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Which came first, E or B?

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Phrak
#37
Jun14-10, 01:15 AM
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Quote Quote by cabraham View Post
How does post #26 prove you right? The E & B relations are symmetrical. E*B, & E^2 - B^2 can be both non-zero, both zero, or neither zero. What is your point?
Nevermind your argument with cmos. I don't really like either of these expressions. E^2 - B^2 implies an invariant Z^2 = E^2 - B^2. What is Z? E*B is a dot product without a metric. Neither looks covariant. Can either be obtained from a contraction of the Farday tensor with itself or its dual?
cmos
#38
Jun14-10, 01:16 AM
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Quote Quote by cabraham View Post
How does post #26 prove you right? The E & B relations are symmetrical. E*B, & E^2 - B^2can be both non-zero, both zero, or neither zero. What is your point?

Claude
Perhaps I should be more clear then. The two quantities I quoted are Lorentz invariants. That is to say that they hold the same quantity in all reference frames. This is a standard topic in upper-level, undergraduate relativity.

For example, take
[tex]E^2-B^2=p[/tex],
where p is a constant, to be a positive quantity in some inertial frame. Then that means that
[tex]E'^2-B'^2=p'=p[/tex]
will be the same quantity in ALL reference frames. Since, for this specific example, we took p to be positive, then that means that there is no inertial frame in which the electric field is zero. QED.
cmos
#39
Jun14-10, 01:39 AM
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Quote Quote by Phrak View Post
Nevermind your argument with cmos. I don't really like either of these expressions. E^2 - B^2 implies an invariant Z^2 = E^2 - B^2. What is Z? E*B is a dot product without a metric. Neither looks covariant. Can either be obtained from a contraction of the Farday tensor with itself or its dual?
Exactly.

Modulo some constants (depending on your system of units) and maybe even a Levi-Civita tensor thrown in there, the first expression is the contraction of the field tensor with itself and the second expression is the contraction of the field tensor with its dual. Perhaps it's apparent at this point, but to answer your first question, Z^2 is just a scalar which happens to be invariant under Lorentz transformations. All this is in the standard references.

At this point I fear that we have hijack this thread. May I be the first to apologize.
mysearch
#40
Jun14-10, 03:31 AM
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Quote Quote by cmos View Post
At this point I fear that we have hijack this thread. May I be the first to apologize.
Thatís fine. I will pursue my remaining questions through a new thread:
http://www.physicsforums.com/showthread.php?t=409967
cabraham
#41
Jun14-10, 07:52 AM
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Quote Quote by Phrak View Post
Nevermind your argument with cmos. I don't really like either of these expressions. E^2 - B^2 implies an invariant Z^2 = E^2 - B^2. What is Z? E*B is a dot product without a metric. Neither looks covariant. Can either be obtained from a contraction of the Farday tensor with itself or its dual?
Fine with me. I hate arguments.

Claude
Phrak
#42
Jun15-10, 04:35 PM
P: 4,512
I'll use Λ as the wedge product operator.

Obects I've found that contain products of E and B and are still antisymmetric are

FΛG = 24(B2-E2) and GΛG = -FΛF = 12(E∑B)


Edit: I was in a hurry to leave. To be precise:

*(FΛG) = (B2-E2) and *(GΛG) = -*(FΛF) = (1/2)(E∑B)

The result is a scalar (density). * is the Hodge duality operator which multiplies the four form by eplison with upper indices and a numerical factor, 4!.
K^2
#43
Jun15-10, 08:05 PM
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The entire EM theory can be written in terms of E field only, with B as relativistic correction. It's a bit awkward in that form, but that and the lack of magnetic charges strongly points as E giving rise to B.
cabraham
#44
Jun15-10, 09:08 PM
P: 1,041
Quote Quote by K^2 View Post
The entire EM theory can be written in terms of E field only, with B as relativistic correction. It's a bit awkward in that form, but that and the lack of magnetic charges strongly points as E giving rise to B.
Nothing but a groundless assumption. The lack of magnetic charges does not mean that E gives rise to B.

The E field can be due to discrete electric charges, or to a time varying B field. E fields due to electric charges have a source & a sink, i.e. they start on a +ve charge, & end on a -ve charge. But E fields due to induction are solenoidal, they have no start or end, identical to a B field.

Einstein's 1905 paper firmly states that E & B mutually coexist, & that "neither one is the seat". The term "seat" means "root, basis, primary, main, fundamental, etc.". He lived 50 yrs. beyond that paper & never modified his position. "E giving rise to B" is not supported by the science community. It is just a belief that some adopt, but has no scientific backing.

