Electron mass increase inside charged sphere?

In summary, the conversation discusses the momentum of an electron inside a hollow dielectric sphere and how it is affected by the presence of a circulating magnetic field and a radial electric field. The momentum density of the electromagnetic field outside the sphere is given by the equation g = \epsilon_0 E \times B and when integrated over all space, it results in a momentum given by p_{field} = -\frac{2}{3} \frac{e V}{c^2} v. The conversation also considers how to ensure that no additional momentum is added to the system when charges are applied to the sphere or when the sphere itself is moving. It is concluded that the charges are assumed to be fixed onto the sphere and thus have no
  • #1
johne1618
371
0
Imagine an electron of mass [itex]m_e[/itex] inside a hollow dielectric sphere.

Assume that the electron is traveling with constant velocity [itex]v[/itex] relative to the sphere.

The momentum of the system [itex]p[/itex] comprises just the momentum of the electron

[itex] p = m_e v [/itex]

There is also a circulating magnetic field [itex]B[/itex] around the electron due to its moving charge [itex]e[/itex].

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

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

Thus there is a momentum density [itex]g[/itex] at every point in the electromagnetic field outside the sphere given by:

[itex] g = \epsilon_0 E \times B [/itex]

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

[itex]p_{field} = -\frac{2}{3} \frac{e V}{c^2} v[/itex]

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

[itex] p = p_e + p_{field} [/itex]

Therefore

[itex] p_e = m_e v + \frac{2}{3} \frac{e V}{c^2} v [/itex]

or

[itex] p_e = (m_e + \frac{2}{3} \frac{e V}{c^2}) v [/itex]

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?
 
Physics news on Phys.org
  • #2
Details of the Electromagnetic Field Momentum Calculation

Assume the electron is at the center of a hollow dielectric sphere with radius [itex]R[/itex].

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

[itex] \large \vec E = \frac{Q}{4 \pi \epsilon_0} \frac{\hat r}{r^2} [/itex]

Now we assume the electron is traveling with constant velocity [itex]\vec v[/itex] relative to the sphere.

The magnetic field at position [itex]\vec r[/itex] due to the electron is given by

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

Now the momentum density [itex]\vec g[/itex] in the electromagnetic field outside the sphere is given by

[itex] \large \vec g = \epsilon_0 \vec E \times \vec B [/itex]

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

[itex] \large p_{field} = \int \vec g \cdot \hat v \ dV [/itex]

Now

[itex] \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} [/itex]

Using the vector formula:

[itex] \vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c) \vec b - (\vec a \cdot \vec b) \vec c[/itex]

We find that

[itex]
\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}
[/itex]

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

[itex] dV = 2 \pi r^2 \sin \theta \ dr \ d\theta [/itex]

Therefore we have

[itex]
\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}
[/itex]

Now we can perform the [itex]\theta[/itex]-integral by changing variables

[itex]
\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}
[/itex]

So we have

[itex]
\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}
[/itex]

Now the potential [itex]V[/itex] on the sphere is given by

[itex] V = \frac{Q}{4 \pi \epsilon_0 R} [/itex]

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

[itex] \large p_{field} = - \frac{2}{3} \frac{e V}{c^2} v [/itex]
 
  • #3
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
mfb said:
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?

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:
  • #5
johne1618 said:
I measure velocities relative to the sphere so by definition the sphere itself is stationary and thus has no momentum.
But then it is pointless to compare "momentum before" and "momentum after" via momentum conservation.

The charges I add or remove are assumed to be fixed onto the sphere and thus have no momentum themselves.
Right, but they don't appear out of nowhere.
 
  • #6
mfb said:
But then it is pointless to compare "momentum before" and "momentum after" via momentum conservation.

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
mfb said:
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?

I believe this to be the correct answer to your query.

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:
  • #8
johne1618 said:
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?


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.

johne1618 said:
I believe this to be the correct answer to your query.

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.


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
Jasso said:
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.

How about this setup?

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:
  • #10
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
Jasso said:
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.

Maybe you're right - I wouldn't know how to do the calculation in detail.
 
  • #12
johne1618 said:
Assume the electron is at the center of a hollow dielectric sphere with radius [itex]R[/itex].

Now we assume the electron is traveling with constant velocity [itex]\vec v[/itex] relative to the sphere.

The magnetic field at position [itex]\vec r[/itex] due to the electron is given by

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

This formula is incorrect at large distances, since there is a retardation effect due to the finite speed of propagation of em fields.
 
  • #13
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 [itex]m[/itex] 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

[itex] \vec{F} = m \vec{a} [/itex]

where [itex]\vec{F}[/itex] is the force generated by the stretched spring.

The accelerating electron produces an electric field [itex]\vec{E}[/itex] 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 [itex]\vec{\Delta F}[/itex] due to this electric field that is proportional to the electron's acceleration and is given approximately by

[itex]\vec{\Delta F} = \frac{e V}{c^2} \vec{a}[/itex]

(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 [itex]\vec{F} + \vec{\Delta F}[/itex].

