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I Lorentz invariant integration - where is the error?

  1. Mar 2, 2017 #1

    Demystifier

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    Let ##j^{\mu}(x)## be a Lorentz 4-vector field in Minkowski spacetime and let ##\Sigma## be a 3-dimensional spacelike hypersurface with constant time of some Lorentz frame. From those I can construct the quantity
    $$Q=\int_{\Sigma} dS_{\mu}j^{\mu}$$
    where
    $$dS_{\mu}=d^3x n_{\mu}$$
    and ##n_{\mu}## is the unit timelike vector normal to ##\Sigma##. The quantity ##Q## is a Lorentz scalar. Since ##j^{\mu}## is a Lorentz vector, it follows that ##dS_{\mu}## must also be a Lorentz vector. But ##n_{\mu}## is also a Lorentz vector, so ##dS_{\mu}## can be a Lorentz vector only if ##d^3x## is a Lorentz scalar. Yet, ##d^3x## is not a Lorentz scalar, leading to a contradiction.

    Where is the error?

    There is a similar "paradox" with 4-momentum defined as
    $$P^{\nu}=\int_{\Sigma} dS_{\mu}T^{\mu\nu}$$
     
    Last edited: Mar 2, 2017
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  3. Mar 2, 2017 #2

    Orodruin

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    ##dS_\mu## is equal to ##n_\mu d^3x## only if ##n_\mu = (1,0,0,0)##, i.e., in the frame where your chosen spatial coordinates are within the surface.
     
  4. Mar 2, 2017 #3

    Orodruin

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    To elaborate on that. If you pick the spatial coordinates of the surface to be coordinates in a frame where ##S## is not a surface of simultaneity, you will get an additional factor that takes care of your invariance. In general, using parameters ##\xi^1##, ##\xi^2## and ##\xi^3## for the surface, we have
    $$
    dS_\mu = \varepsilon_{\mu\nu\sigma\rho} \frac{\partial x^\nu}{\partial \xi^1}\frac{\partial x^\sigma}{\partial \xi^2}\frac{\partial x^\rho}{\partial \xi^3} d^3\xi.
    $$
     
  5. Mar 2, 2017 #4

    Orodruin

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    As an example, take the surface ##t = vx## parametrised with the coordinates ##x##, ##y## and ##z##. This would lead to ##\partial t/\partial x = v## and therefore
    $$
    dS_0 = \varepsilon_{0123} d^3x = d^3x, \quad dS_1 = \varepsilon_{1023} \frac{\partial t}{\partial x} d^3x = -v\, d^3x
    $$
    with the other elements being equal to zero. It should be clear that this is not a unit vector multiplied by ##d^3x##.
     
  6. Mar 2, 2017 #5

    Demystifier

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    Exactly! For that reason, if ##x^{\mu}## is the spacetime coordinate, it is very misleading to write ##d^3x## for integration over ##\Sigma##. Instead, one should write ##d^3q##, where ##q^i## are coordinates on the hypersurface with induced metric
    $$\gamma_{ij}=\frac{\partial x^{\mu}}{\partial q^i} \frac{\partial x^{\nu}}{\partial q^j} g_ {\mu\nu}$$
    With such notation, it is more clear that one should not attempt to make a Lorentz transformation of ##q^i##.
     
  7. Mar 2, 2017 #6

    Orodruin

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    Yes, I would agree with this. But many physicists are notoriously sloppy with notation. I try to write ##dS_\mu## in most cases.

    You actually never need to refer to the induced metric. What you need is the volume form ##\eta## on Minkowski space (which does depend on the metric) and the 3-form to integrate over the surface is ##i_j\eta##. Your defined quantity becomes
    $$
    Q = \int_\Sigma i_j \eta
    $$
     
  8. Mar 2, 2017 #7

    Demystifier

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    I am starting to appreciate the coordinate free notation that mathematicians prefer. With coordinates and indices one can easily be mislead in an attempt to use in 3-space the quantities originally defined in 4-space.
     
  9. Mar 4, 2017 #8
    Why ##n_\mu## has components ##(1,0,0,0)## when the coordinates are within the surface? Would this mean that at the surface both ##x,y## and ##z## are equal to zero? Why?

    EDIT: I just realized that that is because ##n_\mu## is a unit vector, orthogonal to the hyper-surface of constant time.
     
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