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

by kmarinas86
Tags: charges, fields, jerk, magnetic, responsible, velocity
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 Quote by kmarinas86 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
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Quote by DaleSpam
 Quote by kmarinas86 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.

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

 Quote by kmarinas86 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.
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 Quote by DaleSpam 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 $1 - \mathbf{n} \cdot \boldsymbol{\beta}=1$, I can see that the scalar potential $\varphi(\mathbf{r}, t)$ is a function of position (zeroth-order derivative of position) and the time derivative of $\mathbf{A}(\mathbf{r},t)$ is a function of acceleration (second-order derivative of position). The gradient of the scalar potential at $\mathbf{r}$ is proportional to the value of $\varphi(\mathbf{r}, t)$ at $r$. Thus $\mathbf{E} = - \nabla \varphi - \dfrac{\partial \mathbf{A}}{\partial t}$ depends on the zeroth and second-order derivatives of position. The curl of $\mathbf{A}(\mathbf{r},t)$ at $r$ is proportional to $\mathbf{A}(\mathbf{r},t)$ and the sine of the angle between gradient of $\mathbf{A}(\mathbf{r},t)$ at $r$ and the direction of $\mathbf{A}(\mathbf{r},t)$ at $r$. This makes $\mathbf{B} = \nabla \times \mathbf{A}$ dependent on relative velocity (the first-order derivative of position).
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Summary of key questions:

 Quote by kmarinas86 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?
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 Quote by kmarinas86 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 by kmarinas86 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.
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 Quote by DaleSpam 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 $\mathbf{n} - \boldsymbol{\beta}$ appears in the numerators of E and B forms of the Liénard-Wiechert potentials, while the factor $1 - \mathbf{n} \cdot \boldsymbol{\beta}$ appears in the denominators of them. Neither the $- \boldsymbol{\beta}$ term nor the $- \mathbf{n} \cdot \boldsymbol{\beta}$ term "correspond" to the value of E and B. I hope that clears it up.
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 Quote by kmarinas86 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).
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 Quote by DaleSpam 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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