Energy density due to infinite uniform line charges

[Karl86]

Homework Statement

Two infinitely long rods parallel to the $z$-axis have uniform charge density $\lambda$. I need to calculate the energy density $U$ per unit length along the $z$-axis.

Relevant Equations

$\frac{\epsilon_0}{2} \int E^2 d\tau$?

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the point where the rods intersect the $x,y$ plane. I know that on any given point there will be the superpositions of $E_1=\frac{2\lambda}{4\pi \epsilon_0}\frac{1}{(x-x_1)^2+(y-y_1)^2}\hat{r}_1$ and $E_2=\frac{2\lambda}{4\pi \epsilon_0}\frac{1}{(x-x_2)^2+(y-y_2)^2}\hat{r}_2$, which would mean $E^2=\frac{4\lambda^2}{(4\pi\epsilon_0)^2}\frac{1}{(x-x_1)^2+(y-y_1)^2}\frac{1}{(x-x_2)^2+(y-y_2)^2}$. So my thinking was to sum them as vectors and then integrate the square of this sum, which will be a scalar, over the $x,y$ plane, and I though this would give me the energy density per unit length along the $z$-axis. Can you criticize my reasoning? The result should involve a logarithm of the distance between the two rods and that really seems a far cry from what my calculation points to.

 

[PeroK]

Are you sure your relevant equation is valid in this case?

 

[Karl86]

I am not sure it is valid in this case, it can certainly be part of the debate.

 

[PeroK]

I am not sure it is valid in this case, it can certainly be part of the debate.

To get that equation - the integral of $E^2$ - you need to neglect a boundary surface integral, which is not valid if the charge distribution is not bounded.

Do you know another equation for electrostatic energy ?

 

[Karl86]

$\frac{1}{2}\int \lambda V dl?$

 

[Karl86]

My problem with this formula is that I would measure the total energy density due to the rods, not the energy density per unit length along the $z$ axis, as far as I can tell.

 

[PeroK]

My problem with this formula is that I would measure the total energy density due to the rods, not the energy density per unit length along the $z$ axis, as far as I can tell.

The total energy will be infinite. How is energy density related to the integral?

 

[Karl86]

I did not find a paragraph in Griffiths that stated it clearly, but I guess the energy density is actually the integrand?

 

[Karl86]

This was my first, instinctive, idea, but the potential of that configuration is something like
$$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) $$ whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.
Edit: actually I could take $a$ to be #r_2# and that would give $\log (|r_2-r_1|)$. Does it make sense?

 

[PeroK]

This was my first, instinctive, idea, but the potential of that configuration is something like
$$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) $$ whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.
Edit: actually I could take $a$ to be #r_2# and that would give $\log (|r_2-r_1|)$. Does it make sense?

Are you sure that isn't $|\vec{r_2} - \vec{r_1}|$, where these vectors represent the points where the wires cross the $xy$ plane?

PS You could interpret this problem two ways. The way I interpreted it was that you want the energy stored per unit length of wire. The answer you are quoting is consistent with this, as the energy of the configuration is dependent only on the distance between the wires. Note: in this case you need to think about what the variables in the equation for potential of an infinite straight wire actually mean.

Another interpretation would be the energy density of the EM field at points along the z-axis. This may be how you've interpreted the problem. The answer then would depend on the angle between the wires. For example, if the wires were on opposite sides of the z-axis, then the field along the z-axis would be 0.

 

[Karl86]

Are you sure that isn't $|\vec{r_2} - \vec{r_1}|$, where these vectors represent the points where the wires cross the $xy$ plane?

I think it's exactly that, what made you think otherwise? My notation without arrows? I apologise if that's the case. By the way: this expression for the potential confuses me anyway, because my next step is to calculate the electric field of this configuration, assuming the rods are moving in the $x,y$ plane. Should I take the gradient with respect to the $r_1$ or to the $r_2$ coordinates?

 

[PeroK]

I think it's exactly that, what made you think otherwise? My notation without arrows? I apologise if that's the case. By the way: this expression for the potential confuses me anyway, because my next step is to calculate the electric field of this configuration, assuming the rods are moving in the $x,y$ plane. Should I take the gradient with respect to the $r_1$ or to the $r_2$ coordinates?

Moving in the x-y plane?

In any case the field of two rods is just the sum of the two fields by the superposition principle.

 

[Karl86]

Moving in the x-y plane?

