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Partial derivative of inner product in Einstein Notation

  1. Feb 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Can someone please check my working, as I am new to Einstein notation:
    Calculate $$\partial^\mu x^2.$$
    2. Relevant equations
    3. The attempt at a solution

    \partial^\mu x^2 &= \partial^\mu(x_\nu x^\nu) \\
    &= x^a\partial^\mu x_a + x_b\partial^\mu x^b \ \ \text{(by product rule and relabelling indices)} \\
    &=x^a\delta_\mu^a + x_b\delta_\mu^b \\
    I'm not sure is the expression in the second term of the second line is correct, as the partial is with respect to the covariant vector but the argument is a contravariant vector.
  2. jcsd
  3. Feb 4, 2017 #2
    A useful tip to keep in mind is that indices keep their position (upper or lower) unless they are raised or lowered by a metric. So the equality that you wrote,
    [tex]x^a\partial^\mu x_a + x_b\partial^\mu x^b =x^a\delta_\mu^a + x_b\delta_\mu^b,[/tex] is clearly not correct.

    Firstly, ##\partial^{\mu} x_{a} = \delta_{a}^{\mu}## (notice that ##\mu## stays upper and ##a## stays lower).

    Next, you cannot simply perform the derivative ##\partial^{\mu} x^{b}## directly because the derivative is with respect to the contravariant component. So the correct way to go about doing it is to rewrite ##x^{b} = g^{cb} x_{c}##, so that ##\partial^{\mu} x^{b} = g^{cb} \partial^{\mu} x_{c} +x_{c} \partial^{\mu}g^{cb} ##. In a cartesian coordinate system, the metric is just identity and so we get the same result as you do.
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