I'm learning tensor calculus to understand differential geometry. Please verify if I've understood how to employ Einstein's sum convention and index notation correctly.

Suppose that $\varphi := \varphi(x^1,\cdots,x^n)$ is a smooth scalar field, i.e. $\varphi: \mathbb{R}^n \to \mathbb{R}$ is a function that is infinitely differentiable.

The first derivative of $\varphi$ is calculated in this way, by using the chain rule:

$$\dfrac{d}{dt}\varphi = \varphi_i \dfrac{d}{dt}x^i$$

Where $i$ is a dummy variable and it's summed over from $1$ to $n$.

Here's my attempt to calculate the second derivative of $\varphi$: $$\dfrac{d^2}{dt^2} \varphi = \dfrac{d}{dt}(\dfrac{d\varphi}{dt}) = (\varphi_i \dfrac{d}{dt}x^i)_j \dfrac{d}{dt}x^j = \left( \varphi_{ij}\dfrac{d}{dt}x^i+\underbrace{\varphi_i (\dfrac{d}{dt}x^i)_j}_{(*)} \right) \dfrac{d}{dt}x^j $$

But I believe we have $\varphi_i \left(\dfrac{d}{dt}x^i\right)_j= \varphi_i \dfrac{d}{dt} \delta_{ij}$, therefore:

$$\dfrac{d^2\varphi}{dt^2}=\left( \varphi_{ij}\dfrac{d}{dt}x^i+\underbrace{\varphi_i \delta_{ij} \dfrac{d}{dt}}_{(*)} \right) \dfrac{d}{dt}x^j = \varphi_{ij}\dfrac{dx^i}{dt}\dfrac{dx^j}{dt}+\varphi_i \delta_{ij} \dfrac{d^2 x^j}{dt^2} = \varphi_{ij}\dfrac{dx^i}{dt}\dfrac{dx^j}{dt}+\varphi_j \dfrac{d^2 x^j}{dt^2}$$

The reason that I think $(\dfrac{d}{dt}x^i)_j = \dfrac{d}{dt}\delta_{ij}$ holds is this:

$$(\dfrac{d}{dt}x^i)_j = \dfrac{\partial}{\partial x^j}(\dfrac{d}{dt}x^i) = \frac{d}{dt}(\dfrac{\partial}{\partial x_j}x^i)= \dfrac{d}{dt} \delta_{ij}$$

I don't know why we can swap $\dfrac{\partial}{\partial x^j}$ and $\dfrac{d}{dt}$ though.

Are my calculations correct so far? (I want to calculate the third derivative of $\varphi$ as well, but I'll do it under my question as an answer because my question is getting too long).


I'm making the following assumptions in my calculations:

  1. $\dfrac{d}{dt}$ is a derivative operator and it acts to anything that is written right to it. Therefore $\dfrac{d}{dt} \space f = \dfrac{df}{dt}$

  2. I'm assuming that $\dfrac{\partial}{\partial x^j}\dfrac{d}{dt} = \dfrac{d}{dt}\dfrac{\partial}{\partial x^j}$

Using these two assumptions, we have:

$$(\dfrac{d}{dt}x^i)_j = \dfrac{d}{dt}(\dfrac{\partial}{\partial x^j} x^i) = \dfrac{d}{dt} \delta_{ij} = \delta_{ij} \dfrac{d}{dt}$$

The last equality holds because of linearity of $\dfrac{d}{dt}$.

So, we have $(\dfrac{d}{dt}x^i)_j= \delta_{ij} \dfrac{d}{dt}$ as operators. Is that wrong?

