But you have no real motion, if you are looking at just one point t = 0, you may have a differential velocity but this is a limes against 0.

So, the difference between your forces is infinitely small.

Yes, in this situation the law of action and reaction is violated. Ok I know, it's shocking, Newton was wrong, they didn't warn you, but don't worry, you'll get over it. It is possible to save the law of conservation of momentum by adding the field momentum as you say. The simplest way to handle it is to write the EM field in terms of the electric potential phi and the magnetic vector potential A. They form a 4-vector whose value in the two frames is related by the Lorentz transformation. The field momentum is equal to qA. So it turns out that we need relativity to keep electrodynamics consistent! You might also take a look at Feynman II lectures 25,26 and 27 and the example concerning angular momentum in 17.4. 

@Johannes,

The problem is not restricted to a single point but this imbalance of perpendicular component of E extends to a finite intervall delta-x. Consider a circle and an elipse that overlap to envisage the difference between electric fields of a static charge and moving charge. The problem is between the intersection points of both curves.

Adam

Unlike the equation F=ma which is valid only in nonrelativistic limit, the validity of the momentum conservation law is not restricted to Newtonian mechanics but it is related to isotropy of space and it is valid without restriction. See Landau Lipschitz classical mechanics. What is different in relativity in this respect is not that momentum conservation ceases to be valid but it is only that the definition of momentum changes. The chapters you referred in Feynman lectures in electromagnetism give basic general information on the subject. In the thought experiment 17-4, the direction of pointing vector ExB namely the circular flow of energy within the field resolves the paradox but this doesn't help me either with my problem. The direction of ExB in my case is parallel to the velocity of the charge. Thus this doesn't explain the momentum imbalance in direction perpendicular to v. This is why I explicitely added the explanation "considering the arrangement of the charges the momentum of the field doesn't solve the problem either". I added this so that someone who wants to contribute, should first think on the problem instead of simply giving references of chapters on field energy/momentum in general text books in classical electrodynamics.

I know that I must be overlooking something. I just wanted to know whether it has a verbal explanation through field momentum as in the case of resolution of the paradox 17-4 or whether one needs explicite relativistic calculations to resolve it.

Yes, the force on Q1 is larger than the force on Q2 by a factor gamma (the Lorentz factor). But it acts during a shorter time, since the field of Q2 is Lorentz-contracted by the same factor gamma in reference frame S. The net transfer of momentum, which is the time integral of the force, is the same (in magnitude) for both particles.

The temporary imbalance between the two forces is taken by the field. The field momentum density is DxB (see, e.g. eq.(16) of arXiv:physics/0208007). This vector differs from the Poynting vector ExH (energy flux) in matter, but they are equal in vacuum: DxB = ExH = ExB = P.  The field momentum is the space integral of P.

P contains the interference term E{from Q1} x B{from Q2}. Its y-component is                                P_y = E_x{from Q1} times B_z{from Q2}.

Then the time-derivative of the field momentum along y is 

         U = - integral of  dx dy dz E_x{from Q1}(x,y,z)  d B_z{from Q2}(x,y,z) /dt  

Let us take q1 and q2 poisitive. To guess the sign of U at t=0 we divide the integration in 4 regions:

a) region y0 . 

E_x{from Q1} >0 , B_z{from Q2} 0

d) region y>b, x

Thank you for detailed answer Xavier. 

My mistake was considering only the contribution of P on the plane x= 0 which has only  x component and no y component. Of course we must take the space integral.

Again thank you very much.  

Atilla,

Sorry if my previous answer appeared patronizing.  Take a look at the Wikipedia article on the Lienard-Wiechert potential. In the situation you define, beta.n=0 and dbeta/dt =0, wjhere beta=v/c, so the expressions simply a lot. You also need the relativistic expression for the force F=gamma(E+v cross B).

Consider the problem entirely in frame S. Q1 is not moving, so it feels no force from the magnetic field of Q2. It experiences an electric force Q1Q2/(4*pi*eps0*gamma d^2) in the Y direction and  Q1Q2/(4*pi*eps0*gamma d^2) * v/c in the X direction.  Q2 also feels no magnetic force in S, because Q1 is not moving and does not generate a magnetic field. Q2 does feel an electric force Q1Q2/(4*pi*eps0 d^2 in the Y direction, larger by a factor gamma than that felt by Q1, but having the same effect because Q2's mass is dilated by a factor gamma, so the Y forces are in this relativistic sense "equal". However, there is no X force on Q2, this is the part which is balanced by the change in the field momentum  Q1dA/dt.

Adam 

Sorry, Adam, I do not quite understand your conclusion.

I find that in frame S (Q1 at rest) the transverse acceleration of Q1 is

a1 = gamma * Q1Q2 / (4*pi * d^2 * M1) ;

the transverse acceleration of Q2 is

a2 = Q1Q2 / (4*pi * d^2 * gamma * M2)

So a2 i s weaker than a1 by the factor gamma^2.

