Is "E" in Maxwell-Equations really "E"?


by ManDay
Tags: maxwellequations
ManDay
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Apr18-13, 03:14 AM
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Consider a perfectly static and spatially bound magnetic field B ∈ ℝ such that

B ≠ 0 ; ∂/∂t B = 0

further, a continuous, time-stationary, and spatially bound current j ∈ ℝ through that magnetic field

j ≠ 0 ; ∂/∂t j = 0 ; ∇ j = 0

which contributes to B but does not violate ∂/∂t B = 0. The initial charge distribution ρ = 0 for all space. It will therefore remain 0 for all times.

The Lorentz force predicts a force acting on a moving density of charge ρ'. j may be considered such a moving density superimposed by a stationary density -ρ' to satisfy ρ = 0. Therefore, the Lorentz force predicts an (effective) electric field, which is defined by its action on charges - but only on the moving ones, this time! Maxwell Equations, however, say that there is no electric field at all - on moving charges or stationary alike.

How can this be reconciled?
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Apr18-13, 03:24 AM
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Therefore, the Lorentz force predicts an (effective) electric field, which is defined by its action on charges
The force is usually thought of as being due to the magnetic field interacting with the charges. Remember the definition of the electric field vector ##\vec{E}##?
ManDay
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Apr18-13, 03:51 AM
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E is defined as the force per unit of charge on a charge (or density thereof). That's my very point. By that defintion, it should account for the Lorentz force due to a magnetic field as well. But as illustrated, it does not. Maxwell equations which determine E uniquely do not account for the Lorentz force. Therefore, I see a contradiction.

I know my question is not exactly precisely posed, but here is the same contradiction from a different perspective:

Maxwell's equations do not predict any change in electrostatics if j is made larger or smaller. E and the associated potential U will remain exactly the same (namely 0 and a constant respectively).

Now consider a Hall-conductor. Obviously, a different current j will produce a different potential across the perpendicular side! That contradicts Maxwell's prediction if the defintions of E and U coincide.

vanhees71
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Apr18-13, 06:31 AM
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Is "E" in Maxwell-Equations really "E"?


The situation you describe is a bit vague. So let's make it more concrete by the usually solved problem of an infinitely long cylindrical DC conducting wire. There the usual treatment in textbooks is non-relativistic. That's of course fully correct, because the drift velocity of the electrons is very tiny compared to the speed of light, but it's a typical example for the confusion in electromagnetics caused by the tacitly assumed non-relativistic treatment of the matter around.

In fact, here it leads to a contradiction to precisely the point you make concerning the Lorentz force acting on the flowing electrons, which indeed is
[tex]\vec{F}_{\text{L}}=q \left (\vec{E}+\frac{\vec{v}}{c} \vec{B} \right ).[/tex]
In order that the flow is stationary, this force together with the friction force should be 0. The friction force is lumped into the constitutive equation involving the conductivity, and this should read
[tex]\vec{j}=\sigma \underbrace{\frac{1}{\sqrt{1-v^2/c^2}}}_{\gamma} \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]
Here, [itex]\vec{v}[/itex] is the drift velocity of the electrons, and thus
[tex]\vec{j}=n_{\text{cond}} \vec{v}.[/tex]
The friction force on the electrons is compensated by the driving force from the external electric field which is somehow given by the voltage difference between the ends of the wire (which of course in reality is of finite length, but we are looking at a very long wire not to close to the ends). Thus we can make the ansatz
[tex]\vec{v}=v \vec{e}_z,[/tex]
where the [itex]z[/itex] axis is in direction of the wire. Further, assuming [itex]\mu=\epsilon=1[/itex] for simplicity we have the equations
[tex]\vec{\nabla} \cdot \vec{E}=n_{\text{Q}}, \quad \vec{\nabla} \cdot \vec{B}=0,[/tex]
[tex]\vec{\nabla} \times \vec{B} = \frac{1}{c} \vec{j}, \quad \vec{j}=\sigma \gamma \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right).[/tex]
We assume that [itex]n_{\text{cond}}[/itex] is constant in time and thus due to the stationarity condition also [itex]\vec{v}=v_z \vec{e}_z[/itex] and thus finally also [itex]\vec{j}=n_{\text{cond}} \vec{e}_z.[/itex]

