- #1
jostpuur
- 2,116
- 19
What happens to a particle in a scalar potential [itex]U(t,x)[/itex]? I have been living under a belief that the equation of motion would be
[tex]
\frac{d}{dt}\frac{m\dot{x}(t)}{\sqrt{1- \frac{\|\dot{x}(t)\|^2}{c^2}}} = -\nabla_x U(t,x(t))
[/tex]
but I just proved that this isn't Lorentz invariant, and therefore cannot be the answer.
Or to be more precise, I have had some doubts about the previous equation, and in some contexts I have been living under a belief that the equation
[tex]
\frac{d}{dt}\frac{\big(m + \frac{U(t,x(t))}{c^2}\big)\dot{x}(t)}{\sqrt{1- \frac{\|\dot{x}(t)\|^2}{c^2}}} = -\nabla_x U(t,x(t))
[/tex]
would have a better chance of being Lorentz invariant. Now when I looked this more carefully, I was unable to to prove the invariance claim for this to be true or false.
[tex]
\frac{d}{dt}\frac{m\dot{x}(t)}{\sqrt{1- \frac{\|\dot{x}(t)\|^2}{c^2}}} = -\nabla_x U(t,x(t))
[/tex]
but I just proved that this isn't Lorentz invariant, and therefore cannot be the answer.
Or to be more precise, I have had some doubts about the previous equation, and in some contexts I have been living under a belief that the equation
[tex]
\frac{d}{dt}\frac{\big(m + \frac{U(t,x(t))}{c^2}\big)\dot{x}(t)}{\sqrt{1- \frac{\|\dot{x}(t)\|^2}{c^2}}} = -\nabla_x U(t,x(t))
[/tex]
would have a better chance of being Lorentz invariant. Now when I looked this more carefully, I was unable to to prove the invariance claim for this to be true or false.
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