    ...This theory is Lorentz invariant, satisfies the conservation laws, correctly reduces to the Newtonian limit and satisfies the weak equivalence principle.''
     
  5. May 2, 2009 #4
    Thanks for the reply, I am looking into the link and will comment once I attempt the problem again. I agree with you though, the question is rather vague.
     
  6. May 4, 2009 #5
    Okay I think I have this solved. I just went about it the straight forward method by simply crunching the transformations as follows:
    [tex]
    {A}=
    \begin{bmatrix}
    {\frac{\partial^2}{\partial t^2}\frac{1}{c}t}\\
    {\frac{\partial^2}{\partial x^2}x}\\
    {\frac{\partial^2}{\partial y^2}y}\\
    {\frac{\partial^2}{\partial z^2}z}
    \end{bmatrix}

    \begin{bmatrix}
    {\gamma}&{-\beta\gamma}&{0}&{0}\\
    {-\gamma\beta}&{\gamma}&{0}&{0}\\
    {0}&{0}&{1}&{0}\\
    {0}&{0}&{0}&{1}
    \end{bmatrix}
    [/tex]

    Then I dot:
    [tex]{A}\cdot{A'}[/tex]
    Using the primed version of matrix A

    Which I get:
    [tex]\frac{\partial^4}{\partial x^4}xx' + \frac{\partial^4}{\partial y^4}yy' + \frac{\partial^4}{\partial z^4}zz'[/tex]

    After using:
    [tex]\beta= \frac{v}{c},\:\;\: \gamma^2= \frac{c^2}{c^2-v^2}[/tex]
    and canceling out differentials that go to zero

    This answer equals the answer given by:
    [tex]\begin{bmatrix}
    {\frac{\partial^2}{\partial t^2}\frac{1}{c}t}\\
    {\frac{\partial^2}{\partial x^2}x}\\
    {\frac{\partial^2}{\partial y^2}y}\\
    {\frac{\partial^2}{\partial z^2}z}
    \end{bmatrix}
    \cdot
    \begin{bmatrix}
    {\frac{\partial^2}{\partial t^2}\frac{1}{c}t'}\\
    {\frac{\partial^2}{\partial x^2}x'}\\
    {\frac{\partial^2}{\partial y^2}y'}\\
    {\frac{\partial^2}{\partial z^2}z'}
    \end{bmatrix}
    [/tex]

    Therefore the D'Alembertian is invariant under Lorentz Transformation.

    Does this seem like a reasonable way to do this problem?

    Thanks

    BTW should I prime the partial derivatives in the A' matrices? Then my answer would look like this:

    [tex]\frac{\partial^2}{\partial x^2}x\frac{\partial^2}{\partial x'^2}x' + \frac{\partial^2}{\partial y^2}y\frac{\partial^2}{\partial y'^2}y' + \frac{\partial^2}{\partial z^2}z\frac{\partial^2}{\partial z'^2}z'[/tex]

    This wouldn't affect the problem I don't think.
     
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