I The Electric field of a moving charge according to Purcell

[particlezoo]

So it seems those longitudinal electric fields which are not affected by Lorentz boosts per Griffiths are not affected because they happened to be uniform, as in the electric field between capacitor plates. If they are just charges, they do drop in the expected way. The same would apply for any fringe fields leaking from a capacitor. I see what happened now. Thanks you guys for the help!

 

[vanhees71]

My major concern was whether or not the electric flux from a charge is independent of its velocity. If I have a plasma of charged electrons and ions, I don't think the electric flux from negative charges should be dependent on their velocity. Elementary charges are supposed to be discrete. If transverse electric fields are affected by velocity but not longitudinal electric fields, it seems that we have a problem. I don't see anything in the above that suggests that longitudinal electric fields would reduce so as to maintain electric flux conservation. I do know in notable gauges like the Coulomb gauge and the Lorenz gauge, the time-varying magnetic vector potential reduces the longitudinal electric field, however there is also the electric scalar potential, and it too contributes to this electric field. How is it possible for an electrically neutral system of varying temperature with charged species of different masses and velocities to maintain neutrality if electric flux were to change only on its transverse components?Continued: Ok, let me make an attempt here. So in your derivation we have a transformation of the scalar potential of essentially the Lorentz Factor and the reduction of distance |x|. So let's just consider a boost along x. So there is a squared Lorentz facor inside the square root which determines x. So this eliminates Lorentz factor in the numerator. Then we have the quantity inside the square root which is inverse to one of the factors effecting the differential volume element (the other was the Lorentz factor). Ok, so how does this conserve the electric flux of a charge?

I see, so you worry about Gauss's Law. It's of course valid in any inertial frame.

Let's verify this for our example of a uniformly moving point charge, using the explicit solution given in #20. Indeed for $|\vec{x}| \neq 0$, i.e., $\gamma^2 (x^{\prime 1}-v x^{\prime 0})^2+(x^{\prime 2})^2+(x^{\prime 3})^2) \neq 0$ you can show by explicitly taking the derivatives that $\vec{\nabla}' \cdot \vec{E}'=0$. To see that also the singularity comes out right, you have to integrate $\vec{E}'$ over an arbitrary surface containing the singularity. The most simple one is a sphere $S_a$ of arbitrary radius $a$ around $\vec{x}_0'=v x^{\prime 0}$. Then we get with
$$\vec{x}'=\vec{r}+\vec{v} x^{\prime 0}$$
and polar coordinates for $\vec{r}$
$$\int_{S_a} \mathrm{d}^2 \vec{f}' \cdot \vec{E}'=\frac{\gamma q}{4 \pi} \int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta a^3 \sin \vartheta \frac{1}{(\gamma^2 a^2 \cos^2 \vartheta + a^2 \sin^2 \vartheta)^{3/2}}=q,$$
because this implies that indeed
$$\vec{\nabla}' \cdot \vec{E}'=q \delta^{(3)}(x^{\prime 1}-v x^{\prime 0}) \delta(x^{\prime 2}) \delta(x^{\prime 3}),$$
and this is indeed the charge density of a uniformly moving point charge, as also shown via the Lorentz boost in #20.

 

[Dale]

If I separated the x, y, and z components of the electric field into three separate terms, the sum of their divergences should equal the divergence of their sum. If I multiplied the y and z components by -1, I would expect their divergences to change sign. If I doubled them, I would expect their contributions to the divergence to double.

Stop talking about what you expect the math to do and actually do the math.

 

[particlezoo]

Stop talking about what you expect the math to do and actually do the math.

I wanted to make sure I was on the right "math problem" first. Turns out I was not recognizing that the problem with the capacitor plates and uniform field is such that a longitudinal boost does not affect the longitudinal electric fields. That was not a general result.

I see, so you worry about Gauss's Law. It's of course valid in any inertial frame.

