My physics teacher assigned this problem: "Evaluate $\vec\nabla \vec x^2 $ where $\vec x^2 = x^2 + y^2 +z^2$ and $\vec\nabla = \hat{\imath}\frac{\partial}{\partial x}+\hat{\jmath}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$.

I know how to compute the problem. But I am confused about this vector $\vec x^2 = x^2 + y^2 +z^2$. I thought vectors have a norm and direction. But it looks like this is only a norm. I am used to vectors always having components, $\hat{\imath},\hat{\jmath},\hat{k}$.

Could anyone shed some light on the subject? What is this notation?

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    $\begingroup$ It's not a vector, it's the square of its norm. $\endgroup$
    – Javier
    Commented Oct 21, 2016 at 21:09
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    $\begingroup$ $\vec x^2$ is a compact notation for $\vec x \cdot\vec x=|\vec x|^2$. $\vec x^2$ is NOT a vector. $\endgroup$
    – N0va
    Commented Oct 21, 2016 at 21:09
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    $\begingroup$ I just hate physics notations... $\endgroup$
    – Integral
    Commented Oct 21, 2016 at 21:14
  • $\begingroup$ The euclidean inner product in space $\Bbb C^n$ is defined as $$(\vec x|\vec y)=\vec x\cdot\vec y=\sum_{k=1}^n x_k\bar y_k$$ Then I assume that $$(\vec x|\vec x)=\vec x^2=x^2+y^2+z^2$$ is the standard norm in $\Bbb R^3$ where the coordinates are $x,y,z$. $\endgroup$
    – Masacroso
    Commented Oct 21, 2016 at 21:28
  • $\begingroup$ That notation is quite unconventional. More often, we write $\vec x \cdot \vec x$ or $|\vec x|^2$ or $\vec r\cdot \vec r$ or $|\vec r|^2$ to mean $\sum_{i=1}^3 x_i^2$. $\endgroup$
    – Mark Viola
    Commented Oct 21, 2016 at 21:37

2 Answers 2


The notation $\vec x ^2 = x^2 + y^2 +z^2$ seems to be the square of the length of the vector $\vec x$. This is often written as $\|\vec x \|^2$, where $\|\vec x\|$ is the length of the vector $\vec x$.

The symbol $\vec\nabla$ is the gradient operator. We can think of $\vec x^2$ as a function where $$(x,y,z) \longmapsto x^2 + y^2 + z^2$$ The gradient of the function $\vec x^2$ would give $$\vec \nabla \vec x^2 = 2x{\bf i} \ + \ 2y{\bf j} \ + \ 2z{\bf k} $$


The notation $\vec{x}^2$ is presumably being used to mean $(\vec{x})^2$, which is equivalent to $\vec{x}\cdot\vec{x}$. It's not a vector, it's just the square of a norm.

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    $\begingroup$ If $\vec x^2$ is undefined then so is $(\vec x)^2$. There is no difference between the two. $\endgroup$
    – user137731
    Commented Oct 27, 2016 at 20:17
  • $\begingroup$ Yes, that is what I said in my answer. $\endgroup$
    – David Z
    Commented Oct 28, 2016 at 0:27

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