Suppose I have a proton at the origin, and a (stationary) electron on the x-axis at x=r.(adsbygoogle = window.adsbygoogle || []).push({});

Taking q to be the elementary charge, the force acting on each of these particles is:

[tex]F = \frac{q^2}{r^2}[/tex]

Now suppose that the electron is heading towards the proton (in the negative x-direction) at a relativistic speed. Suppose gamma=2. The field generated by the electron will be "squished" in the x-direction, and correspondingly the force experienced by the proton will be weaker than it would have been had the electron been stationary.

The electric field generated by a moving charge (as seen in the "stationary" frame) is given by:

[tex]E = \frac{Q}{r^2}\frac{1-\beta^2}{1-\beta^2\sin^2\theta}[/tex]

where Q is the charge on the moving particle and theta is the angle between the origin of the moving particles coordinate frame and the direction of the particle's motion (in this case, the x-axis).

In our case, we have:

[tex]E = \frac{1}{4}\frac{q}{r^2}[/tex]

and so the force the proton experiences, due to the moving electron, is given by:

[tex]F = \frac{1}{4}\frac{q^2}{r^2}[/tex]

Meanwhile, the force on the (moving) electron, due to the stationary proton, is given by the Lorentz force law:

[tex]F = qE + \frac{1}{c}v \times B[/tex]

Since the proton doesn't generate a magnetic field, the force on the (moving) electron due to the stationary proton, at the moment the electron is passing the point x=r, is:

[tex]F = \frac{q^2}{r^2}[/tex]

In other words, the force the proton imparts to the electron is 4x the force that the electron imparts to the proton.

What happened to Newton's Third Law?

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# Newton's Third Law and Relativity

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