Let's verify this for our example of a uniformly moving point charge, using the explicit solution given in #20. Indeed for $|\vec{x}| \neq 0$, i.e., $\gamma^2 (x^{\prime 1}-v x^{\prime 0})^2+(x^{\prime 2})^2+(x^{\prime 3})^2) \neq 0$ you can show by explicitly taking the derivatives that $\vec{\nabla}' \cdot \vec{E}'=0$. To see that also the singularity comes out right, you have to integrate $\vec{E}'$ over an arbitrary surface containing the singularity. The most simple one is a sphere $S_a$ of arbitrary radius $a$ around $\vec{x}_0'=v x^{\prime 0}$. Then we get with
$$\vec{x}'=\vec{r}+\vec{v} x^{\prime 0}$$
and polar coordinates for $\vec{r}$
$$\int_{S_a} \mathrm{d}^2 \vec{f}' \cdot \vec{E}'=\frac{\gamma q}{4 \pi} \int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta a^3 \sin \vartheta \frac{1}{(\gamma^2 a^2 \cos^2 \vartheta + a^2 \sin^2 \vartheta)^{3/2}}=q,$$
because this implies that indeed
$$\vec{\nabla}' \cdot \vec{E}'=q \delta^{(3)}(x^{\prime 1}-v x^{\prime 0}) \delta(x^{\prime 2}) \delta(x^{\prime 3}),$$
and this is indeed the charge density of a uniformly moving point charge, as also shown via the Lorentz boost in #20.

Thank you.

 

[Dale]

I wanted to make sure I was on the right "math problem" first. Turns out I was not recognizing that the problem with the capacitor plates and uniform field is such that a longitudinal boost does not affect the longitudinal electric fields. That was not a general result.

Thank you.

Any arbitrary EM field follows the transformation equation given by the first set of equations here

https://en.m.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

This transformation is not limited to the special case of a uniform field. It applies to the fields at every event in spacetime.

 

[particlezoo]

Any arbitrary EM field follows the transformation equation given by the first set of equations here

https://en.m.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

This transformation is not limited to the special case of a uniform field. It applies to the fields at every event in spacetime.

Yes I see now. I just wished it didn't take this long for me to realize that the longitudinal fields that "weren't affected" by the boost only applied to a finite region between capacitor plates, and not to the whole volume. I know now that only happens because of the geometry, and not because the fields of a charge preserve their longitudinal components everywhere under a boost.

 

[particlezoo]

Any arbitrary EM field follows the transformation equation given by the first set of equations here

https://en.m.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

This transformation is not limited to the special case of a uniform field. It applies to the fields at every event in spacetime.

It's been a couple weeks so I looked at the page with a fresh set of eyes, and it's pretty clear to me that the page you provided implies that the longitudinal components of the electric field are not affected by Lorentz boosts.

[/PLAIN]
This equation, also called the Joules-Bernoulli equation, considers two inertial frames. [...] In these two frames moving at relative velocity v, the E-fields and B-fields are related by:[2]

\begin{aligned}&\mathbf {{E}_{\parallel }} '=\mathbf {{E}_{\parallel }} \\&\mathbf {{B}_{\parallel }} '=\mathbf {{B}_{\parallel }} \\&\mathbf {{E}_{\bot }} '=\gamma \left(\mathbf {E} _{\bot }+\mathbf {v} \times \mathbf {B} \right)\\&\mathbf {{B}_{\bot }} '=\gamma \left(\mathbf {B} _{\bot }-{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {E} \right)\end{aligned}

[...]

An equivalent, alternative expression is:[3]

\begin{aligned}&\mathbf {E} '=\gamma \left(\mathbf {E} +\mathbf {v} \times \mathbf {B} \right)-\left({\gamma -1}\right)(\mathbf {E} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} \\&\mathbf {B} '=\gamma \left(\mathbf {B} -{\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}}\right)-\left({\gamma -1}\right)(\mathbf {B} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} \\\end{aligned}

where v̂ is the velocity unit vector.

