# Electron mass increase inside charged sphere?

1. Oct 21, 2012

### johne1618

Imagine an electron of mass $m_e$ inside a hollow dielectric sphere.

Assume that the electron is travelling with constant velocity $v$ relative to the sphere.

The momentum of the system $p$ comprises just the momentum of the electron

$p = m_e v$

There is also a circulating magnetic field $B$ around the electron due to its moving charge $e$.

Now let us charge up the dielectric sphere to +V volts.

There is now a radial electric field $E$ outside the sphere due to its surface charge as well as the circulating magnetic field $B$ from the electron.

Thus there is a momentum density $g$ at every point in the electromagnetic field outside the sphere given by:

$g = \epsilon_0 E \times B$

When this momentum density is integrated over all space outside the sphere one finds a resultant momentum given by

$p_{field} = -\frac{2}{3} \frac{e V}{c^2} v$

In order for the total momentum of the system to be conserved we must have

$p = p_e + p_{field}$

Therefore

$p_e = m_e v + \frac{2}{3} \frac{e V}{c^2} v$

or

$p_e = (m_e + \frac{2}{3} \frac{e V}{c^2}) v$

There are no forces acting on the electron inside the sphere to change its velocity so the increase of its momentum must be solely due to an increase in its inertia.

Is this reasoning correct?

2. Oct 22, 2012

### johne1618

Details of the Electromagnetic Field Momentum Calculation

Assume the electron is at the center of a hollow dielectric sphere with radius $R$.

The charge $Q$ on the sphere is much larger than the electron charge so that the electric field at position $\vec r$ outside the sphere is very nearly given by

$\large \vec E = \frac{Q}{4 \pi \epsilon_0} \frac{\hat r}{r^2}$

Now we assume the electron is travelling with constant velocity $\vec v$ relative to the sphere.

The magnetic field at position $\vec r$ due to the electron is given by

$\large \vec B = -\frac{e}{4 \pi \epsilon_0 c^2} \frac{\vec v \times \hat r}{r^2}$

Now the momentum density $\vec g$ in the electromagnetic field outside the sphere is given by

$\large \vec g = \epsilon_0 \vec E \times \vec B$

By symmetry the total momentum in the electromagnetic field is in the direction of the electron's velocity and is given by

$\large p_{field} = \int \vec g \cdot \hat v \ dV$

Now

$\large \vec g \cdot \hat v = - \frac{e Q}{16 \pi^2 \epsilon_0 c^2} \frac{(\hat r \times (\vec v \times \hat r)) \cdot \hat v}{r^4}$

Using the vector formula:

$\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c) \vec b - (\vec a \cdot \vec b) \vec c$

We find that

\begin{align} (\hat r \times (\vec v \times \hat r)) \cdot \hat v &= (\vec v - (\hat r \cdot \vec v) \hat r) \cdot \hat v \\ &= \vec v \cdot \hat v - (\hat r \cdot \vec v)(\hat r \cdot \hat v) \\ &= v - v \ \cos^2 \theta \\ &= v \ \sin^2 \theta \end{align}

We use a volume element that is a ring around the direction of the electron's velocity given in polar co-ordinates by

$dV = 2 \pi r^2 \sin \theta \ dr \ d\theta$

Therefore we have

\large \begin{align} p_{field} &= \int \vec g \cdot \hat v \ dV \\ &= \int_R^\infty \int_0^\pi - \frac{eQ}{16 \pi^2 \epsilon_0 c^2} \frac{v\ \sin^2 \theta}{r^4} 2 \pi r^2 \sin \theta \ dr \ d\theta \\ &= - \frac{e Q v}{8 \pi \epsilon_0 c^2} \int_R^\infty \frac{dr}{r^2} \int_0^\pi \sin^3 \theta \ d\theta \end{align}

