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Question 11.13: In this chapter, several examples of a current density in on...

In this chapter, several examples of a current density in one direction inducing the gradient of an intensive quantity in another direction were shown. These effects are referred to by the name of their discoverers : Righi-Leduc (11.29), Hall (11.75), Nernst (11.85), Ettingshausen (11.80). The latter refers to a temperature gradient induced by an orthogonal electric charge current density. It was pointed out recently that this effect can occur in a crystal, which consists of two types of electric charge carriers and presents a strong crystalline anisotropy in the plane where the heat and electric charge transport take place. No magnetic induction field needs to be applied orthogonally in order to observe this effect. Thematerial has two types of electric charge carriers, electrons (e) and holes (h). Assume that no ‘chemical reaction’ takes place between them. The thermoelectric properties are isotropic, i.e. the same in all directions. Therefore, the Seebeck tensors for the electrons and holes are given by,

∇T = −\Re (j_Q × B)                     (11.29)

 

\nabla \varphi = −H (j_q× B)        (11.75)

 

∇T = −Ε (j_q × B)                               (11.80)

 

\nabla \varphi = - Ν (∇T × B)            (11.85)

 

\varepsilon _e = \Biggl(\begin{matrix} ε_e & 0\\ 0 & ε_e \end{matrix} \Biggr)

 

\varepsilon _h = \Biggl(\begin{matrix} ε_h & 0\\ 0 & ε_h \end{matrix} \Biggr)

However, the conductivities differ greatly in two orthogonal directions. Therefore, the conductivity tensors are given by,

\sigma _e = \Biggl(\begin{matrix} σ_{e,αα} & 0 \\ 0 & σ_{e,bb} \end{matrix} \Biggr)

 

\sigma _h = \Biggl(\begin{matrix} σ_{h,αα} & 0 \\ 0 & σ_{h,bb} \end{matrix} \Biggr)

where a and b label the a-axis and the b-axis that are orthogonal crystalline axes.
Consider an electric charge transport along the x-axis at an angle θ from the α-axis and show that this electric current density j_q induces a heat current density j_Q along the y-axis.
This is the planar Ettingshausen effect. It can be understood by establishing the following facts :

a) Show that the Seebeck tensor for this crystal is given by,

ε = (σ_e + σ_h)^{−1} · (σ_e · ε_e + σ_h · ε_h)

b) Show that the Seebeck tensor for this crystal is diagonal and written as,

\varepsilon =\Biggl(\begin{matrix} ε_{αα} & 0 \\ 0 & ε_{bb} \end{matrix} \Biggr)

where the diagonal component ε_{αα} is different from ε_{bb} in general. The matrix is given here for a vector basis along the crystalline α-axis and b-axis.

c) Write the components of the Seebeck tensor with respect to the coordinate basis (x, y),

\varepsilon = \Biggl(\begin{matrix} ε_{xx} & ε_{xy} \\ε_{yx} & ε_{yy} \end{matrix} \Biggr)

in terms of the diagonal components ε_{αα} and ε_{bb} of the Seebeck tensor with respect to the coordinate basis (a, b).

d) The heat current density j_Q is related to the electric charge current density j_q  by,

j_Q = Π· j_q

which is a local version of the Peltier effect (11.108). The Peltier tensor is related to the Seebeck tensor by,

Π = T ε

j_{QB}− j_{QA} = π_{AB} j_q               (11.108)

In particular, for an electric charge current density j_q = j_q,x \hat{x} , where \hat{x} is a unit vector along the x-axis, show that the component j_Q,y along the y-axis of the heat current density j_Q = j_{Q,x} \hat{x} + j_{Q,y} \hat{y} , where \hat{y} is a unit vector along the y-axis, is given by,

j_{Q,y} = \frac{1}{2} T (ε_{αα} − ε_{bb}) \sin(2 θ) j_{q,x}

Thus, the planar Ettingshausen effect is maximal for an angle θ = π/4.

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a) The electric charge transport equations for the electrons and the holes are given by,

j_{q,e} = −σ_e · ε_e ·∇T − σ_e \nabla \varphi .

 

j_{q,h} = −σ_h · ε_h ·∇T − σ_h \nabla \varphi .

The Seebeck effect is observed when the electric current density vanishes, i.e. j_q = j_{q,e} + j_{q,h} = 0 . Thus,

  j_q = j_{q,e} + j_{q,h} = −(σ_e · ε_e + σ_h · ε_h) ·∇T − (σ_e + σ_h) .\nabla \varphi =0 

According to the Seebeck effect (11.83),

  \nabla \varphi =  −ε (s, n_A, q) ·∇ T                                 (11.83)

