# What does it mean to take the gradient of a vector field?

What does it mean to take the gradient of a vector field? $$\nabla \vec{v}(x,y,z)$$? I only understand what it means to take the grad of a scalar field.

• The short answer is: the gradient of the vector field $\sum v_i(x, y, z)e_i$, where $e_i$ is an orthonormal basis of $\mathbb{R}^3$, is the matrix $(\partial_i v_j)_{i, j=1, 2, 3}$. Jun 11, 2012 at 8:48
• The long answer involves tensor analysis and you can read about it on books such as Itskov, Tensor algebra and tensor analysis for engineers. Jun 11, 2012 at 8:49
• Another possible explanation is that the dot is missing between $\nabla$ and $\vec v$, and the "gradient" is actually divergence.
– user31373
Jun 11, 2012 at 14:27
• Guiseppe Negro's short answer is off, switch his i's and j's and its fixed.
– user78853
May 21, 2013 at 21:10
• @Thomas I've converted your answer into a comment. In the future please only use answers to give answers to the question posed in the original post. I realize that you do not yet have the reputation to comment on other people's posts, but that will come before long if you contribute to the site. May 21, 2013 at 22:12

The gradient of a vector is a tensor which tells us how the vector field changes in any direction. We can represent the gradient of a vector by a matrix of its components with respect to a basis. The $(\nabla V)_{\text{ij}}$ component tells us the change of the $V_j$ component in the $\pmb{e}_i$ direction (maybe I have that backwards). You can check out the Wikipedia article for the details of calculating the components.

To get a physical picture of its meaning we can decompose it into 1) the trace (the divergence) 2) an anti-symmetric tensor (the curl) 3) a traceless symmetric tensor (the shear)

If the vector field represents the flow of material, then we can examine a small cube of material about a point. The divergence describes how the cube changes volume. The curl describes the shape and volume preserving rotation of the fluid. The shear describes the volume-preserving deformation.

It depends on how you define the gradient operator. In geometric calculus, we have the identity $\nabla A = \nabla \cdot A + \nabla \wedge A$, where $A$ is a multivector field. A vector field is a specific type of multivector field, so this same formula works for $\vec v(x,y,z)$ as well.

So we get $\nabla\vec v = \nabla \cdot \vec v + \nabla \wedge \vec v$. The first term should be familiar to you -- it's just the regular old divergence. However the second term is a different type of object entirely (actually, it's a generalization of the familiar $3$D curl $\nabla \times \vec u$ that works in any dimension).

In the same way that a vector field can be though of as associating with every point in your domain an oriented line segment (a vector), $\nabla \wedge \vec v$ associates with every point in your domain an oriented plane segment (which we call bivectors). So $\nabla \wedge \vec v$ is called a bivector field.

So to answer your question, the gradient of a vector field is the sum of a scalar field and a bivector field.

• Interesting! However, how to reconcile this language with the usual one? It seems non trivial: you never add scalars and vectors in "standard" vector analysis. Which is the interpretation? May 25, 2020 at 1:08

Assume the vector $$\vec{\bf F} = (F_1, F_2, F_3)$$ exists in a 3D space with basis $$x_1, x_2, x_3$$, then its gradient is the 3 × 3 matrix: $$\partial_iF_j$$

$$\nabla\vec{\bf F}=\left( {\begin{array}{c} \frac{\partial F_1}{\partial x_1}&\frac{\partial F_1}{\partial x_2}&\frac{\partial F_1}{\partial x_3}\\ \frac{\partial F_2}{\partial x_1}&\frac{\partial F_2}{\partial x_2}&\frac{\partial F_2}{\partial x_3}\\ \frac{\partial F_3}{\partial x_1}&\frac{\partial F_3}{\partial x_2}&\frac{\partial F_3}{\partial x_3}\\ \end{array}} \right).$$

That is, each column is a "usual" gradient of the corresponding scalar component function.

