0
$\begingroup$

Consider a system of charges interacting with one another via their EM fields together with external EM fields applied to the system from outside. Suppose we work out the equations of motion separately for two cases:

  • charges interacting upon one another with the external field turned off
  • external field acting upon the charges with their mutual interaction upon one another turned off

Does the superposition principle allow us to add these two equations of motion to one another to get a valid equation of motion for the total system?

Maxwell's equations are linear and the superposition principle applies, but I still have my doubts that it would work in the above case for reasons I'm ignorant of.

$\endgroup$
1
$\begingroup$

The scheme you purpose is generally not valid. To see why, I shall simplify your case to a example in which an external temporally constant electric field is made up from two sources. As per your reasoning, since the superposition theorem holds for the fields (which it does), the motion of the particle should be the sum of the particles motion under each separate field (incorrect). Note that the correct equation of motion for this case, by the Lorentz law, is:$$\vec F = q\vec E(\vec r(t))$$in which:$$\vec E(\vec r(t)) = {{\vec E}_1}(\vec r(t)) + {{\vec E}_2}(\vec r(t))$$whence by this notation, we mean that while the electric fields are not a function of time in themselves, they would obtain such implicit functionality as per the particles position being a function of time. Your proposition is that solving:$$m\frac{{{d^2}}}{{d{t^2}}}\vec r(t) = q({{\vec E}_1}(\vec r(t)) + {{\vec E}_2}(\vec r(t)))$$could be converted to solving:$$\left\{ \begin{array}{l}m\frac{{{d^2}}}{{d{t^2}}}{{\vec r}_1}(t) = q{{\vec E}_1}({{\vec r}_1}(t))\\m\frac{{{d^2}}}{{d{t^2}}}{{\vec r}_2}(t) = q{{\vec E}_2}({{\vec r}_2}(t))\end{array} \right.$$and then placing:$$\vec r(t) = {{\vec r}_1}(t) + {{\vec r}_2}(t)$$The false assumption in this reasoning should be clear by now, specifically since the electric field is not generally a linear function of the position vector, one may not combine the right hand of the latter system of equations to obtain the original equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.