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Thermodynamics of Electricity

Thermodynamics of Electricity - Physics Forum

Thermodynamics of Electricity - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 01-27-2004, 05:34 PM
The Void
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Default Thermodynamics of Electricity



Okay, so the way I figure it, the value of the electric field E in
each of the x, y, and z directions is a degree of freedom, and the
energy associated with each is proportional to its value squared. So
the equipartition theorem ought to apply. That means the energy of an
infintesimally small volume associated with its electric field in the
x-direction, dV*(1/2)*e_0*E_x^2, on average ought to be equal to
(1/2)kT. But that makes mean(E_x^2)=kT/dV/e_0. But that's infinite!
And it would make the average electric energy density
mean[(1/2)*e_0*(E_x^2+E_y^2+E_z^2)]=(3/2)kT/dV, another infinity.
That would mean that there's an infinite amount of electrical energy
in this room, which is impossible, because not only would it turn the
room into a black hole, it would fry this computer. But I can't find
anything wrong with that analysis. So could someone tell me what the
hell is wrong?
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  #2  
Old 01-27-2004, 08:13 PM
Uncle Al
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Default Thermodynamics of Electricity

The Void wrote:

What makes you think an electric charge is in thermal equilibrium? I
take a hollow thin-walled metal spheres suspended weightless in vacuum
and internally inject a microcoulomb of electrons. I take a hollow
thin-walled fused silica sphere and internally inject a microcoulomb
of electrons. I wait minute hour for things to settle down, and then
measure. I measure again in an hour. I come back in a year and
measure again. What does "thermodynamic equilibrium" mean for these
two charged hollow spheres?

Compare to injecting 100 calories of heat.

Charge is quantized and carries along spin angular momentum. What is
the smallest quanity of heat you can transfer?

--
Uncle Al
[Only registered users see links. ]
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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  #3  
Old 01-27-2004, 08:39 PM
Old Man
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Default Thermodynamics of Electricity

"The Void" <[Only registered users see links. ]> wrote in message
news:d661004a.0401270934.12776f2a@posting.google.c om...

Children should not play with calculus. If one erroneously assumes
that an infinitesimal volume contains finite energy, then the correct
conclusion is that the energy density in that volume is infinite.
(E^2)dV is an infinitesimal. It cannot be set equal kT, which is finite.
To find the average energy density, do the integration over E^2 first,
and then divide that result by the integrated volume. [Old Man]



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  #4  
Old 01-28-2004, 02:28 AM
tadchem
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Default Thermodynamics of Electricity


"The Void" <[Only registered users see links. ]> wrote in message
news:d661004a.0401270934.12776f2a@posting.google.c om...

There is your error. The left side is a differential quantity, the right
side is not.

By dividing by dV, you are trying (in the limit) to divide by 0.

The energy in a differential volume equals the *energy density* times the
volume element, and is itself a differential quantity.

You are assigning to a differential quantity (dV) the total kinetic energy
assigned to a degree of freedom of a gas molecule. The assignment is
unjustifiable.


Tom Davidson
Richmond, VA


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  #5  
Old 01-28-2004, 03:46 PM
Franz Heymann
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Default Thermodynamics of Electricity


"The Void" <[Only registered users see links. ]> wrote in message
news:d661004a.0401270934.12776f2a@posting.google.c om...

Rubbish.
You appear to be totally unfamiliar with the use of infinitesimals.

Franz


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  #6  
Old 01-28-2004, 05:20 PM
The Void
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Default Thermodynamics of Electricity

[Only registered users see links. ] (The Void) wrote in message news:<d661004a.0401270934.12776f2a@posting.google. com>...

You folks aren't answering the question! What does charge being
quantized have to do with it, and where would there be any integration
over E^2 -- and what would that even mean? Maybe I could try to be
more clear:

The electromagnetic field in a region is often in thermodynamic
equilibrium with its surroundings.

The energy associated with the field is proportional to the field
strength squared. Thus the equipartition theorem ought to apply.

The number of degrees of freedom of the electromagnetic field in a
region is infinite.

