Is There a Perpetual Motion Machine Hidden in the Capacitor Paradox?

[Phrak]

The http://en.wikipedia.org/wiki/Four-current" (more correctly the four-current-density) transforms as a four-vector.

Now that you've brought it up, could you clear up a point for me? I hope you don't mind the word doc attachment; the equations were difficult enough as was.

 

[Dale]

My understanding was that assigning units to the tensor itself was the same as assigning units to each of the components. With the caviat that the components are scalars so a tensor with units X cannot be added to a component with the same units. That could be wrong, but it was my understanding.

 

[Phrak]

My understanding was that assigning units to the tensor itself was the same as assigning units to each of the components. With the caviat that the components are scalars so a tensor with units X cannot be added to a component with the same units. That could be wrong, but it was my understanding.

If I have a space-time vector (type 1,0 tensor) whose elements have units (T,D,D,D) and attach units of per-volume, V^-3, to the entire vector I get units (T,D,D,D)V^-3, or (TV^-3, DV^-3, DV^-3, DV^-3).

I don't know if that's quit what you said, but it makes sense that a type(1,0) tensor can have elements with charge densities now you've talked it up. Thanks for helping me over this thing. It's been bothering me for better than a year! And now seems perfectly obvious.

 

[Dale]

Well, usually we multiply the T by c to get units of T L/T = L. This is called making it dimensionally consistent and is really required if you are going to do any summation over the elements. So the standard four-vector has units (T L/T, L, L, L) = (L, L, L, L) = (1,1,1,1) L.

In the case of the four-current, ordinary current density is current/area and current is charge/time, so current density is C/TL². The charge density on the other hand is charge/volume C/L³ which when multiplied by c gives C/L³ L/T = C/TL² which is the same units as current density. So the four-current is also dimensionally consistent with units of current/area.

 

[Naty1]

And so can someone summarize what has been agreed to here and why?? OR what has not been agreed to and why??

It sure would be nice if a Science Advisor would briefly summarize conclusions in long threads like this. Or identify points of remaining contention.

That would not only aid those of us who have not formally studied all the underlying math and theory, but might reduce the number of repeat questions on the same issue...

 

[Dale]

Yeah, but the summary would inevitably wind up being as disputed as the rest of the thread. E.g. I would summarize this thread as follows:

Energy is conserved in Maxwell's equations. You can, of course, ignore Maxwell's equations and then mistakenly conclude that energy is not conserved. One specific example of how to do so is the OP.

Let the objections commence.

 

[DrGreg]

Well, usually we multiply the T by c to get units of T L/T = L. This is called making it dimensionally consistent and is really required if you are going to do any summation over the elements. So the standard four-vector has units (T L/T, L, L, L) = (L, L, L, L) = (1,1,1,1) L.

In the case of the four-current, ordinary current density is current/area and current is charge/time, so current density is C/TL². The charge density on the other hand is charge/volume C/L³ which when multiplied by c gives C/L³ L/T = C/TL² which is the same units as current density. So the four-current is also dimensionally consistent with units of current/area.

What DaleSpam says is true, but note the word "usually". I think I would prefer to say "...often, in special relativity,..."

It's not compulsory to work with "dimensionally consistent" components, indeed in general relativity we often don't. E.g. two of the components could be angles, not lengths, in spherical polar spatial coordinates. The issue of summation over the elements gets handled by the components of the metric tensor in your chosen coordinates.

As an aside, I only recently realized that in (T, L, L, L) Minkowski coordinates the covariant (i.e. not contravariant) 4-momentum comes out in (E, p, p, p) units without any conversion factor required.

Which was nice.

 

[Dale]

Yes, I agree with DrGreg, my comments were the usual convention for special relativity, not general relativity which is much more flexible in that sense and doesn't require dimensional consistency of tensor elements in the same way.

Thanks for clarifying DrGreg.

 

[Privalov]

My equation shows that the stored energy is rotation invarient if E²-B² is the conserved quantity..

Energy in electric field is the potential energy of separated charges. It can change, if the distance between charges changes. In principle, it could have changed in the given problem (it did not though).

Energy in magnetic field is the kinetic energy of moving charges. In this paradox, speed of charges remains the same. Therefore, magnetic field energy remains constant. I believe electromagnetism has nothing to do this particular problem.

 

[Dale]

Energy in magnetic field is the kinetic energy of moving charges.

This is not correct at all. The KE of charges is negligible, typically the electron drift velocity is on the order of 1 mm/s or even less. With so little velocity and so little mass their KE is typically insignificant and in general has nothing to do with the energy of the magnetic field.