Claude
mysearch
#45
Jun16-10, 08:01 AM
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Quote Quote by cabraham View Post
"E giving rise to B" is not supported by the science community. It is just a belief that some adopt, but has no scientific backing.
Claude
I hope you can see by the next quote taken from my own post #2 that I am not looking to argue this point, but rather seek a better understanding of some of the energy mechanisms involved in electrodynamic systems.
Quote Quote by mysearch View Post
Possibly the reference to the ‘chicken and egg’ was not a good idea as I was primarily interested in trying to better understand the source of energy and the nature of its transfer within a system as a whole.
If I could quickly cite 2 examples as a frame of reference. The first is an energy system linked to mass only. If a mass (m) is moving with constant velocity, which is maintained by virtue of the conservation of momentum, it also ‘carries’ kinetic energy [tex]1/2 mv^2[/tex]. It might be thought to also ‘carry’ potential energy [tex]-\frac{GMm}{r}[/tex] by virtue of its changing position with respect to another mass (M). The second example is essentially the same as the first, but now mass (m) carries a charge (q). It would seem that everything said about the first example still applies, but the implication of a moving charge would also suggest the existence of a B-field. In this context (B) seems to be analogous to kinetic energy in that it only exists when (q) has a velocity (v). Likewise, the E-field potential energy [tex]-\frac{KQq}{r}[/tex] might initially be thought of as analogous to the gravitational potential, which also moves with the charge [q] in respect to (Q).

Within the context of the first example, I am not sure anybody would argue that potential energy comes first, simply that it can be converted (or give rise) to kinetic energy, but the opposite is equally true. So what about (E) and (B) specifically?

In part, it seems that both have a relationship with energy, which can be somewhat obscured by the system of units, e.g. SI versus gaussian. While SI is the system I normally use, there does seem to be some advantage to the gaussian system in as much that (E) and (B) end up with the same fundamental units, although they don’t immediately seem to convey any obvious meaning until you square the quantity. At this point, if I have done the conversion correctly, both [tex]E^2 and B^2[/tex] seem to align to energy density [tex]\frac {m^2kg}{s^2}* \frac{1}{m^3}= \frac {kg}{ms^2}[/tex]. The nice thing about this perspective is that if you consider (E) and (B) as field strength amplitudes, then the square of these amplitudes is also analogous to the energy, which is true for mechanical waves. Although I am still left pondering what the energy density means with respect to a photon, which carries energy but can’t specify its volume!

One final point regarding Maxwell’s time dependent equations, which are also presented in the gaussian form, but in a reduced format associated with a plane wave (Ey and Bz) propagating, in vacuum, along the x-axis:

[tex]\nabla\times \vec{E} = \frac{\partial{E_y}}{\partial x} = -\frac {1}{c} \frac{\partial{B_z}}{\partial t} [/tex]

[tex]\nabla\times \vec{B} = -\frac{\partial{B_z}}{\partial x} = \frac {1}{c} \frac{\partial{E_y}}{\partial t} [/tex]

If the equations are correct for the assumptions made, doesn’t it suggest that any change to (E) or (B) at a point in time, i.e. position held constant, requires the complementary change in the other field in the direction of propagation? So can I assume that as (E) changes in time, (B) changes in the direction of propagation and vice versa. Would welcome any clarification or corrections. Thanks
cabraham
#46
Jun16-10, 12:33 PM
P: 1,041
But please reread my most recent post. You keep re-iterating the E field associated with a discrete charge. In wave propagation, the E field is solenoidal. i.e. there are no charges present. Of course there is a time varying B field somewhere due to a time varying current, which means that charges are not only moving with time, but that the rate at which they are moving also varies w/ time.

In Maxwell's eqn from Ampere's law we have:

curl H = J + dD/dt.

J is the conduction current, i.e. charge in motion, but dD/dt is the displacement current. In a charge free region in space, this dD/dt is not charges in motion. I was only clarifying that point.

Whenever someone wants to present their case that E comes before B, they use the existence of electric charge & the non-existence of magnetic charge as proof. But discrete electric charge gives rise to only 1 kind of E field. There is another kind of E field, not due to electric charges. This type of E field, an emf or induction type, has no charge sources or sinks. The absence of magnetic charges does not prove that E comes before B. That is all I'm saying.

We still have much to learn about e/m fields, & QED. We will, with time, unlock secrets. But for now, I really don't think the "which came first, E or B?" question can be answered. The only conclusion that science gives us is that under time varying conditions, they mutually co-exist. Neither exists w/o the other, & neither causes the other, nor is caused by the other.

