From "Introducing Einstein’s Relativity" by Ray D’Inverno page 77-78

enter image description here

In my calculation, the process is

$$\frac{\partial{x^{'a}}}{\partial{x^d}}=\frac{\partial{x^{a}}}{\partial{x^d}}+\frac{1}{2} {Q_{bc}^{a}}{\frac{\partial}{\partial{x^d}}}(x^b x^c)$$

$$=\delta_{b}^{a} + \frac{1}{2}{Q_{bc}^{a}(\delta_{d}^{b}x^c + x^b \delta_{d}^{c})}$$

I think there should be somewhere incorrect, but I cannot find the right way.

for the second one: $$\frac{\partial}{\partial{x^e}}(\delta_{d}^{a}+{Q_{bd}^{a}}x^d)={Q_{bd}^{a}}\delta_{e}^{d}=Q_{be}^{a}$$

which is $Q_{de}^{a}$ in picture

  • $\begingroup$ The author made a mistake -- switching $d$ and $b$ -- and in fact you made the same mistake in a different place. Every term in the first equation you want to derive should have an upper $a$ and a lower $d$ and nothing else other than possibly summed upper/lower pairs that are neither $a$ nor $d$. $\endgroup$
    – user10851
    Apr 30, 2016 at 0:51

1 Answer 1


Your mistake is that you left out the summation over double indices (which would give another delta).

Since you have to sum over all possiblities of e.g. $c$, one of these possibilities is when c=d and only this term remains.

I'll give a simplified example:

$$\frac{\partial}{\partial{x^d}}({Q_{bc}}x^c) = \frac{\partial}{\partial{x^d}}({Q_{b0}}x^0) + \frac{\partial}{\partial{x^d}}({Q_{b1}}x^1) + \frac{\partial}{\partial{x^d}}({Q_{b2}}x^2) + \frac{\partial}{\partial{x^d}}({Q_{b3}}x^3)$$

It doesn't matter what $d$ actually is, you will be left with:

$$\frac{\partial}{\partial{x^d}}({Q_{bc}^{a}}x^c) = \frac{\partial}{\partial{x^d}}({Q_{bd}^{a}}x^d)={Q_{bd}^{a}} \quad, \quad \frac{\partial}{\partial{x^d}}({Q_{bc}^{a}}x^c) = \delta^c_d {Q_{bc}^{a}} $$

Some hints for your calculations:

  • in $\frac{\partial}{\partial{x^d}}({Q_{bc}^{a}}x^b x^c)$ you have to, as you did, differentiate both factors. But keep track of the summation! (You can use deltas to shorten it, it will look similar to what you got, but you might mess it up the first time. The $2$ to cancel out $\frac{1}{2}$ will appear somewhere in this mess).
  • in the result, $\frac{\partial{x^{'a}}}{\partial{x^d}}=\delta_{d}^{a} + Q_{bd}^{a}x^d$, the right $d$ in the sum is NOT the same $d$ as in $\frac{\partial}{\partial{x^d}}$ (since again, the right side is a sum; thats a general rule for this notation), but $\frac{\partial{x^{'a}}}{\partial{x^d}}$ and $\delta_{d}^{a}$ use the same $d$ (and $a$). It can very well be renamed to m: $\frac{\partial{x^{'a}}}{\partial{x^d}}=\delta_{d}^{a} + Q_{bm}^{a}x^m$. Keep that in mind when you stumble upon 4 terms you want to get rid of.
  • for the 2nd derivative, you should start from the first formula, not the intermediate result, as you'll get more terms since you have to keep terms with either $d$ or $e$, not just $d$.

Final remark: the last formula (the 2nd derivative) represents a fundamental concept in how tensor transformation works (when you have to check if something is a tensor you can check if that kind of formula holds), so you will see it again.


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