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  Does more electromagnetic mass mean that more electric field is emitted?

+ 0 like - 4 dislike

 Does a moving charge give more electric field since it has more electromagnetic mass ?

Let I have a moving charge with speed 0.9c. kinetic energy hence gained is $m0.81 c^2$.
Total energy of electron has increased to $1.81mc^2$

Now, let I am moving with 0.9 c. I would see charge at rest. 
But it has $1.81mc^2$ energy. Does this increased energy emit more electric field?

[ I don't know how to use these equations.][2]

  [2]: https://en.m.wikisource.org/wiki/On_the_Electromagnetic_Mass_of_a_Moving_Electron

Closed as per community consensus as the post is not graduate-level
asked May 24, 2016 in Closed Questions by anubhav (-30 points) [ revision history ]
recategorized Jun 3, 2016 by Dilaton

This is not graduate-level and rather unclear, voting to close.

Voting to close.

1 Answer

+ 1 like - 0 dislike

The total energy $E=mc^2/\sqrt{1-v^2/c^2}$ contains the rest energy $mc^2$ and a kinetic part $T(v)=E(v)-mc^2$. In you reference frame there is no kinetic part. (Your formulas are wrong.) And the relativistic formula for $E$ is valid for neutral particles too.

The radiated energy depends on the particle charge, its velocity and acceleration. The notion of electromagnetic mass is misleading. There is a mass defect, which is real for compound "particles", but the rest energy is not of purely electromagnetic nature. Annihilation of electron and positron may occur to a couple of neutral particles other than photons, to a couple of neutrino and antineutrino, for example, i.e., via weak interaction.

Apart from the radiated field, there is a "near" field following the particle. The corresponding formulas for this field for a uniformly moving charge are given in textbooks.

answered May 24, 2016 by Vladimir Kalitvianski (102 points) [ revision history ]
edited May 25, 2016 by Vladimir Kalitvianski

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