The scalar magnitude $S$ of a symmetric 2nd-rank tensor $S_{ij}$ in a given direction having direction cosines $l_i$ is given as: $$\tag{1} S=S_{ij} l_i l_j$$

This result is obtained by starting with Eqn(2) and using the relationship given by Eqn(3): $$\tag{2} S=\frac{|\mathbf{p}|\cos \theta}{|\mathbf{q}|}$$ $$\tag{3} p_i=S_{ij}q_j$$

In Eqn(2), $\cos \theta$ is the angle between the two vectors $\mathbf{p}$ and $\mathbf{q}$. In Eqn(3) the 2nd-rank tensor (or 2nd-order tensor) $S_{ij}$ linearly maps the vector $\mathbf{q}$ to $\mathbf{p}$, each component of $\mathbf{p}$ is a linear function of each component of $\mathbf{q}$:

$$\begin{bmatrix}p_1\\p_2\\p_3 \end{bmatrix}=\begin{bmatrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33} \end{bmatrix}\cdot\begin{bmatrix}q_1\\q_2\\q_3 \end{bmatrix}$$

My question is: Is Eqn(2) an arbitrary definition?

Eqn(2) is stating that the magnitude $S$ characterizing the symmetric tensor property in a given direction is the ratio of the projection of vector $\mathbf{p}$ onto this direction to the length (magnitude) of vector $\mathbf{q}$ laid off (oriented) in the same direction.

But what if instead of stating (2) we were to state (4)? $$\tag{4} S=\frac{|\mathbf{p}|}{|\mathbf{q}|\cos \theta}$$

Unless I'm failing to recognize a mistake in my manipulations (shown below), could we not get a scalar value from combining (4) and (3) in terms of the tensor $S_{ij}$? And if we can, what is wrong with using this scalar value? Or why is the definition of (2) favored over (4)?

$$\tag{5} S=\frac{|\mathbf{p}|}{|\mathbf{q}|\cos \theta}=\frac{|\mathbf{p}|}{\mathbf{q}\cdot\mathbf{p}/|\mathbf{p}|}=\frac{|\mathbf{p}|^2}{\mathbf{q}\cdot\mathbf{p}}=\frac{|\mathbf{p}|^2}{q_i p_i}$$ Substituting (3) into (5), $$\tag{6} =\frac{|\mathbf{p}|^2}{q_i S_{ij}q_j}=\frac{|\mathbf{p}|^2}{S_{ij}q_i q_j}=\frac{|\mathbf{p}|^2}{S_{ij}l_i l_j|\mathbf{q}||\mathbf{q}|}=\frac{|\mathbf{p}|^2}{S_{ij}l_i l_j|\mathbf{q}|^2}$$

After typing this, I would now say that the definition of (2) resulting in (1) is favorable since (1) requires only two inputs: $S_{ij}$ and $l_i$. The definition given by (4) resulting in (6) requires four inputs for (6) to be calculated: $S_{ij}$, $l_i$, $\mathbf{p}$, and $\mathbf{q}$

The motivation for asking about the relation given in Eqn(4):

There are a number of linear phenomenological equations that take the form of Eqn(3), e.g., Darcy’s, Fourier’s, and Ohm’s laws. Initially the anisotropic material property as given by $S_{ij}$ was considered a scalar, i.e., isotropic, and thus the vectors involved, $p_i$ and $q_i$, were assumed co-linear. If this were true, the sample’s geometry and boundary condition have no consequence on the measurement of $S$ in a linear, steady-state, experiment.

Suppose the isotropic material sample being experimented with is in the shape of a right-circular cylinder whose cylindrical surface is insulated (no-flux boundary), but its planar ends are not. The application of $\mathbf{q}$ perpendicular to the planar ends, along the cylinder’s main axis, will cause a flux vector $\mathbf{p}$ oriented along this axis, proportional to the magnitude of $S$. Thus, with the magnitudes of $\mathbf{p}$ and $\mathbf{q}$ that can be measured, one can determine $S$ using: $$\tag{A} S = |\mathbf{p}|/|\mathbf{q}|$$

I emphasized magnitudes since only magnitudes are measured in this experiment. One does not measure the individual components of the vectors $p_i$ and $q_i$, e.g., $q_1,q_2,q_3$, and then from which determines the magnitude using $|\mathbf{q}|=\sqrt{q_1^2+q_2^2+q_3^2}$. Rather, the devices used to measure $\mathbf{p}$ and $\mathbf{q}$ in the experiment simply provide a magnitude (i.e. scalar) value. It is only when the vector (either $\mathbf{p}$ and/or $\mathbf{q}$) is co-linear with the sample’s main axis that the magnitude of the vector measured is equal to the one non-zero component of the vector that is co-linear with the sample’s axis, i.e., $q_{measured} =|\mathbf{q}|=\sqrt{q_1^2+0+0}=q_1$, where $q_1$ is oriented along the sample’s axis (the sample’s axis is taken to be oriented along $x_1$ of the Cartesian $x_1,x_2,x_3$ coordinate system).

