0
$\begingroup$

I'm trying to study tensors. Given a coordinates transformation from cartesian to $u_i$ ones: $$ u_1 = u_1 (x,y,z) \qquad u_2 = u_2 (x,y,z) \qquad u_3 = u_3 (x,y,z) $$ I can write a vector $\mathbf{A}= A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k}$ in at least two different ways $$ C_1 \frac{\partial \mathbf{r}}{\partial u_1} + C_2 \frac{\partial \mathbf{r}}{\partial u_2} + C_3 \frac{\partial \mathbf{r}}{\partial u_3} = c_1 \nabla u_1 + c_2 \nabla u_2 + c_3 \nabla u_3 $$ (where $C_i$ and $c_i$ are appropriate constants) and show that the corresponding coordinates in a different coordinates system $\bar{u}_i$ are related to this ones by (using summation convention) $$ \bar{C}_p = {C}_q \frac{\partial \bar{u}_p}{\partial u_q} \qquad \bar{c}_p = {c}_q \frac{\partial {u}_q}{\partial \bar{u}_p} \qquad $$ I understood proofs until here. If this can help in writing an answer, I say that I understood the meaning of metric tensor too (to do quickly scalar product if we know the contravariant description of the vectors). I can't see how this is useful but on trust I'm going on studying hoping that one day all will be clear. But first of all I don't understand when they started speaking about contravariant and covariant vectors: a vector (a displacement, a speed, an electric field, etc.) is contravariant if I use $\frac{\partial \mathbf{r}}{\partial u_i}$ basis and covariant if I use $\nabla u_i$ basis to describe it? Co or contravariant is not a property of the vector itself but rather a consequence of the choice we do in describing it? In addiction things became very confused when they start speaking about tensors with rank greater than one. I crashed into $$ \bar{A}^{pr} = \frac{\partial \bar{x}^p}{\partial x^q} \frac{\partial \bar{x}^r}{\partial x^s} A^{qs} $$ which is said to be a contravariant tensor of rank two. I suppose this is in some way related to the fact that they used $\frac{\partial \mathbf{r}}{\partial u_i}$ basis in describing a matrix with some physical meaning (say an inertia tensor for example) but I can't see in which way this happens: a matrix is not a linear combination of some basis. When tensor rank become grater than one things look harder and paradoxically books start going faster. How can I proof the last equation I wrote? "This is the definition of a tensor of second rank" shouldn't be an answer. It should be possible to show that if I claim that this entity is the same in both system of coordinates, and if I use the $\frac{\partial \mathbf{r}}{\partial u_i}$ basis in describing it, then coordinates transform in this way. Maybe to be concrete and so clearer, I should play on physical grounds. But anyway I can't do that.

Edit

The ground beyond $ \bar{C}_p = {C}_q \frac{\partial \bar{u}_p}{\partial u_q} $ is $$ C_1 \frac{\partial \mathbf{r}}{\partial u_1} + C_2 \frac{\partial \mathbf{r}}{\partial u_2} + C_3 \frac{\partial \mathbf{r}}{\partial u_3} = \bar{C}_1 \frac{\partial \mathbf{r}}{\partial \bar{u}_1} + \bar{C}_2 \frac{\partial \mathbf{r}}{\partial \bar{u}_2} + \bar{C}_3 \frac{\partial \mathbf{r}}{\partial \bar{u}_3} $$ but I can't see the ground beyond $\bar{A}^{pr} = \frac{\partial \bar{x}^p}{\partial x^q} \frac{\partial \bar{x}^r}{\partial x^s} A^{qs}$.

Edit 2

The first part of my question find an excellent answer in Fleisch statement "It's not the vector itself that is contravariant or covariant, it's the set of components that you form through its parallel or perpendicular projections" (see link given by Void). But I am in trouble whit the second part.

