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Covariant vs. contravariant time component

  1. Oct 4, 2008 #1
    ...of the four-momentum vector.

    Why is the energy of a particle identified with p0 instead of p0? Is there a theoretical basis for this, or was it simply observed that p0 is conserved in a larger set of circumstances?
  2. jcsd
  3. Oct 4, 2008 #2


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    How is the metric written in that book (-+++) or (+---)?

    Covariant and contravariant components are related by the metric.

    Just make sure the energy is positive.
  4. Oct 4, 2008 #3


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    In the more usual metric (+---) of SR, p^0 and p_0 are equal, so it doesn't matter which you use. In any event, although they are equal, it should be p^0 that is the physical energy since the four-momentum is a contravariant vector.
    If you are reading a book that uses the metric (-+++), then everything could be different.
  5. Oct 4, 2008 #4
    I did not have the Minkowski metric specifically in mind. If one uses the Schwarzschild metric -- or any other diagonal metric with [tex]|g_{00}|\neq1[/tex] -- p0 and p0 differ by more than just the sign; they have different magnitudes, so the energy of a particle cannot have both values. It's been my impression that in such circumstances one uses the covariant form instead of the contravariant form, but I don't know why.
  6. Oct 4, 2008 #5

    George Jones

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    If one uses standard Schwarzschild coordinates, then

    [tex]k = \frac{\partial}{\partial t}[/tex]

    is a timelike Killing vector. If [itex]u[/itex] is the 4-velocity of a freely falling particle, then

    [tex]E = g \left( k , u \right)[/tex]

    is constant along the particle's worldline.

    What is the coordinate expression of the above coordinate-free expression?
  7. Oct 4, 2008 #6
    What I had in mind was

    [tex] p^0=m_0 c \frac{dt}{d\tau} [/tex] while [tex] p_0= g_{00}m_0 c \frac{dt}{d\tau}=(1-\frac{r_s}{r})m_0 c \frac{dt}{d\tau}[/tex]

    where [tex]d\tau=ds/c[/tex] and [tex]ds[/tex] is defined using the Schwarzschild metric. Since [tex](1-\frac{r_s}{r})[/tex] won't equal 1 while r is finite, these two terms (p0 and p0) have different values, and I don't know which one (if either) represents the energy of the particle.

    Regarding the coordinate-free expression, when you say that [tex]k = \frac{\partial}{\partial t}[/tex] is a vector, do you mean the partial derivative of the displacement vector with respect to coordinate time? or of a different vector?
    Last edited: Oct 4, 2008
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