• I think that's wrong because the correct result is the transpose of what you wrote. take a look at the page 7 of this document: Apr 20, 2018 at 20:57
• According to Riley, Hobson and Bence (3rd), we have $\nabla\vec{\bf F}=\left( {\begin{array}{c} \frac{\partial F_1}{\partial x_1}&\frac{\partial F_1}{\partial x_2}&\frac{\partial F_1}{\partial x_3}\\ \frac{\partial F_2}{\partial x_1}&\frac{\partial F_2}{\partial x_3}&\frac{\partial F_2}{\partial x_3}\\ \frac{\partial F_3}{\partial x_1}&\frac{\partial F_3}{\partial x_2}&\frac{\partial F_3}{\partial x_3}\\ \end{array}} \right).$ It states on page 937: Apr 9, 2021 at 17:37
• The gradient of a vector. Suppose $v_i$ represents the components of a vector; let us consider the quantities generated by forming the derivatives of each $v_i, i =1,2,3$, with respect to each $x_j, j =1,2,3$ i.e. $$T_{ij} = \frac{\partial v_i}{\partial x_j}\ .$$ Apr 9, 2021 at 17:41
• @ Freshman42 Actually, that PDF is incorrect. If you for example consider a vector field of 2-vectors in 3-space, multiplying the resulting gradient matrix with the 3-vector along which we want to take the directional derivative in order to get the derivative, which is a 2-vector, only works if the matrix is what Mussé Redi describes. Dec 8, 2021 at 19:12
• I believe the discussion about whether this or the transposed is correct comes down to whether numerator or denominator layout is used. The one used in the answer is numerator layout. Mar 21 at 14:29

Gradient of a vector field (or a multi-valued function $f: R^m\to R^n$) is jacobian of the multi-valued function $f$, where each row $r_i$ of the $\text{Jacobian}(f)$ represents the gradient of $f_i$ (remember, each component $f_i$ of the multi-valued function $f$ is a scalar).

• Hope this edit is acceptable. Mar 31, 2013 at 8:24

Gradient of a vector field is intuitively the Flux/volume leaving out of the differential volume dV. Visualise in 2D first. Suppose you have a vector field E in 2D. Now if you plot the Field lines of E and take a particular Area (small area..), Divergence of E is the net field lines, that is, (field line coming out of the area minus field lines going into the area). Similarly in 3D, Divergence is a measure of (field lines going out - field lines coming in). If you mathematically implement this you see you get 3 terms of partial derivatives added, which essentially adds the total net field lines.

For a scalar field(say F(x,y,z) ) it represents the rate of change of F along the the 3 perpendicular ( also called orthonormal ) vectors you defined your system with (say x, y, z ).

The gradient of a vector field corresponds to finding a matrix (or a dyadic product) which controls how the vector field changes as we move from point to another in the input plane.

Details: Let $$\vec{F(p)} = F^i e_i = \begin{bmatrix} F^1 \\ F^2 \\ F^3 \end{bmatrix}$$ be our vector field dependent on what point of space we take, if step from a point $$p$$ in the direction $$\epsilon \vec{v}$$, we have:

$$\vec{F(p+ \epsilon \vec{v})} = F^i(p+ \epsilon v) e_i= F^i(p) e_i + \epsilon \left(v \cdot \nabla F^i(p) \right) e_i$$

But, what is $$( v \cdot \nabla F^i ) e_i= \begin{bmatrix} \nabla F^1 \\ \nabla F^2 \\ \nabla F^3 \end{bmatrix} v= \begin{bmatrix} \partial_1 F^1 & \partial_2 F^1 & \partial_3 F^1 \\ \partial_1 F^2 & \partial_2 F^2 & \partial_3 F^2 \\ \partial_1 F^3 & \partial_2 F^3 & \partial_3 F^3 \end{bmatrix} \begin{bmatrix} v^1 \\ v^2 \\ v^3 \end{bmatrix}$$

And that's what the gradient of a vector field is, a big matrix controls how the gradient vector changes when we move in any direction of the input space.

Note that I have assumed the eulicdean basis (i.e: i,j,k)