I'm not sure if it is better ask here or in math community. I understood rules of general transformations of coordinates for vectors, but I'm trying to give a sense to definition of secon rank tensor. I imagine the simpler way is starting considering metric tensor (in a second step I'll try to understand why what works for $\mathsf{g}$ will work for every matrix whit some physical meaning). Let's consider 2 basis $\{\mathbf{e}_i\}$ and $\{\mathbf{\bar{e}}_i\}$ in $\mathbb{R}^N$. If $a_{ij}$ is $j$ component of vector $\mathbf{\bar{e}}_i$ in not barred base, we have $$ \bar{g}_{ij} \equiv \mathbf{\bar{e}}_i \cdot \mathbf{\bar{e}}_j = (a_{i1} \mathbf{e}_1 + \dots + a_{iN} \mathbf{e}_N ) \cdot (a_{j1} \mathbf{e}_1 + \dots + a_{jN} \mathbf{e}_N ) $$ which lead to (using summation convention) $$ \bar{g}_{ij} = a_{ip} a_{jq} g_{pq} $$ where $ {g}_{pq} \equiv \mathbf{{e}}_p \cdot \mathbf{{e}}_q$. Now, if $\mathbf{e}_i$ and $\mathbf{\bar{e}}_i$ are tangent basis $\frac{\partial \mathbf{r}}{\partial u_i}$ and $\frac{\partial \mathbf{r}}{\partial \bar{u}_i}$ for some coordinate change, components $p$ and $q$ transform in this way (I use $m$ and $n$ as dummy index) $$ {C}_p = \bar{C}_m \frac{\partial {u}_p}{\partial \bar{u}_m} \qquad {C}_q = \bar{C}_n \frac{\partial {u}_q}{\partial \bar{u}_n} $$ so $$ a_{ip} = \bar{a}_{im} \frac{\partial {u}_p}{\partial \bar{u}_m} \qquad a_{jq} = \bar{a}_{jn} \frac{\partial {u}_q}{\partial \bar{u}_n} $$ and then $$ \bar{g}_{ij} = \bar{a}_{im} \bar{a}_{jn} \frac{\partial {u}_p}{\partial \bar{u}_m} \frac{\partial {u}_p}{\partial \bar{u}_n} g_{pq} $$ But $\bar{a}_{im}$ is $i$ component of vector $\mathbf{\bar{e}}_m$ in barred base, so it is zero if $i \neq m$ and $1$ if $i=m$. In other words $\bar{a}_{im} = \delta_{im}$. The same goes with $\bar{a}_{jn}$ so we have $$ \bar{g}_{ij} = \delta_{im} \delta_{jn} \frac{\partial {u}_p}{\partial \bar{u}_m} \frac{\partial {u}_p}{\partial \bar{u}_n} g_{pq} $$ and then $$ \bar{g}_{ij} = \frac{\partial {u}_p}{\partial \bar{u}_i} \frac{\partial {u}_p}{\partial \bar{u}_j} g_{pq} $$ By having used tangent basis $\frac{\partial \mathbf{r}}{\partial u_i}$ and $\frac{\partial \mathbf{r}}{\partial \bar{u}_i}$ I would have expected a completely different rules if transformations (I wouldn't?), the one that books claim is transformation for contravariant tensor: $$ \bar{g}_{ij} = \frac{\partial \bar{u}_i}{\partial {u}_p} \frac{\partial \bar{u}_j}{\partial {u}_q} g_{pq} $$ Bar signs in partial derivatives are inverted. What went wrong?

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/a/131984/52394 and physics.stackexchange.com/a/128063/52394 $\endgroup$ – Void Jan 4 at 15:00
  • $\begingroup$ Fairly sure the tensor equation describes change of coordinate bases. The bases have been left out of the equation. (Remember that the partials end up being position dependent scalar valued factors, and that you actually wrote a system of equations since it is for all p, r) $\endgroup$ – Emil Jan 4 at 15:00
  • $\begingroup$ You write: "a matrix is not a linear combination of some basis". Yes, it is. For example, $$ \begin{pmatrix}a & b \\ c & d\end{pmatrix} = a \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} + b \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} + c \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix} + d \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} $$ so $\{ \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \ \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \ \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \ \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} \}$ is a basis. $\endgroup$ – md2perpe Jan 4 at 18:05
  • $\begingroup$ I was implicitly speaking about basis of vectors, with which I worked in the first part of the question. I can't see at the moment if your observation is fruitful but I'll try to think. $\endgroup$ – Fausto Vezzaro Jan 4 at 21:35
1
$\begingroup$

Let $V$ be a linear space over $K = \mathbb{R}\text{ or }\mathbb{C},$ e.g. $V = \mathbb{R}^3$.

Consider a bilinear function $A : V \times V \to K.$ Given a basis $\{\mathbf{e}_j\}$ of $V$, by the bilinearity of $A,$ if $u = \sum_j u^j \mathbf{e}_j$ and $v = \sum_k v^k \mathbf{e}_k$ we have $$A(u, v) = A(\sum_j u^j \mathbf{e}_j, \sum_k v^k \mathbf{e}_k) = \sum_j \sum_k u^j v^k A(\mathbf{e}_j, \mathbf{e}_k) = \sum_j \sum_k u^j v^k A_{jk},$$ where $A_{jk} = A(\mathbf{e}_j, \mathbf{e}_k).$

