Can Anisotropic Materials Have Asymmetric Thermal Conductivity Tensors?

[dRic2]

Hi, yesterday a professor of mine told me that if you have a temperature gradient along the x-axsis you could have heat flowing in the y direction. Mathematically it is pretty straightforward to find the thermal conductivity tensor required, but in real life can you name some materials that share this property ?

 

[Orodruin]

Essentially most anisotropic materials.

In more technical terms, if the decomposition of the symmetric traceless representation of rotations includes any singlets of the symmetry group of the material, you would have a harder time explaining why that material should not have an anisotropic thermal conductivity than why it should be isotropic. An example that I would expect to have anisotropic thermal conductivity is graphene, but I am no expert on thermal conductivity properties of materials - this is a pure symmetry argument. @Chestermiller ?

 

[Vanadium 50]

Consider even a uniform material.

If I have a long rod with both tips at 0C heat baths the rod will reach 0C. If I replace the ice at one tip with a blowtorch at the center of the tip's surface, a temperature gradient will develop along the rod. The surface of the rod along the long direction will also heat up. That means heat is being transferred radially, not just longitudinally.

Is that an example? If not, why not?

 

[Orodruin]

That means heat is being transferred radially, not just longitudinally.

Is that an example? If not, why not?

Heat in this case is being transferred radially because there will be a radial component of the temperature gradient.

 

[dRic2]

Is that an example? If not, why not?

No because the surface of the rod is initially at 0C and so there is also a radial component of the temperature gradient as the rod warms up that drives the heat flow.

An example that I would expect to have anisotropic thermal conductivity is graphene

Consider a very very thin slab of graphene. Heat will flow poorly through the layers of graphene while it will flow without problems along the layers. If you apply a temperature gradient across the faces of the slab I think you will only see an high resistance to the flow but I don't think you will measure heat flowing in the perpendicular direction. IMHO

PS: As I interpreted the statement of my professor the heat must flow only in the perpendicular direction, there is no heat flow along the direction of the temperature gradient

 

[Orodruin]

Consider a very very thin slab of graphene. Heat will flow poorly through the layers of graphene while it will flow without problems along the layers. If you apply a temperature gradient across the faces of the slab I think you will only see an high resistance to the flow but I don't think you will measure heat flowing in the perpendicular direction. IMHO

You will never get a flow that is exactly orthogonal to the gradient. However, you will get a flow that is not parallel to the gradient (i.e., it has a component that is orthogonal to the gradient) if you apply a temperature gradient that is not along one of the principal directions. In the case of graphene, apply a gradient that is not along the sheets and not perpendicular to the sheets, i.e., taking $\vec n$ to be the direction perpendicular to the sheets and $\vec t$ to be tangent to the sheets, you apply a gradient $\nabla T = T^t \vec t + T^n \vec n$. Conductivity is a linear map from a temperature gradient to a heat flux density and therefore
$$
\vec J = -K(\nabla T) = - K(T^t \vec t + T^n \vec n) = - T^t K(\vec t) - T^n K(\vec n).
$$
The directions $\vec t$ and $\vec n$ are in the principal directions of the material so $K(\vec t) = K^t \vec t$ and $K(\vec n) = K^n\vec n$, where $K^t$ and $K^n$ are material constants. It follows that
$\vec J = - T^t K^t \vec t - T^n K^n \vec n,$
which is not parallel to $\vec J$ unless $K^t = K^n$ or $\nabla T$ is along one of the principal directions. In your example, you took $\nabla T$ to be in a principal direction.

Edit: So, the underlying reason that the heat flow is not parallel to the temperature gradient is that it is much easier for the temperature gradient to drive a flow in a direction along the sheets as compared to driving a flow between the sheets.