  \nabla \varphi = − (σ_e + σ_h)^{−1} · (σ_e · ε_e + σ_h · ε_h) ·∇ T = −ε ·∇ T 

Thus, the Seebeck tensor is given by,

  ε = (σ_e + σ_h)^{−1} · (σ_e · ε_e + σ_h · ε_h)

b) The Seebeck tensor is written in components,

  ε =\Biggl(\begin{matrix} \frac{\sigma _{e,\alpha \alpha } \varepsilon _e +\sigma _{h,\alpha \alpha }\varepsilon _h}{\sigma _{e,\alpha \alpha }+\sigma _{h,\alpha \alpha }} & 0 \\ 0 & \frac{\sigma _{e,bb } \varepsilon _e +\sigma _{h,bb }\varepsilon _h}{\sigma _{e,bb }+\sigma _{h,bb }} \end{matrix} \Biggr)

where the diagonal components are given by,

  \varepsilon _{\alpha \alpha } = \frac{\sigma _{e,\alpha \alpha } \varepsilon _e + \sigma _{h,\alpha \alpha } \varepsilon _h}{\sigma _{e,\alpha \alpha } + \sigma _{h,\alpha \alpha }}

 

  \varepsilon _{bb } = \frac{\sigma _{e,bb } \varepsilon _e + \sigma _{h,bb } \varepsilon _h}{\sigma _{e,bb } + \sigma _{h,bb }}

c) The rotation matrix   \Re_θ that rotates the crystalline α-axis and b-axis by an angle θ in plane on the x-axis and y-axis, and its inverse   \Re_{−θ} , are written in components as,

  \Re_θ = \Biggl(\begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{matrix} \Biggr)

 

  \Re_{−θ} = \Biggl(\begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{matrix} \Biggr)

where   \Re_{−θ} ·\Re_θ = 1 . The coordinates of the electric potential gradient   ∇\varphi in the coordinate basis (x, y) are related to coordinates in the coordinate basis (a, b) by,

  \Biggl(\begin{matrix} \partial_ x\varphi \\ \partial_y \varphi \end{matrix} \Biggr) =\Biggl(\begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{matrix} \Biggr)  \Biggl(\begin{matrix} \partial_ α\varphi \\ \partial_b \varphi \end{matrix} \Biggr)

The coordinates of the temperature gradient ∇T in the coordinate basis (x, y) are related to coordinates in the coordinate basis (a, b) by,

  \Biggl(\begin{matrix} \partial_ x T \\ \partial_y T \end{matrix} \Biggr) =\Biggl(\begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{matrix} \Biggr)  \Biggl(\begin{matrix} \partial_ α T \\ \partial_b T \end{matrix} \Biggr)

Thus, in view of the Seebeck effect (11.83), the coordinates of the Seebeck tensor in the coordinate basis (x, y) are related to coordinates in the coordinate basis (a, b) by,

  \nabla \varphi =  −ε (s, n_A, q) ·∇ T                                 (11.83)

  \Biggl(\begin{matrix} ε_{xx} & ε_{xy} \\ε_{yx} & ε_{yy} \end{matrix} \Biggr) =  \Biggl(\begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{matrix} \Biggr) \Biggl(\begin{matrix} ε_{αα} & 0 \\ 0 & ε_{bb} \end{matrix} \Biggr) \Biggl(\begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{matrix} \Biggr)

which implies that,

  \Biggl(\begin{matrix} ε_{xx} & ε_{xy} \\ε_{yx} & ε_{yy} \end{matrix} \Biggr) =  \Biggl(\begin{matrix} \varepsilon _{\alpha \alpha }\cos ^2\theta +\varepsilon _{bb}\sin ^2\theta & (\varepsilon _{\alpha \alpha }-\varepsilon _{bb})\sin \theta \cos \theta \\ (\varepsilon _{\alpha \alpha }-\varepsilon _{bb})\sin \theta \cos \theta & \varepsilon _{\alpha \alpha }\sin ^2\theta +\varepsilon _{bb} \cos ^2\theta \end{matrix} \Biggr)

d) The Peltier effect is given by,

  j_Q = T ε · j_q

For a heat current density  j_Q = j_{Q,x} \hat{x} + j_{Q,y} \hat{y} and an electric charge current density  j_q = j_q,x \hat{x} , this effect is written in components as,

  \Biggl(\begin{matrix} j_{Q,x} \\ j_{Q,y} \end{matrix} \Biggr) = T \Biggl(\begin{matrix} \varepsilon _{\alpha \alpha }\cos ^2\theta +\varepsilon _{bb}\sin ^2\theta & (\varepsilon _{\alpha \alpha }-\varepsilon _{bb})\sin \theta \cos \theta \\ (\varepsilon _{\alpha \alpha }-\varepsilon _{bb})\sin \theta \cos \theta &\varepsilon _{\alpha \alpha }\sin ^2\theta +\varepsilon _{bb}\cos ^2\theta \end{matrix} \Biggr) \Biggl(\begin{matrix} j_{q,x} \\ 0 \end{matrix} \Biggr)

Thus, the planar Ettingshausen effect is written as,

  j_{Q,y} = \frac{1}{2} T (ε_{αα} − εbb) \sin(2 θ) j_{q,x}

since  \sin (2 θ) = 2 \sin θ \cos θ .

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