 

[Dadface]

I have read through this several times trying to get the gist of it but I am still confused so I apologise in advance if I get it wrong and misinterpret some of which has been written.
Firstly the energy density is given by n=0.5* permittivity*E^2
Secondly it seems to have been suggested that since the volume is the same in each case then so is the energy.This is not so because the capacitance of the capacitor is not proportional to its volume it is proportional to the plate overlap area(A) divided by the plate separation(d).Here I have considered a parallel plate capacitor and I have ignored edge effects.It follows that if A reduced to half of its original value then so would C reduce this resulting in a doubling of the energy stored(Energy=Q^2/2C).I have assumed that the charge (Q) remains constant.If d is reduced to half of its original value C doubles and the energy stored halves.

 

[pallidin]

Take 2 pennies and separate them equal to the thickness to one penny. The air is the dielectric.
Charge those pennies like a capacitor, and note their discharge potential.

Now, take those two pennies and beat them down to such a thickness that it is "paper-thin", such as each penny is now about 6-inches in diameter.

With the same dielectric(air) and dielectric spacing(one penny thickness), charge and note discharge potential.

The second scenario has much greater energy storage potential.

 

[Privalov]

The KE of charges is negligible, typically the electron drift velocity is on the order of 1 mm/s or even less. With so little velocity and so little mass their KE is typically insignificant

Yes; and energy in magnetic field of a typical wire at any given moment is also insignificant. However, any attempt to drain the energy from this magnetic field will result in more energy supplied by the power plant. So, total amount of energy, transferred by wire, can become significant. Kinetic energy of moving electrons remains the energy carrier though.

Anyhow, this seems like terminological dispute. Let me find some material examples.

I believe that accurate calculation will reveal the energy in magnetic field of a moving capacitor will not depend on capacitor orientation (so capacitor experiences no torque). True or false?

It will be hard to prove though (at least based on formulas, which Bob S used).

Let’s say we have a charged particle moving at given velocity. By placing solenoids around it, we can extract energy from its magnetic field. I believe we can not extract more energy, than the kinetic energy of particle. True or false?

it seems to have been suggested that since the volume is the same in each case then so is the energy.This is not so

Not so, if the voltage is the same. Typically, it’s the case. However, in this particular problem (as described in my post #1) there are no constraints on voltage; while charge is constant.

 

[Dadface]

Take 2 pennies and separate them equal to the thickness to one penny. The air is the dielectric.
Charge those pennies like a capacitor, and note their discharge potential.

Now, take those two pennies and beat them down to such a thickness that it is "paper-thin", such as each penny is now about 6-inches in diameter.

With the same dielectric(air) and dielectric spacing(one penny thickness), charge and note discharge potential.

The second scenario has much greater energy storage potential.

So basically you have increased the capacitance by increasing the plate area.It is true that if you take a large and a small capacitor and charge them to the same voltage then the large one will store more energy.With this paradox we can have different situations:
1.BATTERY PERMANENTLY CONNECTED.In this case V will equalise when the capacitance is varied but Q will vary.If C is increased, more charge will flow from the battery and the energy stored will increase.
2.CAPACITOR CHARGED AND BATTERY DISCONNECTED.In this case Q will be constant but V will vary.If C is increased, V will decrease and the energy stored will decrease.
In my post above I was referring to situation two and although both situations give opposite results the paradox still exists.

 

[pallidin]

Perhaps I'm not understanding. What, specifically, is the paradox?

 

[Meir Achuz]

So basically you have increased the capacitance by increasing the plate area.It is true that if you take a large and a small capacitor and charge them to the same voltage then the large one will store more energy.With this paradox we can have different situations:
1.BATTERY PERMANENTLY CONNECTED.In this case V will equalise when the capacitance is varied but Q will vary.If C is increased, more charge will flow from the battery and the energy stored will increase.
2.CAPACITOR CHARGED AND BATTERY DISCONNECTED.In this case Q will be constant but V will vary.If C is increased, V will decrease and the energy stored will decrease.
In my post above I was referring to situation two and although both situations give opposite results the paradox still exists.

I got to this thread late, but just noticed your "paradox".
This is treated in most EM textbooks. A battery connected to keep the capacitor at constant voltage does twice as much work as the negative energy difference at constant Q. This work done by the battery effectively changes the sign. The arithmetic is -1+2=+1.

 

[Dadface]

Perhaps I'm not understanding. What, specifically, is the paradox?

I got to this thread late, but just noticed your "paradox".
This is treated in most EM textbooks. A battery connected to keep the capacitor at constant voltage does twice as much work as the negative energy difference at constant Q. This work done by the battery effectively changes the sign. The arithmetic is -1+2=+1.

I think we are at cross purposes here and it is necessary to read throught the whole thread.The paradox is about the changes that would apparently be observed by different observers if the capacitor was moving at relativistic speeds.In my posts I was mainly pointing out that that the equal volume changes in the two capacitor orientations referred to earlier in the thread do not result in equal capacitance changes.

 

[Dale]

Perhaps I'm not understanding. What, specifically, is the paradox?

I agree with you. There is no real paradox here. Just a typical "apparent paradox" from not using the laws of physics and then being surprised that the result is non-physical.