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I Newtonian analogue for Lorentz invariant four-momentum norm

  1. Jun 22, 2016 #1
    Hi.

    I read that the Lorentz invariance Minkowski norm of the four-momentum
    $$E^2-c^2\cdot \mathbf{p}^2=m^2\cdot c^4$$
    has no analogue in Newtonian physics. But what about
    $$E-\frac{\mathbf{p}^2}{2m}=0\quad ?$$
    It might look trivial by the definition of kinetic energy, but it's still a relation between energy and momentum that's invariant under Galilei transforms.
     
  2. jcsd
  3. Jun 22, 2016 #2
    It's not a relation between energy and momentum but between kinetic energy and momentum. It doesn't work with the total energy.
     
  4. Jun 22, 2016 #3

    haushofer

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    Where did you read it?

    You're right; I think the idea is that the newtonian expression is not the inner product of two four-vectors, since such a product does not really exist in newtonian spacetime; there is no (non-degenerate) metric.
     
  5. Jun 22, 2016 #4

    robphy

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    The natural candidate is ##mu^a## where ##m## is the [rest] mass and ##u^a## is the 4-velocity.
    Presumably, there is a mass-shell in energy-momentum space which would look similar to the [timelike but degenerate] Galilean metric.
    In https://www.desmos.com/calculator/ti58l2sair, set E=0.
    The temporal component would be ##m## (or in standard units of momentum ##mc## where ##c## is a convenient velocity unit with no other significance).
    To get the spatial components ##m\vec v##, one would use the spacelike-but-degenerate Galilean metric.
    (To do this right, one needs to first write down the postulated structure [e.g., (M,##t_a##, ##h^{ab}##, ...) akin to specifying (M,g) for a spacetime] then formulate the dynamics on it.)
    You can do 4-momentum conservation by vector addition.. which amounts to conservation of mass and conservation of spatial-momentum.

    Kinetic energy should really be calculated using the Work-energy-theorem.
     
  6. Jun 30, 2016 #5
    So in the relativistic case, the equation is about total energy? Then I'm running into problems with an answer I got in a different thread:
    Say the left side of
    $$E^2=c^2\cdot \mathbf{p}^2+m^2\cdot c^4$$
    is total energy squared. Consider two identical particles with the same velocity where one is in an electric potential and the other is not. Then their total energies are different, but their momentum is the same. So one of those particles must violate above equation.
     
  7. Jul 1, 2016 #6
    Yes.

    I'm not sure if the term "total energy" makes much sense in this example. It doesn't refer to the total system because the source of the potential is not included.
     
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