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Are both velocity and jerk of charges responsible for magnetic fields?

by kmarinas86
Tags: charges, fields, jerk, magnetic, responsible, velocity
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kmarinas86
#1
Mar7-12, 09:24 PM
P: 1,011
It is commonly believed that changes of electric fields cause contributions to a component of the magnetic field and that changes of magnetic fields cause contributions to a component of the electric field.

In the case where the fields in question are caused by electric charges, do electric fields correspond to even-order derivatives of position, while magnetic fields correspond to odd-order derivatives of position? If so, I would first like to make the statement that this would mean that odd-order derivatives of position of an order greater than one would make contributions to the magnetic field that would still be present in other inertial frames of reference. This would be consistent with the idea that not all magnetic fields can be "Lorentz-transformed" out of the system.

My further analysis is as follows:
______

In AC power, the zeroth, first, second, and third derivatives of position would, I presume, have much greater significance in the total electric and magnetic fields than any higher derivatives. In the case for electric charges, this would correspond to charge, drift current, I'/t', and I''/t' respectively.

charge = Q = linear charge density * length
drift current = Q'/t' = I = linear charge density * velocity
I'/t' = linear charge density * acceleration
I''/t' = linear charge density * rate change of acceleration

From what I understand, the derivative of an acceleration, with respect to time, corresponds to the derivative of force divided by mass, with respect to time. So for a constant mass, I''/t' is proportional to the derivative of force with respect to time. The gradient of the scalar electric potential (in units of force divided by the electric charge) corresponds to a component of the electric field. The derivative of that electric field with respect to time, integrated over the surface area of a conductor and multiplied by electrical permittivity gives the value for displacement current generated in a direction normal to that conductor's surface. Therefore, regardless of the direction the total displacement current, a total that includes the component of displacement current down the length of the conductor, it may ultimately be due to jerk of the charges.

Since jerk is the second-derivative of velocity, there should be cases where the jerk of the charges contribute more to the magnetic field than that provided by the velocity of the charges. While the energy stored in the magnetic field of a solenoid is proportional to the square of velocity, as integration of a variable (v) would imply (i.e. 1/2 * v^2), in the case of the displacement current, we would have a jerk of the electrons, and let's call it k instead of j to not conflict with the imaginary number j used in electrical engineering. Given a constant linear charge density, charge drift velocity v is proportional to drift current, and k is proportional to displacement current.

Knowing that:
sin(t)' = cos(t)
cos(t)' = -sin(t)
-sin(t)' = -cos(t)
-cos(t)' = sin(t)

In the case for a sinusoidal drift current, we have the following relations:

charge / charge_max = -cos(t)
drift current / drift current_max = sin(t)
(I'/t') / (I'/t')_max = cos(t)
(I''/t') / (I''/t')_max = -sin(t)

In this case, the derivative of drift current / drift current_max ((drift current / drift current_max)'=cos(t)) is exactly opposite that of (I''/t') / (I''/t')_max (((I''/t') / (I''/t')_max)'=-cos(theta)). In the case of a lossless LC circuit, changes in drift current are directly opposite of changes of displacement current. However, obviously, this condition does not hold when switching the LC circuit from the off position (open circuit) to the on position (closed circuit), for if the changes of these two currents were directly equal and opposite from one another from the moment of being switched on, then no net change of current, and thus no net change of magnetic field, could develop in the first place. The alternating net magnetic field clearly does exist, even in an AC circuit.

Have I got it right so far?

First, let's stick to the sinusoidal condition:
Now considering that the electromagnetic field produced by a drift current has a stored energy content proportional to v^2, we should analogously expect that the displacement current stores magnetic potential energy proportional to k^2. In a sinusoidal situation, this would imply a total magnetic potential energy storage proportional to x(sin(t))^2 + y(-sin(t))^2, where x and y are proportionality constants. So obviously in that case we would have a situation where the magnetic potential energy goes up and down. Presumably, the electrical potential energy would vary according to, say, x(-cos(t))^2 + y(cos(t))^2, corresponding to position and acceleration, respectively.