Compare this to a prior response received on this thread:

If you look at the equation for the electric field of a moving charge, you should get the same answer as from http://farside.ph.utexas.edu/teaching/em/lectures/node125.html

The field points in the radial direction, at an angle $\theta$ the electric field from <<this link>> is:

$$||E|| = \frac{e}{4 \pi \epsilon_0 r^2} \frac{\gamma}{(1 + \beta^2 \gamma^2 \cos^2 \theta)^\frac{3}{2}}$$

I've taken some liberties with the notation, calculating the magnitude of E (since we know it points in the radial direction), and letting $\beta = v/c$. $\gamma$ is used on the webpage, it's $1/\sqrt{1-\beta^2}$ as usual.

The term
$$\frac{e}{4 \pi \epsilon_0 r^2}$$
is just the Coulomb field. So we see that at $\theta=+/-90$, which is the transverse direction, the field is boosted by a factor of $\gamma$. But when $\theta=0$ or $\theta=180$, in the longitudinal direction, the field is:

$$\frac{\gamma}{(1 + \beta^2 \gamma^2 )^\frac{3}{2}}$$

If we take for example $\beta = .995$ so that $\gamma = 10.01 \approx 10$, we see that the transverse field is about 10 times the columb field of a stationary charge, but the longitudinal field is about $10 / 100^\frac{3}{2} \approx .01$. times the columb field of a stationary charge.

If you're really ambitious, you can set up and do the integral. I think I may have done that once in the very distant past, but it's pretty messy looking. The E-field is normal to the surface of the sphere, but you'll have to do a 3d integral to get the total charge from Gauss' law.

 

[Dale]

It's been a couple weeks so I looked at the page with a fresh set of eyes, and it's pretty clear to me that the page you provided implies that the longitudinal components of the electric field are not affected by Lorentz boosts.

Yes, regardless of the shape of the field.

Compare this to a prior response received on this thread

Yes. You can derive the expression from pervect using Coulomb's law and the transformation equations that I posted.

 

[particlezoo]

Yes, regardless of the shape of the field.

Yes. You can derive the expression from pervect using Coulomb's law and the transformation equations that I posted.

So we take (1: Coulomb's law) then (2: transformation equations you posted) to get (3: the expression from pervect)?

The field points in the radial direction, at an angle $\theta$ the electric field from <<this link>> is:

$$||E|| = \frac{e}{4 \pi \epsilon_0 r^2} \frac{\gamma}{(1 + \beta^2 \gamma^2 \cos^2 \theta)^\frac{3}{2}}$$

So this expression is supposed to be one where the longitudinal components of the electric field are not affected by Lorentz boosts. This requires that r would be function of the Lorentz boosts. I can understand this part. Now, what happens if instead of changing the velocity of the observer, I change the velocity of the source in the frame of the observer, would r be a function of that velocity? I would think that, given enough time, the electric field measured by the observer only depends on the relative velocity between observer and the source charge (ignoring accelerations of course for simplicity). Shouldn't then a change of velocity of the source charge also preserve the longitudinal components of the electric field, given enough time for the fields to update?

 

[pervect]

I'm getting the feeling there is a lack of communication going on here, but I'm not sure exactly what to do about it.

Perhaps asking questions would be better than writing answers. It's worth a shot, going around in circles isn't getting us anywhere.

Suppose we have a stationary charge. There is a red pole sticking out from the stationary charge, pointing in a direction we'll call "north". There is a meter at the end of the pole, and the meter reads "1 volt/meter", and it is calibrated using SI units The red pole has a length of 1 meter - a proper length, that is, a length of 1 meter measured in its own rest frame. We'll also call this frame the "red" frame, so it has the same designation as the pole.

Now consider an observer, moving north (in the same direction the red pole points) at $\beta = v/c = .995$ so that the $\gamma$ factor is 10. We'll call this frame the green frame and the observer the green observer.

The first question is this: Review (or read about for the first time if it isn't familiar) the barn-and-pole paradox of special relativity. What is the length of the red pole according to the moving (green) observer?