Now we can perform the $\theta$-integral by changing variables

\begin{align} \int_0^\pi \sin^3 \theta \ d\theta &= - \int_1^{-1} (1 - \cos^2 \theta) \ d(\cos \theta) \\ &= - \left[ \cos \theta - \frac{\cos^3 \theta}{3} \right]_1^{-1} \\ &= \frac{4}{3} \end{align}

So we have

\large \begin{align} p_{field} &= - \frac{e Q v}{8 \pi \epsilon_0 c^2} \cdot \frac{1}{R} \cdot \frac{4}{3} \\ &= - \frac{e Q v}{6 \pi \epsilon_0 c^2 R} \end{align}

Now the potential $V$ on the sphere is given by

$V = \frac{Q}{4 \pi \epsilon_0 R}$

and so finally the total momentum in the electromagnetic field in terms of the sphere potential is

$\large p_{field} = - \frac{2}{3} \frac{e V}{c^2} v$

3. Oct 22, 2012

### Staff: Mentor

By putting charges on your sphere, you add (or remove) something, which can have momentum. How do you guarantee that you do not add any momentum to the system?
Or maybe the sphere itself is moving afterwards?

4. Oct 22, 2012

### johne1618

I measure velocities relative to the sphere so by definition the sphere itself is stationary and thus has no momentum.

The charges I add or remove are assumed to be fixed onto the sphere and thus have no momentum themselves.

Last edited: Oct 22, 2012
5. Oct 22, 2012

### Staff: Mentor

But then it is pointless to compare "momentum before" and "momentum after" via momentum conservation.

Right, but they don't appear out of nowhere.

6. Oct 22, 2012

### johne1618

Ok.

So I could assume that the frame of reference is independent of the sphere but also that the sphere is very massive.

Would my argument work in that case?

7. Oct 23, 2012

### johne1618

My assumption is that there are no net forces between the electron and the charged shell because of the spherically symmetric distribution of charge.

Therefore the shell does not change momentum when charges are applied to it uniformly. (One could also imagine shrinking a large shell with a fixed amount of charge on it to produce the same increase in potential.)

Thus if the shell has zero momentum initially then it always has zero momentum.

Therefore the shell can be ignored and momentum conservation can be applied to the system comprising the electron and the electromagnetic field alone.

Last edited: Oct 23, 2012
8. Oct 24, 2012

### Jasso

This won't change the fact that the momentum of the sphere changes; it would only mean that the momentum of the sphere is indeterminate.

But the magnetic field generated by the moving charge inside is not spherically symmetric, it curls around the axis of motion of the moving charge. The increase of charge on the shell can be thought of as a radially inward current density, which interacts with the non-radial components of the magnetic field, pushing the charges opposite the motion of the charge inside, imparting an opposing momentum to the shell itself.

9. Oct 24, 2012

### johne1618

Assume the spherical shell is a charged superconductor.

Initially the electron's magnetic field will not penetrate the shell. Eddy currents will occur in the shell but I assume there will not be any net force exerted by the electron on the body of the shell itself.

Now warm up the shell above its superconductor transition temperature so that it becomes a charged dielectric.

By this means we "uncover" the charges on the surface of the sphere without using any currents to get them there.

No momentum will be imparted to the shell in this case.

Last edited: Oct 24, 2012
10. Oct 24, 2012

### Jasso

There would still be an induced momentum. As the magnetic field comes through, it would be a time varying field at the surface. Since -dB/dt = curl(E), there would be an equivalent electric field at the surface pushing the charged shell the other way.

11. Oct 24, 2012

### johne1618

Maybe you're right - I wouldn't know how to do the calculation in detail.

12. Oct 24, 2012

### Dickfore

This formula is incorrect at large distances, since there is a retardation effect due to the finite speed of propagation of em fields.

13. Nov 2, 2012

### johne1618

Hi,

Here is a new argument to show that an electron's inertia is changed by the presence of other charges.

What do people think of it?

