I'm new to tensor calculus: I'm reading a little introductory book whose title is "Quick Introduction to Tensor Analysis", written by R.A Sharipov. I've reached the section called differentiation of tensor fields. There the author explains how the tensors can be differentiated, distinguishing two cases:

  1. with respect to external parameters
  2. with respect to spacial variables (used as arguments of the tensorial function)

while explaining the second case it is told a theorem (which holds only for Cartesian coordinate systems): "for any tensor field $X$ of type $(r, s)$ partial derivatives (21.3) with respect to spacial variables $x^1, x^2, x^3$ in any Cartesian coordinate system represent another field $Y$ of type (r, s+1)":

(21.3) $$ Y^{i_1, \cdots, i_r}_{q, j_1, \cdots, j_s} = {\partial X^{i_1, \cdots, i_r}_{j_1, \cdots, j_s} \over \partial x^q} \Rightarrow Y^{i_1, \cdots, i_r}_{q, j_1, \cdots, j_s} = \nabla_q X^{i_1, \cdots, i_r}_{j_1, \cdots, j_s} $$


$$ \nabla_q = {\partial \over \partial x^q} $$

It is then suggested the following exercise: "Prove the theorem. For this purpose consider another Cartesian coordinate system $\tilde{x}^1, \tilde{x}^2, \tilde{x}^3$ related to $x^1, x^2, x^3$ via $x^i = S^i_j \tilde{x}^j + a^i$ and $\tilde{x}^i = T^i_j x^j + \tilde{a}^i$ (where $a^i$ is the ith component of the translation vector from $O$ to $\tilde{O}$ and $\tilde{a}^i$ is the ith component of the translation vector from $\tilde{O}$ to $O$, $O$ and $\tilde{O}$ being the origins of the two coordinate systems). Then in the new coordinate system consider the partial derivatives $$ \tilde{Y}^{i_1, \cdots, i_r}_{q, j_1, \cdots, js} = {\partial {\tilde{X}^{i_1, \cdots, i_r}_{j_1, \cdots, js}} \over \partial \tilde{x}^q} $$ and derive relationships binding this above partial derivative and (21.3)"

I've struggled trying to find a solution to this exercise, but I'm not even sure I've really understand what it is asking. Anyone could help me out?

P.S. Another side question: could I think of $ \nabla_q X^{i_1, \cdots, i_r}_{j_1, \cdots, j_s} $ as the product of two tensors, one being $\nabla_q$?


1 Answer 1


The tensor expression has different realizations in different coordinate systems. For example, the expression $T_{ij}$ has the transformation rule:

$$ T_{i' j'} = T_{ij} J_{i'}^i J_{j'}^j$$

And similarly the derivative also has similar transformation rule, take for example a function given in one coordinates as $f(x,y)$ then in another coordinates as $f(r(x,y) , \theta(x,y) )$ then:

$$ \frac{\partial f(x(r,\theta) , y(r ,\theta) )}{\partial r} = \frac{\partial x}{\partial r} \partial_x f + \frac{\partial y}{\partial r} \partial _y f$$

You see the rule to take derivative also transforms. In general:

$$ \frac{\partial}{\partial x^{i' } } =J_{i'}^i \frac{\partial }{\partial x^i}$$

The prime denotes the new 'indexed' coordinates. So, all the exercise is asking you to do is plug in the expression for the transformed tensor expression w.r.t cartesian coordinates and evaluate the derivative.

Note: $J_{i'}^i$ is denoting the Jacobian matrix. If the index is upper, there is a different jacobian transformation rule.

I will use Pavel's notation to work this out, we begin with the equations given:

$$ x^{i'}= T_j^i x^j + a^{i'}$$

I assume the displacement vector is constant, and I differentiated both sides with $x^p$ assuming $T_j^i$ is a constant:

$$ \frac{\partial x^{i'} }{\partial x^p} = T_j^i \frac{\partial x^j}{\partial x^p}$$

Now, $\frac{\partial x^j}{\partial x^p} = \delta_p^j$ and with contraction on the kronceker:

$$ \frac{\partial x^{i'} }{\partial x^p} = T_p^i$$

Relabelling indices $( p \to i)$and recalling the definition of jacobian:

$$J_{i}^{i'} = T_{i}^{i'}$$

Similarly we can find the inverse jacobian $J_{i'}^i$.. now it is simply a matter of plugging the jacobian terms in for the transformation:

$$Y_{i_1',i_2',...i_n'}^{j_1',j_2'...,j_n'} = Y_{i_1,i_2...,i_n}^{j_1,j_2...,j_n} J_{i_1'}^{i_1} J_{i_2'}^{i_2}...J_{i_n'}^{i_n} J_{j_1'}^{j_1}... J_{j_n'}^{j_n} $$

However I do wish to make the same remark the pdf does here:

Warning 21.1. Theorem 21.1 and the equality (21.5) are valid only for Cartesian coordinate systems. In curvilinear coordinates things are different.

You may find this question by me and pavel's approach in here easier

  • $\begingroup$ My phrasing may have been poor.. however, if you can, try check out Pavel Grinfeld's Tensor calculus book. He explains it better than I do $\endgroup$
    – Babu
    Commented Jun 23, 2021 at 12:43
  • $\begingroup$ Thank you for your time and effort in answering me. I will try to use your explanation to prove the theorem and I will let you know in case of other problems. One more thing: I am an high school student who is intrested in understanding upper level physics, would you recommend to buy the book you've mentioned to understand more about tensors, is it at a temptable level for an highschooler? $\endgroup$
    – Luke__
    Commented Jun 23, 2021 at 17:42
  • $\begingroup$ Hi friend, I too was in a similar background when I started and I will say yes! There are lectures to go along with the book. @Luke__ on YouTube . There is another book which was recommended to me (but I haven't started) that I thought good it's called John Baez Gauge fields knots and gravity. I skimmed it and it seemed qute simpled Feel free to drop in my chat if you want more details. $\endgroup$
    – Babu
    Commented Jun 23, 2021 at 18:04
  • $\begingroup$ This youtube series is also nice @Luke__ $\endgroup$
    – Babu
    Commented Jun 23, 2021 at 18:04
  • $\begingroup$ Thank again for your support! I've tryed to go on through the problem using your suggestion and, what I think I've managed to prove is that the new object obtained by differentiating the tensor, is actually a new tensor: $\tilde{Y}^{i_1, \cdots, i_r}_{q, j_1, \cdots, j_s} = (T^{i_1}_{h_1} \cdots T^{i_r}_{h_r} S^{k_1}_{j_1} \cdots S^{k_s}_{j_s} S^p_q) Y^{h_1, \cdots, h_r}_{p, k_1, \cdots, k_s}$. This, I think, prove that the nature of $Y$ is tensorial, but how is it possible to prove that $Y$ has a valency of $(r, s+1)$, while $X$ is of type $(r, s)$ ? $\endgroup$
    – Luke__
    Commented Jun 23, 2021 at 19:13

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