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The Lorentz theory of radiation reaction in the non relativistic limit is critically examined from the principles of symmetry. In Newtonian motion the applied force and the consequent momentum change generally always have a direction. In the case of a charged body this momentum change is accompanied by the radiation of photons. The question that needs to be answered is: does this radiated field carry any mo mentum? The following result of investigations seems to indicate this fact contrary to the prediction of Lorentz theory which states that the field carries no net momentum.

Electromagnetic radiation reaction entails both the conservation of energy and momentum. When a charged particle moves in a straight line under the action of an external force directed along its line of motion it is expected to radiate electromagnetic energy in the shape of photons. Now these photons also carry away with them momentum and it cannot be assumed that the net momentum carried out by the electromagnetic field will be zero purely from considerations of symmetry for this highly asymmetrical problem. Suppose that the line of accelerated motion of the charged particle is along the z axis then from symmetry consideration one can say that there will be no net momentum carried out by the photons which will have x and y components but certainly that will not be true for the z component of momentum. In fact we have tried to show in our earlier publication [

The following equation for the acceleration a ( T ) of a charged particle as a function of time T was derived by us in [

a ( T ) = f 0 m [ e T / τ 0 ( 1 − u ( T ) ) + u ( T ) ] − f 0 m [ e ( T − t ) / τ 0 ( 1 − u ( T − t ) ) + u ( T − t ) ] (1)

where u ( T ) is the unit step function. The equation of motion m a = f e x t + m τ 0 d a d t can be written in the integrated form as

m v ( T ) | − ∞ t = ∫ 0 t f e x t d T + m τ 0 a ( T ) | − ∞ t , (2)

where v ( T ) = ∫ a ( T ) d T . The net momentum carried out by the radiated photons is given by the term m τ 0 a ( T ) | − ∞ t and it is easily seen to be zero as both a ( t ) = 0 and a ( − ∞ ) = 0 . This means that the photons carry +z and -z components of momentum in equal amounts although the particle gains +z component of momentum only in the case of motion along the positive z direction. In our opinion this is incorrect from the considerations of the symmetry of the problem itself. Even in an inertial frame of reference in which the charged particle is momentarily at rest there is a direction in which it is accelerated and so some z directed momentum is transferred to the particle by the force which is acting on it and hence is it not to be accepted that the radiated electromagnetic field will be imparted with a z directed net momentum too? We think that the answer to this question will be in the affirmative but then there will be a distinction between the natures of the momentum transfer by the force to the charged particle on one hand and by the force to the electromagnetic field on the other.

The loss rate of momentum P L , was worked out by us for a charged particle having a velocity v and acceleration a in the z direction and this is expressed as (see Equation (5) of [

P L = u ^ z q 2 a 2 v 5 π c 5 ε 0 (3)

where u ^ z is the unit vector in the z direction; q is the charge carried by the particle. In addition we have the fundamental constants of nature c the velocity of light and ε 0 the permittivity of free space. We had stated that this is consistent with the energy loss rate in a certain frame of reference which for the moment we just call a privileged inertial frame. A return to the discussion of the nature of this frame will be made shortly. For the moment we would like to remind the reader that the laws of Newtonian mechanics are unchanged under a Galilean transformation that is the momentum and energy conservation are both observed in different inertial frames. The momentum conservation follows directly from m d v d t = f e x t in some inertial frame which becomes in another inertial frame having relative velocity V with respect to the first frame m d ( v − V ) d t = f e x t after the transformation is applied. The two are the same equations of motion as d V d t = 0 . Similarly for energy we have the two equations m d v d t ⋅ v = f e x t ⋅ v and m d ( v − V ) d t ⋅ ( v − V ) = f e x t ⋅ ( v − V ) which are also consistent with each other. The situation changes when dissipation is present for example in the case of a viscous drag in a fluid and this was mentioned by the author earlier [

We would like to thank our institute for the financial support towards the payment of publication charges.

The author declares no conflicts of interest regarding the publication of this paper.

Roy, R. (2019) A Note on Momentum Conservation in the Non-Relativistic Lorentz Theory of Radiation Reaction. Open Access Library Journal, 6: e5534. https://doi.org/10.4236/oalib.1105534