Electric Field Directly Ahead of or Behind a Moving Charge

[shinobi20]

TL;DR Summary

I am browsing through E&M by Purcell and Morin Chapter 5.6 and have some confusion.

Since it is stated that ##E'_x = E_x##, I am going to set a special case where ##z' = z = 0##, ##E_x## in (5.10) reduces to,

##E_x = \frac{1}{4 \pi \epsilon_0}\frac{Q}{x^2}##

However, ##E'_x## in (5.13) reduces to,

##E'_x = \frac{1}{4 \pi \epsilon_0}\frac{Q}{\gamma^2 x'^2}##

There is an extra ##\gamma^2## factor in the denominator, why is that so?

It makes sense that both ##E'_x## and ## E_x## are equal directly ahead (or behind) of the charged particle, but the above calculation shows otherwise. Any comments?

 

[pervect]

Summary:: I am browsing through E&M by Purcell and Morin Chapter 5.6 and have some confusion.
Since it is stated that ##E'_x = E_x##, I am going to set a special case where ##z' = z = 0##, ##E_x## in (5.10) reduces to,

##E_x = \frac{1}{4 \pi \epsilon_0}\frac{Q}{x^2}##

However, ##E'_x## in (5.13) reduces to,

##E'_x = \frac{1}{4 \pi \epsilon_0}\frac{Q}{\gamma^2 x'^2}##

There is an extra ##\gamma^2## factor in the denominator, why is that so?

It makes sense that both ##E'_x## and ## E_x## are equal directly ahead (or behind) of the charged particle, but the above calculation shows otherwise. Any comments?

x' is not equal to x. Recall that (t,x) and (t', x') are related by the Lorentz transform.

 

[shinobi20]

x' is not equal to x. Recall that (t,x) and (t', x') are related by the Lorentz transform.

Yeah, so I have to apply the Lorentz contraction ##x' = x/\gamma## in order to get the same result.

Also, I have a similar confusion when reading the same topic from a different book, from Ohanian.

You can see in the last sentence of the second paragraph, he stated that ##E_x## must be,

##E_x = q\frac{1}{(x-x_q)^2}##

which I think is missing a ##\gamma^2## factor in the denominator as can be deduced from (48). Is that a typo or I missed something? I have consulted the Errata sheets for this book and that is not listed as one.

 

[PeterDonis]

so I have to apply the Lorentz contraction ##x/\gamma##

No. You have to use the Lorentz transformation to convert ##t## and ##x## to ##t'## and ##x'## and vice versa. You can't just look at ##x'## and ##x## in isolation, because of relativity of simultaneity. The book you quote in the OP gives the transformations as equations 5.11 and 5.12. You have to use them; note that neither of those transformations contains ##x' = x / \gamma##.

 

[shinobi20]

No. You have to use the Lorentz transformation to convert ##t## and ##x## to ##t'## and ##x'## and vice versa. You can't just look at ##x'## and ##x## in isolation, because of relativity of simultaneity. The book you quote in the OP gives the transformations as equations 5.11 and 5.12. You have to use them; note that neither of those transformations contains ##x' = x / \gamma##.

In the sentence above (5.13), it is stated that the case considered is ##t'=0##, meaning the case considered is that both frames coincide.

 

[Pencilvester]

Yeah, so I have to apply the Lorentz contraction ##x' = x/\gamma## in order to get the same result.

I don’t see what your confusion is. You’re right, they’re only evaluating ##E_x## at ##t’ =0##. So plug that into your equation ##E’_x = \frac{1}{4 \pi \epsilon_0} \frac{Q}{\gamma^2 x’^2}## and what do you get?

 

[PeterDonis]

In the sentence above (5.13), it is stated that the case considered is ##t'=0##, meaning the case considered is that both frames coincide.

"Both frames coincide" is not a correct description of what ##t' = 0## means. It means you are only considering events in the hyperplane ##t' = 0##. Events in this hyperplane do not all have the same value of ##t##. The particular event where the charge ##Q## passes the origin of the primed frame has ##t = 0## and ##t' = 0##, and also has ##x = 0## and ##x' = 0##. But this is the only event where things "coincide" in this manner.

 

[shinobi20]

I don’t see what your confusion is. You’re right, they’re only evaluating ##E_x## at ##t’ =0##. So plug that into your equation ##E’_x = \frac{1}{4 \pi \epsilon_0} \frac{Q}{\gamma^2 x’^2}## and what do you get?

Yes, that slipped through me but I get that now.