The equipartition theorem would then tell us that the expected energy
of the field in the region is equal to fkT/2, with f the number of
degrees of freedom.

But that would make the energy in the region infinite, contradicting
obvious empirical facts. So where is the error?
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  #7  
Old 01-28-2004, 08:26 PM
zigoteau
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Default Thermodynamics of Electricity

[Only registered users see links. ] (The Void) wrote in message news:<d661004a.0401270934.12776f2a@posting.google. com>...

Hi Void,

I think that none of the answers so far has hit the spot, so I'll have
my twopence worth.



You have rediscovered a problem which was known and worried about
roughly 100 years ago. I think it was Jeans who did the calculation.
At the time it was called the ultraviolet catastrophe.

Yes, according to classical statistical thermodynamics, each degree of
freedom of a system should be excited by thermal fluctuations to have
an average energy of kT/2; each mode an average energy of kT. A field
has infinitely many degrees of freedom. If you have a system with a
finite number of particles, then when equilibrium is reached all the
energy ends up in the field.

The ultraviolet catastrophe involves black-body radiation, the closest
approximation to which is achieved experimentally by making a closed
oven with a small hole - the so-called cavity (German: Hohlraum)
radiation. Statistical mechanical calculations, of the sort first
carried out by Maxwell, predicted that the brightness at frequency f
should be proportional to f^2, i.e. a black body should be incredibly
bright at X-ray wavelengths and an even brighter source of gamma rays.
Clearly this is a nonsense. Quite a few experimenters investigated the
problem, including Planck. Experimentally, the black-body spectrum
does nothing of the sort, but has maximum brightness at a particular
wavelength which depends on the temperature.

Planck came up with a formula which fitted the experimental data
extremely well. His formula had one constant, h. Planck conjectured
that his formula meant that energy was only exchanged in packets, or
quanta.

The rest, as they say, is history, and the solution to the problem was
quantum mechanics. In quantum statistical thermodynamics, the average
energy in a mode is only kT per mode for low-frequency modes. Very
high-frequency modes are hardly excited at all, so that at equilibrium
at 300K there is not very much energy in the radiation field after
all, and you may sleep soundly in your bed.

Cheers,

Zigoteau.
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  #8  
Old 01-29-2004, 12:10 AM
Old Man
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Default Thermodynamics of Electricity


"The Void" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ] m...
news:<d661004a.0401270934.12776f2a@posting.google. com>...

The ambiguity here originates with "The Void".


This is ambiguous. Give an example. Specify the boundary conditions.
Better to speak of the charge momentum distribution as being in thermal
equilibrium. To Nth order, static electromagnetic fields do not interact
with dynamic electromagnetic waves: superposition rules.


Bullshit. The field has amplitude as well as direction. Quantization
is required, wherein the field is represented in phase space as a
distribution of virtual photons. The equapartition of energy applies
to each photon, not being limited to the three degrees of freedom
of the field.


The error is in "The Void's" presumption of competence. Erroneous
calculus and erroneous physics are the tools of untutored simpletons.
In any case, as is well known, a competent treatment reveals classical
electrodynamics to be inadequate here. [Old Man]


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  #9  
Old 01-29-2004, 09:45 PM
Franz Heymann
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Default Thermodynamics of Electricity


"The Void" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ] m...

[snip]


Please be more explicit.
What is "often" in this context?
When is this equilibrium achieved and when is it not achieved?
By what processes of energy exchange is this equilibrium achieved?

Franz


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  #10  
Old 01-29-2004, 09:45 PM
Franz Heymann
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Default Thermodynamics of Electricity


"zigoteau" <[Only registered users see links. ]> wrote in message
news:9da9cba1.0401281226.60f93978@posting.google.c om...
news:<d661004a.0401270934.12776f2a@posting.google. com>...

Sorry Zigoteau, but it is not obvious that that is what The Void wqs on
about. He should be worried about the simple nonsense of an equation which
has an infinitesimal on one side and not on the other.


You have been discussing a topic far more erudite than the level at which
The Void will be able to comprehend.

Franz


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