But what if we do not have a sinusoidal condition? Then the derivative of the drift current / drift current_max is not necessarily equal and opposite to the derivative of (I''/t') / (I''/t')_max. Then the idea that displacement current was necessarily converted from the drift current becomes invalid.

Maybe the displacement current is a manifestation of the jerk of electrons or other kinds of charge, such as virtual electron-positron pairs in the vacuum? If the magnetic potential energy stored in the field of an inductor were proportional to (I^2 + b^2*(I''/t')^2) (where b is another constant) instead of simply being proportional to I^2, all one would have to do to create a magnetic field is to jerk the electrons in the circuit. The rate at which you could turn the field on and off would in turn be proportional to jounce of the electrons, as opposed to the acceleration of the electrons, as in the case of passing alternating drift current in an inductor.

While capacitance relates to stored charge, resistance relates to drift current (I), and inductance relates to I'/t', there does not appear to be any conventionally taught terms relating to I''/t' and I'''/t'.

We would have an effective current equal to I + b*(I''/t'). The integral of this with respect to time would equal to the "effective" charge passed through. The integral of I with respect to time (=Q) is the amount of electrical charge that had passed through a surface cutting through the conducting path. The integral of (I''/t') with respect to time is (I'/t'), which equals linear charge density * acceleration. The electric flux divided by surface of that conductor gives the electric field, and the charge subject to that field will have an acceleration proportional to it. Therefore, the increase of electric flux through that surface is proportional to the acceleration of the electrons at that surface. When the acceleration is ceased, as is the case for DC, then this additional electric flux component is eliminated. That residual electric flux could thus be switched on and off simply by applying and removing an external E-field onto to the electrons in the circuit. The stored energy of that electric flux would fluctuate at a rate that can easily be disproportional to drift current, as it would instead be proportional to the second derivative of drift current with respect to time.... For, in the case of non-sinusoidal drift currents, particularly in situations where the circuit is driven at a frequency higher than its natural one (off-resonant), the rise of displacement current produced is "decoupled" from the fall in drift current, resulting in what may lead to a displacement current greater than the peak value of I.

As more charge is withdrawn from the (-) source, the acceleration of the charges plummet exponentially as time passes, as does the displacement current. Some of this is acceleration down the length of the conductor, and some of it is deceleration with respect to the surface of the conductor. What amount of time does it take for the E-field to be created? Very little time, limited by the propagation speed of the electromagnetic wave down the conductor and the length of that conductor. Thus, simply applying the electric field polarizes the medium interface consisting of the wire and the space out of it much faster than electrons flow through it. The energy stored in the polarization of the medium consisting of that interface is therefore proportional to the deceleration of electrons with respect to surface of the conductor. The power of the alternation between states of polarization is therefore proportional to the jerk of the electrons with respect to the surface of the conductor, for it is that component of acceleration contributing to the displacement current through that surface. Another way to look at it is that this component of power is proportional to the slew rate of the electric flux, which for a fixed area, is proportional to the slew rate of the electric field, and which for a fixed length, is proportional to the slew rate of the voltage. Therefore, for a wire element of fixed dimensions, the power of the alternation between states of polarization is proportional to the rate change of the voltage. The maximum of that power occurs just after the very moment the circuit is switched on, which is also when the drift current is at a minimum.

_____

In contrast to the above, for magnetic dipoles, we would have the opposite relationship, in which case electric fields correspond to odd-order derivatives of position and magnetic fields correspond to even-order derivatives of position.

The above would be implied if variations of electric and magnetic fields were based on derivatives of positions of electric monopoles and magnetic dipoles. Therefore, I ask this:

Are both velocity and "jerk" of charges responsible for magnetic fields?