"Both frames coincide" is not a correct description of what ##t' = 0## means. It means you are only considering events in the hyperplane ##t' = 0##. Events in this hyperplane do not all have the same value of ##t##. The particular event where the charge ##Q## passes the origin of the primed frame has ##t = 0## and ##t' = 0##, and also has ##x = 0## and ##x' = 0##. But this is the only event where things "coincide" in this manner.

I understand what you're saying, I think a better term is "origins coincide". All that in correspondence to what the book is implying where it set a special case of ##t' = 0##.How about my second question in post #3?

 

[vanhees71]

Purcell seems to systematically confuse the issue of relativistic electrodynamics and with that the students.

The most simple way to get the fields for the uniformly moving point charge is to start from the Lorenz-gauge vector potential in the rest-frame of the particle (inertial frame ##\Sigma'##), where
$$A^{\prime 0}=\frac{q}{4 \pi r'}, \quad \vec{A}'=0.$$
Now the only building blocks to build this expression in covariant form is the four-velocity of the particle ##u^{\mu}=\gamma (1,\vec{\beta})## and ##x^{\mu}##. In the rest frame ##u^{\prime \mu}=(1,0,0,0)## thus in covariant terms we have
$$A^{\mu}=\frac{q}{4 \pi r'} u^{\mu},$$
and
$$r'=\sqrt{(u_{\mu} u_{\nu} -\eta_{\mu \nu}) x^{\mu} x^{\nu}}.$$
Then the field components are simply given by
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
The same can be very easily derived also using the retarded Green's function for the Lorenz-gauge potential. For details and explicit expressions, see Sec. 3.6 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Pencilvester]

How about my second question in post #3?

The equation you’re questioning is immediately preceded by “a charge at rest”. Anything considered in its own rest frame has gamma equal to one.

 

[shinobi20]

The equation you’re questioning is immediately preceded by “a charge at rest”. Anything considered in its own rest frame has gamma equal to one.

No, ##E'_x## is the electric field in the rest frame of the particle and ##E_x## is the electric field of the particle viewed in the lab frame. The image in post #3 is just saying that the electric field ahead (or behind) of the charge "looks the same" as that if the particle were at rest, BUT that only applies to points which are AHEAD or BEHIND the particle, anywhere else the E-field is different. So I am thinking that there should be a ##\gamma^2## in the denominator as can "easily" be derived from (48).

So ##E_x## is the electric field of the moving particle as viewed in the lab frame, ##\gamma## cannot be exactly 1 there, just canceled if the circumstance requires.

 

[shinobi20]

Purcell seems to systematically confuse the issue of relativistic electrodynamics and with that the students.

The most simple way to get the fields for the uniformly moving point charge is to start from the Lorenz-gauge vector potential in the rest-frame of the particle (inertial frame ##\Sigma'##), where
$$A^{\prime 0}=\frac{q}{4 \pi r'}, \quad \vec{A}'=0.$$
Now the only building blocks to build this expression in covariant form is the four-velocity of the particle ##u^{\mu}=\gamma (1,\vec{\beta})## and ##x^{\mu}##. In the rest frame ##u^{\prime \mu}=(1,0,0,0)## thus in covariant terms we have
$$A^{\mu}=\frac{q}{4 \pi r'} u^{\mu},$$
and
$$r'=\sqrt{(u_{\mu} u_{\nu} -\eta_{\mu \nu}) x^{\mu} x^{\nu}}.$$
Then the field components are simply given by
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
The same can be very easily derived also using the retarded Green's function for the Lorenz-gauge potential. For details and explicit expressions, see Sec. 3.6 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

I also find Ohanian's treatment much clearer, starting from the argument of generalizing the electrostatic field equations to incorporate the "mysterious force" in the rest frame of the moving particle and satisfying some relativistic principles, then arriving at the EM field tensor, poof, E and B-fields expressions just drop out.
However, I just wanted to consult Purcell just to enforce my understanding of the topic, but it seems there are paragraphs where it really needs extra effort to understand.

Anyways, my lingering problem aside from the OP is post #3. Thanks for the notes @vanhees71 ! I greatly appreciate it coming from you, I will carefully read it.

 

[Pencilvester]

No, ##E'_x## is the electric field in the rest frame of the particle and ##E_x## is the electric field of the particle viewed in the lab frame. The image in post #3 is just saying that the electric field ahead (or behind) of the charge "looks the same" as that if the particle were at rest, BUT that only applies to points which are AHEAD or BEHIND the particle, anywhere else the E-field is different. So I am thinking that there should be a ##\gamma^2## in the denominator as can "easily" be derived from (48).

