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Lab Book_2014_05_10 Special Relativity, Rotating Frames, and Quantum Mechanics

Lab Book 2014_05_10     Hamilton Carter

A seemingly simple question about why magnetic forces act at right angles to their associated field lines led me to derive that the transverse forces on a charged particle moving in a circular path have no gamma terms associated with special relativity.  This seems to tie nicely into why the quantum mechanical number operator predicts no spectrum of Fulling-Unruh radiation from a particle moving in a circular path, but a Fourier decomposition of the wave solution does as shown by Letaw and Pfautsch.  L and P left the lack of spectrum predicted by the number operator as an open question.

Someone asked an interesting question on stackexchange regarding why the Lorentz force from a magnetic field acts at right angles to the direction of the magnetic field.  The simple offhanded answer is that in the fame of the moving particle, the magnetic field transforms into an electric field that is parallel to the direction of the produced force.  Deriving this at length though brought up another interesting point related to the theoretical work that I’m doing.  A byproduct of the derivation is that it points out that there are no special relativistic effects due to forces acting at right angles to the direction of the particle’s moving frame.  Let me say that much more precisely.  If my particle is moving in a circular path, it can be shown that the centripetal force that keeps it in the circular path produces no additional special relativistic effects.
OK, so what does this have to do with my theory research?  I’m looking at one aspect of a problem posed by Letaw and Pfautsch in 1980.  They showed that the spectrum of particles associated with Fulling-Unruh radiation due to circular motion is different than what the quantum mechanical number operator N predicts.  The number operator predicts that ,unlike the case of tangential acceleration which produces a thermal spectrum of particles, in the case of perpendicular acceleration, (circular motion), there should be no spectrum of particles at all.  In their paper, this is left as an unaddressed oddity.
What I figured out today is outlined below.  In all fairness, it might only be a semantic rewording of something that was obvious, but it also might not be.  Here are the points.
1.  Quantum mechanics from the point of view of matter-waves, as developed by DeBroglie, was constructed on top of the special relativistic four momentum vector.
2.  The number operator is constructed on top of quantum mechanics.  See any quantum book for this derivation.  Nieto and Carruthers present a particularly nice derivation for my purposes.
3.  Special relativity produces null results for transverse accelerations, as in circular motion.
4.  It stands to reason that if special relativity doesn’t ‘know about transverse accelerations’, then neither does the number operator, N, constructed on top of it.  Consequently, N never had a chance of producing a non-null spectrum in the problem investigated by Letaw and Pfautsch.

The EM transformation mentioned above follows in pdf.  It utilizes two expressions due to Karapetoff’s oblique angle and hyperbolic treatments of special relativity. 

EM Transformation work in wikiTex
Thanks for asking an interesting question!  One way to think about the answer is using special relativity.  In short, the particle itself sees the magnetic field in, (say the z direction), in our stationary frame transformed into an electric field in the -y direction in its moving frame.  This electric field has the correct magnitude to create a force on the charge equal to the Lorentz force from the magnetic field in our frame.  Just as time and space are actually each part of space-time and transform between each other as the velocity between the moving and laboratory frames change, the electric and magnetic fields are part of the electromagnetic field and transform in a similar manner. 

For a frame, $S^{\prime}$, moving with our charged particle along the x axis with respect to electric, $E$, and magnetic $H$ fields in the laboratory frame, Karapetoff[1][2] gives the following Lorentz transform for the $E'$ and $H'$ fields.

$E_y' = E_y cosh\left(u\right)-H_z sinh\left(u\right)$,

$H_y' = E_y cosh\left(u\right)-H_z sinh\left(u\right)$,

$E_z' = E_z cosh\left(u\right)-H_y sinh\left(u\right)$, and

$H_z' = H_z cosh\left(u\right)-E_y sinh\left(u\right)$,

where $u$ is the rapidity of the frame that moves with the particle. As a matter of reference, u can be expressed as

$v/c = tanh\left(u\right)$,

where v is the speed of the particle in the laboratory frame, but we won't make use of this.

There are a few interesting things to note here.  First, the $E$ and $H$ fields transform via a rotation matrix in a hyperbolic space.  Second, there's an interesting contrast to the Lorentz transform for space-time.  Whereas in space-time only lengths parallel to the direction of motion are changed, when the electromagnetic field is Lorentz transformed, only quantities perpendicular to the direction of motion are changed.

Getting back to the original question, "Why does the magnetic force act at right angles to the magnetic field", let's look at what the moving particle sees in its own frame, S'.  Due to the Lorentz transform shown above, in the $S'$ frame, there is now an electric field in the y direction. 

$E_y' = E_y cosh\left(u\right)-H_z sinh\left(u\right)$,

but  $E_y$ in the laboratory frame is zero so

$E_y' = -H_z sinh\left(u\right)$

There is still an $H$ field in the z direction equal to $H_z cosh\left(u\right)$.  This threw me for a bit, because I was worried about the Lorentz force due to this field.  There is however no need for concern.  Since we're in the frame of the moving charged particle, its relative velocity is zero.  Consequently the magnetic field produces no force in this frame.

Now we'll use the formula

$sinh\left(u\right) = \frac{v}{\sqrt{1-v^2/c^2}} = v\gamma$

to get,

$E_y' = -H_z v\gamma$

We're interested in the force on the particle, $F=qE_y'$, so we write down

$E_y' q = -q H_z v\gamma$

This is starting to look pretty good.  The E field producing a force on the moving charged particle is proportional to the magnetic field times the velocity of the particle.  There's still one problem.  We have a $\gamma$ in the expression which doesn't show up in the Lorentz force law.  Force can also be expressed as $F=ma$.  Since we're working in special relativity, however, we need to express the (transverse in this case) force using relativistic mass so we get $F=\gamma m a$.  This finally gives us

$F = \gamma m a = -q H_z v\gamma$

The $\gamma$'s cancel out to give the correct answer for the magnitude of the force:

$F = -qH_zv$.  In this case, however, the force was created by an electric field in the same direction.

There's one last interesting thing to note here, at least for my purposes.  The force acting in a direction perpendicular to the motion of the charged particle doesn't need to be adjusted for special relativity.  There are no factors of $v/c$ in the final, exact, (as opposed to a low speed approximation), answer.


1.  Karapetoff, V., "Restricted Theory of Relativity in Terms of Hyperbolic Functions of Rapidities", American Mathematical Monthly, **43**, (1936), 70



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