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We know that relativistic mass of a object is given by: $$m=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}.$$

So the mass of an object will become very large if it travels with a speed near to the speed of light.

My question is that electricity travels with a speed which is near to the speed of light and the electricty flows in a conductor due to the flow of electrons. So the speed of electrons in the case will also be near to the speed of light so the mass of electrons should become very large because: $$m=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}.$$

The value of $\frac{v^2}{c^2}$ will be near about $1$ (since $v^2$ is nearly equal to $c^2$).

So $\sqrt{1-\frac{v^2}{c^2}}$ will be some about close to 0.

Therefore the relativistic mass of electron at that speed should also become very large but that never happens.

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  • $\begingroup$ Also in this question brainly.in/question/163300 how do I apply the law of conservation of mass as electron is travelling at a speed close to that of light so its mass will change. $\endgroup$
    – The Mathemagician
    Commented Jun 10, 2018 at 5:42
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    $\begingroup$ Why do you think this does not happen? Particle accelerators are a good experiment to show that.momentum behaves like.you described. $\endgroup$
    – lalala
    Commented Jun 10, 2018 at 6:21
  • $\begingroup$ @lalala How can this happen if it happens then the mass of electron becomes very large so it will attract earth and other objects towards it due to gravitational force. $\endgroup$
    – The Mathemagician
    Commented Jun 10, 2018 at 6:24
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    $\begingroup$ It's best to avoid relativistic mass - it can be misleading. Modern treatments of relativity just use rest mass. Anything that's legitimate to do with relativistic mass can be done instead with relativistic momentum and kinetic energy formulas. $\endgroup$
    – PM 2Ring
    Commented Jun 10, 2018 at 6:56
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    $\begingroup$ In general relativity, the gravitational mass is still being the rest mass. When we using four-vectors, we can abandon the term of "relativistic mass". See another answer here. $\endgroup$
    – Ng Chung Tak
    Commented Jun 10, 2018 at 7:17

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So the speed of electrons in the case will also be near to the speed of light

At the Wikipedia article Drift velocity, find a numeric example for the rather ordinary case of $1\,\mathrm{A}$ current through a copper wire of $1\,\mathrm{mm}$ radius. The calculated value is $2.3\times10^{-5}\,\frac{m}{s}\lll c$

For what value of current would the drift velocity be say, a tenth of the speed of light?

$$I = \frac{c}{10}nAe = (3\times10^7)(8.5\times10^{28})\cdot(\pi\cdot 0.001^2)\cdot(1.6\times10^{-19}) = 1.28\times10^{12}\,\mathrm{A}$$

So, for any reasonable value of current, the drift velocity of the mobile electrons is in fact much, much smaller than $c$.

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The electrons don't move with the speed of light, rather it 'drifts'. And there ain't one electron that it'll gain mass and become infinite. Rather, there are millions and millions of electrons that collide with each other and the average velocity is taken between the free electrons.

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The actual velocity is negligible compared to that of the Electric field and/or of light in the medium, and can actually be made to almost vanish on average. The extra Electric field flux and or Electric energy flux due to the motion and interactions is the point.

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Although the electrons move slowly, a perturbation of the electric field moves at a sizeable fraction of the speed of light. An analogy is when you push a brick with a shock on one edge. The delay with which the opposite edge starts moving equals the distance between the edges divided by the sound speed in the rock. So the perturbation travels at a speed of the order of a 1000 m/s, while the whole rock moves much slower.

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