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Warning: You might be wondering why this isn’t in Math Stack Exchange. In fact, it is. I asked the same question there a few days ago but got no answer, and since I think that this question is more directed towards the “physicist POV” of tensors and is related to Einstein’s Convention, I’ll leave the question here.

Original Question Below

I'm trying to self study tensor calculus. I was trying to derive the notation for covariant and contravariant indexes of a linear transformation matrix ($(1,1)$ type tensor).

So I did the following: Try for the $2 \times 2$ case, and then try to find a pattern.

For covectors (covariant) $x_i$: (I shall first assume that both indexes of the matrix are up as contravariant, just for simplicity of notation. I'll later "correct" this according to what i have found).

We have:

$$ \begin{bmatrix} x_1' & x_2 ' \end{bmatrix} = \begin{bmatrix} x_1 & x_1 \end{bmatrix} \begin{bmatrix} a^{11} & a^{12} \\ a^{21} & a^{22} \end{bmatrix}$$ And so, for $x_j'$, I'll have: $$x_j' = a^{ij}x_i$$ (Summation convention here). Here, I am summing on the first index of the matrix. For the contravariant case, I have: $$x^i ‘= a^{ij}v^j$$ Hence summing on the second index of the matrix. So, I tought about writing $a$ as $a_{i}^j$, calling then $i$ as the covariant index and $j$ as the contravariant index. Does this make any sense? The first problem I see is that this goes against the summation convention, who states that the indexes must be summed up when they are at different positions (ex: $a^iv_i$ would mean a summation in $i$, but $a_ iv_i$ would not).

This construction I've made would be equivalent to: $$a = a_ib^j \mathbf{e}_i\otimes \mathbf{e}^j$$

My other question is if it is possible arriving to the following:

$$a = a^ib_j \mathbf{e}^i\otimes \mathbf{e}_j$$

Using matrix algebra? Is that possible?

I'm really confused. I've seen 2nd order tensors being written as $a^i_j$ and as $a_i^j$. What is the difference between them? Do they act the same way on vectors?

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  • $\begingroup$ This type of tensor manipulation is known as Ricci Calculus en.wikipedia.org/wiki/Ricci_calculus $\endgroup$ – Triatticus Jun 6 '18 at 0:23
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    $\begingroup$ there, in the comments I suggest math.stackexchange.com/q/1047994 as an observation of how in math this is settle, but now I support to Vitor expecting someone could elaborate a bit more here with applications viewpoint. Going to upvote cuz the braveness asking now in PhysSE. $\endgroup$ – janmarqz Jun 6 '18 at 1:17
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  • $\begingroup$ @AccidentalFourierTransform I don’t believe this is a duplicate for 1) I don’t speak specifically about Lorentz Transformations 2) I’m also interested in knowing if I can name these indices as covariant or contravariant and putting them in this order. 3) The cited question also doesn’t answer the question as to whether I can arrive to both constructions using matrix algebra. If i can’t, why not? $\endgroup$ – Vitor C Goergen Jun 9 '18 at 16:46
  • $\begingroup$ I just answered a very similar question here. Long story short, it's very easy to make a mistake while treating tensors as matrices, and I don't really think it's worth doing at all, but you can make it work. $\endgroup$ – knzhou Jun 10 '18 at 21:47
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First of all, I have to say that matrices are not tensors or vice versa. Matrices are just arrays of numbers while tensors are invariant objects having co-variant or contra-variant components up to some basis.

By the way, the equalities you have written are not proper. If you have on one side a contra-variant tensor, then you need to have the other side contra-variant, as well. For instance, $x'_j = a^{ij} x_i$ should be $x'_j = a {}^i {}_j x_i$, or alike.

First, I would like to clarify some aspects of co-variant and contra-variant transformations of a tensor before I write the answer to the question about the order of indices, which would be trivial in the end.