If the answer is yes, then I also ask the following:

In this situation, we have capacitance formed between a wire in a closed circuit and the surrounding environment, causing lines of electric flux to concentrate near the - end and + end of the wire. Presumably, this causes displacement current to flow out of the wire at some acute angle to the surface of the wire depending on the weakening of the electric field from the - end to the + end of the circuit. This would result in displacement current possessing a net longitudinal component outside of the wire along its course between the - end and the + end. Thus, it would act as a displacement current in parallel to the drift current, and it would be capable of producing a magnetic field that is generally in the same direction as that which would be produced by the flow of the electrons (with minor deviations near the - and + ends of the wire). In other words, it would reinforce the magnetic field produced by the electrons, in a way similar, but different, to that of a capacitor hooked in parallel.

Is it conceivable that for non-sinusoidal waves generated by pulsing far beyond the natural frequency of the circuit, such as electrical impulses, that the relationship, between excess charge on the conductor and the number of charges passing through a conductor at any given time, breaks down so much that electrons that move down the wire account only for up to 1% of the magnetic field in the wire?

If that is conceivable, this would mean that the drift current can be arbitrarily small, as if the circuit were still open, and yet the same desired magnetic field can be produced, and propagating near the speed of light, if circuit parameters are optimized to manifest this condition, though that component of the magnetic field (due to displacement current) would only exist so as long as I''/t' is not zero.

Thus, it would be logical to think that the number of electrons displaced from the (-) end of the circuit can be made arbitrarily small compared to the fields produced surrounding the conductor. Is that not a paradox? What is missing here?

If all the above is the case for non-sinusoidal pulses, in a non-sinusoidal waveform, where does the energy stored as a magnetic field (resulting from the displacement currents flowing between the negative end to the positive end of the circuit) come from, if it is arbitrarily disproportional with the amount of charge displaced from the (-) end of the circuit?
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DaleSpam
#2
Mar7-12, 10:07 PM
Mentor
P: 17,543
Quote Quote by kmarinas86 View Post
In the case where the fields in question are caused by electric charges, do electric fields correspond to even-order derivatives of position, while magnetic fields correspond to odd-order derivatives of position?
No.

See: http://en.wikipedia.org/wiki/Li%C3%A...hert_potential
kmarinas86
#3
Mar7-12, 11:34 PM
P: 1,011
Quote Quote by DaleSpam View Post
Quote Quote by kmarinas86 View Post
In the case where the fields in question are caused by electric charges, do electric fields correspond to even-order derivatives of position, while magnetic fields correspond to odd-order derivatives of position?
No.

See: http://en.wikipedia.org/wiki/Li%C3%A...hert_potential
If so, then it is wrong to say that some component of the magnetic field can depend on the derivative of an electric field, and that some component of the electric field can depend on the derivative of the magnetic field. So, is that wrong to say?

BTW: You do realize that some formulas which do not have higher derivative terms can be expanded to an infinite series with those higher derivatives... Right? This is why I find your answer incomplete.

DaleSpam
#4
Mar8-12, 06:16 AM
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P: 17,543
Are both velocity and jerk of charges responsible for magnetic fields?

Quote Quote by kmarinas86 View Post
BTW: You do realize that some formulas which do not have higher derivative terms can be expanded to an infinite series with those higher derivatives... Right? This is why I find your answer incomplete.
You are certainly welcome to do such an expansion for the LW fields, but whatever it may do at 3rd order and higher derivatives it is clear that the E field depends on the 1st derivative of position and the B field depends on 0th and 2nd order derivatives. That is in direct contradiction to your assumption.
kmarinas86
#5
Mar8-12, 09:17 AM
P: 1,011
Quote Quote by DaleSpam View Post
You are certainly welcome to do such an expansion for the LW fields, but whatever it may do at 3rd order and higher derivatives it is clear that the E field depends on the 1st derivative of position and the B field depends on 0th and 2nd order derivatives. That is in direct contradiction to your assumption.
From what I understand:
The zeroth derivative of position is just position.
The first derivative of position is velocity.
The second derivative of position is acceleration. Etc.

