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Science Forum Index » Physics - Electromagnetic Forum » Problem with Poynting vector.
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| Army1987 |
 Posted: Mon Feb 04, 2008 1:06 pm |
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In a homework assignment, I was asked to calculate the flux of Poynting
vector on the surface of a cylindrical conductor of constant resistivity,
with uniform electric field within it, parallel to the cylinder's axis. It
asked to compare it with the power dissipated (by Joule effect in it).
The Poynting vector is supposed to be the superficial density of power
flowing through a point, but I found a *negative* flux, i.e. the vector
pointing *into* the conductor, as if the energy was coming from outside
into it.
I've checked carefully whether I committed a sign error, but it can't find
one.
Any thoughts?
--
Army1987 (Replace "NOSPAM" with "email") |
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| Timo Nieminen |
| Posted: Mon Feb 04, 2008 8:58 pm |
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On Mon, 4 Feb 2008, Army1987 wrote:
Quote: In a homework assignment, I was asked to calculate the flux of Poynting
vector on the surface of a cylindrical conductor of constant resistivity,
with uniform electric field within it, parallel to the cylinder's axis. It
asked to compare it with the power dissipated (by Joule effect in it).
The Poynting vector is supposed to be the superficial density of power
flowing through a point, but I found a *negative* flux, i.e. the vector
pointing *into* the conductor, as if the energy was coming from outside
into it.
I've checked carefully whether I committed a sign error, but it can't find
one.
Any thoughts?
The Joule heating is _inside_ the conductor. Electromagnetic energy is
being converted to heat. Since it's all in steady-state, electromagnetic
energy must be moving _into_ the conductor to replace the loss.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Army1987 |
| Posted: Tue Feb 05, 2008 2:50 am |
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Timo Nieminen wrote:
Quote: On Mon, 4 Feb 2008, Army1987 wrote:
In a homework assignment, I was asked to calculate the flux of Poynting
vector on the surface of a cylindrical conductor of constant resistivity,
with uniform electric field within it, parallel to the cylinder's axis. It
asked to compare it with the power dissipated (by Joule effect in it).
The Poynting vector is supposed to be the superficial density of power
flowing through a point, but I found a *negative* flux, i.e. the vector
pointing *into* the conductor, as if the energy was coming from outside
into it.
I've checked carefully whether I committed a sign error, but it can't find
one.
Any thoughts?
The Joule heating is _inside_ the conductor. Electromagnetic energy is
being converted to heat. Since it's all in steady-state, electromagnetic
energy must be moving _into_ the conductor to replace the loss.
I think I get the point. But it appears that the energy passes through the
lateral surface of the cylinder, rather than through the ends. Does it
mean that the flux of the vector through a closed surface is the power,
but its direction and magnitude in a given point needn't?
--
Army1987 (Replace "NOSPAM" with "email") |
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| Benj |
| Posted: Tue Feb 05, 2008 11:50 am |
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On Feb 5, 7:50 am, Army1987 <army1...@NOSPAM.it> wrote:
Quote: Timo Nieminen wrote:
On Mon, 4 Feb 2008, Army1987 wrote:
In a homework assignment, I was asked to calculate the flux of Poynting
vector on the surface of a cylindrical conductor of constant resistivity,
with uniform electric field within it, parallel to the cylinder's axis. It
asked to compare it with the power dissipated (by Joule effect in it).
The Poynting vector is supposed to be the superficial density of power
flowing through a point, but I found a *negative* flux, i.e. the vector
pointing *into* the conductor, as if the energy was coming from outside
into it.
I've checked carefully whether I committed a sign error, but it can't find
one.
Any thoughts?
The Joule heating is _inside_ the conductor. Electromagnetic energy is
being converted to heat. Since it's all in steady-state, electromagnetic
energy must be moving _into_ the conductor to replace the loss.
I think I get the point. But it appears that the energy passes through the
lateral surface of the cylinder, rather than through the ends. Does it
mean that the flux of the vector through a closed surface is the power,
but its direction and magnitude in a given point needn't?
