When F’ = F = 0

G.R.Dixon, May 18, 2007

It is generally accepted that Newton’s laws are valid in all cases and in all inertial frames, provided the dependence of inertial mass on speed is taken into account. That is, if a mass’ speed is u, and to the extent

, (1)

then the following (Newton’s 2nd law) is generally valid:

. (2)

Since the Lorentz transformation provides formulas for transforming u into u’, etc., the transformation of F to F’ is essentially an exercise in algebra. For example, assuming inertial frame K’ moves in the positive x direction of K with speed v, the full force transformations are:

, (3a)

, (3b)

. (3c)

A special case is when the particle being acted upon has no net force acting on it. According to Newton’s 1st law, in such a case the particle moves with a constant velocity (including zero velocity) in an inertial frame. And according to the Lorentz transformation, any particle moving with a constant velocity in one inertial frame moves with constant velocities in all inertial frames. In brief, If F=0 in one inertial frame, then F’=0 in every inertial frame.

A celebrated example is depicted in Fig. 1. A test charge is at rest beneath an uncharged line current. The current consists of a resting negative line charge, of line charge density l-, and a superimposed positive line charge, of density l+ = -l- and moving to the left at speed v.

Figure 1

Test Charge at Rest Beneath Current

Since E=0 everywhere and since the test charge is at rest, it experiences no Lorentz force: F=0. The uncharged line current is just I=l+v. Thus the magnetic field at the test charge has magnitude

. (5)

Let us now view things from frame K’, moving in the positive x direction of K at speed v. The force transformation requires that F’=0 also. But now q moves to the left at speed v, and B’ is not zero. In particular, the general field transformations specify that

. (6)

Evidently the test charge experiences an upward magnetic force of magnitude qvBz’ in K’. This must be exactly balanced by a downward electric force if F’total is to equal zero. And indeed according to the general field transformations

. (7)

(The nonzero Ey’ in K’ owes to the fact that l’ = l+’ + l-’ is not zero in K’.) Note that this relativistic effect is manifest at all values of v, in particular values of v<<c.

In general the Lorentz force, F=q(E + v x B), always transforms in accordance with Eqs. 3a through c … a requirement for any force that can be equated to d(mv)/dt.

Let us, for example, consider Fig. 2. In frame K a positive test charge, q, is at rest beneath an unbalanced negative line charge, l-. At q’s location the line charge’s E points upward and has magnitude l-/2pLeo.

Figure 2

Test Charge at Rest Beneath Negative Line Charge

Since q is at rest, it must be that a downward pointing force cancels out the upward pointing electric force, so that Ftotal = 0. Note in the figure that the downward pointing force is provided by a "black box" … it could be any of a number of different agents.

Now in frame K’, according to the field transformations, the Lorentz force on q is

. (8)

Thus F’Lorentz is again quite as the general force transformation requires. Since F’total = Ftotal = 0, the force exerted by the black box must also transform like Felec does, regardless of the physical source of that force.

The reader can substitute his/her own pet forces for the black box in Fig. 2. Obviously if the black box is another negative line charge, at distance L beneath the test charge, then both F’total and Ftotal will equal zero. Other possibilities for the black fox are a spring, magnetism (assuming a magnet is mechanically linked to q), gravity (assuming a neutral mass is mechanically linked to q), the force exerted by a piston with a partial vacuum beneath it, etc.