So if a charged particle has no velocity relative to the observer (i.e. first-order derivative of position x'=0), how can there be no electric field?

Just by looking at the equations, in the case that we assume that [itex]1 - \mathbf{n} \cdot \boldsymbol{\beta}=1[/itex], I can see that the scalar potential [itex]\varphi(\mathbf{r}, t)[/itex] is a function of position (zeroth-order derivative of position) and the time derivative of [itex]\mathbf{A}(\mathbf{r},t)[/itex] is a function of acceleration (second-order derivative of position). The gradient of the scalar potential at [itex]\mathbf{r}[/itex] is proportional to the value of [itex]\varphi(\mathbf{r}, t)[/itex] at [itex]r[/itex]. Thus [itex]\mathbf{E} = - \nabla \varphi - \dfrac{\partial \mathbf{A}}{\partial t}[/itex] depends on the zeroth and second-order derivatives of position. The curl of [itex]\mathbf{A}(\mathbf{r},t)[/itex] at [itex]r[/itex] is proportional to [itex]\mathbf{A}(\mathbf{r},t)[/itex] and the sine of the angle between gradient of [itex]\mathbf{A}(\mathbf{r},t)[/itex] at [itex]r[/itex] and the direction of [itex]\mathbf{A}(\mathbf{r},t)[/itex] at [itex]r[/itex]. This makes [itex]\mathbf{B} = \nabla \times \mathbf{A}[/itex] dependent on relative velocity (the first-order derivative of position).
kmarinas86
#6
Mar8-12, 09:20 AM
P: 1,011
Summary of key questions:

Quote Quote by kmarinas86 View Post
In the case where the fields in question are caused by electric charges, do electric fields correspond to even-order derivatives of position, while magnetic fields correspond to odd-order derivatives of position?

(True or False?) [T]he derivative of an acceleration, with respect to time, corresponds to the derivative of force divided by mass, with respect to time. So for a constant mass, I''/t' is proportional to the derivative of force with respect to time. The gradient of the scalar electric potential (in units of force divided by the electric charge) corresponds to a component of the electric field. The derivative of that electric field with respect to time, integrated over the surface area of a conductor and multiplied by electrical permittivity gives the value for displacement current generated in a direction normal to that conductor's surface. Therefore, regardless of the direction [of] the total displacement current, a total that includes the component of displacement current down the length of the conductor, it may ultimately be due to jerk of the charges.

(True or False?) In the case for a sinusoidal drift current, we have the following relations:
charge / charge_max = -cos(t)
drift current / drift current_max = sin(t)
(I'/t') / (I'/t')_max = cos(t)
(I''/t') / (I''/t')_max = -sin(t)
In this case, the derivative of drift current / drift current_max ((drift current / drift current_max)'=cos(t)) is exactly opposite that of (I''/t') / (I''/t')_max (((I''/t') / (I''/t')_max)'=-cos(theta)). In the case of a lossless LC circuit, changes in drift current are directly opposite of changes of displacement current. However, obviously, this condition does not hold when switching the LC circuit from the off position (open circuit) to the on position (closed circuit), for if the changes of these two currents were directly equal and opposite from one another from the moment of being switched on, then no net change of current, and thus no net change of magnetic field, could develop in the first place. The alternating net magnetic field clearly does exist, even in an AC circuit.

But what if we do not have a sinusoidal condition?

Maybe the displacement current is a manifestation of the jerk of electrons or other kinds of charge, such as virtual electron-positron pairs in the vacuum?

Are both velocity and "jerk" of charges responsible for magnetic fields?