Good for you! You've just discovered where one of the dead bodies is
buried in Electromagnetics! People to find it convenient to think of
the Poynting vector as some kind of instantaneous power density. And
yes, sometimes that works, but Plonsey and Colin in their textbook
note: "It is not possible to state where the energy is located. Only
the total energy associated with a given field has a physical
meaning." So yes, it does appear as if the power is flowing into the
resistor from outer space! But it's really as you surmise just one
more case where you have to throw out the result as "non-physical".
You'll find quite a few like that in EM. In spite of EM being around
for a great many years, it's not as neatly tied a package as most of
us pretend! |
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| Army1987 |
| Posted: Tue Feb 05, 2008 2:52 pm |
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Timo Nieminen wrote:
Quote: On Tue, 5 Feb 2008, Army1987 wrote:
[Why does Poynting's vector point *into* a current-carrying conductor?]
I think I get the point. But it appears that the energy passes through the
lateral surface of the cylinder, rather than through the ends.
The current doesn't carry energy - it's the source of losses. The energy
is carried by the fields outside the wire. You can think of the wire as a
waveguide. For some fun, see what happens when you make the resistance of
the wire -> zero. (Also fun to look at the AC case. This might not be easy
- I don't know what kind of course you're doing, what level, or what
textbook you're using.)
I'm a second year physics student, and my professor is perverse enough to
ask problems like that on exams.
I'm trying the AC case. The flux of S is the negative of the Joule
power, plus a term which looks like the time derivative of the e.m. energy
inside the cylinder. That term is actually rho * epsilon * dWe / dt, where
rho is resistivity and We is the electric energy. The formula for the
magnetic energy is somewhat complicated and I'll take a look at it
tomorrow.
Quote: Anyway, as R -> 0, the E field along the wire -> 0, and the inward
Poynting vector goes to zero. If there is a resistive load further along
the wire, there must be power transport along the wire, which means
there must be an outward electric field - the wire must have a surface
charge over its outside. The wire on the far side of the resistance must
also have a surface charge, and together these produce the electric
field along the resistance.
Yes, I'm aware of the issue. Surface charges on discontinuities in
resistivity (as well as volume charges on continuously nonuniform
resistivity) are among my professor's favorite topics.
[...]
Quote: However, I think it's useful to remind yourself that electric circuits
are fundamentally not about electrons moving in wires, but about
electric and magnetic fields around the wires. The movement of electrons
is almost just a side-effect. That's why there's no problem with
electrical signals travelling along wires at the speed of light, while
the electrons crawl along at some mm/s.
Indeed. A problem on the transmission of a signal along a transmission
line was giving me nonsensical results because in one place I had confused
the "direction" of the current and the direction of propagation of the
signal.
Quote: Does it
mean that the flux of the vector through a closed surface is the power,
but its direction and magnitude in a given point needn't?
The relevant conservation law is
du/dt + J.E + div(S) = 0,
where u = electromagnetic energy density, J.E is the density of energy
loss through Joule heating, and S is the Poynting vector.
Here's the problem. In my notes there wasn't the J.E term in the equation.
This because my professor was considering propagation of e.m. waves
through a perfect insulator (J = 0) while deriving it.
[...]
Quote: This leads to fun cases such as "Is there an energy flow in the static
field when, say, you put a charged capacitor between two magnetic
poles?"
[...]
So, to answer your question, I think your conclusion is correct, but in
the case you base this on, the Poynting vector does actually give you
the direction and magnitude at each point.
Ok, thanks.
--
Army1987 (Replace "NOSPAM" with "email") |
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| Timo Nieminen |
| Posted: Tue Feb 05, 2008 7:00 pm |
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On Tue, 5 Feb 2008, Army1987 wrote:
Quote: Timo Nieminen wrote:
On Mon, 4 Feb 2008, Army1987 wrote:
In a homework assignment, I was asked to calculate the flux of Poynting
vector on the surface of a cylindrical conductor of constant resistivity,
with uniform electric field within it, parallel to the cylinder's axis. It
asked to compare it with the power dissipated (by Joule effect in it).