In any case the field of two rods is just the sum of the two fields by the superposition principle.

Yes, meaning the two rods move (while staying perpendicular to the $x,y$ plane) and thus the intersection points move. I have to find differential equations for $E_x,E_y$ and show that the component $E_z$ of the electric field satisfies:
$$ \left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0$$
but $E_z$ should be zero, because the potential $C\cdot \log(|r_2-r_1|)$ does not depend on z. It just seems too trivial. The field I get by taking the gradient wrt both coordinates and adding is:
$$ \mathbf{E}=\left( \frac{x_1-x_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}, \frac{y_1-y_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}},\\ \frac{z_1-z_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}\right)$$
but we are on the $x,y$ plane so the last component is $0$.
P.S. The energy density should be understood as the energy in the field between the planes $z=a$ and $z=a+1$ which means our interpretation was correct, right?

 

[PeroK]

The potential you calculated was only valid when the rods were static.

You need to look for the potential formulation for electrodynamics!

Hint: Lorentz Gauge.

 

[Karl86]

We have not done those yet, so I believe there must be a way to prove that $E_z$ obeys that equation and to find equations for $E_x,E_y$ without calculating them explicitly. Do you think that's possible? If you think I should start another thread for this, let me know.

 

[Karl86]

Are you sure that isn't $|\vec{r_2} - \vec{r_1}|$, where these vectors represent the points where the wires cross the $xy$ plane?

PS You could interpret this problem two ways. The way I interpreted it was that you want the energy stored per unit length of wire. The answer you are quoting is consistent with this, as the energy of the configuration is dependent only on the distance between the wires. Note: in this case you need to think about what the variables in the equation for potential of an infinite straight wire actually mean.

Another interpretation would be the energy density of the EM field at points along the z-axis. This may be how you've interpreted the problem. The answer then would depend on the angle between the wires. For example, if the wires were on opposite sides of the z-axis, then the field along the z-axis would be 0.

Actually, the fact that it should be interpreted as the energy stored in the field between the planes of equations $z=a$ and $z=a+1$ for any $a$ makes me think that there needs to be an integration, $\frac{1}{2} \lambda U$ alone is a local quantity. But then I don't understand how it can be so close to the result, this is truly confusing.

 

[PeroK]

Actually, the fact that it should be interpreted as the energy stored in the field between the planes of equations $z=a$ and $z=a+1$ for any $a$ makes me think that there needs to be an integration, $\frac{1}{2} \lambda U$ alone is a local quantity. But then I don't understand how it can be so close to the result, this is truly confusing.

I thought we had solved the first question?

I'm sorry I have not had much time to look at this. In the second part are the rods moving in the z direction?

 

[Karl86]

hi, no, just $x,y$ and they stay parallel to the $z$ axis. That generates an E and a B which I don't need to write explicitly, just to write equations for. For the $z$ component of $E$ I just want to show that
$$
\left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0
$$
so basically I already have the equation.

 

[PeroK]

hi, no, just $x,y$ and they stay parallel to the $z$ axis. That generates an E and a B which I don't need to write explicitly, just to write equations for. For the $z$ component of $E$ I just want to show that
$$
\left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0
$$

The problem I see is that symmetry in the z direction should result in $E_z =0$.

 

[Karl86]

That is weird for me too, although the equation would be satisfied (but in a completely trivial way). And if they were indeed moving in the $z$ direction?

 

[Karl86]

Actually: why does the fact that it doesn't move in the z direction imply that $E_z=0$? It should imply that $E_z$ doesn't depend on $z$.

 

[PeroK]

Actually: why does the fact that it doesn't move in the z direction imply that $E_z=0$? It should imply that $E_z$ doesn't depend on $z$.

Because the rod is infinite, we can take any point to have $z=0$. I.e. there's an infinite line of charges above the point mirrored by an infinite line of charges below the point.

We can pair these charges off and study the fields for a pair of charges, the same distance above and below the x-y plane.

If the field at any point has $E_z > 0$, then why is this? What distinguishes the "up" direction? Someone looking from below might see the charges moving to their right, say, and the field point away from the nearer charge. But, someone looking from above (and "upside down") would also see the charges moving to the right, but the field pointing towards the nearer charge.

This is physically incompatible. Hence the field must have $E_z = 0$ everywhere, for every pair of charges, hence for the whole rod.

