Suppose I have a scalar field $\varphi\left(\vec r(t), t\right)$ where $\vec r$ is a vector in $\mathbb R^3$ that depends on the parameter $t$.

I'd like to figure out what $\dfrac{\mathrm d}{\mathrm dt}\nabla \varphi$ is.

I start as follows, $$\nabla \varphi = \partial_i \varphi \hat{e}^i $$ (which I attempt to write in Einstein notation)

From there, I take the time derivative component-wise and using the directional derivative I get $$\dfrac{\mathrm d}{\mathrm dt}\nabla \varphi = \left( v_j \partial _j \partial_i \varphi +\partial _t \partial _i \varphi \right)\hat{e}^i$$ where $v^\nu = \dfrac{\mathrm d r^\nu }{\mathrm dt} $.

I can rewrite this as (where $^T$ denotes the transpose)$$\dfrac{\mathrm d\nabla\varphi }{\mathrm dt} = \pmatrix{\nabla \partial _x\varphi^T \\ \nabla \partial _y \varphi ^T \\ \nabla \partial_z \varphi ^T }\vec v+\dfrac{\partial}{\partial t}\nabla \varphi$$ and am wondering if there is a way for me to simplify this further, in particular if there is a notation or name for the $\pmatrix{\nabla \partial _x\varphi^T \\ \nabla \partial _y \varphi ^T \\ \nabla \partial_z \varphi ^T }$ matrix.

  • 1
    $\begingroup$ I think you mean e.g. $\nabla\varphi=\partial_\mu\hat{e}^\mu$ or, even better, $\nabla\varphi=\partial_i\hat{e}^i$. $\endgroup$
    – J.G.
    Commented Jun 11, 2022 at 22:08
  • $\begingroup$ @J.G. yes, thank you $\endgroup$
    – user256872
    Commented Jun 11, 2022 at 22:21
  • $\begingroup$ Sorry, I meant $\partial_i\varphi\hat{e}^i$. You still don't want an index on $\varphi$. $\endgroup$
    – J.G.
    Commented Jun 11, 2022 at 22:23
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    $\begingroup$ Hi! Let me know if you want more details/ more notation help. $\endgroup$ Commented Jun 11, 2022 at 22:31
  • $\begingroup$ @J.G. ah yes, it's a scalar -- for some reason I was thinking of $\varphi^i$s as components of the gradient :/ $\endgroup$
    – user256872
    Commented Jun 11, 2022 at 22:32

1 Answer 1


Here is your answer:

$$ \frac{d}{dt} \nabla \phi = \left[ \nabla_v + \partial_t \right] \nabla \phi$$

It maybe confusing to see $\nabla_v$ acting on a vector here. The meaning of the notation is that you act the operator on all the component of the vector.

  • 1
    $\begingroup$ Btw you can actually think about $\nabla$ acting twice on a vector. This is known as the vector gradient. $\endgroup$ Commented Jun 11, 2022 at 22:31
  • $\begingroup$ ^^ Twice on a scalar. Or once on a vector. Confused myself. $\endgroup$ Commented Jun 11, 2022 at 22:38

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