The moving (green) observer also carries an E-field meter. When the green observer's green e-field meter is at the same location as the red observer's e-field meter, what does the green observers E-field meter read? And to ask the first question in a different way, when the green meter is read, how far away is it from the charge in the green frame?

Eventually, we'll need to introduce more poles, and talk about Gauss's law for that matter, but at this point it'd be too confusing I think.
We'll see how this works out first in any event.

 

[Dale]

So we take (1: Coulomb's law) then (2: transformation equations you posted) to get (3: the expression from pervect)?

Yes.

Now, what happens if instead of changing the velocity of the observer, I change the velocity of the source in the frame of the observer,

The fields are different in these two scenarios, but all measurements are symmetric. This is, in fact, the point made by Einstein in the beginning of his 1905 paper

 

[particlezoo]

Yes.

The fields are different in these two scenarios, but all measurements are symmetric. This is, in fact, the point made by Einstein in the beginning of his 1905 paper

That must be because the distance differs from one scenario than other. I was essentially taking the distance to be a certain value prior to changing one of the velocities, but depending on whether I decided to change the velocity of observer, or just the source, determines whether r is affected, such that electric field observed is different in each case.

 

[particlezoo]

I'm getting the feeling there is a lack of communication going on here, but I'm not sure exactly what to do about it.

Perhaps asking questions would be better than writing answers. It's worth a shot, going around in circles isn't getting us anywhere.

Suppose we have a stationary charge. There is a red pole sticking out from the stationary charge, pointing in a direction we'll call "north". The meter reads "1 volt/meter", and it is calibrated using SI units The red pole has a length of 1 meter - a proper length, that is, a length of 1 meter measured in its own rest frame. We'll also call this frame the "red" frame, so it has the same designation as the pole.

Now consider an observer, moving north (in the same direction the red pole points) at $\beta = v/c = .995$ so that the $\gamma$ factor is 10. We'll call this frame the green frame and the observer the green observer.

The first question is this: Review (or read about for the first time if it isn't familiar) the barn-and-pole paradox of special relativity. What is the length of the red pole according to the moving (green) observer?

The moving (green) observer also carries an E-field meter. When the green observer's green e-field meter is at the same location as the red observer's e-field meter, what does the green observers E-field meter read? And to ask the first question in a different way, when the green meter is read, how far away is it from the charge in the green frame?

Eventually, we'll need to introduce more poles, and talk about Gauss's law for that matter, but at this point it'd be too confusing I think.
We'll see how this works out first in any event.

Length of the red pole according to the green observer = 10 cm

I'm assuming that the red E-field meter is one meter away from the charge in the red rest frame. So the distance of the charge from the green observer in the green observer's frame = 10 cm.

Now for the part that I am not sure about. I take it that since I am boosting from the red frame to the green frame, that the longitudinal component of the electric field is the same as measured by both meters, which are presumably identical and only measuring the longitudinal electric field. So it would measure the same 1 volt/meter.

 

[particlezoo]

Length of the red pole according to the green observer = 10 cm

I'm assuming that the red E-field meter is one meter away from the charge in the red rest frame. So the distance of the charge from the green observer in the green observer's frame = 10 cm.

Now for the part that I am not sure about. I take it that since I am boosting from the red frame to the green frame, that the longitudinal component of the electric field is the same as measured by both meters, which are presumably identical and only measuring the longitudinal electric field. So it would measure the same 1 volt/meter.

Now since the distance is 10x closer according to the green frame, that would be a factor of 100 times the force if it weren't for the fact that the electrical field measured in this frame is reduced for a different reason, i.e. the movement of the charge according to this frame. The net effect is that the measured field is the same.

 

[Dale]

That must be because the distance differs from one scenario than other.

There is no such thing as "the distance" since the distance changes over time. It is any distance at some point in time.

Any measurement that you can come up with will be the same in the two different scenarios.

 

[pervect]

Length of the red pole according to the green observer = 10 cm

I'm assuming that the red E-field meter is one meter away from the charge in the red rest frame. So the distance of the charge from the green observer in the green observer's frame = 10 cm.