If we take an example of a magnetic field, say, that points clockwise about the direction of motion for a single charge, then the field for two charges moving in the same direction and the same distance above and below a plane will cancel on the plane.

 

[Karl86]

Because the rod is infinite, we can take any point to have $z=0$. I.e. there's an infinite line of charges above the point mirrored by an infinite line of charges below the point.

We can pair these charges off and study the fields for a pair of charges, the same distance above and below the x-y plane.

If the field at any point has $E_z > 0$, then why is this? What distinguishes the "up" direction? Someone looking from below might see the charges moving to their right, say, and the field point away from the nearer charge. But, someone looking from above (and "upside down") would also see the charges moving to the right, but the field pointing towards the nearer charge.

This is physically incompatible. Hence the field must have $E_z = 0$ everywhere, for every pair of charges, hence for the whole rod.

If we take an example of a magnetic field, say, that points clockwise about the direction of motion for a single charge, then the field for two charges moving in the same direction and the same distance above and below a plane will cancel on the plane.

Ok, and equation-wise what can one say? Taking the curl of one of Maxwell's equations I get
$$ \nabla \times \nabla \times \mathbf{E} = -\frac{\partial}{\partial t} \nabla \times \mathbf{B}$$
from which
$$ \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 E = - \mu_0 \frac{\partial \mathbf{J}}{\partial t} -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} $$
but $\mathbf{J}$ should be zero (correct me if I'm wrong) and the gradient of $\nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0}$ should also be zero (although I'm not sure about this, perhaps you can tell me if it is true). So would $E_x, E_y$ satisfy the same equation as $E_z$ or different ones?

 

[PeroK]

Where did you get this problem?

You can, of course, get differential equations for the fields in a vacuum. But, these are always the same equations, independent of the charges and currents elsewhere.

I assume the equations you are looking for are for the specific case of charged rods?

Maybe not?

I'm going offline now. I might have a chance to help later.

 

[Karl86]

Where did you get this problem?

You can, of course, get differential equations for the fields in a vacuum. But, these are always the same equations, independent of the charges and currents elsewhere.

I assume the equations you are looking for are for the specific case of charged rods?

Maybe not?

I'm going offline now. I might have a chance to help later.

Thank, appreciate your help. At this point I am convinced that it's just about getting those differential equations for the generic fields. But I think that this question could also shed light on the matter: if we double the charge density on the rods ($2\lambda$ instead of $\lambda$) how do $E$ and $B$ change?

 

[PeroK]

Ok, and equation-wise what can one say? Taking the curl of one of Maxwell's equations I get
$$ \nabla \times \nabla \times \mathbf{E} = -\frac{\partial}{\partial t} \nabla \times \mathbf{B}$$
from which
$$ \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 E = - \mu_0 \frac{\partial \mathbf{J}}{\partial t} -\mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} $$
but $\mathbf{J}$ should be zero (correct me if I'm wrong) and the gradient of $\nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0}$ should also be zero (although I'm not sure about this, perhaps you can tell me if it is true). So would $E_x, E_y$ satisfy the same equation as $E_z$ or different ones?

You have derived a general equation for $\vec{E}$ in any region where there is no charge or current. In this case, we can see that $\vec{E}$ is independent of $z$ - although, in this case, we can also see that $E_z = 0$.

That's about all there is to it.

 

[Karl86]

This was my first, instinctive, idea, but the potential of that configuration is something like
$$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) $$ whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.
Edit: actually I could take $a$ to be #r_2# and that would give $\log (|r_2-r_1|)$. Does it make sense?

Actually, one last thing. Here I somehow convinced myself I would get $\log (|r_2-r_1|)$ but I actually don't, do I? $$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) = -\frac{\lambda}{2\pi\epsilon_0}\left(\log(r_2) - \log(r_1) \right)$$ which is quite different from $\log{|\vec{r_2}-\vec{r_1}|}$. Can you see how I would get $\log{|\vec{r_2}-\vec{r_1}|}$ or did you just trust me when I wrote it :D?

 

[PeroK]

Actually, one last thing. Here I somehow convinced myself I would get $\log (|r_2-r_1|)$ but I actually don't, do I? $$ -\frac{\lambda}{2\pi\epsilon_0}\left(\log\frac{r_1}{a}+\log\frac{r_2}{a}\right) = -\frac{\lambda}{2\pi\epsilon_0}\left(\log(r_2) - \log(r_1) \right)$$ which is quite different from $\log{|\vec{r_2}-\vec{r_1}|}$. Can you see how I would get $\log{|\vec{r_2}-\vec{r_1}|}$ or did you just trust me when I wrote it :D?

