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Where is potential energy in relativistic formula of energy?

  1. Jul 24, 2011 #1
    Please teach me this:
    The relativistic energy E=m.square(c)/squareroot(1-square(v)/square(c)) is determined by momentum p(because of ''square''(4-p)=square(m)).Then what is the role of potential in relativistic energy?When we consider the interaction between particles,how can we express the kinetic energy plus potential energy in the same formula?
    Thank you very much in advance.
     
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  3. Jul 24, 2011 #2

    bcrowell

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    You may be more likely to get helpful responses to your posts if you make them easier to read by marking the math up in LaTeX. Here is an example: [itex]\sqrt{x^2+y^2}=1[/itex]. To see how I did that, click the QUOTE button on my post.
     
  4. Jul 24, 2011 #3
    The question has to be more specific: what kind of force? Because if it is gravity, you need to do some GR and it's not exactly a potential anymore. If it's EM, the potential is a 4-vector: [itex](\phi, \mathbf{A})[/itex], where [itex]\phi[/itex] is the scalar potential and [itex]\mathbf{A}[/itex] the vector potential. The Hamiltonian in this case is:
    [tex]H = \sqrt{m^2 c^4 + \left(\mathbf{p} - e\mathbf{A}\right)^2 c^2} + e\phi[/tex]
     
  5. Jul 24, 2011 #4
    If it's gravity, then both potential and kinetic energy are generalised as (letting c=1):
    [itex]-m\frac{d\tau}{dt}[/itex]​
    where
    [itex]d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}[/itex]​
    We can see this clearly when we assume spherical symmetry and consider the Newtonian limit:
    [itex]-m\frac{d\tau}{dt}=-m\sqrt{\frac{g_{tt}dt^2+g_{mm}(dx^m)^2}{dt^2}}[/itex]​
    [itex]=-m\sqrt{g_{tt}-\dot{x}^2}[/itex]​
    [itex]=-m\sqrt{1-\frac{2GM}{r}-\dot{x}^2}[/itex]​
    [itex]\approx-m+\frac{GMm}{r}+\frac{1}{2}m\dot{x}^2[/itex]​
    And given that the action in a gravity well is
    [itex]-m\int{d\tau}=0[/itex]​
    then we recover the non-relativistic
    [itex]E=\Delta{U}+KE[/itex]​
     
    Last edited: Jul 24, 2011
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