So ##E_x## is the electric field of the moving particle as viewed in the lab frame, ##\gamma## cannot be exactly 1 there, just canceled if the circumstance requires.

It looks like you’re comparing eq. 48 to the equation at the end of the paragraph. At y=0 and for a charge at rest, eq. 48 reduces to the equation at the end of the paragraph. You’re welcome to add gamma’s in, but when it’s said “at rest”, the implication is that gamma equals one.

Again, I’m not sure what’s causing the confusion. Is it from the fact that when using the charge’s primed coordinates you simply use x’ and in the lab frame you have to subtract the coordinate of the position of the charge?

 

[shinobi20]

It looks like you’re comparing eq. 48 to the equation at the end of the paragraph. At y=0 and for a charge at rest, eq. 48 reduces to the equation at the end of the paragraph. You’re welcome to add gamma’s in, but when it’s said “at rest”, the implication is that gamma equals one.

Again, I’m not sure what’s causing the confusion. Is it from the fact that when using the charge’s primed coordinates you simply use x’ and in the lab frame you have to subtract the coordinate of the position of the charge?

Based on my understanding in the unprimed frame, the charge is moving, and so it CANNOT be at rest. What the paragraph is saying is that if ##y=0##, then the electric field perceived by the observer in the unprimed frame is "THE SAME" as with the electric field in the primed frame (charge is at rest), BUT the charge IS NOT at rest in this unprimed frame, the electic field JUST LOOKS the same as with the rest frame of the charge.

 

[Pencilvester]

in the unprimed frame, the charge is moving, and so it CANNOT be at rest.

Does it specify that ##v \neq 0## for the charge in the lab frame? I was under the impression that they were not restricting the velocity in that way, especially since they say “for a charge at rest,” and then give the reduced equation for a charge at rest in the lab frame.

I think you’re over-complicating this. They’re offering up eq. 48, then pointing out that if the charge is at rest, then along y=0, the x component equation reduces to the familiar inverse square law equation.

 

[TSny]

My 2 cents isn't worth much. But my interpretation of Ohanian is the same as @shinobi20. Ohanian appears to me to be saying that for points on the ##x## axis in the unprimed frame, the field of the moving charge is the same as if the charge were at rest. But, of course, this is not true.

 

[vanhees71]

Anyways, my lingering problem aside from the OP is post #3. Thanks for the notes @vanhees71 ! I greatly appreciate it coming from you, I will carefully read it.

You are welcome. The field (both electric and magnetic components) are given there. I don't know, where the problem still is, or is there something wrong? I'm always thankful for hints at typos ;-).

 

[TSny]

I don't know, where the problem still is, or is there something wrong?

The question still lingering is the question in post #3 regarding the following excerpt from Ohanian:

There is disagreement on the interpretation of the last sentence. In the underlined sentence, is Ohanian still referring to (48) as applied to a charge in motion, or is he now assuming that the charge is at rest?

@shinobi20 and I believe that he is still assuming that the charge is in motion. We see him claiming that for points in front of (or behind) the moving charge, the E field is the same as if the charge were at rest. This claim is false as can be easily seen by examination of the first equation in (48). It's probably just an oversight on Ohanian's part.

Others believe that in the last sentence he has switched to assuming that the charge is actually at rest. But if so, why would he bother to include the phrase "..directly ahead of (or behind)..". Surely if the charge is actually at rest, then (48) reduces to the field of a point charge at rest at all points, not just points "ahead of (or behind)".

 

[Pencilvester]

In the underlined sentence, is Ohanian still referring to (48) as applied to a charge in motion, or is he now assuming that the charge is at rest?

Eq. 48 is a general equation. It is still applicable even if the velocity of the charge is zero in the lab frame. What makes you think otherwise?

in the last sentence he has switched to assuming that the charge is actually at rest.

Again, it is a general formula—there is no “switching”. Why would he need to “switch” for ##v=0##, but not for any other velocity?

why would he bother to include the phrase "..directly ahead of (or behind).."

I agree, the word “directly” is a little misleading here.

 

[TSny]

Eq. 48 is a general equation. It is still applicable even if the velocity of the charge is zero in the lab frame. What makes you think otherwise?

I never meant to imply that (48) is not applicable to v = 0. The question is whether or not in the last sentence of the paragraph Ohanian is assuming that v is zero.

Again, it is a general formula—there is no “switching”. Why would he need to “switch” for ##v=0##, but not for any other velocity?

