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I read "The Feynman Lectures on Physics Vol II The New Millennium Edition" and the equation (1.2) stated that we can combine electrical force with the equation of motion and get:

$$\frac{\text{d}}{\text{d}t}\left(\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}v\right)=q(E+v\times B)$$

My question is: why has the relativistic mass been used here? Is it just to generalize or is it supposed to indicate something?

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The expression in big brackets whose time derivative you are taking is the relativistic expression for momentum. If you wish to think of it as product of two factors, think of it as the invariant $m$ (that used to be called rest mass' and is now simply called 'mass') multiplied by the kinematic factor $\frac{\vec v}{\sqrt{(1-v^2/c^2)}}$.

You are strongly advised not to think of it as the product of $\vec v$ and a factor $\frac{m}{\sqrt{(1-v^2/c^2)}}$, especially if you regard the latter factor as some sort of speed-dependent mass. What's wrong with doing this? (a) It doesn't help, for example substituting $\frac{m}{\sqrt{(1-v^2/c^2)}}$ for $m$ in the Newtonian formula for kinetic energy won't give you the right relativistic formula for kinetic energy. (b) $\frac{m}{\sqrt{(1-v^2/c^2)}}$ is actually much more closely related to energy than to mass.

But to understand the equation that you have quoted, simply think of the left hand side as the rate of change of momentum of the charged particle.

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  • $\begingroup$ Thank you! So, if I understood right -- it is a moveton to think about $\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}$ as a relativistic mass $m_0$? $\endgroup$
    – KeepKolmogorov
    Commented Aug 20, 2021 at 16:23
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    $\begingroup$ @KeepKolmogorov It is outdated and bad practice to use the concept of relativistic mass. There is only one quantity of mass which is the "rest mass". In other words, mass is just mass. It's just that in relativity the momentum is $\gamma mv$ instead of $mv$. Small particles (such as ions, protons and electrons) are so light that they easily approach a significant fraction of light speed in a typical electric field, so the relativistic formula must be used. $\endgroup$
    – Vincent Thacker
    Commented Aug 20, 2021 at 16:34
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    $\begingroup$ I've learnt a new word: 'moveton'! Yes, 'relativistic mass' is an unhelpful and unnecessary and potentially misleading term, and the factor $\frac{m}{\sqrt{(1-v^2/c^2)}}$, is of no help in understanding relativistic momentum. $\endgroup$
    – Philip Wood
    Commented Aug 20, 2021 at 16:55
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    $\begingroup$ At least it was somehow useful for you too! (I think we use it in Russian widely because of the French influence in 18-19 centuries - originally "mauvais ton" - fr. means bad tone) $\endgroup$
    – KeepKolmogorov
    Commented Aug 20, 2021 at 17:44
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Using the relativistic momentum expression makes sense when you realize that it's quite easy, using EM fields, to accelerate fundamental particles (e.g. electrons) to relativistic velocities.

As an example, in a CT (Xray) tube, you might accelerate electrons through 140 kV; given that their rest mass is 511 keV, it's obvious that 140 keV is "substantial", and the change in mass can no longer be ignored.

Incidentally, this is why, in particle physics experiments, the "velocity" of a particle is often given in keV / MeV / GeV - there comes a point where their actual velocity is very close to the speed of light, and only their mass / energy is changing. So that becomes a sensible metric.

And it all starts with the equation that Feynman gave.

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  • $\begingroup$ thank you for the useful answer! I don't have enough rep but you will obtain your vote as soon as I have it! $\endgroup$
    – KeepKolmogorov
    Commented Aug 20, 2021 at 16:40

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