Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Kind of an extension to this question:

If you heat up an object, and put it in contact with a colder object, in an ideal insulated box, the heat from one will transfer to the other through thermal conduction and they will eventually reach an equilibrium temperature at the midpoint, correct?

Now if you have a hot resistor (electrical component) and a cold resistor, and connect them by their leads, so that they make a circuit:

hot and cold resistors

there will be the same conduction and radiation heat transfers. But also, the hotter resistor will have a larger noise current, right? So will there additionally be a transfer of electrical energy from one resistor to the other? Would completing the circuit allow them to reach equilibrium temperature faster than if they were just touching through an insulator with the same thermal conductivity?

share|improve this question
    
I think it would be helpful if you separated the concepts of thermal resistance and electrical resistance in your mind, because as I see it your question is unanswerable because one can only guess what thermal resistances for said connections would be. BTW, you can, in fact, have a thermal circuit where heat is conducted over a wire and several components do things like resistance and energy conversion. But an electrical resistor is mostly irrelevant to that. –  Alan Rominger May 23 '11 at 19:00
    
@Zassounotsukushi I'm not sure they're even separable. The thermoelectric effect involves heat currents carried by electric charge, for instance. I've tried to clarify what I'm asking about, though. The thermal resistance should be irrelevant. –  endolith May 23 '11 at 19:04

2 Answers 2

A resistor at a temperature T has a fluctuating voltage. This is a consequence of the fluctuation dissipation theorem which you can use to calculate the spectrum of the voltage. The wikipedia article on the Fluctuation Dissipation Theorem has a section on resistor thermal noise. When measured over a bandwidth $\Delta\nu$, the average squared voltage is: $$\langle V^2\rangle = 4Rk_BT\Delta\nu$$ where $k_B= 1.38\times 10^{-23}$J/K is Boltzmann's constant.

Suppose we have a resistor $R$ maintained at a temperature T, and hooked up to a resistor $R_0$ initially at absolute zero. The thermal noise of the warm resistor will be as given above. When you apply a voltage $V$ to a resistor $R_0$, the dissipation (in watts) will be given by $IV = V^2R_0$, so the watts applied to the resistor initially at absolute zero, over a bandwidth $\Delta\nu$ will be $$\langle V^2\rangle = 4R_0Rk_BT\Delta\nu.$$

As that resistor warms up to temperature $T_0$, it will apply a fluctuating voltage on the warm resistor. Following the above, but with the two resistors swapped, the power applied to the warm resistor by the colder resistor at temperature $T_0$ will be: $$\langle V^2\rangle = 4RR_0k_BT_0\Delta\nu.$$

The system will be in balance when the above two powers are equal. This happens algebraically when $T=T_0$.


A possible source of paradoxical confusion is that the above calculation was done over a limited bandwidth range. But the calculation does not depend on frequency; instead the power transmitted is simply proportional to the range of bandwidths.

For the usual physical system, we consider frequencies that run from 0 to infinity. Thus the total bandwidth is infinite. This suggests that the power flow in the above should be infinite. This paradox is avoided by noting that physical resistors have a limited bandwidth. There is always a parasitic capacitance so that the bandwidth is limited on the high side. Thus the power transfer rate depends on how ideal your resistors are.


As an example calculation, suppose that a resistor has a maximum frequency of 100 GHz $= 10^{11}$ Hz, a (room) temperature of 300K, and a resistance of 1000 ohms. Then the power transfer rate is: $$ 4\times 1000 \times 1.38\times 10^{-23} 10^{11} \times 300 = 1.66\;\;\textrm{uWatts}$$ Given the heat capacity of the resistor, you can compute the relaxation time with which the colder resistor exponentially approaches an equal temperature.

share|improve this answer
    
Wait, so the equilibrium point occurs when both resistors are at absolute zero? :( –  endolith Jun 27 '11 at 21:16
    
No, the equilibrium occurs when the power generated by the two resistors (and dissipated by the other) are equal. I.e. $4RR_0k_bT_0\Delta_v = 4R_0Rk_bT_1\Delta_v$ and therefore $T_0=t_1$. –  Carl Brannen Jun 29 '11 at 3:15
1  
Dissipation is $V^2/R$, not $V^2R$. –  Retarded Potential Mar 1 '13 at 17:04

Actually, that's exactly what happens: heat is transported by freely moving electrons. - but you wouldn't call it noise current, you would call it heat transfer, regardless of whether it's phonons or electrons which carry the heat.

The whole thing really has nothing to do with the resistors, it would work equally well with any passive component (proof: otherwise you could use that to build a 2nd order perpetuum mobile).

share|improve this answer
    
"you wouldn't call it noise current" But it is the same thing as noise current, isn't it? "heat is transported by freely moving electrons" Meaning it's carried by their vibrations pushing against each other, or by electric field waves, not by a net DC flow in one direction? –  endolith May 24 '11 at 15:08
    
"regardless of whether it's phonons or electrons which carry the heat" But phonon and electron heat transfer are independent of each other, no? Metals are good thermal conductors/electrical conductors because of free electrons, but diamond is an electrical insulator and good thermal conductor because of phonons in the crystal structure? –  endolith May 24 '11 at 15:11
    
Yes, you just can't see this from the outside. –  leftaroundabout May 24 '11 at 16:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.