Understanding E-Field Transformation in Feynman II_26

In summary, the conversation discusses the topic of Special Relativity and the transformation of electric fields in moving frames of reference. Feynman's Lectures are mentioned as a source for learning, specifically in Chapter II_26 where he gives the transformation for the E-field of a moving charge and discusses the reduction of the field strength in the stationary frame. Later on, in Sec 26-3, a different derivation is presented that directly relates the E-field in the stationary and moving frames, leading to the conclusion that the E-fields in the x direction are equal. However, there is a discrepancy between Eqn 26.11 and Table 26-2 in terms of the coordinates being used. The conversation then delves into a diagram that
  • #1
Verne42
3
0
I have been learning SR from various sources. Most of the time from Feyman's Lectures but that's not the only place.

In II_26 he gives the transformation for the E-field of a moving charge in the x direction under a standard Lorentz configuration. In Eqn 26.11 he derives a formula for the Ex strength (at the x coordinate of a point to the right of the charge as observed in the stationary frame which sees the E-field moving) and states the strength is reduced by 1/gamma compared to what the strength would be as seen by the frame of the charge, (that is, E'x.) This derivation came by first transforming the potentials, and then taking derivatives.

Then later, in Sec 26-3 he does a different sort of derivation where he derives the E-field transformations directly from the E fields themselves. That is, he derives the relationship between E and E'. He arrives at the formulas given in Table 26.2 which clearly show that the primed and unprimed versions of the E-field in the x direction are equal. That is, Ex = E'x.

A double-take ensues. Are the two versions of the E field in the x direction equal or not? Eqn 26.11 says no. Table 26-2 says yes. I suspect it's some kind of misunderstanding about the coordinates being used. For example, in the Table26-2 equation for Efield transform Ex = E'x, is it really implying Ex(x, y, z, t) = E'x(x', y', x', t'). I assume that's what it means, yet Eqn 26.11 is not giving it that way, it's saying Ex(x, y, z, t) = blahblah(x, y, z, t). In other words, no primed coordinates on either side. Is that the discrepancy between the two formulas?

I've spent a lot of time going over this material, and admit I don't fully understand the 4-vector stuff though I get the gist of it. And I suspect the aforementioned usages of coordinates is the cause of my confusion. I concocted a diagram that illustrates the E field in both frames of reference (2D only), for an arbitrary amount of squishiness (read: velocity of the charge) and I look at the value of the E-field strength at an arbitrary point on the X-axis and give that strength a value of 17 V/m just because I like the number (it's made up, in other words). I wanted to visually see how it tracks between the two frames of reference. I do the same thing in the Y direction, picking a point on the equipotential line of the first point.

electric field II.png


Does this diagram show things correctly? To wit:

(1) Are the equipotential lines on the squished E-field correctly proportioned?

(2) I (presumably) understand how the x direction is getting squished. Just plain ol' length contraction. But why is the Y direction expanded? Yes, I know it's due to the 4-vector transformations, but that doesn't really explain it. In fact, I can't find anywhere on the web, in books, etc., where this is explained, even though many places say that it happens. UPDATE: I see Feynman says in the same chapter that it's something to do with the ratio of the energy of the particle to its rest mass -- don't know how I missed that. My intuition on the diagram -- before I saw the Feynman statement -- says that it's due to a conservation of charge thing, kind of like squeezing a ball laterally and having the transverse sides pop out, aka conservation of volume on the part of the ball. Is this a good intuition? Does it roughly fit with Feynman's statement? Is it conservation of charge or conservation of the charge's momentum? Why doesn't anyone (besides Feynman in one short sentence) talk about this? Is it given in lectures and just not preserved in books and web articles? I unfortunately was never privy to such lectures when in undergrad and grad school in mechanical engineering way back in the 70s and 80s.

3) This stretching of the Y direction is implying that if a test charge were placed vertically above the moving charge, and that test charge was moving with same velocity, that from the stationary frame's point of view, the test charge would feel an extra upward force, beyond what Coulomb's law would say given their vertical distance. And that this is in addition to a magnetic force which ends up being directed downward. I assume that in general, these two additional effects do not entirely cancel. Is all of this correct?

Thanks for any help. I hope I explained the problem clearly enough.
 

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  • #2
Verne42 said:
3) This stretching of the Y direction is implying that if a test charge were placed vertically above the moving charge, and that test charge was moving with same velocity, that from the stationary frame's point of view, the test charge would feel an extra upward force, beyond what Coulomb's law would say given their vertical distance. And that this is in addition to a magnetic force which ends up being directed downward. I assume that in general, these two additional effects do not entirely cancel. Is all of this correct?

Extra Coulomb force? No. A static charge sees a moving charge's field as contracted, and feels some extra Coulomb force. If the static test charge started to accelerate after the moving charge, then the test charge would see the field un-contracting, and the Coulomb force approaching normal.

But maybe a static observer could say that there is extra Coulomb force between two co-moving charges? Hmmm ... it's much better if the static observer does not say such thing, because that just gets very confusing. And the force (total net force) between the charges gets smaller in the static frame as the speed increases.
 