  • $\begingroup$ note that $\frac{d}{dt}\delta_{ij}=0$ and you make $\frac{d}{dt}$ and $\delta_{ij}$ "commute" in the r.h.s. of the first equality of the final expression for the second derivative. $\endgroup$
    – Avitus
    Jul 31, 2014 at 10:49
  • $\begingroup$ @Avitus: I think $\dfrac{d}{dt}(\delta_{ij} f)$ could be interpreted to be the same as $\delta_{ij} \dfrac{d}{dt}(f)$ because $\delta_{ij}$ is constant and $\dfrac{d}{dt}$ is a linear operator. $\endgroup$
    – math.n00b
    Jul 31, 2014 at 10:53
  • $\begingroup$ @Avitus: Unfortunately my differential geometry book hasn't explained tensor notation very well. :( So, I'm just trying to figure out how it works on my own and it's a pretty time-consuming and mistake-prone job. :| $\endgroup$
    – math.n00b
    Jul 31, 2014 at 10:55
  • $\begingroup$ I like the fact you are trying to explain notation on your own, well done :) Btw, let us go on slowly: are you convinced with notation in my answer? That is the "standard" approach. $\endgroup$
    – Avitus
    Jul 31, 2014 at 10:57
  • $\begingroup$ @Avitus: Yes, your standard approach makes perfect sense. $\endgroup$
    – math.n00b
    Jul 31, 2014 at 10:58

1 Answer 1


I do not fully understand your notation, when you write $(\varphi_i\frac{dx_i}{dt})_j$; using a more standard notation and $\varphi=\varphi(x_1(t),\cdots,x_n(t))$ I get

$$\frac{d\varphi}{dt}=\sum_i \frac{d\varphi}{dx_i}\frac{dx_i}{dt} $$ and $$\frac{d^2\varphi}{dt^2}= \sum_i \frac{d}{dt}\left(\frac{d\varphi}{dx_i}\right)\frac{dx_i}{dt}+\sum_i \frac{d\varphi}{dx_i}\frac{d^2x_i}{dt^2}= \sum_i\sum_j \frac{d^2\varphi}{dx_idx_j}\frac{dx_j}{dt}\frac{dx_i}{dt}+\sum_i \frac{d\varphi}{dx_i}\frac{d^2x_i}{dt^2}. $$

A consideration on the notation $(\cdots)_j$ in the OP (add on).

In the OP $(\cdot)_j$ denotes $\frac{\partial}{\partial x_j}$. Then

$$(\frac{dx_i}{dt})_j =0$$ as $\frac{dx_i}{dt}$ does not depend on $x_j$, for all $i$' and $j$, but only on the parameter $t$ if $i\neq j$.

If $i=j$, then $g_j:=\frac{dx_j}{dt}$, with $g_j=g_j(x_j,t)$. We have

$$\frac{dg_j}{dt}=\frac{\partial g_j}{\partial x_j}\frac{dx_j}{dt}, $$

or $$\frac{\partial g_j}{\partial x_j}=\frac{dg_j}{dt}\frac{1}{\frac{dx_j}{dt}}= \frac{d^2x_j}{dt^2}\frac{1}{\frac{dx_j}{dt}}$$

if $\frac{dx_j}{dt}\neq 0$ for all $t$. In summary

$$(\frac{dx_i}{dt})_j =\delta_{ij}\frac{d^2x_i}{dt^2}\frac{1}{\frac{dx_i}{dt}}. $$

  • $\begingroup$ I see now: then $\partial_j \frac{dx_i}{dt}=0$ because $\frac{dx_i}{dt}$ does not depend on $x_j$ (but only on $t$). $\endgroup$
    – Avitus
    Jul 31, 2014 at 10:34
  • $\begingroup$ I edit my answer accordingly $\endgroup$
    – Avitus
    Jul 31, 2014 at 10:36
  • $\begingroup$ I consider the function $g_j=g_j(x_j(t))$ and I apply the chain rule as we did for $\varphi(x_1(t),...,x_n(t))$. $\endgroup$
    – Avitus
    Jul 31, 2014 at 11:47
  • $\begingroup$ The second you wrote. $\endgroup$
    – Avitus
    Jul 31, 2014 at 11:54
  • $\begingroup$ And how do you get your second equation? Do you just take $\dfrac{dx_j}{dt}$ to the other side? $\endgroup$
    – math.n00b
    Jul 31, 2014 at 12:01

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