If a body becomes charged by losing electrons for example, it will experience a braking force when it is accelerated due to radiation called Bremsstrahlung radiation. Part of the energy used to accelerate the charged body will be emitted as radiation. It should therefore take more energy to accelerate the charged mass than the energy required to accelerate the body without the charge.

The Larmor formula calculates the non-relativistic power radiated by the acceleration. There is also a relativistic derivation.

The charged body will appear to be more massive due to this effect. It can be explained by the fact that part of the energy is going to kinetic energy and part to radiation. Nevertheless, F=ma should imply a higher mass so are we dealing with a more massive body?

  • $\begingroup$ The self energy for elementary particles is a slightly different issue. The mass of the electron and the contribution from the electric field for example is still not fully understood. In the case of a charged body that loses electrons, if you put it on a scale (balance or spring) it will measure the same force of gravity as the uncharged body. It will however exhibit more resistance to acceleration. It will also fall slower in a gravitational filed than an uncharged body. $\endgroup$ – Peter R Nov 24 '15 at 15:26
  • $\begingroup$ That is exactly what is happening, $\endgroup$ – Peter R Nov 24 '15 at 16:34
  • $\begingroup$ if you remove an electron from an uncharged body, has it's mass increased? If you put it on a scale, will it show more mass? $\endgroup$ – Peter R Nov 24 '15 at 17:01
  • $\begingroup$ Perhaps you should rephrase your question in those specific terms $\endgroup$ – docscience Nov 24 '15 at 20:24
  • $\begingroup$ While taking the electron away will reduce the mass of the atom, the atom will be of positive charge, but I don't believe that would increase the ion's mass, or would it since it's energy is potentially increased. $\endgroup$ – docscience Nov 24 '15 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.