In S' (Q2 at rest at t=0=t'), the transverse acceleration a'2 of Q2 can be calculated from a2 :

a'2 = d^2 y2 / dt'^2 = (d^2 y2 / dt^2) * [dt/dt'(y'=0)]^2 = a'2 * gamma^2 = a1, 

where I have used [dt/dt'(y'=0)]  = gamma.

Xavier

If I stick to the very clear Attila's notation

You were right to say : it turns out that we need relativity to keep electrodynamics consistent!"

Adam,

1. The force q1 experiences due to the electric field of q2 is by factor gamma larger then the force  q2 experiences due to electric field of q1.Thus if your explanation with relativisitic mass increase were correct it would make the problem even worse.  Greater force acts upon the body with less mass namely q1.

2. The transversal mass is not effected by the motion, because an infinitesimal change of direction does not have any effect on the magnitude of the velocity. 

This is why the term "relativistic mass increase by factor gamma" is somewhat misleading.  If we continue to define relativistic mass through F=ma (this what we do implicitely if we use a term like mass increase) then mass becomes a tensor. Thus the rate of increase depends on the angle between acting force and velocity.  Direction of force and acceleration may differ in general.

By the way this is also the simplest way to understand why there is no rotation in Trouton Noble experiment if one considers the problem in the reference frame where the capacitor is in motion.

Adam, 

Let me add the following last remark:

Even if your suggestion  about mass increase compensating force increase leading to equal and opposite acceleration were correct,  this would only ensure the conservation of total transverse velocity but not conservation of transversal momentum since the masses would be different. 

Gentlemen. I'm afraid that much of what I said in my previous posts on this subject was wrong. Thinking about it over the weekend, at last, now I understand. The key point is that the Lienard-Wiechert potentials are retarded i.e. they are not calculated using the position of the source now, but its position when it emitted the EM field that is being measured now. For source  Q1, it makes no difference, because it is stationary (in frame S). For source Q2, it is simple to show that the field felt now by Q1, was emitted by Q2 at time d/ sqrt (c^2-v^2) , when Q2 was at vector position  (-gamma d v/c, d).

Using the formulae in the Wikipedia article, the unit vector from the source                   n= (v/c ,-1/gamma), and 1-n.beta s = 1/gamma ^2. Substitute in to the formula given,   E(r,t) = -Q2 gamma/ (4 pi eps d^2), giving a force exerted by Q2 on Q1 *exactly equal and opposite* to the force exerted by Q1 on Q2. (Note that  here   F = Q1E simply, because Q1 is stationary).

The reason why Sir Isaac is right, in this particular case, is that dA/dt =0: all the electric field is due to Del phi, so there is no field momentum, and mass dilation is only involved in the sense that the instantaneous acceleration of Q2 is lower by a factor gamma than that of Q1, when measured in S, contrary to what I said before!

All this applies to objects with charge, but no intrinsic magnetic moment. For things like electrons, which have  one, things are different.

Thanks for a very interesting discussion (and apologies for my earlier nonsense!)

Adam

Adam, 

1.There is no need to apologize for mistakes one makes while thinking, mistakes are natural part of thinking process. That I asked the question at all was possible only because I couldn't see things in all clarity. So this is normal. 

2.Regarding your explanation, I must think on Xavier's solution and your solution to the problem before I can decide. As far as I know the uniform motion is a unique case where one does not need to consider the retardation because the whole field moving symetrically together with the charge is the exact solution of Maxwell equations for uniform motion. However I must first think myself. Anyway thanks for the contribution. 

Adam,

   Atilla's problem is much more interesting that I first tought, because we still not agree !

 - You write  " E(r,t) = -Q2 gamma/ (4 pi eps d^2)  " - I agree

Then " giving a force exerted by Q2 on Q1 *exactly equal and opposite* to the force exerted by Q1 on Q2. " - I disagree, because of the the gamma factor above.

 - You write " there is no field momentum " - I disagree (see my first contribution)

- You write "the instantaneous acceleration of Q2 is lower by a factor gamma than that of Q1, when measured in S " - I disagree; I found a factor gamma^2 instead of gamma (see my second contribution).

This gamma^2 factor is in accordance with Lorentz time dilation :  In frame S, let us calculate the time Delta t1 needed for Q1 to make a small displacement                 Delta y

Xavier, I don't understand what you mean. In frame S, since q2 is moving, it experiences a force F = gamma (E+v X B).The electric field on q2 due to q1, measured in frame S, is  q1q2/(4pi eps 0 d^2),  because q1 is stationary, there is no retardation. There is also no magnetic field from q1. So the force on q2 due to q1, measured in frame S, is  trivially q1q2gamma/(4pi eps 0 d^2) .

Adam, why do you put a gamma factor in your expression for F in the second line of your answer ?

Atilla,

I think I treat exactly this problem in Ch 9 of my book, available in full-text here:

Chapter Motional Consequences - A new approach to Special Relativity

Section 9.4.3

Greetings,

Kjell