Understanding the Seebeck effect

  • #1
fluidistic
Gold Member
3,896
232
I'm trying to understand the Seebeck effect in an open circuit of a single material. In other words the system consists of a metal whose ends are kept at different temperature, and we wait long enough for the steady-state to establish.

Apparently we can think of the (quasi)electrons in a semi-classical manner, where they have well defined speed and position in the metal. Also, rather than being an equilibrium state where the electrons are static, it is question of a non equilibrium steady-state where there are two different forces acting on the electrons, whose magnitude is equal but whose direction is opposite, so that the net force is zero. These forces are the one due to the temperature gradient, and the one due to the charge distribution creating an electric field thanks to the Seebeck effect.

What I completely fail to swallow is that, despite having 0 force on the electrons in average, we can (and we must) think of them as "some are pushed solely due to the Seebeck E field in a particular direction, whilst others are pushed in the opposite direction due to the temperature gradient". So there are two currents, that cancel out exactly in the steady-state. But in classical mechanics, when the net force acting on a particle vanishes, the particle has a constant velocity. In our case I would expect that velocity to be zero, else charges would accumulate on a side of the metal and we wouldn't have been in steady-state. But this is not the case. There is clearly a dynamics of electrons, even though there is no net motion in average.

So my question can be reworded as "how is that possible to have a steady-state while the net force on any particle, in average, vanishes?".
 

Answers and Replies

  • #2
SpinFlop
35
38
So my question can be reworded as "how is that possible to have a steady-state while the net force on any particle, in average, vanishes?".

A simple harmonic oscillator has a force ##F = -kx## with the steady state solution ##x = A\cos(\sqrt{k/m} t)##. Thus, the time average is $$\bar F = \lim_{t_{f} \rightarrow \infty} \frac{\int_0^{t_{f}}-Ak\cos(\sqrt{\frac{k}{m}} t)\, dt}{t_{f} - 0} = -\sqrt{mk}\lim_{t_{f} \rightarrow \infty}\frac{\sin(\sqrt{\frac{k}{m}} t_{f})}{t_{f}} = 0$$
So this is a very simple example of how you can have a steady state with a vanishing average force. In your Seebeck material the two opposing forces act effectively as a single restoring force. Namely, when an electron gets to close to one end, then it is pushed back in the other direction. Just like the above harmonic oscillator, the time average of the force vanishes, ie: over a long enough period of time it is pushed to the left just as much as it is pushed to the right.
 
  • #3
fluidistic
Gold Member
3,896
232
A simple harmonic oscillator has a force ##F = -kx## with the steady state solution ##x = A\cos(\sqrt{k/m} t)##. Thus, the time average is $$\bar F = \lim_{t_{f} \rightarrow \infty} \frac{\int_0^{t_{f}}-Ak\cos(\sqrt{\frac{k}{m}} t)\, dt}{t_{f} - 0} = -\sqrt{mk}\lim_{t_{f} \rightarrow \infty}\frac{\sin(\sqrt{\frac{k}{m}} t_{f})}{t_{f}} = 0$$
So this is a very simple example of how you can have a steady state with a vanishing average force. In your Seebeck material the two opposing forces act effectively as a single restoring force. Namely, when an electron gets to close to one end, then it is pushed back in the other direction. Just like the above harmonic oscillator, the time average of the force vanishes, ie: over a long enough period of time it is pushed to the left just as much as it is pushed to the right.
Thanks a lot, I hadn't realized a SHM was a steady-state system. That was very enlightening.
I'm still entirely blank regarding the Seebeck effect in general. If I understand correctly, in the case you describe where an electron goes too close to an end, say the hotter end, would it come back to near it should thanks to the two forces (temperature gradient and electrostatic force)? And not solely thanks to a single force.

I still have other blind spots with the Seebeck effect. For instance, would this effect modify the temperature distribution in the metal?
On one hand, it shouldn't, because it has no impact on Fourier's law and therefore the temperature distribution should be linear.
On the other hand, the Seebeck effect modifies the charge distribution, by Poisson equation. Using the steady-state condition which links voltage to temperature ($\nabla V = -S\nabla T$). And so a link between temperature and charge distribution is established. So the temperature distribution results affected by the Seebeck voltage... which contradicts the fact that Fourier's law is "Seebeck effect free".

What am I missing?
 
  • #4
DrDu
Science Advisor
6,258
906
What I completely fail to swallow is that, despite having 0 force on the electrons in average, we can (and we must) think of them as "some are pushed solely due to the Seebeck E field in a particular direction, whilst others are pushed in the opposite direction due to the temperature gradient". So there are two currents, that cancel out exactly in the steady-state.
This is wrong. There are no two kinds of electrons, ones which are only pushed by the field and others pushed by the temperature gradient. Any decomposition of the net flux (which is zero in steady state) into components is an artificial device. Anyhow the temperature gradient is not an actual physical force. The point here is that the electric field gradient is not the only force acting on the electrons. Rather there are the continuous pushes due to collisions with atomic cores which changes their speed, direction and energy. An electron with a higher velocity will spent less time in a given volume. Hence the hotter regions of the metal will contain less electrons on average.

There are other factors contributing, and I am not sure about their relative importance.
For example, the motion of the electrons is hindered by the presence of other electrons, the electrons being Fermions and hence the Pauli principle in place. At higher temperatures, there are more states accessible to an electron, hence the chance to find an electron in a region of higher temperature increases. You can see that this effect is somehow complementary to the one I described before.
 
