# (Self) Inductance of a straight current carrying wire, vol. 2

1. Aug 18, 2015

### QuasiParticle

This thread is continuation to:

https://www.physicsforums.com/threads/self-inductance-of-a-straight-current-carrying-wire.461622/

It's a really old thread, but I happened to come across the same problem a while back and since the old thread does not give the correct answer, I decided to create continuation for it. The solution was actually a little bit surprising to me.

If one considers a single infinitely long straight wire with radius $R$ and computes the energy from $1/2 \int B^2/\mu_0$, it indeed gives infinity for the reason already mentioned in the old thread. However, if one calculates the total energy using the vector potential: $E = 1/2 \int \vec{A} \cdot \vec{J} d^3r$, the energy (and inductance) is finite*.

It turns out that the energy density $B^2/2\mu$ is not correct in the case of an infinite wire. The reason I was surprised was that I had always thought $E=1/2\int B^2/\mu$ to be exact in all cases. When deriving this result for the energy, the integration is extended to all space and a surface integral term appearing in the expression is neglected. The term in question is

$$-\frac{1}{2\mu} \oint (\vec{A} \times \vec{B}) \cdot d\vec{S}$$
Usually this can be neglected, when integrating over all space. For a dipole field, for example, $B\sim1/r^3$ and $A\sim1/r^2$, so the surface integrand is $\sim1/r^3$. The infinite wire, however, is a very special case, and the surface integral term becomes infinite.

This also demonstrates how difficult it is to localize electromagnetic energy. The integration of $B^2/2\mu$ must be extended over all space to give the correct total energy. This leads me to state that $B^2/\mu$ does not represent any true energy density of the field, even though it has units $[\mathrm{J/m}^3]$.

*There remains a constant in the vector potential, which I cannot determine and which is a bit strange. The vector potential is:

$$A_z=\left\{ \begin{array}{l l} \frac{\mu_0I}{4\pi}\left(C-\frac{r^2}{R^2} \right) & r\leq R\\ \frac{\mu_0I}{4\pi}\left(C-1-\ln \frac{r^2}{R^2} \right) & r\geq R \end{array}\right.$$
I think the correct answer is found, when we set $C=1$, but I have no justification for this. Help?

Last edited by a moderator: Aug 19, 2015
2. Aug 20, 2015

### Jano L.

Do you mean total energy comes out finite or energy per unit length of wire comes out finite?

3. Aug 21, 2015

### QuasiParticle

Sorry, per unit length, of course.

4. Aug 21, 2015

### Jano L.

Infinite wire is a nice idealization of real wires when calculating magnetic field via the Biot-Savart formula, but its unphysicality manifests when we try to calculate the energy necessary to have infinitely long wire with electric current.

The magnetic energy density formula $\frac{1}{2\mu_0}B^2$ is based on the work-energy interpretation of the Poynting theorem, which is mathematically valid even in case of infinite wire. It gives infinite energy per unit length of wire. That is not really an implausible result considering that to establish current even in the smallest part of the wire, one needs to overcome opposing electromotive forces (self-inductance effect) of an infinitely long wire.

I do not think there is a more valid formula for the energy per unit length giving finite value. Could you elaborate on why do you think that standard formula does not apply (other than the energy comes out infinite)? Why would another formula you mentioned be valid?

Last edited: Aug 21, 2015
5. Aug 21, 2015

### QuasiParticle

Let's begin with the total work (against the EMF) required to create the currents of a system:

$$W=\frac{1}{2} \int \vec{A}\cdot\vec{J}d^3r$$
If you plug in the vector potential of an infinite wire (given in the first post) with a current density

$$\vec{J}=\frac{I}{\pi R^2}\hat{e}_z$$
then the energy per unit length is finite and (if I'm not mistaken with that one constant) equals:

$$W/ \Delta z=\frac{\mu I^2}{16\pi}$$
To see why $\int B^2/2\mu$ is wrong in this case, we write the current density in terms of the field:

$$\vec{J}=(\nabla \times \vec{B})/\mu$$
and get

$$W=\frac{1}{2\mu}\int (\nabla \times \vec{B})\cdot \vec{A}d^3r$$
Now, using $(\nabla \times \vec{B})\cdot \vec{A}=B^2-\nabla \cdot (\vec{A} \times \vec{B})$ and the divergence theorem, the work can be written in the form

$$W=\frac{1}{2\mu}\int B^2 d^3r- \frac{1}{2\mu} \oint (\vec{A} \times \vec{B})\cdot d\vec{S}$$
Then usually the integration volume is extended over all space and the surface integral is neglected, as explained in the first post. However, in the case of an infinite wire, this surface term becomes infinite, and thus cannot be neglected. It cancels the infinity coming from the $B^2$-integral.

