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Consider a proton in uniform circular motion in a plane perpendicular to a magnetic field. The Lorentz force acts as the required centripetal force.

At a low speed, we have $\frac{m\cdot v^2}{r}=B\cdot q\cdot v\to$ $$v=\frac{B\cdot q\cdot r}{m}\qquad(1)$$ I wonder how this changes at high speeds when the rest energy is not yet negligible compared to the kinetic energy.
It is tempting to add a factor of gamma to the expression for the centripetal force and leave the expresssion for the Lorentz force unchanged:
$$v=\frac{B\cdot q\cdot r}{m_0}\cdot\sqrt{1-\frac{v^2}{c^2}}$$ $$v=\frac{1}{\sqrt{(\frac{1}{c})^2+(\frac{m_0}{B\cdot q\cdot r})^2}}\qquad(2)$$ Is this correct for a proton moving at a speed of the order of $c/2$, when the rest energy is not yet negligible compared to the kinetic energy?

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2 Answers 2

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Your equations are correct. The square root comes from the relativistic equation for momentum: $p=mv\gamma=mv/\sqrt{1-v^2/c^2}$.

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  • $\begingroup$ My derivation of equation (2) begins in terms of force. There isn't much talk of forces in special relativity, perhaps because forces don't transform nicely. Can you give me a hint how to derive equation (2) in a more appropriate way, e.g. in terms of momentum? $\endgroup$
    – gamma1954
    Commented May 27, 2022 at 21:03
  • $\begingroup$ It is better to avoid talking about 'force' in special relativity, because force means different things to different people. In your case, it is better to use $\frac{\bf dp}{dt}=\frac{d(m{\bf v}\gamma)}{dt}=q{\bf v\times B}$. $\endgroup$ Commented May 28, 2022 at 10:38
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This is precisely what is seen in a relativistic particle accelerator. https://en.wikipedia.org/wiki/Cyclotron#Relativistic_considerations

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