  • #3
Verne42 said:
I suspect it's some kind of misunderstanding about the coordinates being used.
I believe you are correct - there's coordinate jiggery-pokery going on. (Incidentally - link to the relevant chapter: http://www.feynmanlectures.caltech.edu/II_26.html)

If I sit in a lab with an EM sensor, and you walk through the lab with identical equipment, there are four things you can define:
  1. E(x,y,z,t), B(x,y,z,t) - my EM sensor readings at an event defined with my clocks and rulers.
  2. E'(x,y,z,t), B'(x,y,z,t) - your EM sensor readings at an event defined with my clocks and rulers.
  3. E(x',y',z',t'), B(x',y',z',t') - my EM sensor readings at an event defined with your clocks and rulers.
  4. E'(x',y',z',t'), B'(x',y',z',t') - your EM sensor readings at an event defined with your clocks and rulers.

If I understand correctly, the relationships in table 26-2 are between (1) and (2). The relationship in 26.11 is between (1) and (4).
 
  • #4
Verne42 said:
(2) I (presumably) understand how the x direction is getting squished. Just plain ol' length contraction. But why is the Y direction expanded?
There is more field at the middle, because the field concentrated towards the middle, when the field contracted.

That's how things contract: Stuff making up the thing moves towards the center of the thing. I mean towards the center line. A football field becomes a narrow rectangle, and so on.

I admit that electric field may be different to a football field.

But, if we draw some lines radially from the center point of a football field, those lines transform like electric field lines, right?
 
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  • #5
Verne42 said:
I have been learning SR from various sources. Most of the time from Feyman's Lectures but that's not the only place.

In II_26 he gives the transformation for the E-field of a moving charge in the x direction under a standard Lorentz configuration. In Eqn 26.11 he derives a formula for the Ex strength (at the x coordinate of a point to the right of the charge as observed in the stationary frame which sees the E-field moving) and states the strength is reduced by 1/gamma compared to what the strength would be as seen by the frame of the charge, (that is, E'x.) This derivation came by first transforming the potentials, and then taking derivatives.

Then later, in Sec 26-3 he does a different sort of derivation where he derives the E-field transformations directly from the E fields themselves. That is, he derives the relationship between E and E'. He arrives at the formulas given in Table 26.2 which clearly show that the primed and unprimed versions of the E-field in the x direction are equal. That is, Ex = E'x.

A double-take ensues. Are the two versions of the E field in the x direction equal or not?

Consider a moving charge (moving in the x-direction), a moving E-field meter (moving in the same direction and velocity as the charge), and a stationary E-field meter.

Suppose the two E-field meters (the moving and stationary ones) are at the same location. Feynman is saying that the x-component of the E-field (we won't worry about the other components unless you develop an interest) are the same when the two meters are in the same location at the same time.

If you draw a space-time diagram, though, you'll see that the distance between the moving charge and the intersection of the world-lines of the two E-field meters (the point in space-time when they are at the same location at the same time) isn't the same in the moving frame and the stationary frame. This is often described as Lorentz contraction.

I could try to draw a space-time diagram for you, but it's better if you can draw your own.

Basically, one of your calculations implicitly uses one distance, the other calculation implicitly uses the other distance. But the two distances don't have the same value.
 
  • #6
Verne42 said:
(1) Are the equipotential lines on the squished E-field correctly proportioned?
Don't know - what equation did you use to draw them? And what velocity?
Verne42 said:
3) This stretching of the Y direction is implying that if a test charge were placed vertically above the moving charge, and that test charge was moving with same velocity, that from the stationary frame's point of view, the test charge would feel an extra upward force, beyond what Coulomb's law would say given their vertical distance. And that this is in addition to a magnetic force which ends up being directed downward. I assume that in general, these two additional effects do not entirely cancel. Is all of this correct?
Directly above the charge the increased electrostatic repulsion ought to cancel with the magnetic attraction, since the relativistic force transformation law in that case simplifies to F'y=Fy. (Edit: Mucked up the cancellation (twice!), as noted by jartsa below. Directly above the charge, the force on the test charge changes by a factor of ##\gamma## - see the link in the struck-out text - which you can verify from Feynman's table 26-2 in the case ##E_y\neq 0##, ##E_x=E_z=0## and B=0).
 
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  • #7
Verne42 said:
the equipotential lines on the squished E-field

The strictly correct response to this is that, in a frame in which the source is moving, the field is not a pure static E-field and does not have equipotential surfaces; those only work for a pure static E-field.
 
  • #8
Ibix said:
Directly above the charge the increased electrostatic repulsion ought to cancel with the magnetic attraction, since the relativistic force transformation law in that case simplifies to ##F'_y=F_y##. (Edit: Indeed, you can verify this trivially from Feynman's table 26-2 in the case ##E_y\neq 0##, ##E_x=E_z=0## and B=0).
No. Internal transverse forces inside moving objects, like for example co-moving charges, are reduced by gamma.

F'y=Fy/gamma

I guess you miscalculated something.
 
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  • #9
jartsa said:
No. Internal transverse forces inside moving objects, like for example co-moving charges, are reduced by gamma.

F'y=Fy/gamma

I guess you miscalculated something.
Indeed. Corrected above - thanks.
 
