1
$\begingroup$

I wanted to know if the example I came up with for tensors is valid, or completely wrong. So here is the setup. A force, $F$ is applied onto a block, at a certain angle from the ground, and pushes it for a certain distance, $S$. Here is a picture (made by myself):enter image description here

The work done on that block, which is a scalar, has value of $W=\vec{F}\cdot\vec{S}$. If I were to calculate this now, I simplify it by calculating the dot product of the vectors to yield $W=F^xcos(\theta)S$. Since $\vec{S}$ only points in the "x" axis, I just put the $S$ letter in, but on the force vector, I specified the x component of the force. Instead of doing that, is it valid to write this as a tensor, describing all the possible component combinations for the amount of work done on certain axes? A bit like this: \begin{align} W^{ab}&=F^aS^b\\ W^{ab}&= \begin{bmatrix} F^xS^x & F^xS^y & F^xS^z \\ F^yS^x & F^yS^y & F^xS^z \\ F^zS^x & F^zS^y & F^xS^z \end{bmatrix} \end{align} I am just starting to learn tensors, so please, if I am wrong, tell me and please show me what would be a valid example of tensor use and explanation. Thanks!

$\endgroup$
3
  • $\begingroup$ Well ask yourself is your expression coordinate system independent? $\endgroup$ Mar 16, 2018 at 6:00
  • $\begingroup$ As I understand from your equations, isn't it the work $W$ = $\delta^{aa} W^{ab}$ = $W^{aa}$ so that it is a scalar, independent of coordinate transformation. $\endgroup$
    – user187456
    Mar 16, 2018 at 6:54
  • $\begingroup$ What would be the meaning of $F^x S^y$? In what physical situation is that quantity useful? Of course taking the outer product of two vectors produces a mathematically valid tensor, but whether it's useful for physics is another matter. $\endgroup$ Mar 16, 2018 at 17:26

3 Answers 3

1
$\begingroup$

As mentioned in the comments, work must be a scalar, which is invariant under the change of coordinate system. However, a 2-form tensor is not, so this tensor formulation of work makes things unnecessarily complicated.

Nevertheless, I think the diagonal elements do make sense on some level; in fact, if you take the trace of this matrix, you will find that it gives the work done on the block. So on some intuitive level, you can think of the $F_xS_x$ term as the contribution of force along the $x$ component, and similarly for $F_yS_y$ and $F_zS_z$, so the total work is just the addition of the three.

The off-diagonal elements are not very physical, and I can't think of a physical measure as to what they would represent.

$\endgroup$
1
$\begingroup$

In physics, objects are almost always defined by the virtue of how it transforms under some particular rule. Personally, I like to think tensors as objects defined in a convenient way to denote how they transform under a particular transformation. The details of how the object transforms is `encoded', so to speak, in the tensor. Just like how transformations come in varieties, tensors come in varieties.

Take a scalar, for example. Scalar is just a number. No matter how you transform your set-up, the scalar doesn't change. (Temperature is a good example. If you tilt your head and see the room, the temperature at each point in the room remains the same as before you tilted your head.) These objects are invariant under a prescribed transformation (it needn't always be plain rotation, it can be Lorentz boosts, and all kinds of other transformations). This kind of object is an one element object. We say that scalar is a tensor of rank zero.

Vectors are objects which change when you change the space (or coordinate system) in which it lives. Vectors are objects that are inherently sensitive to sense of 'directions' in the space (unlike scalars). To put it mathematically, vectors are objects that transform linearly when you rotate the coordinate system. If you have a vector $\vec{A} = x_0\hat{x}$ that is purely along the $\hat{x}$-direction, and you rotate your coordinate system by 45$^0$ about the $z$-axis in the $xy$-plane, the object is now not strictly along $\hat{x}$-direction anymore! It has turned into some $A\hat{x}+B\hat{y}$. That is, the rotational transformation has now made it a linear combination of $x_1\hat{x}$ and $y_1 \hat{y}$. These objects are conveniently written in the form of a column matrix. The number of elements in the column matrix would denote the dimension of the space the vector lives in. In Euclidean space, you would have three elements for a vector. Since you need one column to describe a vector, it is a tensor of rank one.

To up the game a little bit, you can define objects that are not made of just one column matrix like in the case of vectors but an $N \times N$ matrix! This type of object can have slightly weird connotations, but it is an object no stranger than, say, a vector. Since this object takes columns and rows, this is a tensor of rank two.

Similarly, you can create a tensor of rank three, but it would be hard to write it on paper as the third array would come out of the page, i.e., an $N \times N \times N$ matrix.

Now, there is no point in defining these objects unless we have some kind of composition rule that makes them interact with one other. This is where we introduce the concept of a dual. Each of these objects, be it a scalar or a vector or a tensor, live in their own space. But there is also a dual (or a twin, if you like) for each of these elements which live in another space which we call the 'dual space'. Why do we need this dual space? Because we insist that any rule of composition can be made only between objects and its counterpart in the dual space. No two objects from the same space can 'compose'!

For example, the dual of a column vector would be a row vector! And when you insist on the composition rule being matrix multiplication (row vector multiplied by column vector), we then magically get a scalar! We call this the 'dot product'. Of course, you can also get a vector by composing two vectors together, like you would in the case of cross-products.