  • #5
fluidistic
Gold Member
3,896
232
First of all, thanks a lot for your reply.

This is wrong. There are no two kinds of electrons, ones which are only pushed by the field and others pushed by the temperature gradient. Any decomposition of the net flux (which is zero in steady state) into components is an artificial device.
Ok, great.


Anyhow the temperature gradient is not an actual physical force. The point here is that the electric field gradient is not the only force acting on the electrons. Rather there are the continuous pushes due to collisions with atomic cores which changes their speed, direction and energy. An electron with a higher velocity will spent less time in a given volume. Hence the hotter regions of the metal will contain less electrons on average.
May I know what are some differences between a real and a non real physical force? It's not obvious at all to me that the temperature gradient is not an actual force.

Then, are you talking about phonon drag, i.e. collisions or better say, interactions between phonons and electrons leading to an enhancing of the Seebeck coefficient? I understand that there's a sort of diffusion (though I am not sure at all there really is a diffusion equation involved at all. I couldn't derive it from Boltzmann transport equation. There was no way on Earth I could make a term with $\nabla _x ^2$ appear.), and I think this diffusion should be present regardless of the presence of the phonon drag effect. All is required is the temperature gradient and the Seebeck effect to take place (hence a modified charge distribution leading to a non zero E field).

There are other factors contributing, and I am not sure about their relative importance.
For example, the motion of the electrons is hindered by the presence of other electrons, the electrons being Fermions and hence the Pauli principle in place. At higher temperatures, there are more states accessible to an electron, hence the chance to find an electron in a region of higher temperature increases. You can see that this effect is somehow complementary to the one I described before.
I see.
 
  • #6
fluidistic
Gold Member
3,896
232
If the temperature gradient does not represent a real force, is the following reasoning correct?
In a material with non homogeneous temperature distribution and in which the chemical potential is temperature dependent, in such way that any tiny region can be considered as having a well defined temperature and chemical potential, and adjacent regions possesses a different chemical potential, then the gradient of the chemical potential does not vanish, and electrons will be driven to establish an equilibrium state by moving across the material as to equate the chemical potential everywhere. This "motion" can be described by a diffusion equation and is not driven by a "real force" per se. Instead, it results simply due to the non homogeneous chemical potential.
Note that by assuming a temperature dependent chemical potential, we made the underlying assumption that the Seebeck coefficient does not vanish.

I'm really not sure about what I just wrote at all.
 
  • #7
DrDu
Science Advisor
6,258
906
The Seebeck effect falls within the realm of linear irreversible thermodynamics. There, a stationary state is characterized by minimal entropy production.
The driving forces ##F_i## are the gradients of the inverse temperature 1/T and of ##\mu/T## where ##\mu## is the electrochemical potential.
They are related to the currents of heat and particle number by the relation ##J_i =\sum_j L_{ij} F_j##. In the stationary state of the Seebeck effect, the heat current is different from 0 while the particle current vanishes. With the definition of the forces given above, the thermal current will be due to a temperature gradient and even for a diagonal matrix L, a difference of the electrochemical potential must arise so that ##\mu/T## is constant in all of the material. Hence this does not necessarily mean that the chemical potential ##\mu_0=\mu+e\Phi## varies at all.
 
  • #8
DrDu
Science Advisor
6,258
906
Maybe it would be interesting to first consider a simpler problem, namely that of two boxes containing a charged ideal gas which can be hold at different temperatures and different potential. The electrochemical potential is known and the currents can be calculated.
 
  • #9
fluidistic
Gold Member
3,896
232
The Seebeck effect falls within the realm of linear irreversible thermodynamics. There, a stationary state is characterized by minimal entropy production.
The driving forces ##F_i## are the gradients of the inverse temperature 1/T and of ##\mu/T## where ##\mu## is the electrochemical potential.
They are related to the currents of heat and particle number by the relation ##J_i =\sum_j L_{ij} F_j##. In the stationary state of the Seebeck effect, the heat current is different from 0 while the particle current vanishes. With the definition of the forces given above, the thermal current will be due to a temperature gradient and even for a diagonal matrix L, a difference of the electrochemical potential must arise so that ##\mu/T## is constant in all of the material. Hence this does not necessarily mean that the chemical potential ##\mu_0=\mu+e\Phi## varies at all.
Thanks, it is a bit clearer now to me. Essentially you have assumed that Onsager reciprocity relations hold. I think this implies that the system possesses time reversal symmetry (assuming we change B to -B). I thought this would mean no entropy production, but I was wrong on this and you are absolutely right. I was lead to confusion because it is often claimed that thermoelectric effects are thermodynamically reversible. For example on the wikipedia page, they justify this claim by stating that any device working at Carnot efficiency is thermodynamically reversible. In the case of thermoelectrics, this means an infinite ZT value.
 

Suggested for: Understanding the Seebeck effect

  • Last Post
Replies
0
Views
558
Replies
0
Views
582
  • Last Post
Replies
1
Views
670
  • Last Post
Replies
2
Views
1K
Replies
1
Views
271
Replies
4
Views
2K
  • Last Post
Replies
6
Views
683
Replies
0
Views
672
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
6
Views
1K
Top