Claude
cmos
#47
Jun17-10, 03:39 PM
P: 367
Quote Quote by K^2 View Post
The entire EM theory can be written in terms of E field only, with B as relativistic correction. It's a bit awkward in that form, but that and the lack of magnetic charges strongly points as E giving rise to B.
Quote Quote by cabraham View Post
Nothing but a groundless assumption. The lack of magnetic charges does not mean that E gives rise to B.
What, I believe, K^2 is talking about is Purcell's example of calculating "magnetic forces" without ever recognizing the existence of the magnetic field. This is covered quantitatively in Purcell's book as well as Sec. 12.3.1 of the 3rd Ed. of Griffith's book, among others. Alternately, do a web search for Hans de Vries; he has written a nice free article about this.

Just to outline the argument for anybody interested:
1) Place a current carrying, but charge neutral, wire next to a moving test charge. Because we already know E&M, we know there will be a magnetic force; however, our goal is to calculate the force without ever recognizing the magnetic field.

2) Lorentz transform to the rest frame of the test charge. In the wire, however, positive and negative charges will be Lorentz contracted differently (because they moved in different directions in the original frame to give a net current). You can assume that in the present frame, there is no current flow in the wire. A result, however, of the different Lorentz contractions is that a net charge builds up in the wire. You can now calculate the electric force between the wire and the test charge. Let's call this F'.

3) You can then go back to the original frame and use the rules of Lorentz transformations on F' and transform it to F, the force between the wire and the test charge in the original frame. If we did not know about magnetism, then F would be a purely relativistic phenomenon. If you invoke the magnetic force, however, your result will still be F.

My view on the matter:
From a classical perspective (since we are in the classical physics forums), you can claim that there are only electric forces; however, those forces must be heavily scrutinized, with the tools of relativity, to accurately describe it in an arbitrary reference frame. Note that, with this specific point of view, we are necessarily talking about an action-at-a-distance theory. It is NOT a field theory. A proper field theory would require the recognition of both E and B; e.g. how conventional covariant E&M recognizes both E and B as components of the field tensor.
cabraham
#48
Jun17-10, 07:34 PM
P: 1,041
Quote Quote by cmos View Post
What, I believe, K^2 is talking about is Purcell's example of calculating "magnetic forces" without ever recognizing the existence of the magnetic field. This is covered quantitatively in Purcell's book as well as Sec. 12.3.1 of the 3rd Ed. of Griffith's book, among others. Alternately, do a web search for Hans de Vries; he has written a nice free article about this.

Just to outline the argument for anybody interested:
1) Place a current carrying, but charge neutral, wire next to a moving test charge. Because we already know E&M, we know there will be a magnetic force; however, our goal is to calculate the force without ever recognizing the magnetic field.

2) Lorentz transform to the rest frame of the test charge. In the wire, however, positive and negative charges will be Lorentz contracted differently (because they moved in different directions in the original frame to give a net current). You can assume that in the present frame, there is no current flow in the wire. A result, however, of the different Lorentz contractions is that a net charge builds up in the wire. You can now calculate the electric force between the wire and the test charge. Let's call this F'.

3) You can then go back to the original frame and use the rules of Lorentz transformations on F' and transform it to F, the force between the wire and the test charge in the original frame. If we did not know about magnetism, then F would be a purely relativistic phenomenon. If you invoke the magnetic force, however, your result will still be F.

My view on the matter:
From a classical perspective (since we are in the classical physics forums), you can claim that there are only electric forces; however, those forces must be heavily scrutinized, with the tools of relativity, to accurately describe it in an arbitrary reference frame. Note that, with this specific point of view, we are necessarily talking about an action-at-a-distance theory. It is NOT a field theory. A proper field theory would require the recognition of both E and B; e.g. how conventional covariant E&M recognizes both E and B as components of the field tensor.
That's a fair presentation. E & B as components of a tensor is what modern physics uses. Of course you mentioned "placing a test charge" near the wire. I presume that said charge is in motion so that v x B is non-zero. Otherwise there is no force acting on said test charge.

My view is the following. In his 1905 paper "OTEOMB", AE stated that "neither is the seat". He lived another 50 yrs., & 55 yrs. have elapsed since then. Nobody has ever been able to topple his position. I accept AE's thesis that neither E nor B can be taken as the "seat". It has withstood scrutiny for 105 yrs. & still stands like Gibraltar. I cannot argue with such an immutable finding.

They mutually co-exist. You brought some good points to the discussion which I appreciate.

Claude


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