Poynting's theorem relates the work done on a charge distribution with the energy change in the electromagnetic field.

6. Aug 21, 2015

### Jano L.

Your math seems OK, but why do you assume that energy is given by

$$W=\frac{1}{2} \int \vec{A}\cdot\vec{J}d^3r$$

and not by

$$W= \int \frac{1}{2\mu_0} B^2\,d^3r$$ ? After all, the latter is a standard expression inferred from the Poynting theorem.

From what I know, I would assume the latter is correct and use your steps backwards to show the former formula does not apply for infinite wire, since the surface integral of $A\times B$ does not vanish.

7. Aug 23, 2015

### QuasiParticle

I think the vector potential version of the energy is more fundamental. It can be derived with some rather general arguments. And then from that one finds the $B^2$-energy by neglecting that one term.

I'm not convinced that the $B^2$-energy term can be derived easily from Poynting's theorem; without making the same trick as in the vector potential case.

8. Aug 24, 2015

### Jano L.

I do not think so. Could you post a reference to source that does that?
The magnetic energy is defined as the $B^2$-energy expression; it is not derived. The definition is based on the work-energy interpretation of the Poynting theorem. Check out Panofsky, Melba Phillips, Classical electricity and magnetism,1962 section 10-5 for its derivation.

9. Aug 24, 2015

### QuasiParticle

See, for example:

Griffiths, Introduction to Electrodynamics (1999), section 7.2.4
Vanderlinde, Classical Electrodynamic Theory (2005), section 4.1.4
Jackson, Classical Electrodynamics (1999), section 5.16

The $B^2$-term might be defined as the magnetic energy term, but it gives the correct total energy only if the already-mentioned surface term can be neglected.

If you look at section 10-1 in Panofsky & Phillips, they derive the $B^2$-energy term and they have to make the same assumption about the surface term. Consequently they proceed to derive the vector potential energy expression from this. In the process they again neglect a surface integral term; I think they are basically restoring the term neglected earlier.

With Poynting's theorem I have doubts as to what is the total energy of a system with some given currents. Poynting's theorem states

$$\frac{dW}{dt}=-\frac{d}{dt} \int \frac{1}{2}\left(\epsilon_0 E^2+B^2/\mu_0 \right) d^3r-\oint (\vec{E} \times \vec{B})/\mu_0 \cdot d\vec{S}$$
I'm not sure how to derive the total energy from this. I suspect the surface term might give a contribution equivalent to the one often neglected.

10. Aug 25, 2015

### Jano L.

Now I think both formulae are OK and give the same infinite energy per unit length.

Check out Griffiths' section 7.2.4. He derives the formula 7.31 for a wire in form of a closed circuit under the assumption that magnetic flux $\Phi$ is finite and can be expressed as
$$\Phi = \oint \mathbf A\cdot d\mathbf l.$$

Now we can take the circuit and expand its size to infinity, and the formula is valid throughout the process. The inductance of the circuit $L$ goes as $R(\ln \frac{8R}{a} - 2)$ where $R$ is size of the circuit and $a$ is radius of the wire circular cross-section.This was first calculated by Maxwell I think, and can be found for example in the document

Self inductance of a wire loop as a curve integral
R. Dengler
http://arxiv.org/pdf/1204.1486.pdf

Vector potential in the wire thus depends on $R$ as $\ln \frac{8R}{a}$.

This means its magnitude diverges into infinity as the wire changes into the infinitely long straight wire via the limiting process. This makes sense, as the vector potential in this quasi-static process is given by the Coulomb-like integral

$$\mathbf A(\mathbf x)=\frac{\mu_0}{4\pi}\int \frac{\mathbf j}{|\mathbf x - \mathbf x'|} d^3\mathbf x'.$$

This vector potential, evaluated inside the wire, is finite for finite sized circuit but infinite for infinitely large circuit, due to infinite contribution of distant parts of the wire. It is the same behaviour as with electric potential of an infinitely long charged wire; inside the wire, it is infinitely high due to distant parts of the wire.

Thus the net energy per unit length of wire is infinite, per both formulae.

Generally, the $B^2$ variant of magnetic energy expression is better in the sense it follows naturally from Poynting's theorem which is valid even in highly violent processes, when the fields change quickly like near radiating antenna or radiating particle beams. In such processes, the assumptions behind derivation of the variant 7.31 do not hold and this latter formula cannot be relied upon.

11. Sep 12, 2015

### QuasiParticle

I agree that a loop (or two parallel wires) in the limit of infinite radius (distance) has an infinite inductance per unit length. However, I think the case of a single infinitely long wire is different.

