Why do metals like iron, cobalt, and nickel show a negative Thomson's effect? What is the reason for other metals like copper and zinc to show a positive Thomson's effect? Is there any reasonable idea explaining this other than experimental observation?
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@Lamichhane88, is my answer helpful, or is there anything I'm missing? – auden Aug 02 '16 at 21:29
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@heather, You made an statement: In certain metals the effect reverses sign as the structure of it is altered or the temperature is increased.But I wish to know how does the structure alters leading to two different types of effects. – Lamichhane88 Aug 03 '16 at 14:45
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@Lamichhane88, I cannot seem to find any information for how that particular aspect of it works, but I do know that the first part of the explanation (the one relevant to the question) is correct. – auden Aug 04 '16 at 15:26
1 Answers
I'm just going to quickly go through the different thermoelectric effects, and then explain the answer.
Seebeck Effect
The Seebeck effect is the conversion of heat directly into electricity at the junction of different types of wire (for example, at the intersection of a piece of copper wire and a piece of gold wire). This occurs because the electron energy levels in each type of wire shift differently, which creates a voltage difference between the two wires, an electric current, and therefore a magnetic field around the wires. It was discovered in 1821 by Thomas Seebeck, though he didn't realize electric current was involved; Hans Oersted corrected the oversight and coined the term thermoelectricity. The Seebeck effect is a classic example of an electromotive force (emf). The Seebeck effect is generally described locally by the creation of an electromotive field
$E_{emf} = -S \nabla T$,
where $S$ is the Seebeck coefficient and $\nabla T$ is the gradient in temperature $T$. For ordinary materials at room temperature, the Seebeck coefficient could vary from −100 μV/K to +1,000 μV/K.
Peltier Effect
The Peltier effect is the presence of heating or cooling at the electrified junction of two different conductors. It was discovered in 1834 by Jean Peltier. This is described by
$\dot{Q} = \left( \Pi_\mathrm{A} - \Pi_\mathrm{B} \right) I$,
where $\dot{Q}$ is Peltier heat generated at the junction per unit time, $II_A(II_B)$ is the Peltier coefficient of the conductor, and $I$ is the electric current. The total heat generated may also be influenced by Joule heating.
Thomson Effect
In many materials, the Seebeck coefficient is not constant in temperature, and so a spatial gradient can result in a gradient in the coefficient. If a current is then driven through this gradient, a continuous version of the Peltier effect will occur. This is the Thomson effect, predicted and observed by Lord Kelvin in 1851. It describes the heating or cooling of a conductor with a temperature gradient. It is described as
$\dot q = -\mathcal K \mathbf J \cdot \boldsymbol \nabla T$
where $\nabla T$ is the temperature gradient, $\mathcal K$ is the Thomson coefficient, $\mathbf J$ is the current density, $\dot q$ is the heat production per unit volume. It should be noted that the relation between the Thomson coefficient and the Seebeck coefficient is $\mathcal K = T \frac{d S}{d T}$.
Thomson Relations
In 1854, Lord Kelvin found relationships between the Seebeck, Thomson, and Peltier coefficients, implying that the three effects were really different examples of one effect, which is characterized by the Seebeck coefficient. The first Thomson relation can be expressed as
$\mathcal K \equiv {d\Pi \over dT} - S$,
where ${\displaystyle \scriptstyle T}$ is the absolute temperature, $\mathcal K$ is the Thomson coefficient, $\Pi$ is the Peltier coefficient, and $S$ is the Seebeck coefficient. The second Thomson relation is
$\Pi = TS$
(It should be noted that if the material in question is placed in a magnetic field, the second relation is not as simple as shown.) It is also worth noting that the Thomson coefficient is unique among the three coefficients in that it is directly measurable for individual materials; the other two coefficients can only be easily determined for pairs of materials.
Your Question
As far as I can tell, the direction of the Thomson emf (which is, of course, directly related to the Thomson coefficient, which is what you are referencing as positive or negative) is such that it supports a current from cold to hot regions of the metal. In certain metals the effect reverses sign as the structure of it is altered or the temperature is increased.
This has to do with the structure of the metal, namely, because of subtle differences in electron band structure. This book (google book format; open to the correct page) elaborates further.
I'll be updating this as I find more information. In the meantime, this website is very informative for the basics. Hope this helps!
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@Lamichhane88, I'd be glad to address what I've missed. What do I need to explain? Thank you for the bounty, by the way. – auden Aug 08 '16 at 15:13
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,only if you can :how does the structure alters leading to two different types of effects? – Lamichhane88 Aug 08 '16 at 15:19
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@Lamichhane88, so you want to know how the structure of the metals leads to the positive/negative effect? – auden Aug 08 '16 at 15:20
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yes indeed......and note that bounty was provided with the expectation of answer required... – Lamichhane88 Aug 08 '16 at 15:23
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I downvoted for a few innacuracies here and there and the lack of answering the question. – untreated_paramediensis_karnik Apr 28 '18 at 15:42
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