# Is an EMF observed by a constantly accelerating loop the moment it is stationary to an electric charge?

Suppose I had a positive charge $$+q$$ fixed at the origin $$(x,y,z)=(0,0,0)$$. The coordinate system is that of the charge's rest frame.

Now suppose I had a circular loop with a radius of 1 whose axis is in the positive z direction. Let the position of the center of the circular loop be a function of time t such that its position is $$(x,y,z)=((1/2)at^2,2,0)$$.

At $$t=-\epsilon$$, the circular loop saw a magnetic field through it in the $$+z$$ direction, and at $$t=+\epsilon$$, the circular loop will see a magnetic field in the $$-z$$ direction. So in the non-inertial frame of reference of the loop, there will be a changing magnetic field. However, at $$t=0$$ the loop sees the charge as stationary, so the charge does not show a magnetic field at $$t=0$$. The Lorentz transformation depends on the observer's velocity, not the observer's acceleration. Therefore, at $$t=0$$ the perceived electric field of the charge should be curl free, not dependent on the observer's acceleration. For that reason, the acceleration of the loop should not be directly responsible for an EMF.

So what's going on? The magnetic field seen by the loop depends on its velocity $$\mathbf{v}$$ and the electric field it is subject to.

$$\mathbf{B} = -\frac{1}{c^2}\mathbf{v}\times\mathbf{E}$$

Since the velocity of the loop is not constant, there are two ways the electric field as seen by the loop can change with respect to time, as revealed by applying the product rule:

$$\frac{d\mathbf{B}}{dt} = -\frac{1}{c^2}\frac{d\mathbf{v}}{dt}\times\mathbf{E} -\frac{1}{c^2}\mathbf{v}\times\left(\mathbf{v}\cdot\nabla\right)\mathbf{E}$$

The average change in the magnetic field through the loop as seen by the loop over the period from $$t=-\infty$$ to $$t=+\infty$$ is zero. Therefore:

$$\int_{-\infty}^{+\infty} -\frac{1}{c^2}\frac{d\mathbf{v}}{dt}\times\mathbf{E}\ dt = \int_{-\infty}^{+\infty} \frac{1}{c^2}\mathbf{v}\times\left(\mathbf{v}\cdot\nabla\right)\mathbf{E}\ dt$$

The integrals on both sides are non-zero. To see that this is the case, suppose that instead of a constant acceleration, the loop had constant velocity until $$t=-\epsilon$$ when it suddenly changes to an equal and opposite velocity at $$t=+\epsilon$$ with acceleration $$\mathbf{v}/ \epsilon$$.

So we have a dilemma. Either:

1. The time-averaged emf in the loop is zero because the curl of the electric field seen by the observer is a function of that observer's acceleration.

or

1. Maxwell's equations should not be modified for accelerating observers, and as result, the time-averaged emf into the loop is not zero from the loop's point of view.

Option 1 implies a non-trivial modification to Maxwell's equations for accelerating observers, whereas Option 2 implies that it would be possible to induce a time-averaged direct current into a loop over an unbounded period of time without an unbounded change in magnetic fields. Is an Option 3 possible? If so, what is it? Some ideas below:

1. Notions such as "the curl of the electric field" do not make sense for an accelerated observer, even though a person standing at sea level in Earth's gravity is subject to such acceleration in curved spacetime.

2. The effective masses of positive and negative charges in a loop decrease and increase, respectively, in the presence of charge $$+q$$'s electric potential such that an EMF results at $$t=0$$ despite the curl of $$+q$$'s electric field being zero, resulting in an uncompensated increase of the loop's magnetic field in the $$+z$$ direction at $$t=0$$.

3. The effective masses of positive and negative charges in a loop are modified in opposite ways as the loop moves along the gradient of charge $$+q$$'s potential such that the loop does not experience an EMF at any point along the way (despite all evidence to the contrary).

## 1 Answer

The solution requires recognizing that:

1. In the general case, force is not exactly equal to mass times acceleration.
2. In the rest frame of the loop, the magnetic Lorentz force acting on the loop is zero, while the electric field is relativistically transformed.
3. Due to the time dilation of the relatively moving loop as viewed in the rest frame of the external charge, the acceleration of an internal charge as observed by the rest frame of the external charge is reduced by the Lorentz factor derived from the observed velocity $$v$$.

Therefore, while the Lorentz force is:

$$\mathbf{F} = q(\mathbf{E}+\mathbf{v} \times \mathbf{B})$$

The following is not generally true:

$$\mathbf{F} = m \mathbf{a}$$

Instead, we have:

$$\frac{\mathbf{qE}'}{\gamma} = m a = q\left(\left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left (\frac{{\gamma-1}}{\gamma} \right ) \mathbf{E}_\parallel \right)\\$$

Thus:

$$\mathbf{F} = m \mathbf{a} + q\left (\frac{{\gamma-1}}{\gamma} \right ) \mathbf{E}_\parallel\\$$

This equation predicts that as the Lorentz factor $$\gamma$$ approaches infinity, the acceleration component parallel to the velocity vector $$\mathbf{a}_\parallel$$ approaches zero, as is observed in particle accelerators.

The effective EMF induced into the loop is therefore generally a combination of transformer EMF, motional EMF, and an apparently overlooked "relativistic" EMF:

$$\mathcal{E}=\oint_{C} \left(\mathbf{E} + \mathbf{v} \times \mathbf{B} - \left (\frac{{\gamma-1}}{\gamma} \right ) \mathbf{E}_\parallel\right) \cdot \mathrm{d} \boldsymbol{ \ell } \ ,$$

In the example given in the above post, the first two contributions to the EMF are completely absent in the rest frame of the external charge, with only the third contribution remaining. The only time when the third term does not contribute either is at the precise moment when the loop is stationary, which means the answer to the question I posed:

"Is an EMF observed by a constantly accelerating loop the moment it is stationary to an electric charge?"

...is "no".

Interestingly enough, the term $$- \left (\frac{{\gamma-1}}{\gamma} \right ) \mathbf{E}_\parallel$$ in the above formula predicts that the EMF induced the loop for $$t < 0$$ possesses the same sign as the EMF induced into the loop for $$t > 0$$ since $$\gamma$$ remains positive even if velocity is reversed, and thus a time-averaged net current may be produced inside the loop despite that:

The average change in the magnetic field through the loop as seen by the loop over the period from $$t=-\infty$$ to $$t=+\infty$$ is zero.