It is true that in deriving the vector potential form for the energy, one considers the flux through a loop. However, I still feel that the energy expression is valid even for an infinite wire, even though the loop in the derivation may not be perfectly defined. If the vector potential expression for the energy is not correct in the case of an infinite wire, then I don't know how to compute the energy, since the $B^2$ expression is not correct.

I still disagree that the $B^2$ version follows naturally from Poynting's theorem; additional assumptions has to be made (e.g. neglecting surface integral terms).

12. Sep 13, 2015

### Jano L.

The surface integral in the Poynting equation vanishes if the surface containing the circuit is expanded to infinity and the process of current increase is quasi-static. Both energy formulas are valid, and both give +infinity.

13. Sep 13, 2015

### QuasiParticle

Right, but my suspicion is that the surface integral cannot be neglected for the infinite wire.

I don't understand. By using the vector potential of an infinite wire given in the first post, the energy (per unit length) is clearly finite.

14. Sep 14, 2015

### Jano L.

Infinite wire does not exist, it is only a result of a limiting process where circuit size expands to infinity. If the enclosing surface is much greater than the circuit in the process, then the surface term vanishes.

The vector potential is infinite in the wire due to its distant parts. I think the formulae you gave in the first post do not obey the condition that $\mathbf A$ is given by the standard Coulomb-like integral, which I think is necessary if the derivation of the A.j formula for energy is to hold (the vector potential needs to decay with distance).

15. Sep 22, 2015

### QuasiParticle

So are you saying that we should not even try to compute the inductance (or energy) of an infinite wire?

I disagree that an infinite wire is the limiting case of a large circuit. In the infinite circuit case the vector potential at the other wire due to the other becomes infinite (like you mention). This doesn't happen for the infinite wire.

Wouldn't an infinite wire be approximated well by a very long wire with an initial charge imbalance between the ends? This would not form a loop at all.

Not sure I'm following your argument. The vector potential for an infinite wire is obtained directly using Maxwell's equations (see e.g. Vanderlinde p. 78).

By the way, in Vanderlinde the case of two infinite parallel wires is calculated. He says about the resulting inductance that "The constant term, $\mu_0 l/4\pi$, represents the self inductance of the current in each wire on itself." This is precisely the result one obtains for the single infinite wire using the vector potential energy expression and the vector potential of the first post.

16. Sep 22, 2015

### Jano L.

No, I am not; we may try to compute energy per unit length of an infinite wire, but we will find that the integral diverges and the only reasonable result we can assign it is +infinity.

Infinite straight wire is not necessarily limit of a circuit, but limiting finite circuit to infinity it's the only physically sensible way to evaluate the infinite case. Laws of electromagnetism are experimentally confirmed for finite bodies, everything about hypothetical infinite bodies is extrapolation.

Wire with charge imbalance would be hard to force into obeying the condition of quasistaticity for which the A.j formula was derived.

His vector potential is fine as far as the role of potential for $\mathbf B$ goes. However, derivation of the A.j formula is based on the idea of finite-sized circuits where loop integrals $\oint \mathbf{A}\cdot d\mathbf{l}$ have finite values. For infinite wire, there is no loop and even if we tried to put it there, the integral would be infinite. There are no grounds to use this diverging A in the A.j formula.

Moreover, his potential diverges as $r\rightarrow\infty$. I think this may make it invalid to use in the general derivation of the A.j formula.

Last edited: Sep 22, 2015
17. Sep 24, 2015

### QuasiParticle

I think more or less static situation could be created with an appropriate RC circuit. But in any case, we can try to compute the result based on the electromagnetic theory, regardless of whether it is possible to construct an infinite wire in reality.

Here is an argument for the vector potential energy expression without reference to a circuit. Let's start with the Lorentz force
$$\vec{F} = q\left( \vec{E}+\vec{v}\times\vec{B} \right)$$
Only the electric field can do work on the charges. Let the work done against the EMF of a group of charges $\sum q_i$, which are displaced by $\vec{r}_i=\vec{v}_i dt$ be
$$dW = -\sum\vec{F}_i\cdot d\vec{r}_i = -\sum q_i\vec{E}\cdot \vec{v}_i dt$$
Let's replace the charges with the current density $\vec{J}(\vec{r})=\sum q_i \vec{v}_i \delta(\vec{r}-\vec{r}_i)$:
$$dW = -\int d^3r ( \vec{J}\cdot\vec{E}) dt$$
The electric field in terms of the vector potential is $\vec{E}=-\partial \vec{A}/\partial t$. Let's integrate to get the total work:
$$W = \int d^3r \int_{0}^{\vec{A}} \vec{J}\cdot d\vec{A}$$
By using the linearity between the current density and the vector potential, this integral yields
$$W = \frac{1}{2} \int \vec{J}\cdot\vec{A}~d^3r$$
I believe it is completely valid to use this energy expression for the infinite wire. And since the vector potential of an infinite wire is finite, the energy per unit length is also finite.

