# Why particles don't lose mass when they radiate

Charged particles radiate when accelerated: in the rest frame of the particle moving with acceleration $\textbf{a}$ the amount $dW$ of radiated energy over time $dt$ is $$dW \propto \textbf{a}^2 dt$$ so the particle loses energy $dE = - dW$ and does not lose momentum $d\textbf{P} = \textbf{0}$. So it seems like it should also lose mass $dm = dE/c^2 = -dW/c^2$.

At first I thought that the external field which produces the acceleration $\textbf{a}$ does work on the particle and the energy $dW$ of the emitted radiation is actually absorbed from the field. But then I realised that work done by any force $\textbf{F}$ is $(\textbf{vF})$ and so it should be zero in the rest frame of the particle.

So now I'm confused.

• Your premise that the accelerated particle has a constant momentum is false simply because it is accelerated. Further, the Bremsstrahlung you describe is a conversion of kinetic energy into radiation.
– user73762
Apr 5, 2015 at 19:23
• @pyramids It has zero momentum in the associated rest frame. so the change in the mass is $m dm = E dE - 0 = m dE \neq 0$
– xaxa
Apr 5, 2015 at 21:43
• You're not allowed to assign a "rest frame" to an accelerated particle. Or, you can, but you have to be very careful how you jump from that frame to the external "universal' frame. Apr 5, 2015 at 21:48
• @xaxa Your rest frame changes. Hence what may be conserved in one (constant, unaccelerated) frame of reference is not necessarily conserved in the changing rest frame you adopt.
– user73762
Apr 5, 2015 at 21:49
• @CarlWitthoft I'm not assigning a single rest frame for the whole motion of the charge. This is just a frame that "happened" to be at rest relative to the charge at the present moment. In this frame the charges seem to be losing energy.
– xaxa
Apr 5, 2015 at 21:50

## 3 Answers

I'd like to explore the question in two parts: first considering only fundamental particles and secondly considering composite particles and other system.

For fundamental particles their mass is a part of their basic identity. It is not variable, so the reaction $$X \rightarrow X + \gamma \,, \tag{*}$$ (where $X$ is any fundamental particle) in free space and without outside interactions would violate energy conservation.

That really only leaves systems composed of more than one particle. An excited nucleus can participate in a reaction like (*) though if we're being careful with our notation we ought to have a mark for excitation which will be lost or changed on the right-hand side.

When you are talking about Bremsstrahlung, you are necessarily invoking a compound system composed of the particle and the field which causes it to accelerate. That system can lose energy without altering the particle's mass, which is what is actually happening.

• I'm OK with composite particles losing their mass in radiation due to the change in internal structure. But could you give more details on how the energy loss is compensated in case of truly elementary particles like electron? Please see my comment on why I can't figure out how it works physics.stackexchange.com/questions/174390/… Thanks!
– xaxa
Apr 6, 2015 at 6:33

To start with a particle loses kinetic energy, and therefore momentum, when radiating electromagnetic energy in some electric field, it is the basic reason why the planetary model of an atom cannot work.

Brehmstrahlung, "braking radiation" or "deceleration radiation") is electromagnetic radiation produced by the deceleration of a charged particle when deflected by another charged particle, typically an electron by an atomic nucleus. The moving particle loses kinetic energy, which is converted into a photon, thus satisfying the law of conservation of energy.

But could you give more details on how the energy loss is compensated in case of truly elementary particles like electron?

A free electron does not accelerate or decelerate. A field is needed for any interaction to happen and this is supplied by the spill over field of the electrons around the nucleus of an atom. Or there could be a macroscopic magnetic field and when the electron interacts with it, it will be radiating away some of its kinetic energy.

Kinetic energy A of electron +field energy (before) kinetic energy of electron B plus photn energy C : A=B+C

Sitting at the rest system of the electron it will be the field that will be changing and giving off electromagnetic radiation, i.e. the atom will be accelerating as far as the electron knows. More complicated equations, but consistent with energy conservation.