One has now to solve for the various unknowns in this equations, which are the em. field [itex](\vec{E},\vec{B})[/itex] and [itex]n_Q[/itex], which is the total charge density in the wire. It's easy to see that the non-relativistic approximation [itex]n_Q=0[/itex] is inconsistent:

Taking the divergence of Ohm's Law gives
[tex]\vec{\nabla} \cdot \vec{j}=\vec{\nabla} (\vec{\nabla} \times \vec{B})=0=\sigma \gamma \vec{\nabla} \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]
Using [itex]\vec{v}=\text{const}.[/itex] leads to
[tex]\vec{\nabla} \cdot \vec{E}=n_Q=\frac{v^2}{c^2} n_{\text{cond}}=\text{const}.[/tex]

Solving the system further leads to a tiny electric-field component in radial direction (i.e., [itex]\perp \vec{e}_z[/itex] in addition to the electric field in [itex]z[/itex] direction. This is determined by the demand that the total force on an electron in radial direction must be 0 in order to keep the corresponding current component 0. Thus the force on this electron by the magnetic field produced by all other flowing electrons must be compensated by precisely this electric field, which is the (static) Hall effect. Thus you have a tiny negative charge density within the wire, producing this field.

In reality for any "household current" these effects are of course practically 0, because [itex]v \lesssim 10^{-3} \text{m/s}.[/itex] The non-relativistic approximation is thus practically exact, but in principle there is this tiny Hall effect, making the whole scheme consistent with the full treatment of the Lorentz force on a conducting electron in the wire. This nicely demonstrates the connection between relativity and the necessity for the [itex]\vec{v}\times \vec{B}[/itex] term of the Lorentz force, i.e., the force on a particle moving in a magnetic field.
Philip Wood
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Apr18-13, 08:21 AM
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Quote Quote by ManDay View Post
Therefore, the Lorentz force predicts an (effective) electric field, which is defined by its action on charges - but only on the moving ones
When a force acts on a moving charge but not on stationary charge, the charge is situated in a magnetic field. There is no electric field. If there were an electric field, then even a stationary charge would experience a force.

What I'm doing here is defining electric and magnetic fields using the Lorentz force. This is what the elementary definitions of E and B amount to: the quantities in F = qE + qvxB which will give rise to the observed force F on a 'test' charge q.
ZapperZ
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Apr18-13, 08:24 AM
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Quote Quote by ManDay View Post
E is defined as the force per unit of charge on a charge (or density thereof). That's my very point. By that defintion, it should account for the Lorentz force due to a magnetic field as well. But as illustrated, it does not. Maxwell equations which determine E uniquely do not account for the Lorentz force. Therefore, I see a contradiction.
But you are completely ignoring the fact that based on the dimensions of E alone (i.e. V/m), it actually is more clearly defined as the gradient of the electrostatic potential!

Zz.
cabraham
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Apr18-13, 08:48 AM
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Quote Quote by ZapperZ View Post
But you are completely ignoring the fact that based on the dimensions of E alone (i.e. V/m), it actually is more clearly defined as the gradient of the electrostatic potential!

Zz.
Not at all. Two flaws in that argument. E = -grad V is derived from the integral V = integral E*dl, by working backwards. Hayt's text Electromagnetic Fields for Engineers details this proof. Also, if B is time varying, then E is a non-conservative field, i.e. E fields associated with Faraday induction, cannot be represented as a scalar potential gradient.

E = -grad V does not apply for this non-conservative type of E field. This type of E field is described by E = -dA/dt. Did I help?

Claude
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Apr18-13, 08:57 AM
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Quote Quote by cabraham View Post
Not at all. Two flaws in that argument. E = -grad V is derived from the integral V = integral E*dl, by working backwards. Hayt's text Electromagnetic Fields for Engineers details this proof. Also, if B is time varying, then E is a non-conservative field, i.e. E fields associated with Faraday induction, cannot be represented as a scalar potential gradient.