Increase of Electron Inertia inside a Charged Dielectric Sphere (version 2)

Imagine an electron of mass $m$ at the center of a very massive hollow dielectric sphere charged to a potential of +V volts.

Assume that the electron is attached to the inside wall of the sphere by a stretched spring.

As soon as the electron is released at the center of the sphere it accelerates such that

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

where $\vec{F}$ is the force generated by the stretched spring.

The accelerating electron produces an electric field $\vec{E}$ at the surface of the sphere which itself is assumed to be virtually stationary due to its large mass.

The sphere feels a resultant force $\vec{\Delta F}$ due to this electric field that is proportional to the electron's acceleration and is given approximately by

$\vec{\Delta F} = \frac{e V}{c^2} \vec{a}$

(see Appendix for calculation)

As the total force on the system must be zero this extra force must be balanced by an opposite force in the spring.

Thus the spring must pull the electron and the sphere together with a total force $\vec{F} + \vec{\Delta F}$.

In order to maintain consistency the electron must experience the same acceleration $\vec{a}$ with this increased force. Therefore its mass must increase to $m^\ast$ such that:

$\vec{a} = \frac{\vec{F}}{m} = \frac{\vec{F}+\vec{\Delta F}}{m^\ast}$

$m^\ast \vec{a} = m \vec{a} + \frac{eV}{c^2} \vec{a}$

Therefore

$m^\ast = m + \frac{eV}{c^2}$

(If the electron mass did not increase then the increased force in the spring would lead to an increased acceleration of the electron. The increased acceleration would lead to an increased electric force on the sphere which, being transmitted to the electron by the spring, would further increase the electron's acceleration ad infinitum.)

Appendix: Force on Charged Sphere due to Central Accelerating Electron

The electric field at the sphere is given by

$\vec{E} = -\vec\nabla{\phi} - \frac{\partial \vec{A}}{\partial t}$

For an exact calculation one should use the Lienard-Wiechert potentials for a moving charge. For an approximate calculation we can use the following expressions for the scalar potential $\phi$ and vector potential $\vec{A}$ at a radial distance $r$ from an electron with charge $e^-$

$\phi = - \frac{e}{4 \pi \epsilon_0 r}$

$\vec{A} = - \frac{e \vec{v}}{4 \pi \epsilon_0 c^2 r}$

The electric field at the spherical dielectric of radius $R$ is then given by

$\vec{E} = - \frac{e \hat{r}}{4 \pi \epsilon_0 R^2} + \frac{e}{4 \pi \epsilon_0 c^2 R} \frac{d \vec{v}}{dt}$

The first part is symmetric over the sphere so it does not impart any net force on it. The resultant force on the sphere with charge $+Q$ due to the accelerating electron is therefore given by

$\vec{\Delta F} = \frac{e Q}{4 \pi \epsilon_0 c^2 R} \vec{a}$

If the potential on the sphere is $+V$ volts then the force is finally given by

$\vec{\Delta F} = \frac{e V}{c^2} \vec{a}$

Last edited: Nov 2, 2012
14. Nov 2, 2012

Staff Emeritus
This thread is very close to being locked. Repeating the same errors in slightly different configurations is not progress.

15. Nov 2, 2012

### Dickfore

This vector potential would create a magnetic field:
$$\begin{array}{lcl} \mathbf{B} & = & \nabla \times \mathbf{A} \\ & = & -\frac{e}{4\pi \, \varepsilon_0 \, c^2} \nabla \left(\frac{1}{r}\right) \times \mathbf{v} \\ & = & -\frac{e}{4\pi \, \varepsilon_0 \, c^2} \, \left(-\frac{1}{r^2} \, \frac{\mathbf{r}}{r}\right) \times \mathbf{v} \\ & = & \frac{\mu_0}{4\pi} \frac{e (\mathbf{r} \times \mathbf{v})}{r^3} \end{array}$$
where I used $1/\varepsilon_0 \, c^2 = \mu_0$, and $\nabla r = \mathbf{r}/r$.