Equation (48) is for any v < c. I believe Ohanian is still assuming the charge has any v < c in the last sentence. That is, he hasn't suddenly "switched" to assuming the charge has v = 0. Rather, he seems to be saying that while the charge is moving at any velocity v, the field in front of (and behind) the charge is the same as for a charge at rest.

 

[TSny]

Here's a way to make sense of what Ohanian says.

Suppose we attach a thin rod of rest-length ##L_0## to the charge. The rod is oriented along the ##x## axis to the right of the charge. In Ohanian’s primed frame, the charge and rod are at rest. The field ##E’_x## at the far end of the rod in this frame is ##E’_x = \frac{q}{L_0^2}##. In the unprimed frame where the charge and rod are in motion, the rod is contracted to length ##\Delta x = \frac{L_0}{\gamma}##. From (48), the field at the far end of the rod is $$E_x = \frac{q}{(\gamma\Delta x)^2} =\frac{q}{(\gamma L_0/\gamma)^2} = \frac{q}{L_0^2} = E’_x$$

So, the field at the far end of the rod is the same in both frames. That is, the field at a point in front of the moving charge is the same as the field in front of the charge when it is at rest if you choose the distance in front of the moving charge to be the Lorentz contraction of the distance in front of the charge in the rest frame.

If that's what Ohanian means, then I'm OK with it. But it still appears to me that his equation ##E_x = \frac{q}{(x-x_p)^2}## at the end of the last sentence should be ##E_x = \frac{q}{\gamma^2(x-x_p)^2}##.

Originally, it seemed to me that he was saying that if you go to the unprimed frame where the charge is in motion and you measure the field, say, 1 meter in front of the charge (measured in the unprimed frame) then you will get the same field that would exist 1 meter from the charge as measured in the rest frame of the charge. That wouldn't be right.

Whew.

[EDIT: A picture might help]

 

[Pencilvester]

... if you choose the distance in front of the moving charge to be the Lorentz contraction of the distance in front of the charge in the rest frame.

I’m disinclined to believe this is what he meant simply because relativity is all about distinguishing between frame dependent quantities and frame independent ones. The electric field is frame dependent, so it doesn’t really make sense to say that it’s independent as long as you change what you mean by x=1 depending on your frame.

But upon re-reading the paragraph, I was suddenly able to see the source of confusion. I still think he’s just wanting to point out how the equation reduces to the familiar inverse square law equation, but I’m less sure than I was.

I think we can all agree now, however, that it’s a weird and confusing sentence—the important thing is that we do understand relativity well enough to figure out ways of forcing it to make sense.

 

[TSny]

It is interesting that if you have a region of electric field and if some event occurs in that region, then the x-component of the electric field at that event will have the same value for any frames moving relative to each other in the x-direction: ##E'_x = E_x##.

This fact is related to the picture I posted in my last post, where ##P## and ##P'## can be thought of as marking the position of the same event.

 

[shinobi20]

Does it specify that ##v \neq 0## for the charge in the lab frame? I was under the impression that they were not restricting the velocity in that way, especially since they say “for a charge at rest,” and then give the reduced equation for a charge at rest in the lab frame.

I think you’re over-complicating this. They’re offering up eq. 48, then pointing out that if the charge is at rest, then along y=0, the x component equation reduces to the familiar inverse square law equation.

The original situation is that there is a charge ##q## moving with velocity ##v## in the lab frame (unprimed frame) and we can set up a moving frame (primed frame) also with velocity ##v##. So yes, ##v \neq 0## and again, the charge is NOT at rest in the lab frame which corresponds to ##E_x##, thus ##\gamma \neq 1##.

It is my impression from the books that I've read that the usual convention is that "lab frame" views objects moving and we set up a moving frame with the same velocity as that object and claim that as the rest frame of the object, maybe this was not clear to you and I apologize for that. But then, if we are to proceed with your statement in accordance to post #3, it would not make sense because there will be no point of writing ##E'_x## (equation above (48) which is clear to be the E-field for a charge at rest) and another more general field ##E_x## which as you claim reduces to the "charge at rest", why write prime and unprime then if the other is just a special case of the more general, also knowing that we use prime and unprime for different frames?

Again, ##E'_x## and ##E_x## are fields viewed from different frames so you cannot just compare it the way you compare it.

My 2 cents isn't worth much. But my interpretation of Ohanian is the same as @shinobi20. Ohanian appears to me to be saying that for points on the ##x## axis in the unprimed frame, the field of the moving charge is the same as if the charge were at rest. But, of course, this is not true.

Exactly. This is where the discussion must begin.