Tensors and their components

Tensors are not matrices or array of numbers. Tensors are objects on a vector space that does not change, as a whole, under a coordinate transformation. Vectors are one example for tensors, a vector is the same vector even if you express it in a different coordinate system. Only the components are transforming since the bases are changing, but not the whole thing.

So, expressing a tensor in terms of array of numbers implies some specific transformation properties on the components, depending on the transformation from one basis to another. These properties consist of co-variant and contra-variant transformations.

Let's assume we work in coordinate bases, $dx^i$ and $\frac{\partial}{\partial x^i}$. So, tensors $\mathbf{A}$ and $\mathbf{B}$ could be written in different coordinate systems as follows: \begin{align} \tag{1.a} \mathbf{A} &=A_i \, dx^i \\ \tag{1.b} &= A'_i \, dx'^i \end{align} and \begin{align} \tag{2.a} \mathbf{B} &= B^i \frac{\partial}{\partial x^i} \\ \tag{2.b} &= B'^i \frac{\partial}{\partial x'^i} \end{align} where summation convention is used. I have just written the same tensors in different coordinate systems, primed or unprimed.

So, the primed components could be written in terms of unprimed components, respectively, as follows:

$$ \tag{co-variant} A'_i = A_i \frac{\partial x^i}{\partial x'^i} $$ $$ \tag{contra-variant} B'^i = B^i \frac{\partial x'^i}{\partial x^i} $$

where one needs to track the primes according to chain rule, when using the equality between (.a)'s and (.b)'s, respectively. Anything obeys these component-basis transformations which leaves the object unchanged is called a tensor.

Now, you can also express the tensor in any basis, even those non-coordinate ones, i.e., $\mathbf{e}^a = e {}^a_i dx^i$ and $\mathbf{e}_a = (e^{-1}) {}_a^i \frac{\partial}{\partial x^i}$ where the invertible map, $e {}^a_i (x)$, is called tetrad field or vierbein (in 4D). So, the new components would be $A_a = A_i (e^{-1}) {}_a^i$ and $B^a = B^i e {}^a_i$, respectively.

If you consider the proper version of the equality you have written, $v'_j = a {}^{i} {}_j v_i$, in this context I explained above, you can see how it becomes valid if you explicitly write the coordinate transformations.

Order of the indices and the metric tensor

On the other hand, the relation between the co-variant and contra-variant components is handled via a map called metric: $$ A_i = g_{ij} A^j $$ and you can easily show that $g_{ij}$ are also components of a tensor, since both sides of this identity must transform co-variantly.

Now, let's take a second rank tensor, $T_{ij}$. The order of indices are important if the tensor is asymmetric, i.e., $T_{ij} \neq T_{ji}$. If you raise up the first or the second index via the metric, respectively, $$ T {}^i {}_j = g^{ik} T_{kj} \\ T {}_j {}^i = g^{ik} T_{jk} $$ it becomes obvious that they can only be equal if and only if $T_{ij}$ is symmetric in its indices.

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  • $\begingroup$ Thanks for your answer. 1: About my equalities being incorrect wrt the indexes, I've stated that I would correct them later. I can't quite understand your answer. Could you explain why choosing $dx^i$ and $\frac{\partial}{\partial x^i}$ as basis? And not some other basis vectors? I also don't understand what you mean by "prime" components. Another thing: In the last paragraph (which I think is supposed to answer the question afterall) It seems to me that you've explained the difference between $T^i_j$ and $T_j^i$. I'm sorry, I cant relate that to the difference between $T^i_j$ and $T_i^J$. $\endgroup$ – Vitor C Goergen Jun 18 '18 at 20:25
  • $\begingroup$ As for the bounty that expired, I will start a new one, since I was unable to access the site lately. $\endgroup$ – Vitor C Goergen Jun 18 '18 at 20:25
  • $\begingroup$ OK, thanks for the feedback, I will edit the answer. $\endgroup$ – Oktay Doğangün Jun 19 '18 at 10:34

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