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

[itex] \vec{a} = \frac{\vec{F}}{m} = \frac{\vec{F}+\vec{\Delta F}}{m^\ast} [/itex]

[itex] m^\ast \vec{a} = m \vec{a} + \frac{eV}{c^2} \vec{a} [/itex]

Therefore

[itex] m^\ast = m + \frac{eV}{c^2} [/itex]

(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

[itex] \vec{E} = -\vec\nabla{\phi} - \frac{\partial \vec{A}}{\partial t}[/itex]

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 [itex]\phi[/itex] and vector potential [itex]\vec{A}[/itex] at a radial distance [itex]r[/itex] from an electron with charge [itex]e^-[/itex]

[itex] \phi = - \frac{e}{4 \pi \epsilon_0 r} [/itex]

[itex] \vec{A} = - \frac{e \vec{v}}{4 \pi \epsilon_0 c^2 r} [/itex]

The electric field at the spherical dielectric of radius [itex]R[/itex] is then given by

[itex] \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} [/itex]

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 [itex]+Q[/itex] due to the accelerating electron is therefore given by

[itex] \vec{\Delta F} = \frac{e Q}{4 \pi \epsilon_0 c^2 R} \vec{a} [/itex]

If the potential on the sphere is [itex]+V[/itex] volts then the force is finally given by

[itex] \vec{\Delta F} = \frac{e V}{c^2} \vec{a} [/itex]
 
Last edited:
  • #14
This thread is very close to being locked. Repeating the same errors in slightly different configurations is not progress.
 
  • #15
johne1618 said:
[itex] \vec{A} = - \frac{e \vec{v}}{4 \pi \epsilon_0 c^2 r} [/itex]
This vector potential would create a magnetic field:
[tex]
\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}
[/tex]
where I used [itex]1/\varepsilon_0 \, c^2 = \mu_0[/itex], and [itex]\nabla r = \mathbf{r}/r[/itex].

But, this is the same magnetic field that you had stated and to what I objected in my previous post.
 
  • #16
Dickfore said:
This formula is incorrect at large distances, since there is a retardation effect due to the finite speed of propagation of em fields.

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

[tex]
\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}
[/tex]

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

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

[tex]
\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}
[/tex]

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

I think this advanced formula gives the electromagnetic field in an arbitrarily moving non-inertial frame due to a stationary remote charge [itex]q[/itex]. 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 [itex]P[/itex] changes before the position of the remote charge [itex]q[/itex] changes (unlike in the retarded formula describing an inertial frame where the charge position changes before the unit vector).
 
Last edited:
  • #17
johne1618 said:
The Heaviside-Feynman formula is a relativistically correct expression for the electromagnetic field at point [itex]P[/itex] due to an arbitrarily moving charge [itex]q[/itex]:

[tex]
\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}
[/tex]

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

johne1618 said:
I think this advanced formula gives the electromagnetic field in an arbitrarily moving non-inertial frame due to a stationary remote charge [itex]q[/itex]. 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 [itex]P[/itex] changes before the position of the remote charge [itex]q[/itex] changes (unlike in the retarded formula describing an inertial frame where the charge position changes before the unit vector).

I'm afraid I don't have an idea what you're talking about.
 
  • #18
Dickfore said:
First of all, I don't know where the minus sign in the expression for the electric field comes from?!

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.

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 assuming that the sphere is a dielectric so there shouldn't be any conducting boundaries.
 
  • #19
johne1618 said:
I'm assuming that the sphere is a dielectric so there shouldn't be any conducting boundaries.

But, there are dielectric boundaries, which render the solution inapplicable.
 
  • #20
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.
 
  • #21
johne1618 said:
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.

Electromagnetic forces are not inertial, so no.
 
  • #22
Dickfore said:
Electromagnetic forces are not inertial, so no.

Does an accelerated charge q_1 feel *any* sort of electromagnetic force due to the presence of a stationary charge q_2 (apart from Coulomb attraction/repulsion)?
 
Last edited:
  • #23
Coulomb's Law is valid only for two charges at rest. As soon as you start moving charges around, they start emitting em waves, and there may be magnetic fields. Thus, you need to explicitly take the degrees of freedom of the em field into consideration, and the concept of force due to one charge on another looses its meaning. But, there is an approximate Darwin Lagrangian that contains terms up to order [itex]O(v^2/c^2)[/itex].
 
  • #24
Dickfore said:
they start emitting em waves.

Ok.

By the way according to Feynman Lectures in Physics vol II 28-7 an electron only emits electromagnetic radiation if its acceleration changes with time.
 

1. What is electron mass increase inside a charged sphere?

Electron mass increase inside a charged sphere is the phenomenon of electrons gaining mass while inside a charged sphere. This is due to the influence of the electric field on the electrons.

2. Why does electron mass increase inside a charged sphere?

This increase in mass is a result of the interaction between the electrons and the electric field. The electric field exerts a force on the electrons, causing them to gain energy and therefore, mass.

3. How does the electric field affect the mass of electrons inside a charged sphere?

The electric field exerts a force on the electrons, causing them to accelerate. According to Einstein's theory of relativity, as an object's velocity increases, its mass also increases. Therefore, the electric field causes the electrons to gain mass while inside the charged sphere.

4. Does the mass of the charged sphere also increase?

No, the mass of the charged sphere remains the same. The increase in mass is only observed in the electrons inside the sphere.

5. Is electron mass increase inside a charged sphere significant?

The increase in mass is very small and is usually only observed in extremely high electric fields. In everyday situations, the increase in mass is too small to be measured. However, it is an important concept in the field of physics and is crucial for understanding the behavior of charged particles in electric fields.

Similar threads

Replies
2
Views
856
Replies
1
Views
869
  • Electromagnetism
Replies
1
Views
675
Replies
11
Views
2K
Replies
11
Views
740
Replies
6
Views
647
Replies
7
Views
907
  • Electromagnetism
Replies
4
Views
855
  • Electromagnetism
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
327
Back
Top