The question still lingering is the question in post #3 regarding the following excerpt from Ohanian:

View attachment 256358
There is disagreement on the interpretation of the last sentence. In the underlined sentence, is Ohanian still referring to (48) as applied to a charge in motion, or is he now assuming that the charge is at rest?

@shinobi20 and I believe that he is still assuming that the charge is in motion. We see him claiming that for points in front of (or behind) the moving charge, the E field is the same as if the charge were at rest. This claim is false as can be easily seen by examination of the first equation in (48). It's probably just an oversight on Ohanian's part.

Others believe that in the last sentence he has switched to assuming that the charge is actually at rest. But if so, why would he bother to include the phrase "..directly ahead of (or behind)..". Surely if the charge is actually at rest, then (48) reduces to the field of a point charge at rest at all points, not just points "ahead of (or behind)".

If he is assuming that the charge is now at rest with respect to the lab frame, he would have stated so as what usual textbooks do, so yes, there is no change to your original perception of the situation. Besides, that would abruptly interrupt the flow of the discussion unless stated otherwise. Based on the last sentence, it will be false depending if there is a typo in the equation in that sentence, which is my original inquiry.

If he actually switched to assuming that the charge is now at rest, there wouldn't be any point to it right? Assuming he already established a rest frame for the charge, which is so.

Eq. 48 is a general equation. It is still applicable even if the velocity of the charge is zero in the lab frame. What makes you think otherwise?

It is a general equation for the E-field as viewed in the lab frame where the charge is moving, which was the original setup.

I agree, the word “directly” is a little misleading here.

I don't think the word directly is misleading, I totally get what he wants to say there, it is the equation in the last sentence in combination to what he wants to say that is misleading.

I never meant to imply that (48) is not applicable to v = 0. The question is whether or not in the last sentence of the paragraph Ohanian is assuming that v is zero.

Equation (48) is for any v < c. I believe Ohanian is still assuming the charge has any v < c in the last sentence. That is, he hasn't suddenly "switched" to assuming the charge has v = 0. Rather, he seems to be saying that while the charge is moving at any velocity v, the field in front of (and behind) the charge is the same as for a charge at rest.

Exactly my point to @Pencilvester

Here's a way to make sense of what Ohanian says.

Suppose we attach a thin rod of rest-length ##L_0## to the charge. The rod is oriented along the ##x## axis to the right of the charge. In Ohanian’s primed frame, the charge and rod are at rest. The field ##E’_x## at the far end of the rod in this frame is ##E’_x = \frac{q}{L_0^2}##. In the unprimed frame where the charge and rod are in motion, the rod is contracted to length ##\Delta x = \frac{L_0}{\gamma}##. From (48), the field at the far end of the rod is $$E_x = \frac{q}{(\gamma\Delta x)^2} =\frac{q}{(\gamma L_0/\gamma)^2} = \frac{q}{L_0^2} = E’_x$$

So, the field at the far end of the rod is the same in both frames. That is, the field at a point in front of the moving charge is the same as the field in front of the charge when it is at rest if you choose the distance in front of the moving charge to be the Lorentz contraction of the distance in front of the charge in the rest frame.

If that's what Ohanian means, then I'm OK with it. But it still appears to me that his equation ##E_x = \frac{q}{(x-x_p)^2}## at the end of the last sentence should be ##E_x = \frac{q}{\gamma^2(x-x_p)^2}##.

Originally, it seemed to me that he was saying that if you go to the unprimed frame where the charge is in motion and you measure the field, say, 1 meter in front of the charge (measured in the unprimed frame) then you will get the same field that would exist 1 meter from the charge as measured in the rest frame of the charge. That wouldn't be right.

Whew.

[EDIT: A picture might help]

View attachment 256364

This confirmation is what I'm waiting for, so can I say that the equation in the last sentence has a typo, that is, it must have a ##\gamma^2## in the denominator? Actually I think this must not complicate further as I think it is just a plain typo, I totally understand what is happening until this suspected typo. BTW, how did you draw that diagram? Thanks for the effort!

 

[TSny]

This confirmation is what I'm waiting for, so can I say that the equation in the last sentence has a typo, that is, it must have a ##\gamma^2## in the denominator? Actually I think this must not complicate further as I think it is just a plain typo, I totally understand what is happening until this suspected typo.

Yes, that's how I see it.

BTW, how did you draw that diagram?

I used Microsoft Paint.