  • #10
Jartsa: The two charges are moving together.

Ibix: I'll have to chew on your first answer some more. As far as your second answer, I think you are also confirming what I said, the y direction gets strength by a factor of gamma

pervect: Your first answer is basically confirming what I was saying ... I think.

PeterDonis: Thanks for the strict interpretation.

Thanks all of you for replying. I'll chew on this a while longer.
 
  • #11
Verne42 said:
Jartsa: The two charges are moving together.

Ibix: I'll have to chew on your first answer some more. As far as your second answer, I think you are also confirming what I said, the y direction gets strength by a factor of gamma

pervect: Your first answer is basically confirming what I was saying ... I think.

PeterDonis: Thanks for the strict interpretation.

Thanks all of you for replying. I'll chew on this a while longer.

I can expand on my previous answer a bit.

Consider a system of a charge, a meter stick, and an E-field meter, all at rest. The E-field meter will have a constant reading directly proportioanl to the charge, and inversely proportional to the distance from the charge, which I will call L. L is the proper distance between the meter and charge. We can then write

$$(1)\quad E = \frac{q}{4\pi\epsilon_0 L^2}$$

Now, let's view the system from a moving frame, where the motion is in the same direction as the stick. We could either rigidly accelerate the charge, stick, and meter, or we could just change our viewpoint to that of a moving frame.

The meter-stick Lorentz contracts, so if the proper length of the meter stick was L in the meter-stick frame, the distance between the charge and the meter in the moving frame, d, will be Lorentz contracted. This gives us:
$$(2) \quad d = L / \gamma$$

In 26.11 of http://www.feynmanlectures.caltech.edu/II_26.html, Feynman calculates the E-field reading of a stationary meter. We will repeat the calculations using the relationship ##E_{x'}=E_x## from Feynman's table 26.2. This equation tells us that the field in the direction of motion, in this case the x-component of the field, is unchanged by the motion. As we see from the table, this equality is not true in general, but it is true in this case of interest. Thus, we can find the reading of the stationary field meter by finding the reading of a co-moving E-field meter. To find the reading of the co-moving E-field meter, all we need to calculate this is the proper distance between the charge and meter, then we can apply my equation (1)

We know the value of d, which is equal to Feynamn's (x-vt). We know this because in th moving frame, the x-coordinate of the meter is x, and the x-coordinate of the charge at time t is vt, thus the distance between the charge and the meter is (x-vt). We compute that the proper distance, L, between the co-moving E-field meter and the charge is ##L = \gamma \, d = \gamma \, (x-vt)## by using my eq (2) to solve for L.

Applying my eq (1), we then get

$$E = \frac{q}{4\pi\epsilon_0 \, \gamma^2 (x-vt)^2}$$

Substituting ##\gamma = \frac{1}{\sqrt{1-v^2}}## we get Feynman's 26.11

$$E = \frac{q(1-v^2)}{4\pi\epsilon_0 \, (x-vt)^2}$$

Thus we've re-derived 26.11 using a different method, i.e by knowing the reading of a co-moving field meter and transforming the reading to a stationary frame.
 
  • #12
Verne42 said:
Jartsa: The two charges are moving together.
Yes, I understood that. It just did not occur to me that the magnetic field of a fast moving point charge is contracted to a narrow and strong magnetic field just like the electric field of a fast moving point charge is contracted to a narrow and strong electric field. This formula tells us that:

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

And that magnetic field should produce a Lorentz-force that is larger than the Coulomb force, as the net force should decrease like this:

$$F'=\frac {F}{ \gamma}$$
 
  • #13
jartsa and pervect: I'll take and chew on these awhile. Thanks for the time.
 

Related to Understanding E-Field Transformation in Feynman II_26

1. What is the purpose of studying E-Field Transformation in Feynman II_26?

The purpose of studying E-Field Transformation in Feynman II_26 is to gain a better understanding of how electric fields behave and transform in different scenarios. This knowledge can then be applied in various fields such as physics, engineering, and technology.

2. How does Feynman II_26 contribute to our understanding of E-Field Transformation?

Feynman II_26 is a mathematical approach to understanding E-Field Transformation based on the principles of quantum mechanics. It provides a framework for predicting and analyzing the behavior of electric fields in different situations, leading to a deeper understanding of this phenomenon.

3. Can E-Field Transformation be observed in real life?

Yes, E-Field Transformation can be observed in real life through various experiments and applications. For example, the transformation of electric fields can be seen in the behavior of lightning, the functioning of electronic devices, and the behavior of charged particles in magnetic fields.

4. What are some practical applications of understanding E-Field Transformation in Feynman II_26?

The understanding of E-Field Transformation has numerous practical applications, including the design and optimization of electronic circuits, the development of new technologies such as wireless charging and particle accelerators, and the study of natural phenomena such as lightning and auroras.

5. Is Feynman II_26 the only approach to understanding E-Field Transformation?

No, there are other mathematical approaches and theories that can also help in understanding E-Field Transformation. However, Feynman II_26 is a widely recognized and extensively studied approach that has contributed significantly to our understanding of this phenomenon.

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