18. Sep 25, 2015

### Jano L.

This formula gives work done by electromotive forces (e.g. in a battery) against electric forces.

This is not a valid transformation of variables, because $\mathbf J(\mathbf x)$ is not a function of $\mathbf A(\mathbf x)$ (in words, current density at a point is not a function of vector potential at this point).

19. Sep 25, 2015

### QuasiParticle

But the work done to produce the system is exactly the energy of the system.

Which transformation of variables are you talking about? Replacing the electric field with the derivative of the vector potential? What does this have to do with the current density?

20. Sep 26, 2015

### Jano L.

I agree with that. My point was that the work of electromotive force (chemical, mechanical or due to temp. gradient) is stored into the EM energy of the system, not the work of electric force.

The transformation where the integral over time is replaced by the integral over components of $\mathbf A$. This is not a valid transformation, because, while $\mathbf J(\mathbf x)$ is a function of $t$, it is not a function of $\mathbf A(\mathbf x)$.

21. Sep 26, 2015

### QuasiParticle

There is a 1-to-1 correspondance between the magnitudes of the current density and the vector potential. This is enough and the transformation of variables is perfectly valid. One might even make a stronger statement that, given some boundary conditions, there is a 1-to-1 correspondance between the current density and the vector potential, though I'm not absolutely sure about this.

The same form for the energy in terms of an integral over $\vec{A}$ is given in many textbooks, for example, in Jackson and Zangwill [1,2]. I'm quite convinced that it is correct.

By the way, I just noticed that Schwartz uses the exact same argument for the vector potential energy expression as the one given in the post above [3]. So, he also derives it without any reference to a circuit. Furthermore, he also continues from this expression to derive the $B^2$-expression, making the same approximation mentioned in the first post. I think it is quite clear that the vector potential expression is "more fundamental" in the sense that the $B^2$-expression is obtained from it after making an approximation. The infinite wire represents an illuminating case, where this approximation cannot be made. The energy per unit length of an infinite wire is finite, and therefore also the inductance is finite.

[1] Jackson, Classical Electrodynamics (1999), Ch. 5.16.
[2] Zangwill, Modern Electrodynamics (2013), Ch. 12.6.
[3] Schwartz, Principles of Electrodynamics (1987), Ch. 4-7.

22. Sep 26, 2015

### Jano L.

I agree it is true in case of quasistatic (slow enough) changes in the current, which is the case here.

I also agree that the A.j formula can be derived without any reference to a circuit, Schwartz does that for quasistatic processes. However, this derivation is valid only if the system is finite (bounded in space).

Why?

The expression

$$I = \int_V \mathbf A\cdot\mathbf j \,dV$$

depends on the vector potential, which is arbitrary; there is infinity of different vector potentials that give the same magnetic field $\mathbf B$. It also depends on the region $V$, which we assume to contain all the system. But the work done against the electric field during establishing the current in the system is a definite number. There is a possibility that different choice of $\mathbf A$ will result in a different value of the integral and thus fail to give correctly the work done.

So, in order for any formula to give the value of the work done in the system, we require that its value does not depend on the choice of vector potential. From this it follows that we require that the above integral is gauge-invariant.

That is not always the case. In particular, it is not the case for the integral over infinitely long space, which is necessary to include the work done inside the infinite wire.

Why?

Let us consider some vector potential $\mathbf A$ and introduce another one
$$\mathbf A + \nabla \chi$$
where $\chi$ is some function of coordinates. Let us calculate $I$ for this potential:

$$I' = \int_V (\mathbf A + \nabla \chi)\cdot \mathbf j \,dV$$

If $I$ is finite, we may express this as
$$I' = I + \int_V \nabla \chi \cdot \mathbf j\, dV$$
The difference
$$D=\int_V\nabla \chi \cdot \mathbf j\, dV$$
can be expressed as
$$\int_V\nabla \cdot (\chi \mathbf j) - \chi \nabla\cdot\mathbf j\, dV.$$
Since we're considering quasistatic process, $\nabla\cdot\mathbf j = 0$ everywhere so
$$D = \int_V\nabla \cdot (\chi \mathbf j)\, dV.$$

Using Stokes' theorem, we can transform the volume integral into surface integral
$$D = \oint_\Sigma \chi \mathbf j \cdot \, d\boldsymbol{\Sigma}.$$
where $\Sigma$ is boundary of the region $V$.