 

[dRic2]

You will never get a flow that is exactly orthogonal to the gradient. However, you will get a flow that is not parallel to the gradient (i.e., it has a component that is orthogonal to the gradient) if you apply a temperature gradient that is not along one of the principal directions. In the case of graphene, apply a gradient that is not along the sheets and not perpendicular to the sheets, i.e., taking $\vec n$ to be the direction perpendicular to the sheets and $\vec t$ to be tangent to the sheets, you apply a gradient $\nabla T = T^t \vec t + T^n \vec n$. Conductivity is a linear map from a temperature gradient to a heat flux density and therefore
$$
\vec J = -K(\nabla T) = - K(T^t \vec t + T^n \vec n) = - T^t K(\vec t) - T^n K(\vec n).
$$
The directions $\vec t$ and $\vec n$ are in the principal directions of the material so $K(\vec t) = K^t \vec t$ and $K(\vec n) = K^n\vec n$, where $K^t$ and $K^n$ are material constants. It follows that
$\vec J = - T^t K^t \vec t - T^n K^n \vec n,$
which is not parallel to $\vec J$ unless $K^t = K^n$ or $\nabla T$ is along one of the principal directions. In your example, you took $\nabla T$ to be in a principal direction.

Edit: So, the underlying reason that the heat flow is not parallel to the temperature gradient is that it is much easier for the temperature gradient to drive a flow in a direction along the sheets as compared to driving a flow between the sheets.

That is obvious: it is just the vector nature of the flow. I was asking the other thing but you said it was impossible.

For example can a conductivity tensor be like this:
$$ \mathbf k = \begin {bmatrix}
0 & k_{xy} \\
k_{yx} & 0 \\
\end {bmatrix}
$$ ?

In this way if you take a temperature gradient only on the x direction

$$\mathbf Q = \mathbf k \cdot ∇ T =
\begin {bmatrix}
0 & k_{xy} \\
k_{yx} & 0 \\
\end {bmatrix} \cdot \begin{bmatrix} \partial_x T \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ k_{yx} \partial_x T \end{bmatrix}
$$

 

[Orodruin]

That would imply a negative eigenvalue and therefore heat could flow from lower to higher temperature, which breaks the laws of thermodynamics. The conductivity tensor must be positive definite.

 

[Orodruin]

Hi, yesterday a professor of mine told me that if you have a temperature gradient along the x-axsis you could have heat flowing in the y direction.

That is obvious: it is just the vector nature of the flow.

If you think it is obvious I don't understand your confusion. The statement, as you have recounted it, does not say that the flux is exclusively in the y-direction.

 

[Chestermiller]

If you have a composite material, the thermal conductivity can typically be anisotropic. For example, if you have rebar in cement, the preferred direction of heat conduction will be in the long direction of the rebar. This is analogous to the anisotropic mechanical properties of a composite.

 

[Andy Resnick]

Hi, yesterday a professor of mine told me that if you have a temperature gradient along the x-axsis you could have heat flowing in the y direction. Mathematically it is pretty straightforward to find the thermal conductivity tensor required, but in real life can you name some materials that share this property ?

Anisotropic crystalline materials:

https://arxiv.org/ftp/arxiv/papers/1809/1809.04762.pdf
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.66.064525
https://dspace.mit.edu/bitstream/handle/1721.1/4445/rle-tr-377-04740983.pdf?sequence=1

 

[dRic2]

If you think it is obvious I don't understand your confusion. The statement, as you have recounted it, does not say that the flux is exclusively in the y-direction.

I think I solved it. Probably I misunderstood what the professor said: tomorrow I'll talk to him.

Can you tell me more about the problem of negative eigenvalues because I didn't get it very much.

Thanks to everyone for the replies. :D

 

[Orodruin]

Can you tell me more about the problem of negative eigenvalues because I didn't get it very much.

In essence, if you have a conductivity on the form you suggested, then it would have negative eigenvalues. This would mean that there is a direction in which you can apply a temperature gradient and get heat to flow in the same direction as the gradient, i.e., from low to high temperature. If you can do this you break the laws of thermodynamics.

 

[dRic2]

In essence, if you have a conductivity on the form you suggested, then it would have negative eigenvalues. This would mean that there is a direction in which you can apply a temperature gradient and get heat to flow in the same direction as the gradient, i.e., from low to high temperature. If you can do this you break the laws of thermodynamics.

I got it now. I have one more question : Does the tensor need to be symmetric ? I'm a bit confused over this one... Is this property shared by anisotropic materials linked to the symmetry of the tensor ?

 

[Chestermiller]

I got it now. I have one more question : Does the tensor need to be symmetric ? I'm a bit confused over this one... Is this property shared by anisotropic materials linked to the symmetry of the tensor ?

Even if the material is anisotropic, the thermal conductivity will typically be symmetric, with 3 real principal- directions and magnitudes.