E = -grad V does not apply for this non-conservative type of E field. This type of E field is described by E = -dA/dt. Did I help?

Claude
I didn't realize this was not an electrostatic case. The post that I replied to appears to indicate that it is.

Still, how does that change the fact that, per the dimension of E alone, it really is the change in potential per unit length, and not solely defined in terms of charge?

Zz.
cabraham
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Apr18-13, 09:50 AM
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Quote Quote by ZapperZ View Post
I didn't realize this was not an electrostatic case. The post that I replied to appears to indicate that it is.

Still, how does that change the fact that, per the dimension of E alone, it really is the change in potential per unit length, and not solely defined in terms of charge?

Zz.
The 2 are equivalent, i.e. an E field intensity is equivalent to a change in V per unit length. No argument there. But I only wished to point out that "V" is defined in terms of work per unit charge, work is the integral of force over path, but force is E/q, where F is Coulomb's law force.

The definitions quickly become circular. Coulomb's force law is, I believe, an axiom. We "feel" force between 2 charged bodies. But changes in a body's position require a finite time to be felt by the other body, hence the field concept was born. E = F/q, is a basic definition, derived only from Coulomb's law and the fact that changes in force due to charge motion require a finite propagation time. The E field due to charged bodies w/o time varying B present are conservative. If we define potential V, by integrating the force over the path, divided by unit charge, we get V = integral E*dl, where "dl" is incremental path length.

This type of field is path independent. So Hayt in his text works backwards and start w/ the integral, ending up with E as a function of V, hence E = -grad V. The integral where V is defined by E is the starting point. In other words E is defined then V derived. V does not define E.

This is supported by the case w/ time varying B. When fields are induced due to Faraday induction, i.e. curl E = -dB/dt, the relations hold as I described for V = integral E*dl. However in this case E is non-conservative and is not the gradient of a scalar potential. So V is always defined as the integral of E*dl, under all conditions.

If E is conservative, we can say E is the -grad V. If E is non-conservative, V is still integral E*dl, but E is NOT -grad V. I hope I did not make matters worse. Did this help? BR.

Claude
cabraham
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Apr18-13, 11:10 AM
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Maxwell does not predict zero E field, rather zero curl of E field. There is a difference. Maxwell per Faraday is curl E = -dB/dt. So a zero value of dB/dt simply means that the E field can be non-zero or zero, with NO ROTATION. I.e. curl E = 0. In the op case, Ohm's law applies as always, E = rho*J. Also, Ampere's law states that curl H = J + dD/dt. For static case, dD.dt = 0, so that curl H = J. But J = sigma*E, where sigma = 1/rho, so we have curl H = sigma E. In this case H is static (non-varying with time), and J is current density. As long as sigma is finite there will be non-zero E. For a SC, E=0 even if J is finite.

Did I help at all?

Claude
Simon Bridge
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Apr18-13, 04:03 PM
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Quote Quote by ManDay
E is defined as the force per unit of charge on a charge (or density thereof). That's my very point.
You missed out the very important requirement that the charge be stationary at the time the force is measured. Force on a moving charge is not due to a static electric field - by the definition of the field.

@cabraham: how does all that answer OPs original question?
Is the force on a moving charge evidence of a special electric field that only acts on moving charges?
Do we really need to postulate the existence of such a field to explain the motion?
cabraham
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Apr18-13, 05:16 PM
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Quote Quote by Simon Bridge View Post
@cabraham: how does all that answer OPs original question?
Is the force on a moving charge evidence of a special electric field that only acts on moving charges?
Do we really need to postulate the existence of such a field to explain the motion?
Referring to the OP original question, he stated that the Lorents force moves charges, which it does. He then discusses the accompanying E field, which we know from Ohm's law E = rho*J. He then asks how this can be in light of Maxwell per Faraday since dB/dt = 0 implies "no E field"? I pointed out that no E field rotation or curl, does not imply zero E field, just that E fields due to charged bodies are conservative, i.e. ir-rotational.

Thus Lorentz' force law, Ohm;s law, & Maxwell-Faraday law are in accord. That was the point I was making. If I need to explain further let me know. BR.

Claude


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