 

[vanhees71]

I'm not sure, whether I understand this enigmatic sentence by Ohanian either. Note that in my notation
$$u \cdot x=\gamma (c t-\vec{v} \cdot \vec{x}).$$
Thus in the denominator in my [1] equation (3.6.9) you have
$$(u \cdot x)^2-x \cdot x=\gamma^2 (ct-\vec{\beta} \cdot \vec{x})^2-c^2 t^2 + \vec{x}^2.$$
"Directly ahead or before" the charge, I'd interpret as at a position ##\vec{x}=(v t+\epsilon) \vec{\beta}/\beta##
Then the expression in the denominator gets
$$(u \cdot x)^2 -x \cdot x = \gamma^2 \epsilon^2$$
Then indeed
$$\vec{E}=\frac{q \gamma}{4 \pi} \frac{\vec{x}-\vec{v} t}{[(u \cdot x)^2-x \cdot x]^{3/2}}=\frac{q}{4 \pi \gamma^2 \epsilon^2} \frac{\vec{\beta}}{\beta}.$$
Thus there's an additional factor ##1/\gamma^2## compared to the expression at rest.

The electric field of a charge moving with velocities close to ##c## is sharply peaked in transverse direction (with the longitudinal component suppressed by the said factor ##1/\gamma^2=1-\beta^2##.

[1] https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Pencilvester]

there will be no point of writing ##E'_x## (equation above (48) which is clear to be the E-field for a charge at rest) and another more general field ExExE_x which as you claim reduces to the "charge at rest", why write prime and unprime then if the other is just a special case of the more general, also knowing that we use prime and unprime for different frames?

##E’_x## is defined to be the E field in the charge’s own frame, using the charge’s own coordinates. So it doesn’t matter whether or not the charge is moving in the lab because it can always consider itself to be at rest. Eq. 48 does not reduce to the equation at the top because you’re using different coordinates—you’d need a coordinate transformation for that (which happens to be a simple translation if you assume ##v=0##). So the reason they write prime and unprime is to distinguish which frame you are considering the E field in.

Again, ##E'_x## and ##E_x## are fields viewed from different frames so you cannot just compare it the way you compare it.

You’re right, in post #10 I was sloppy with my phrasing. I though I fixed that before hitting Post, but I guess I didn’t. It should have said “Anything at rest in a given frame has gamma equal to one.”

If he actually switched to assuming that the charge is now at rest, there [would be no] point to it right? Assuming he already established a rest frame for the charge, which is so.

Not necessarily. ##v=0## for a given particle in a given frame does not imply that the coordinate transformation to get from said frame to the particle’s frame is the identity matrix. There is still the possibility of translation and rotation.

By the way, I’m not meaning for what I’m saying here to support Ohanian’s weird words or even my own forced interpretation of what he may be trying to say. I’m just trying to reveal some subtle assumptions you may be making. I do agree now, whatever Ohanian was trying to convey, he did it poorly.

so can I say that the equation in the last sentence has a typo, that is, it must have a γ2γ2\gamma^2 in the denominator?

At this point, I don’t think it matters how you make sense of it—there are going to be problems no matter what. If you just assume a simple typo of a missing gamma squared, then you have the problem of trying to explain the phrase “... the electric field is the same as that for a charge at rest; that is...” One, there’d be no need for gammas in an equation for something at rest, and two, that whole statement is just plain wrong. Assuming non-zero velocity, they’re not the same. In my case, I have trouble explaining ”the electric field is the same as”. You’re right in that he makes it sound as though he is referring to the electric field around a charge with possibly non-zero speed (or definitely non-zero, depending on your assumptions), and again, that would be wrong.

 

[shinobi20]

##E’_x## is defined to be the E field in the charge’s own frame, using the charge’s own coordinates. So it doesn’t matter whether or not the charge is moving in the lab because it can always consider itself to be at rest. Eq. 48 does not reduce to the equation at the top because you’re using different coordinates—you’d need a coordinate transformation for that (which happens to be a simple translation if you assume ##v=0##). So the reason they write prime and unprime is to distinguish which frame you are considering the E field in.

You’re right, in post #10 I was sloppy with my phrasing. I though I fixed that before hitting Post, but I guess I didn’t. It should have said “Anything at rest in a given frame has gamma equal to one.”

Not necessarily. ##v=0## for a given particle in a given frame does not imply that the coordinate transformation to get from said frame to the particle’s frame is the identity matrix. There is still the possibility of translation and rotation.

By the way, I’m not meaning for what I’m saying here to support Ohanian’s weird words or even my own forced interpretation of what he may be trying to say. I’m just trying to reveal some subtle assumptions you may be making. I do agree now, whatever Ohanian was trying to convey, he did it poorly.