In your question, however, you have defined 'work' as a tensor, but work is not that kind of an object. Work is, by definition, a scalar, and thus it cannot be denoted by an $N \times N$ matrix. By definition, it is the dot product of two vectors, $\vec{F}$ and $\vec{s}$, where former is the force and latter is the path taken by the object in lieu of the force. Since work is a scalar, it has to just be a number. So $\vec{F}.\vec{s}$ would be: \begin{align} \begin{bmatrix} F_x F_y F_z \end{bmatrix} \begin{bmatrix} s_x \\ s_y \\ s_z \end{bmatrix} = F_x s_x + F_ys_y + F_zs_z = W \end{align}

Writing out the matrix each and every time can be a bit tedious, so we have the wonderful index notation, whereby repeated indices are summed over. For instance, the above matrix multiplication can be succinctly written as $F_\mu s^\mu$, where $\mu$ in our above case runs from 1 to 3, for each $x$, $y$, and $z$. In this way of writing, you make sure that a lower-indexed object always get clubbed with the upper indexed object. That is because $a_\mu$ is from one space and $a^\mu$ is its twin from the dual space! You can multiply objects only with its duals. (It happens so that Euclidean is self-dual, so normal cartesian vectors don't need that much of a distinction.)

You can also have matrices defined by this index notation, and that quantity would be hovered by two indices, one index to denote the row and another to denote the column ($M^{21}$ would denote the element in $2^{nd}$ row and $1^{st}$ column in matrix $M$). If I have a vector $\vec{A}$ (denoted by column matrix) and I act a rotation matrix $M$ on $A$ to rotate it to some other vector $B$, matrix-wise I would write as

\begin{align} \begin{bmatrix} M^{11} M^{12} M^{13} \\ M^{21} M^{22} M^{23} \\ M^{31} M^{32} M^{33} \\ \end{bmatrix} \begin{bmatrix} A_1 \\ A_2 \\ A_3 \end{bmatrix} = \begin{bmatrix} B_1 \\ B_2 \\ B_3 \end{bmatrix}= \begin{bmatrix} A_1 M^{11} + M^{12}A_2 + M^{13}A_3 \\ A_1 M^{21} + M^{22}A_2 + M^{23}A_3 \\ A_1 M^{31} + M^{32}A_2 + M^{33}A_3 \end{bmatrix} \end{align}

Instead of writing it out like that, I can simply write the same transformation as $M^{\mu\nu}A_\nu = B_\mu$, where $\mu$ runs from 1 to 3, and the repeated indices are summed over. You can compare and see for yourself, it works out neatly! It becomes especially very handy if you are going to write down transformations that are very high in dimensions and the rank of the tensors also becomes high. (You cannot write a 3D matrix on paper, but with index notation, sky is the limit!) I hope the answer was of some use to you. Cheers!

$\endgroup$
1
$\begingroup$

Work done by a force is defined to be a scalar, so it's not a good start to learn tensors. Work is just a scalar mapping of two vectors. Actually, it's using a rank 2 tensor since it implies a metric, which is trivially the identity matrix of elements $\delta_{ij}$ in cartesian coordinates : the scalar product in 3D euclidien space. For some infinitesimal displacement (of cartesian components $ds_i$), you could write \begin{equation}\tag{1} dW_F = \vec{\boldsymbol{\mathrm{F}}} \cdot d\vec{\boldsymbol{\mathrm{s}}} \equiv \sum_{i, j} \delta_{ij} \, F_i \, ds_j. \end{equation}

Now consider the electric current $\vec{\boldsymbol{\mathrm{J}}}$ and the external electric field applied $\vec{\boldsymbol{\mathrm{E}}}$. In an linear, homogeneous and isotropic material, the relation is very simple (it's the microscopic law of conduction): \begin{equation}\tag{2} \vec{\boldsymbol{\mathrm{J}}} = \sigma \, \vec{\boldsymbol{\mathrm{E}}}. \end{equation} In cartesian components form: \begin{equation}\tag{3} J_i = \sigma \, E_i. \end{equation} Here, $\sigma$ is a simple constant (the material is homogeneous), interpreted as the material conductivity. Now, if the matierial is homogeneous, but not isotropic (like in a cristal), you need to change the previous relation as this: \begin{equation}\tag{4} J_i = \sum_{j = 1, 2, 3}\sigma_{ij} \, E_j. \end{equation} Here, the cartesian components $\sigma_{ij}$ defines the conductivity matrix elements. Conductivity becomes a rank 2 tensor, which linearly applies the electric field to the current density. You may then write this tensorial expression (the tensor product $\otimes$ is a book-keeping device to reminds you the order of indices): \begin{equation}\tag{5} \boldsymbol{\sigma} = \sum_{i, j} \sigma_{ij} \, \boldsymbol{\mathrm{e}}_i \otimes \boldsymbol{\mathrm{e}}_j, \end{equation} where $\boldsymbol{\mathrm{e}}_i$ are the three ($i = 1, 2, 3$) cartesian unit vectors associated to your cartesian axes.

You could also consider the relation between the angular momentum $\vec{\boldsymbol{\mathrm{L}}}$ of some solid shape and its angular velocity $\vec{\boldsymbol{\mathrm{\omega}}}$ as another example. In general $L_i \ne I \, \omega_i$, where $I$ is the moment of inertia around some axis. You could write this: \begin{equation}\tag{6} L_i = \sum_{j = 1, 2, 3}I_{ij} \, \omega_j, \end{equation} where $I_{ij}$ are the cartesian components of the inertia tensor (a symetric 3 by 3 matrix). This is another example of a rank 2 tensor. It's linearly mapping the angular velocity of rotation (around some axis) to the angular momentum of the rotating object.

A bit related to your problem, you could consider the pressure (or tension, or more generally stress) exerted in a fluid or an elastic solid. It's a linear mapping from the infinitesimal area vector (of cartesian components $dA_i$) to the infinitesimal force vector applied on that area : \begin{equation}\tag{7} dF_i = \sum_{i, j} p_{ij} \, dA_j. \end{equation}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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