Is it conceivable that for non-sinusoidal waves generated by pulsing far beyond the natural frequency of the circuit, such as electrical impulses, that the relationship, between excess charge on the conductor and the number of charges passing through a conductor at any given time, breaks down so much that electrons that move down the wire account only for up to 1% of the magnetic field [surround]in[g] the wire?

Thus, it would be logical to think that the number of electrons displaced from the (-) end of the circuit can be made arbitrarily small compared to the fields produced surrounding the conductor. Is that not a paradox? What is missing here?

If all the above is the case for non-sinusoidal pulses, in a non-sinusoidal waveform, where does the energy stored as a magnetic field (resulting from the displacement currents flowing between the negative end to the positive end of the circuit) come from, if it is arbitrarily disproportional with the amount of charge displaced from the (-) end of the circuit?
DaleSpam
#7
Mar8-12, 11:04 AM
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P: 17,543
Quote Quote by kmarinas86 View Post
From what I understand:
The zeroth derivative of position is just position.
The first derivative of position is velocity.
The second derivative of position is acceleration. Etc.
Correct.

Quote Quote by kmarinas86 View Post
So if a charged particle has no velocity relative to the observer (i.e. first-order derivative of position x'=0), how can there be no electric field?
I said that the E field depends on the 1st derivative (contradicting your statement that it corresponded to even derivatives). I never said that the E field depends ONLY on the 1st derivative. Similarly for the B field.
kmarinas86
#8
Mar8-12, 12:07 PM
P: 1,011
Quote Quote by DaleSpam View Post
Correct.

I said that the E field depends on the 1st derivative (contradicting your statement that it corresponded to even derivatives). I never said that the E field depends ONLY on the 1st derivative. Similarly for the B field.
When I say "correspond", I say this to mean "having a dependent degree of proportionality" (a less elegant way of saying "correspond"). For example, if:

a = c*(1-b)/(1-d)

a corresponds to c but not either to b nor d.

The degree of proportionality depends on b or d, but neither b nor d are proportional to a.

The factor [itex]\mathbf{n} - \boldsymbol{\beta}[/itex] appears in the numerators of E and B forms of the Liénard-Wiechert potentials, while the factor [itex]1 - \mathbf{n} \cdot \boldsymbol{\beta}[/itex] appears in the denominators of them. Neither the [itex]- \boldsymbol{\beta}[/itex] term nor the [itex]- \mathbf{n} \cdot \boldsymbol{\beta}[/itex] term "correspond" to the value of E and B. I hope that clears it up.
DaleSpam
#9
Mar8-12, 12:45 PM
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P: 17,543
Quote Quote by kmarinas86 View Post
When I say "correspond", I say this to mean "having a dependent degree of proportionality" (a less elegant way of saying "correspond"). For example, if:

a = c*(1-b)/(1-d)

a corresponds to c but not either to b nor d.

The degree of proportionality depends on b or d, but neither b nor d are proportional to a.
Then by that very strange definition of "corrsepond" the E field does not "correspond" to any derivatives of the position (also contradicting your assumption).
kmarinas86
#10
Mar8-12, 01:31 PM
P: 1,011
Quote Quote by DaleSpam View Post
Then by that very strange definition of "corrsepond" the E and B fields do not "correspond" to any derivatives of the position (also contradicting your assumption).
a=c*(1-b)/(1-d)

c is a factor in that equations. So are (1-b) and (1-d). -b and -d are the second terms of each factor, respectively. (1-b)/(1-d) "corresponds" to a as well, but both factors (1-b) and (1-d) are dimensionless. a and c are not necessarily dimensionless. The point anyway is not the dimensionality, but rather the fact that -b and -d are terms, not the factors themselves. It is the factors which may have proportionality, not the terms within them.

I defined current as:

current = linear charge density * velocity

linear charge density and velocity are factors in this equation. They both "correspond" to current. On the other hand, say we had the following equation:

current = linear charge density * (speed of light - some other velocity)

"some other velocity" does not correspond to the current, even if "velocity = speed of light - some other velocity".


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