Differentiation of scalar fields using tensor notation

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).

EDIT:

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?

• 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. Jul 31, 2014 at 10:49
• @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. Jul 31, 2014 at 10:53
• @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. :| Jul 31, 2014 at 10:55
• 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. Jul 31, 2014 at 10:57
• @Avitus: Yes, your standard approach makes perfect sense. Jul 31, 2014 at 10:58

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

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

• 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 $t$). Jul 31, 2014 at 10:34
• I edit my answer accordingly Jul 31, 2014 at 10:36
• I consider the function $g_j=g_j(x_j(t))$ and I apply the chain rule as we did for $\varphi(x_1(t),...,x_n(t))$. Jul 31, 2014 at 11:47
• The second you wrote. Jul 31, 2014 at 11:54
• And how do you get your second equation? Do you just take $\dfrac{dx_j}{dt}$ to the other side? Jul 31, 2014 at 12:01