The variable, $r$, in the potential for a charged rod is the distance from the rod.
$$V(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln(\frac{r}{a})$$

This potential, however, does not go to $0$ at $\infty$. You need to be careful, therefore, if you use this potential to find the energy of an electrostatic configuration. What you need is the potential difference between $\infty$ and the point at which you want to evaluate the potential. That's one issue.

But, this potential difference is infinite for an infinite wire. What you want is the energy per unit length. So, you need to find the energy per unit length for a finite wire and take the limit as the length of the wire goes to infinity.

There might be a neat trick to use the logarithmic potential directly, but I don't see it.

 

[Karl86]

The variable, $r$, in the potential for a charged rod is the distance from the rod.
$$V(r) = -\frac{\lambda}{2\pi \epsilon_0} \ln(\frac{r}{a})$$

This potential, however, does not go to $0$ at $\infty$. You need to be careful, therefore, if you use this potential to find the energy of an electrostatic configuration. What you need is the potential difference between $\infty$ and the point at which you want to evaluate the potential. That's one issue.

But, this potential difference is infinite for an infinite wire. What you want is the energy per unit length. So, you need to find the energy per unit length for a finite wire and take the limit as the length of the wire goes to infinity.

There might be a neat trick to use the logarithmic potential directly, but I don't see it.

But that gives something like $C\cdot \log\frac{a + \sqrt{x^2+a^2}}{-a+\sqrt{x^2+a^2}}$ assuming a symmetric rod, times some constant, $1/2 \lambda$ and that limit is still infinite for $a \rightarrow \infty$...

 

[PeroK]

But that gives something like $C\cdot \log\frac{a + \sqrt{x^2+a^2}}{-a+\sqrt{x^2+a^2}}$ assuming a symmetric rod, times some constant, $1/2 \lambda$ and that limit is still infinite...

Yes, it's tricky. One idea is to go back to your original idea of using the electric field!

 

[Karl86]

Yes, it's tricky. One idea is to go back to your original idea of using the electric field!

Those integrals were infinite too :(

 

[PeroK]

Those integrals were infinite too :(

Where did you get this problem?

 

[Karl86]

It was simply suggested in class I have literally 0 reference on it. And the energy density was defined as the energy in the field between the plane z=a and z=a+1. But I cannot find a way to calculate that that doesn't end up being infinite.

 

[PeroK]

It was simply suggested in class I have literally 0 reference on it.

You've been quoting a finite answer, though. Where did that come?

whereas the expression I need is something like $C \cdot \log (|r_2-r_1|)$.

It's possible that the question is ill-defined. You've got two "infinite" wires and they start "infinitely" far apart.

You could interpret that different ways. Do they start "far apart" compared to their length? Or, are they "long" compared to the distance they start apart.

The answer you quote, if you take $C$, just to be some constant of proportionality can be naively obtained from the the potential formula (*). But, that presupposes the problem has a unique, well-defined numerical answer.

(*) I hadn't thought any further than that until you forced me to think about it more clearly.

I suspect if you put different proportions into the ratio of lengths of wires and starting distances you would get different answers, potentially infinite if you take the limit in all cases.

For any finite problem (long wires, far apart), there will be a value of $C$ and an equation related to the log of the final distance apart. Approximately.

 

[Karl86]

You've been quoting a finite answer, though. Where did that come?
It's possible that the question is ill-defined. You've got two "infinite" wires and they start "infinitely" far apart.

You could interpret that different ways. Do they start "far apart" compared to their length? Or, are they "long" compared to the distance they start apart.

The answer you quote, if you take $C$, just to be some constant of proportionality can be naively obtained from the the potential formula (*). But, that presupposes the problem has a unique, well-defined numerical answer.

(*) I hadn't thought any further than that until you forced me to think about it more clearly.

I suspect if you put different proportions into the ratio of lengths of wires and starting distances you would get different answers, potentially infinite if you take the limit in all cases.

For any finite problem (long wires, far apart), there will be a value of $C$ and an equation related to the log of the final distance apart. Approximately.