But, this is the same magnetic field that you had stated and to what I objected in my previous post.

16. Nov 7, 2012

### johne1618

The Heaviside-Feynman formula is a relativistically correct expression for the electromagnetic field at point $P$ due to an arbitrarily moving charge $q$:

$$\begin{array}{rcl} \mathbf{E} & = & \frac{-q}{4 \pi \varepsilon_0} \left\{ \left[ \frac{\mathbf{\hat{r}}}{r^2} \right]_{ret} + \frac{\left[ r \right]_{ret}}{c} \frac{\partial}{\partial t}\left[\frac{\mathbf{\hat r}}{r^2}\right]_{ret} + \frac{1}{c^2} \frac{\partial^2 \left[ \mathbf{\hat r} \right]_{ret}}{\partial t^2} \right\} \\ c\ \mathbf{B} & = & -\left[ \mathbf{\hat r} \right]_{ret} \times \mathbf{E} \end{array}$$

where $[\mathbf{\hat r}]_{ret}$ is the unit vector and $[r]_{ret}$ is the distance from the field point $P$ at time $t$ to the retarded position of the charge $q$ at time $t - [r]_{ret}/c$ (see Feynman lectures on Physics vol I and II).

But I think there must be another Heaviside-Feynman formula - the advanced version given by

$$\begin{array}{rcl} \mathbf{E} & = & \frac{-q}{4 \pi \varepsilon_0} \left\{ \left[ \frac{\mathbf{\hat{r}}}{r^2} \right]_{adv} + \frac{\left[ r \right]_{adv}}{c} \frac{\partial}{\partial t}\left[\frac{\mathbf{\hat r}}{r^2}\right]_{adv} + \frac{1}{c^2} \frac{\partial^2 \left[ \mathbf{\hat r} \right]_{adv}}{\partial t^2} \right\} \\ c\ \mathbf{B} & = & -\left[ \mathbf{\hat r} \right]_{adv} \times \mathbf{E} \end{array}$$

where $[\mathbf{\hat r}]_{adv}$ is the unit vector and $[r]_{adv}$ is the distance from the field point $P$ at time $t$ to the advanced position of the charge $q$ at time $t + [r]_{adv}/c$.

I think this advanced formula gives the electromagnetic field in an arbitrarily moving non-inertial frame due to a stationary remote charge $q$. This electromagnetic field produces an electromagnetic inertial force on a co-moving test charge that is proportional to the frame's acceleration. This inertial force appears at the instant that the test charge is accelerated as it should do if it is a true inertial force. This is because in the advanced formula that describes a non-inertial frame the unit vector at the field point $P$ changes before the position of the remote charge $q$ changes (unlike in the retarded formula describing an inertial frame where the charge position changes before the unit vector).

Last edited: Nov 7, 2012
17. Nov 7, 2012

### Dickfore

First of all, I don't know where the minus sign in the expression for the electric field comes from?!

Second, this formula is valid in vacuum, when there are no conducting boundaries. If you want to take these into account, the problem becomes rather complicated.

I'm afraid I don't have an idea what you're talking about.

18. Nov 7, 2012

### johne1618

The minus sign comes about because I have defined the unit vector at the field point to point towards the source charge q rather than away from it.

I'm assuming that the sphere is a dielectric so there shouldn't be any conducting boundaries.

19. Nov 8, 2012

### Dickfore

But, there are dielectric boundaries, which render the solution inapplicable.

20. Nov 10, 2012

### johne1618

My basic question is this:

Does an accelerated charge q_1 feel an electromagnetic inertial force due to the presence of a stationary charge q_2?

I argue that in the frame of q_1 the charge q_2 has an opposite "fictitious" acceleration that induces an electric field and therefore an inertial "fictitious" force on q_1 that is proportional to its acceleration.