Now for the part that I am not sure about. I take it that since I am boosting from the red frame to the green frame, that the longitudinal component of the electric field is the same as measured by both meters, which are presumably identical and only measuring the longitudinal electric field. So it would measure the same 1 volt/meter.

Yes, that's what happens. When Purcell says the longitudinal component of the E-field is the same, he means that the red meter measures the same value as the green meter.

Note that this is only true in the longitudinal direction, it's not generally true, and Purcell points this out in the transformation law for the transverse component, which is different.

The remaining step as I see it is to discuss how we get the enclosed charged via Gauss's law.
The procedure is to calculate the flux $\Phi_E$ of the electric field, as per <<wiki link>>. Then $Q = \epsilon_0 \Phi_E$, where $\epsilon_0$ is a constant, the permitivity of free space, and $\Phi_E$ is the flux of the electric field through the surface.

To do this easily, we want to construct a spherical array of rods, because to calculate the electic flux, we need to compute the normal component of the field, and the field turns out to be normal to the surface of a sphere, but not normal to other shapes.

First we do this in the red frame. If we take a 1 meter sphere, the field at any point is 1. The field is everywhere normal to the surface of the sphere of radius 1, which has a surface area of $4 \pi$. So the value of the flux is $4 \pi$, and the enclosed charge is $4 \pi \epsilon_0$.

Next, we try another sphere, a 10cm sphere. The field is now 100, not 1, as it follows an inverse square law. But the surface area of the sphere is proportional to the square of the radius, so it's only $.04 \pi$. Thus the value of the enclosed charge is still $4 \pi \epsilon_0$, regardless of the radius of a sphere.

What if we use an ellipsoid, rather than a sphere? This is a perfectly legitimate thing to do, but it's more involved. Now the field is no longer normal to the surface of the enclosing shape, and we need to compute the normal component of the field as discussed in wiki. It will turn out that Gauss's law says we get the same answer for the enclosed charge regardless of the shape the enclosure - as long as we do it correctly. I'm not going to get into more detail, other than to say that before we can compute the value of the moving charge from Gauss's law, we need to know how to compute the value of a non-moving charge from Gauss's law.

To compute the value of the electric flux $\Phi_E$ in the green frame, we need to define what shape we are doing the integral over. The first possibility is to construct a sphere in the green frame. This way the field turns out to be normal to the surface - something that isn't particularly obvious, by the way, but it turns out that way.

Because the charge is moving, the electric field will be varying with time in the green frame, we also need to pay attention to simultaneity issues. Simultaneity is not the same in the red and green frames. The fortunate simplification we can make is that the field doesn't vary with time in the red frame. The field on the green sphere varies with time, but we'll concentrate on the time t=0 as defined in the green frame.

If we imagine more than one red rod in the red frame, all one meter long and radiating away from the charge, we can see right off that the shape traced out by the ends of the rods in the red frame will not be spherical. So if we want to do the flux integral over the surface of a sphere, we need to correct for this, which means finding the shape that isn't spherical in the red frame, but transforms so that it's spherical in the green frame. This winds up with our distorted sphere being 10x longer in the north-south direction in the example I used with $\gamma=10$ and the longitudinal direction being north.

Because of this, the field at the northern tip of the sphere in the green frame (at t=0 in the green frame) is only .01, the same as the field on the distorted, non-spherical shape in the red frame.

The field on the green sphere varies then from about 100 or so, in the direction transverse to the motion around the equator, to .01, at the north and south pole. The condition for the charge to be constant (which turns out to be true) is that the average value of the normal component of the field is 1, and we can see that this is certainly possible, as the field varies from 100 to .01, though we haven't worked out the details by any means.

The field on the green sphere is giving more conveniently by the field expressed in my post #13 <<link>>.

This is more convenient to my way of thinking than doing all the Lorentz contractions and transformations needed, but certainly both approaches should give identical answers.