At this point, I don’t think it matters how you make sense of it—there are going to be problems no matter what. If you just assume a simple typo of a missing gamma squared, then you have the problem of trying to explain the phrase “... the electric field is the same as that for a charge at rest; that is...” One, there’d be no need for gammas in an equation for something at rest, and two, that whole statement is just plain wrong. Assuming non-zero velocity, they’re not the same. In my case, I have trouble explaining ”the electric field is the same as”. You’re right in that he makes it sound as though he is referring to the electric field around a charge with possibly non-zero speed (or definitely non-zero, depending on your assumptions), and again, that would be wrong.

I think what Ohanian is saying in the last sentence is correct, it is just that the equation in there lacks the ##\gamma^2##.

This is also taken from Ohanian,

So the fields parallel to the direction of motion are indeed the same with respect to Lorentz transformation, which is what he is saying in the last sentence.

 

[vanhees71]

Well, but what he forgets is that you have to transform both, the field components themselves (the fields are of course invariant under Lorentz transformations as any tensor) and the spacetime argument. In covariant form
$$F_{\mu \nu}'(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x)={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\hat{\Lambda}^{-1} x').$$
The electric field of a moving point charge with speeds close to ##c## is almost transverse to the direction of the particle's velocity, i.e., it is very different compared to the isotropic Coulomb field of a charge at rest.

The lesson is: Don't make to many words about formulae which speak for themselves!

 

[pervect]

This is also taken from Ohanian,

View attachment 256475

That's a very standard result, but some care most be taken in it's interpreation.

The cleanest way to interpret the above is that if one has two E and B field meters at the same event in space-time, the above relationship tells you how the meter readings transform.

To use this to find the field of a moving charge takes a bit of thought. One way of doing this is to imagine a moving charge, and a co-moving E-field meter. They are in some inertial frame, that I'll call the red frame. We'll call the lab frame the green frame.

In the red frame, the charge has a coulomb field, that we can write down. And that's what the red meter will read.

Knowing the above transform law, we can find the fields in the green frame, but we have to be careful to do it correctly.

In order to fiend the field in the green frame, as a function of the green coordinates, though, we can start by considering some green coordinates (t_g, x_g). Then we convert these coordinates via the Lorentz transform to the red frame, (t_r, x_r). We've already noted that columb's law gives the meter readings in the red frame. The last step is to convert these meter readings using the above transformation law to get the green meter readings.

This is essentailly what Vanhees wrote above, in tensor notation, which is the easiest way to do it. If one is not familiar with the tensor notation, one can work through it carefully, but the tensor approach is easier.

Not that the distance from the charge to the meter will be different in the red and green frames. The simultaneity conventions will also be different in the red and green frames. So to avoid errors and get all the correct factors of gamma as the textbook does, one has to be careful not to skip any steps, and work through the above procedure exactly, using the Lorentz transforms.

 

[shinobi20]

Then indeed
$$\vec{E}=\frac{q \gamma}{4 \pi} \frac{\vec{x}-\vec{v} t}{[(u \cdot x)^2-x \cdot x]^{3/2}}=\frac{q}{4 \pi \gamma^2 \epsilon^2} \frac{\vec{\beta}}{\beta}.$$
Thus there's an additional factor ##1/\gamma^2## compared to the expression at rest.

The electric field of a charge moving with velocities close to ##c## is sharply peaked in transverse direction (with the longitudinal component suppressed by the said factor ##1/\gamma^2=1-\beta^2##.

Just a clarification, there's an additional factor ##1/\gamma^2## compared to the expression at rest with respect to the unprimed frame, i.e.,

##E_x = q\frac{1}{\gamma^2 (x-x_q)^2}~## compare with ##~E_x = q\frac{1}{(x-x_q)^2}~## (E-field of a charge at rest in the unprimed frame)

if there is a charge at rest in the unprimed frame (although in this case the charge is at rest in the primed frame).

Compared to the expression at rest with respect to the primed frame (which is the case in Ohanian) THEY ARE THE SAME, i.e.,

##E_x = q\frac{1}{\gamma^2 (x-x_q)^2} = q\frac{1}{(x')^2}~## compare with ##~E'_x = q\frac{1}{(x')^2}##

since ##x' = \gamma (x - x_q)##.

This is correct right?