How do the values $A_{jk}$ change if we switch to another basis $\{\mathbf{e}_j'\}$? Assume that $\mathbf{e}_j = \sum_p a_j^p \mathbf{e}_p'.$ Then, $$ A_{jk} = A(\mathbf{e}_j, \mathbf{e}_k) = A(\sum_p a_j^p \mathbf{e}_p', \sum_q a_k^q \mathbf{e}_q') = \sum_p \sum_q a_j^p a_k^q A(\mathbf{e}_p', \mathbf{e}_q') = \sum_p \sum_q a_j^p a_k^q A_{pq}', $$ where $A_{pq}' = A(\mathbf{e}_p', \mathbf{e}_q').$

When we have a field on some manifold (e.g. space-time) covered by a curvilinear coordinate system then we take $\mathbf{e}_j = \partial\mathbf{r}/\partial x_j$ and different coordinate systems thus give different bases. The transformation coefficients $a_j^p$ then are given by partial derivatives of one coordinate system with respect to the other coordinate system: $$ \mathbf{e}_j = \frac{\partial\mathbf{r}}{\partial x_j} = \frac{\partial x_p'}{\partial x_j} \frac{\partial\mathbf{r}}{\partial x_p'} = \frac{\partial x_p'}{\partial x_j} \mathbf{e}_p'. $$

Thus, $$ A_{jk} = \sum_p \sum_q \frac{\partial x_p'}{\partial x_j} \frac{\partial x_q'}{\partial x_k} A_{pq}'. $$

In physics the above transformation rule is taken as the definition of a tensor with 2 lower indices and 0 upper indices.

$\endgroup$
  • $\begingroup$ After studying again on the problem I see that, if we use tangent basis $ \frac{\partial \mathbf{r}}{\partial \bar{u}_i}$, metric tensor change in this way $g_{ij} = \frac{\partial \bar{u}_p}{\partial u_i} \frac{\partial \bar{u}_q}{\partial u_j} \bar{g}_{pq}$, which agree with the formula that you derived but books say that this is formula for co-variant tensor: how can this be true if I used tangent basis? Shouldn't I have gotten the formula for contra-variant 2 tensor? $\endgroup$ – Fausto Vezzaro Jan 12 at 12:28
  • $\begingroup$ When we use tangent basis for 1-tensor (vector) and we study how the description of this object change when we change coordinates, we get $\bar{C}_p = {C}_q \frac{\partial \bar{u}_p}{\partial u_q}$ which is very different from the formula we obtained for 2-tensors (still using tangent basis): the bar in partial derivatives are inverted. What went wrong? $\endgroup$ – Fausto Vezzaro Jan 12 at 12:32
  • $\begingroup$ @FaustoVezzaro. Why should anything be wrong? $g_{ij}$ is covariant in both its indices, while $C_q$ is contravariant in its only index. A contravariant index is usually written raised: $C^q.$ $\endgroup$ – md2perpe Jan 12 at 20:15
0
$\begingroup$

The last equation you wrote is a definition: you don’t need to prove definitions;

Brutally, a tensor (contravariant or covariant) is something that transforms in a particular way under change of coordinates (namely, its index are “rotated” through the Jacobian or through the inverse Jacobian); the number of indices (high or low) simply tell you how many matrices of each kind you need for that tensor.

The books goes faster because once you spent time studying the transformation properties of the prototypes of contravariancy and covariancy (respectively, the differential $dx^i$ and the partial derivative $\frac\partial{\partial x^i}$) you simply extend these transformation laws defining a tensor as said above;

$\endgroup$
  • $\begingroup$ The definition is in $\bar{A}^{pr} = \frac{\partial \bar{x}^p}{\partial x^q} \frac{\partial \bar{x}^r}{\partial x^s} A^{qs}$ or in the fact that the objects we call "tensors" is the same in both coordinates system? I suppose both this are definitions. My problem is that I see the link between this two view with vectors but I can't see it with matrices. $\endgroup$ – Fausto Vezzaro Jan 4 at 21:43
  • $\begingroup$ The usual definition is the first, while the second is (if I understand what you mean) is the statement that a tensor/vector exists independently of the coordinate system you choose, so its components must transform in a particular way, that is, with the inverse of the matrix of the change of basis. As you can write a vector as $v=v^a e_a$ and a tensor as $T=T^{ab} e_a \otimes e_b$, if $e_a\rightarrow M^c\,_a e_c$ you must have $v’^a=(M^{-1})^a\,_c v^c$ for the vector, but also $T’^{ab}=(M^{-1})^a\,_c (M^{-1})^d\,_b T^{cd}$ for the tensor (and similarly for higher rank tensor). $\endgroup$ – Francesco Bernardini Jan 4 at 22:04
  • $\begingroup$ Ooooh can’t believe this, I hope you can dive between the latex. In my example the matrix $M^{-1}$ is the Jacobian of the coordinate change, so $\frac{\partial \bar x^i}{\partial x^j}$ $\endgroup$ – Francesco Bernardini Jan 4 at 22:06

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.