Now, if the material property is anisotropic, then careful consideration needs to be made with regards to the sample’s geometry and boundary conditions when measuring the scalar magnitude of $S_{ij}$ in a particular direction. In general, $\mathbf{p}$ and $\mathbf{q}$ will not be co-linear.

Consider two special cases of geometry for the cylindrical sample being tested, one case where the diameter is much greater than its length (i.e., a “thin disc”), the other case where the length is much greater than its diameter (i.e., a “long rod”). One can perform the same experiment as was done for the isotropic sample to these two special-case anisotropic samples. Further, consider that these two samples were cut from a homogeneous anisotropic sample (anisotropic with respect to the property $S$) in a way such that none of the three principal values of $S_{ij}$ are oriented along the sample’s main axis.

For the thin-disc case, the vector $\mathbf{q}$ will be oriented along the sample’s axis (ignoring edge and end “effects”) and the vector $\mathbf{p}$ will be oriented along some other direction. The magnitudes of $\mathbf{p}$ and $\mathbf{q}$ measured in the experiment will be $p_{measured} =p_1=|\mathbf{p}|\cos⁡ \theta$ and $q_{measured} =q_1=|\mathbf{q}|$. Following the mathematical relationship given in Eqn(A), one would calculate the scalar magnitude of $S_{ij}$ in the direction of $q$:

$$\tag{2} S=|\mathbf{p}|\cos \theta/|q|$$

In the tensorial framework given by Eqn(3), this scalar magnitude of $S_{ij}$ determined corresponds to its $S_{11}$ coefficient.

For the long-rod case, the flux vector $\mathbf{p}$ will be compelled down the axis of the rod due to its cylindrical surface being insulated. Thus, its orientation is along the sample’s axis (again, ignoring edge and end effects). The vector $\mathbf{q}$ will be oriented along some other direction. The magnitudes of $\mathbf{p}$ and $\mathbf{q}$ measured in the experiment will be $p_{measured} =p_1=|\mathbf{p}|$ and $q_{measured} =q_1=|\mathbf{q}| \cos⁡ \theta$. Following the mathematical relationship given in Eqn(A), one would calculate the scalar magnitude of $S_{ij}$ in the direction of $\mathbf{p}$: $$\tag{4} S=|\mathbf{p}| /(|\mathbf{q}| \cos \theta)$$

It is better to analyze the measured $\mathbf{p}$ and $\mathbf{q}$ magnitudes obtained from the long-rod case using the reciprocal form of Eqn(3): $$\tag{B} q_i=R_{ij} p_j;\ \ \mathbf{R}=\mathbf{S}^{-1}, $$

where the scalar magnitude of $R_{ij}$ determined corresponds to the $R_{11}$ coefficient.

However, I want to try to understand the “mistake” one is making if one chooses to analyze the long-rod data in accordance to Eqn(4) rather than in accordance to Eqn(B). Or perhaps, I am trying to understand the mathematical relationship between the scalar magnitude of $S_{ij}$ in the direction of $\mathbf{p}$ and scalar magnitude of $S_{ij}$ in the direction of $\mathbf{q}$.

  • $\begingroup$ What are $\mathbf{p}$ and $\mathbf{q}$ here? $\endgroup$
    – Javier
    May 15 '20 at 2:55
  • $\begingroup$ @Javier My apologies for not clarifying. $\mathbf{p}$ and $\mathbf{q}$ are vectors and $\cos \theta$ is the angle between them. I've updated the post to include this clarification. $\endgroup$
    – Armadillo
    May 15 '20 at 13:37

First, a small comment. In the first definition, $\mathbf{l}$ and $\mathbf{q}/|\mathbf{q}|$ are the same thing: you're writing the same definition in two ways. This is because equation $(2)$ can be rewritten as

$$S = \frac{\mathbf{p}\cdot\mathbf{q}}{|\mathbf{q}|^2} = \frac{q_i}{|\mathbf{q}|} S_{ij} \frac{q_j}{|\mathbf{q}|}.$$

In words, this just says that to define $S$ you pick a unit vector (which $q_i/|\mathbf{q}|$ is) and contract it with both indices of $S_{ij}$. This is precisely what equation $(1)$ says, only with a different name for the vector. Therefore, in the second definition (equation $(6)$), it doesn't make a lot of sense to have $\mathbf{q}$ and also $\mathbf{l}$: the scalar magnitude is defined using a single vector, not two.