 

[TSny]

Here's how Ohanian described it in the first edition of his text:

 

[vanhees71]

I don't understand this. In any textbook you find the important result that the electric field for a relativistically moving point charge is almost transverse. The important point is that you must use all quantities in one frame. In my notation the charge is at rest in the primed frame, and there you have
$$\vec{E}'(t',\vec{x}')=\frac{q}{4 \pi r^{\prime 3}} \vec{e}_{3}'.$$
The field in the original frame, where it moves with velocity ##\vec{v}=v \vec{e}_1## is given by (written in 3D-notation)
$$\vec{E}(t,\vec{x}) = \frac{q}{4 \pi} \frac{\gamma (\vec{x}-v t \vec{e}_1)}{[\gamma^2(x^1-v t)^2+(x^2)^2+(x^3)^2]^{3/2}}.$$
The magnetic field turns out to be
$$\vec{B}=\vec{\beta} \times \vec{E}.$$
This you get either by Lorentz boosting the fields (and writing everything, i.e., the spacetime arguments and the fields with the components in the respetive frame) or just using the Lienard-Wiechert potentials for the uniformly point charge (as given in my SRT FAQ [1]).

Note that the above given result is identical with (3.6.9) in my FAQ since the expression in the denominator is
$$(u \cdot x)^2-x \cdot x=\gamma^2 (ct-v x^1)^2-(c t)^2+\vec{x}^2=\gamma^2(x^1-v t)^2 + (x^2)^2 + (x^3)^2.$$

The text by Ohanian is misleading, because he doesn't take into account that he as to express what he calls "proper distance" in terms of the (in his notation) unprimed coordinates!

[1] https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[pervect]

Just a clarification, there's an additional factor ##1/\gamma^2## compared to the expression at rest with respect to the unprimed frame, i.e.,

##E_x = q\frac{1}{\gamma^2 (x-x_q)^2}~## compare with ##~E_x = q\frac{1}{(x-x_q)^2}~## (E-field of a charge at rest in the unprimed frame)

if there is a charge at rest in the unprimed frame (although in this case the charge is at rest in the primed frame).

Compared to the expression at rest with respect to the primed frame (which is the case in Ohanian) THEY ARE THE SAME, i.e.,

##E_x = q\frac{1}{\gamma^2 (x-x_q)^2} = q\frac{1}{(x')^2}~## compare with ##~E'_x = q\frac{1}{(x')^2}##

since ##x' = \gamma (x - x_q)##.

This is correct right?

Depending on what you are comparing, I believe it's correct. Let us specify exactly what you are comparing. You are comparing the field one unit distance (say 1 meter) away from the charge as measured in the unprimed frame to the field, one meter away, from the charge as measured in the primed frame.

The relationship ##E_{\parallel} = E'_{\parallel}## is also a perfectly correct relationship in the correct context. You don't seem to be understanding correctly the context in which this relationship is correct, because you're more focused on a different context. The context in which ##E_{\parallel} = E'_{\parallel}## is correct is that we are comparing the electric field as the same event as seen by two different observers.

Now we ask, "is that an event 1 unit distance unit (say 1 meter) away from the charge q in the unprimed frame is not the same event as an event 1 meter away from the charge q in the primed frame.". The answer is no, they are different events.

When you suggested that "Lorentz contraction" was the explanation for the factor of gamma discrepancy, way back near the beginning of the post, you were basically on the right track. The distance changes by a factor of gamma due to Lorentz contraction, and the field is inversely proportional to the square of the distance, so there is your factor of gamma^2.

You seem to be trying to convince yourself there is a typo in your textbook, rather than figuring out where you went wrong :(.

Getting the details right and understanding the direction of the correction factor of ##\gamma^2## - is it multiplicative, or does it divide? - takes some more attention , which is what I tried to outline in my post #30, and other posters have outlined other mathematical approaches.

There's a graphical approach based on the "field lines" drawn by Ohanian as quoted by another poster in post #32. With the graphical approach, the field strength is proportional to the density of the "field lines" in the transverse direction. See for instance https://en.wikipedia.org/wiki/Field_line.

An individual field line shows the direction of the vector field but not the magnitude. In order to also depict the magnitude of the field, a selection of field lines can be drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to the magnitude of the vector field at that point.

It's hard to get the numbers from the field line approach, especially since you'd need a 3-d diagram as wiki also mentions, but the field line expression gives the correct qualitative answer.

 

[vanhees71]

Again, please make sure to also write down the arguments. The complete notation of the above statement is
$$E_{\parallel}'(x')=E_{\parallel}(x),$$
i.e. with the Lorentz transformation matrix ##\hat{\Lambda}##, ##x'=\hat{\Lambda} x## you have
$$E_{\parallel}'(x')=E_{\parallel}(\hat{\Lambda}^{-1} x').$$