• "when radiating electromagnetic energy in some electric field" - I think this is incorrect. A particle will radiate EM waves regardless of the source of its acceleration (be it external EM field or smth else). "at the rest system of the electron it will be the field that will be ... giving off ... radiation" - yes, but won't the field in turn produce a recoil of electron? This is fine, the problem is that via Larmor's formula, the total loss of momentum of electron is zero in the rest frame while total loss of energy is not. This is what I don't understand. Does electron lose mass?
– xaxa
Jul 26, 2015 at 15:47
• @xaxa The energy comes from the potential energy that the field supplies when an electron is accelerating or decelerating in it. When the electron is at rest because one has transformed to the rest system of the electron, the electromagnetic radiation is coming from the changes in the field which is also transformed to the rest system of the electron. For another pov look at this philpapers.org/rec/HARRFA Jul 26, 2015 at 19:23
• Can the charge change because of radiation? Apr 29, 2022 at 19:55
• @Filippo no electromagnetic radiation is neutral, photons have no charge. Apr 30, 2022 at 3:47

in the rest frame of the particle moving with acceleration $\textbf{a}$ the amount $dW$ of radiated energy over time $dt$ is $$dW \propto \textbf{a}^2 dt$$ so the particle loses energy $dE = - dW$

This is not entirely correct if taken literally and taking it as such leads to paradox - if the particle is at rest and its mass does not change, how could it possibly lose energy?

The resolution is in understanding what the "energy per unit time" in the Larmor formula actually means. It has a technical meaning, which I'll explain below. But first, it is important to realize that the Larmor formula was derived from the work-energy interpretation of the Poynting theorem. This interpretation is valid only when the expression $\mathbf j \cdot \mathbf E$ is integrable at the locations of the particles which it is if all charged particles have charge distribution that is not too singular in space. It does not apply to point particles, where the above product has no sense. Consequently, we cannot apply the Larmor formula to point particles.

Extended particles consist of charged parts that repel each other, so the whole particle has some internal electrostatic energy. When such particle is accelerated, its parts may move with respect to each other (how much, depends on the internal forces keeping them together) and their electrostatic energy can change.

We have no evidence of non-zero size or internal energy change of electrons and this is consistent with the idea that they are points that have no parts. If so, the Larmor formula does not apply to them. If they have some small dimensions and their charge is distributed in a non-singular way, the Larmor formula may apply. But then the particle has internal degrees of freedom and its internal energy can change.

Now let us consider your scenario, assuming the particle has some small but finite dimensions so Larmor formula is valid. The particle is being accelerated by external force, and we look on it in its instantaneous comoving frame. The energy radiated per unit time at time $t'$ is defined roughly this way: we consider how much EM energy will fly through a distant spherical surface at time $t'$, the sphere being centered at the particle position; according to Larmor, this energy is given by the particle acceleration at a previous time $t'-R/c$, where $R$ is radius of the sphere. This radiated energy at time $t'$ has no direct implication for energy change of the particle itself at time $t'-R/c$; only to energy change of the big sphere containing the particle. So, while energy inside the big sphere is decreasing with rate $ka^2$' at the time $t'$, energy of the particle inside may be either increasing, decreasing or staying the same at the time $t'-R/c$. It depends on the details of interaction between the parts of the particle and the external forces accelerating it at time $t'-R/c$.

The sphere has to be big enough so that the EM field on its surface can be accurately approximated by the "wave part" of the field of the particle only. If we use too small a sphere, the surface is too close, the field is more complicated: the external field accelerating the particle has its say, the field of the particle itself is much more complicated that the "wave-part" and integrating the Poynting vector over the sphere surface would be much more difficult and not result in such a simple expression as Larmor formula has. The value of energy loss per time would not be simply function of acceleration squared.

If the accelerated motion is periodic and stable, the energy of the particle will oscillate and will have constant average value. This is what happens to particles in a radiating antenna: they radiate energy but this energy is constantly being resupplied by the power source the antenna is fed with. In a steady operation, the charged particles do not lose their internal energy over time, but just shape the energy flow from localized linear flow near the power lines to more angularly spread pattern of radiation propagating out from the metal bodies to surrounding space.