Now, starting from $(2)$ seems much stranger to me than starting from $(1)$. $(2)$ does certainly look a bit arbitrary, but $(1)$ is clear: it's the result of applying the tensor $S_{ij}$ to a certain direction. To understand why this makes sense, we need basically two facts:

  • Tensors do things by being applied to vectors and returning vectors or numbers. That is, the point of the nine numbers $S_{ij}$ is that you can take a vector $\mathbf{q}$ and transform it to the vector $S_{ij} q_j$, or that you can take two vectors $\mathbf{a}$ and $\mathbf{b}$ and calculate the number $S_{ij} a_i b_j$. Of course, the interpretation of this depends on the particular tensor, but these are the typical operations one does with a tensor.

  • A symmetric tensor is completely determined by values of the form $S_{ij} q_i q_j$. Proof:

$$2 S_{ij} a_i b_j = S_{ij}(a_i+b_i)(a_j+b_j) - S_{ij}(a_i-b_i)(a_j-b_j)$$

Therefore, it makes sense to put a name to things of the form $S_{ij}l_i l_j$. It's hard to give a geometric interpretation without lots of examples, but they basically tell you how strong the tensor is in each direction. If you take all the unit vectors $\mathbf{l}$ you can plot an ellipsoid where the "radius" in direction $\mathbf{l}$ is $S_{ij} l_i l_j$; the axes of the ellipsoid then correspond to the eigenvectors of $S_{ij}$.

And as a last comment, your alternative definition can be rewritten as

$$\frac{q_i S_{ij} S_{jk} q_k}{q_i S_{ij} q_j}.$$

You can see that it's really a combination of simpler quantities: the magnitude of $S$ squared divided by the magnitude of $S$. The definition $(1)$ is clearly simpler, and more suited to be used as a building block in other situations.

  • $\begingroup$ Great answer! Some clarifications & questions I have. When you say, “First, a small comment. In the first definition,…”. Here, are you referring to my Eqn(1) as the “first definition”? I just want to make sure we are not talking about my Eqn(2). Assuming “yes”, I’d like to check the following with you concerning Eqn(1)... $\endgroup$
    – Armadillo
    May 15 '20 at 16:28
  • $\begingroup$ ...We have: $\mathbf{l}=l_i=\frac{q_i}{|q_i|}=\frac{\mathbf{q}}{|\mathbf{q}|}$. Therefore, can we make the following statement for Eqn(1)? $S=S_{ij}l_i l_j=\mathbf{S}\frac{\mathbf{q}}{|\mathbf{q}|}\frac{\mathbf{q}}{|\mathbf{q}|}=\mathbf{S}\frac{\mathbf{q}\cdot\mathbf{q}}{q^2}=\mathbf{S}\frac{qq\cos(0)}{q^2}=\mathbf{S}\frac{q^2}{q^2}=\mathbf{S}, \ \therefore S=\mathbf{S}$. Is that right? Should I include some sort of symbolic notation to emphasize “magnitude” to the relation above, e.g., $S=|\mathbf{S}|$? $\endgroup$
    – Armadillo
    May 15 '20 at 16:28
  • 1
    $\begingroup$ By the first definition I mean (1) and (2) together, that is, the original definition and not the alternative you propose. Your derivation is not correct; you can't pass from $\mathbf{q} \mathbf{S} \mathbf{q}$ to $\mathbf{S} \mathbf{q}\cdot \mathbf{q}$. The equation doesn't even make sense because the LHS is a scalar and the RHS a tensor. That's why you can't say $S = \mathbf{S}$; that, and the fact that $S$ depends on a direction. $\endgroup$
    – Javier
    May 15 '20 at 17:23
  • $\begingroup$ Ah, OK, I failed to realize the error of my way, going from indicial summation notation to vector-matrix notation, as you said, $S=S_{ij}l_il_j=\mathbf{l^tSl}$. Continuing to inquire about your first comment, “In the first definition [in which you mean Eqns (1) and (2) together], $\mathbf{l}$ and $\mathbf{q/|q|}$ are the same thing..." I’m having trouble understanding where the term $\mathbf{q/|q|}$ is found in Eqns (1) & (2) and the point you are making with it... $\endgroup$
    – Armadillo
    May 15 '20 at 18:49
  • 1
    $\begingroup$ @Armadillo I edited my answer, please tell me if it helps. Still, I don't think the details are what really matter; the point is that the standard definition is simple and is a quantity that is often used, while yours isn't. $\endgroup$
    – Javier
    May 15 '20 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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