This integral vanishes for all $\chi$ if no current passes through anywhere on the boundary. This is most easily enforced by making the boundary large enough, so it contains the whole system; then $\mathbf j = 0$ everywhere on the boundary.

For infinite wire with current, no such boundary exists, $D$ does not vanish and the A.j integral depends on the choice of vector potential. It thus seems the A.j formula is appropriate only for bounded systems and cannot be applied to infinite systems.

23. Sep 27, 2015

### QuasiParticle

The vector potential energy expression must be gauge invariant, because $\vec{E}$ is gauge invariant in
$$\frac{dW}{dt}=\int_V \vec{E}\cdot \vec{J}$$
from which the expression is derived. The gauge transformation does not only transform the vector potential, but also the electric potential. The integral remains unchanged.

The vector potential energy formula is, by the way, used in many textbooks to compute the energy of infinite systems (e.g. Vanderlinde, Landau-Lifshitz).

Vanderlinde, Classical Electrodynamic Theory (2005), section 4.1.4.
Landau, Lifshitz, Electrodynamics of Continuous Media (1960), section §33.

24. Sep 27, 2015

### Jano L.

The quoted part is true, but the integral in question is
$$I = \int_V \mathbf A\cdot\mathbf j \,dV$$
over finite region, which is not gauge-invariant if current passes through its boundary (if you do not agree, check my previous post where this is derived). If the integral is over infinite region, where $\mathbf j$ extends to infinity, it is not possible to make this argument without limiting finite region to infinite region. However, we can still investigate gauge invariance for a particular system.

If we return to your expressions for the vector potential in the post #1, we see that there is infinity of different vector potentials for each value of $C$, all giving the same magnetic field. This is because of the gauge freedom in the choice of $\mathbf A$. The integral $I$ for infinite wire is thus gauge-dependent.

This happens because the derivation in Schwartz's book assumes $\mathbf E = -\partial_t \mathbf A$ instead of the general relation

$$\mathbf E = -\partial_t \mathbf A -\nabla \phi$$
which you mentioned ($\phi$ being electric potential). This is correct only in particular class of gauges, so it is no surprise that the derived integral $I$ is gauge dependent.

The calculation in Vanderlinde is unsound, because he assumes without any basis that the integration constants $D'$ of the two vector potential functions have the same value. Because the currents are of equal magnitude, he ends up with total $\mathbf A$ function that has no gauge freedom. But in general, the two constants $D'$ may have any value and the sum of functions will depend on their difference. The valid way to do this is to preserve the integration constant in the total vector potential, so it has gauge freedom just as the single-wire potentials have.

Landau&Lifshitz derive the A.j formula from the H.B formula, so their derivation and usage of the A.j formula is only valid for systems where the H.B integral converges. It cannot be used for single current carrying wire.
vanishes

25. Sep 27, 2015

### QuasiParticle

I'm sorry, I don't understand how the integral can go from gauge independent to gauge dependent in the process. I think your gauge transformation is not correct. You are forgetting the transformation to the electric potential.

If we go back to the form
$$\frac{dW}{dt}=-\int_V \vec{E}\cdot\vec{J}$$
we both agree that this is gauge independent. Earlier I did not include the electric potential or the gauge transformation explicitly, but more generally, we should use, as you say, the form
$$\vec{E}=-\nabla\phi-\partial_t\vec{A}$$
The above power integral becomes
$$\frac{dW}{dt}=-\int_V (-\nabla\phi-\partial_t\vec{A})\cdot\vec{J}$$
and the gauge transformation is
$$\left\{ \begin{array}{l} \vec{A}' =\vec{A}+\nabla\Lambda\\ \phi' =\phi-\partial_t\Lambda \end{array}\right.$$
Clearly the form of the integral before and after the gauge transformation is the same, as it should (we set $\phi=0$). I.e.
$$\frac{dW}{dt}=\int_V (\partial_t\vec{A})\cdot\vec{J}$$
So it is gauge independent.

Actually, when determining the vector potential in post #1, the gauge has already been fixed to the Coulomb gauge. The equation relating the vector potential and the current density depends on the gauge.

Yes, this constant business is a little strange, and it appears also in post #1. I have to think about this. But note that the constants do not render the energy integral infinite in any case. You can add any constant to the vector potential and the resulting energy per unit length is still finite. Note also that Vanderlinde also fixed the gauge before he started to determine the vector potential. Also, regardless of their derivation, Landau and Lifshitz are using the expression for an infinite system.

Last edited: Sep 27, 2015
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