The answer is given: it is $-\frac{2\lambda^2}{4\pi\epsilon_0} \log(|r_2-r_1|)$. What I am asked is to prove that it can be written like that. How is it you got that expression naively? If I get right the $\log(|r_2-r_1|)$ bit I am basically done.

 

[PeroK]

The answer is given: it is $-\frac{2\lambda^2}{4\pi\epsilon_0} \log(|r_2-r_1|)$. What I am asked is to prove that it can be written like that. How is it you got that expression naively?

You start with one "infinite" wire. You bring a unit of line charge in from "infinity" - and don't worry that the potential is infinite at infinity - and you have the energy to create the configuration, per unit length of the second wire.

But, now it seems very clear, if you released that unit line charge, it has infinite potential energy. So, there is no finite answer.

If, however, you consider the wire long but finite, then eventually the logarithmic potential will cease to hold. The PE of the unit length of charge is, therefore, finite for any finite wire, but tends to infinity for an infinite wire, so the problem is not well-defined.

 

[Karl86]

You start with one "infinite" wire. You bring a unit of line charge in from "infinity" - and don't worry that the potential is infinite at infinity - and you have the energy to create the configuration, per unit length of the second wire.

But, now it seems very clear, if you released that unit line charge, it has infinite potential energy. So, there is no finite answer.

If, however, you consider the wire long but finite, then eventually the logarithmic potential will cease to hold. The PE of the unit length of charge is, therefore, finite for any finite wire, but tends to infinity for an infinite wire, so the problem is not well-defined.

You are seeing it in terms of energy to get a certain configuration, but the wires are fixed in a constant position. There are no assumptions on how close or far they are, not infinitely far though, one intersects the xy plane in a point r_1 and the other in a point r_2. What the question seems to point to is they generate a field and this field carries some energy. Calculate it between the planes z=a and z=a+1. Is the way you are trying to see it equivalent to this?

 

[PeroK]

You are seeing it in terms of energy to get a certain configuration, but the wires are fixed in a constant position. There are no assumptions on how close or far they are, not infinitely far though, one intersects the xy plane in a point r_1 and the other in a point r_2.

Nevertheless, they store infinite energy per unit length.

The analysis I did is necessary to calulate the energy stored, either explicitly or through a previously derived formulas, that uses the "bring a charge in from infinity" approach. Note also:

To get that equation - the integral of $E^2$ - you need to neglect a boundary surface integral, which is not valid if the charge distribution is not bounded.

Do you know another equation for electrostatic energy ?

I overlooked that the same objection applies to the other formula using the potential.

In short, you cannot define the energy stored in the field of an infinite wire, because the potential is not finite at infinity.

The moral is that you need to be careful with these hypothetical charge distributions that go to infinity.

 

[Karl86]

Ok so you stand by your analysis that it seems not well defined, but what did you mean when you said that $C\log(|r_2-r_1|)$ can be deduced from the configuration? How?

 

[PeroK]

Ok so you stand by your analysis that it seems not well defined

What happens if you release a positive point charge in the field of an infinite positively charged wire?

What is the final KE of the charge?

 

[Karl86]

What happens if you release a positive point charge in the field of an infinite positively charged wire?

What is the final KE of the charge?

The field is constant and pushes it away from the wire, the charge never stops, so the KE is never 0. IS that right?

 

[PeroK]

The field is constant and pushes it away from the wire, the charge never stops

It's not a question of whether it stops or even whether the force is never zero - those apply to a point charge source as well - the issue is that the force drops off as $1/r$ and the potential is logarithmic, so that there is no final asymptotic KE "at infinity". The KE increases without bound.

Therefore, the PE of the point charge is not well defined.

The infinite wire configuration, although useful for calculating the potential difference between two finite points, as an approximation of a long finite wire, is ultimately a purely hypothetical configuration, with infinite energy density.

 

[Karl86]

Yes, I'm convinced. The discussion that arose from a seemingly innocent problem was remarkable though :D

 

[PeroK]

PS The only question now is whether this question was a deliberate "sandbag" or a genuine misconception on the part of whoever gave it to you!

 

[Karl86]

It was not a deliberate sandbag, it is meant to be answered correctly, so as things stand I am failing to answer it properly, and yet I communicated it to you correctly, and I am persuaded the discussion is correct.

 

[Karl86]

Oh, ok I asked my instructor and I was told to set the reference point not at infinity but at a finite distance from the wire. Do you think that can help?