The Poynting vector is supposed to be the superficial density of power
flowing through a point, but I found a *negative* flux, i.e. the vector
pointing *into* the conductor, as if the energy was coming from outside
into it.
I've checked carefully whether I committed a sign error, but it can't find
one.
Any thoughts?
The Joule heating is _inside_ the conductor. Electromagnetic energy is
being converted to heat. Since it's all in steady-state, electromagnetic
energy must be moving _into_ the conductor to replace the loss.
I think I get the point. But it appears that the energy passes through the
lateral surface of the cylinder, rather than through the ends.
The current doesn't carry energy - it's the source of losses. The energy
is carried by the fields outside the wire. You can think of the wire as a
waveguide. For some fun, see what happens when you make the resistance of
the wire -> zero. (Also fun to look at the AC case. This might not be easy
- I don't know what kind of course you're doing, what level, or what
textbook you're using.)
Anyway, as R -> 0, the E field along the wire -> 0, and the inward
Poynting vector goes to zero. If there is a resistive load further along
the wire, there must be power transport along the wire, which means there
must be an outward electric field - the wire must have a surface charge
over its outside. The wire on the far side of the resistance must also
have a surface charge, and together these produce the electric field along
the resistance.
There've been occasional papers on this topic in American Journal of
Physics (which is a physics education journal). The best I've seen is J.
D. Jackson, AJP 64(7), 855-870 (1996).
While this gives the link between electromagnetic theory and circuit
theory (which is just a convenient approximate method of solving circuit
problems for "electrically small" circuits), it isn't necessary if all one
wants to know is the _effects_ of/on the circuit, such as what the
currents, potentials, and losses of energy in the circuit are. So don't
expect it to be mentioned in a circuit theory course.
However, I think it's useful to remind yourself that electric circuits are
fundamentally not about electrons moving in wires, but about electric and
magnetic fields around the wires. The movement of electrons is almost just
a side-effect. That's why there's no problem with electrical signals
travelling along wires at the speed of light, while the electrons crawl
along at some mm/s.
Quote: Does it
mean that the flux of the vector through a closed surface is the power,
but its direction and magnitude in a given point needn't?
The relevant conservation law is
du/dt + J.E + div(S) = 0,
where u = electromagnetic energy density, J.E is the density of energy
loss through Joule heating, and S is the Poynting vector.
It's converted to the flux-through-a-surface version using the the
divergence theorem. This doesn't mean that S is necessarily the energy
flux. J. Slepian had a paper in J. Applied Physics in 1942 (vol 13, pp
512-518) on this; this paper has a respectable number of citations so it's
hardly unknown.
This leads to fun cases such as "Is there an energy flow in the static
field when, say, you put a charged capacitor between two magnetic poles?"
For the cases I've looked at where (a) E and H are produced by the same
source, or (b) H is produced entirely by currents resulting from E, or (c)
E is produced entirely by charge densities resulting from H, the Poynting
vector is a perfectly reasonable energy flux. For cases where different
sources contribute (like the magnet+capacitor above, or an electromagnetic
wave superimposed on a static electric field), it looks downright silly to
interpret the Poynting vector as the energy flux.
So, to answer your question, I think your conclusion is correct, but in
the case you base this on, the Poynting vector does actually give you the
direction and magnitude at each point.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html |
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| Army1987 |
| Posted: Wed Feb 06, 2008 2:04 am |
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Army1987 wrote:
Quote: Timo Nieminen wrote:
On Tue, 5 Feb 2008, Army1987 wrote:
I'm trying the AC case. The flux of S is the negative of the Joule
power, plus a term which looks like the time derivative of the e.m. energy
inside the cylinder. That term is actually rho * epsilon * dWe / dt, where
rho is resistivity and We is the electric energy.
Anyway, rho * epsilon is on the attosecond scale for metals, and anyway
very small for any reasonably conducting material. So, unless I'm passing
megahertz frequency AC through water, or I'm studying the leak conduction
of capacitors, I can neglect that term.
--
Army1987 (Replace "NOSPAM" with "email") |
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