Differentiating vector fields

6.1 A fixed arrow with changing components#

Draw an arrow pointing east on a flat sheet. At each point of a circle around the origin, describe that same eastward direction using two local unit arrows: one pointing outward and one pointing around the circle. At the rightmost point, east is outward. At the top, east is opposite the direction of increasing angle. The direction stayed fixed, but its components changed.

A vector field assigns a vector to each position. A velocity field, for example, tells you the direction and speed of flow at every location. Its arrows are instructions about motion at those locations; they are not a photograph of separate particles. Start with the uniform eastward field below. Move P and Q around the circle and compare their component readouts. Then bring the two physical arrows to the same origin, where their difference is easy to see.

GEOMETRY LAB / 02

Compare two nearby arrows.

Each arrow gives the velocity at its tail. Compare two positions, P and Q. The thin gold directions point outward and around the circle at each position.

Velocity arrows share one scale. Gold arrows mark unit directions, not velocities.

Bring both arrows to one origin

At PAt QDifference

Actual change: (0.00, 0.00) m/s

At P · radial, tangential(0.85, -0.52) m/s

At Q · radial, tangential(0.41, -0.91) m/s

The pink and blue velocity arrows coincide when brought to one origin. The field is unchanged. The two pairs of polar components differ because “outward” and “around” point in different directions at P and Q.

Separate the component change from the frame change

First change the two component numbers while keeping P’s frame. Then turn that frame into Q’s frame, keeping the new numbers. The gold and blue arrows below show these two contributions, added head to tail. Their sum is the actual change in velocity.

ComponentsFrame

For a uniform field, the two contributions cancel. Try rotation: its polar components are constant, so the entire change comes from the turning frame.

(-0.17, -0.56) + (0.17, 0.56) = (0.00, 0.00) m/s

Reduce the separation to approach a derivative. We will next divide these changes by the angular step and let that step approach zero.

To check the construction, collect the two unit directions as the columns of a matrix $B$, and the two component numbers into a column $c$. Then $V=Bc$. Adding and subtracting $B_Pc_Q$ gives the exact finite identity

$$V_Q-V_P=B_P(c_Q-c_P)+(B_Q-B_P)c_Q$$

The second term uses Q’s component numbers because the first step has already changed the numbers. No small-step approximation is needed for this decomposition.

The fields and their limits

The uniform flow has velocity $(1,0)$ metres per second. The other fields are $V=k(x,y)$ for expansion, $V=k(-y,x)$ for rotation, and $V=k(y,0)$ for shear, with $k=0.4\,\mathrm{s}^{-1}$. P and Q lie on a circle of radius 1.8 metres. Every velocity arrow in the field uses the same length scale; the comparison plot has its own labeled velocity axes.

These are prescribed, steady velocity fields in a flat plane. The arrows are samples of the field, not moving particles, and no fluid force law is being solved. Bringing them to one origin uses ordinary translation in a Euclidean plane. On a curved surface we must specify how to transport a vector first.

The gold frame consists of the unit vectors $\hat e_r,\hat e_\theta$. The polar coordinate basis is different: its angular vector is $\partial_\theta=r\hat e_\theta$. The readouts here use the unit frame, so both component numbers have velocity units.

The uniform field and the rotating field give opposite surprises. A fixed physical arrow can have changing components. A turning physical arrow can have constant components. In both cases, the missing information is how the local frame changes. Our derivative must keep track of the whole arrow.

We can check the eastward example with the unit basis from §4.8. For the algebra, let $V$ denote the unit direction of the eastward flow. In fixed Cartesian components,

The eastward unit vector $V=(1,0)$ can therefore be written as

Differentiating its two coefficients gives $-\sin\theta$ and $-\cos\theta$. If we stop there, we predict a changing vector. But the basis vectors also change:

Apply the product rule to the whole expression for $V$:

The cancellation agrees with the drawing. A useful derivative must account for changes of basis as well as changes of components.

This example used unit vectors. The polar coordinate basis is $\partial_r=\hat e_r$, $\partial_\theta=r\hat e_\theta$: its angular vector also changes length with radius. We now work in coordinate bases and use the chain rule to identify the general problem.

Let a vector field have components $V^\alpha$ in coordinates $x^\mu$. Under new coordinates $x'^\alpha(x)$, define the Jacobian and its inverse by

Then $V'^\alpha=J^\alpha{}_{\beta}V^\beta$. Apply the chain rule and the ordinary product rule:

The first term is exactly how a tensor with one upper and one lower index should transform. The second term is the trouble: it involves derivatives of the coordinate transformation itself. A tensor transformation changes the description of an object using the Jacobian at the point; it does not need second derivatives of the map.

Notice when the trouble disappears. A Cartesian rotation or a Lorentz transformation has a constant Jacobian, so the extra term vanishes. For these transformations, differentiating the components already gives a tensor. A position-dependent change of coordinates requires the extra correction.

The solution is to differentiate the geometric vector, accounting for the changing local basis. We call the resulting operation the covariant derivative.

6.2 What a derivative must be able to compare#

There is a subtlety even before the algebra. A vector at event $p$ belongs to $T_pM$, the tangent space at $p$. A vector at a neighboring event $q$ belongs to $T_qM$. These are different vector spaces. Subtracting their component lists does not, by itself, define a geometric subtraction.

On the plane we could compare arrows using one fixed Cartesian basis. A general manifold does not come with that common basis. A connection supplies a rule for comparing vectors at neighboring points. Repeating the comparison along a path will let us transport a vector from one point to another; different paths can give different results.

In a coordinate basis $e_\nu=\partial/\partial x^\nu$, define connection coefficients by

Here $\nabla_\mu$ means differentiation along coordinate direction $\mu$ using the chosen comparison rule. The coefficients $\Gamma^\rho{}_{\mu\nu}$ describe the resulting change of basis vector $e_\nu$. They generalize the basis derivatives we just calculated on the plane.

Apply the product rule to $V=V^\nu e_\nu$:

Therefore

The first term measures changing components. The second corrects for the comparison of local bases. Neither term separately has to be a tensor; their sum does.

The derivative index $\mu$ is a lower index because differentiation asks for a direction as its input. If $X^\mu$ specifies that direction, then $\nabla_XV=X^\mu\nabla_\mu V$ is a vector. Before choosing $X$, $\nabla V$ is a tensor with an extra covector slot waiting to receive it.

WORKED EXAMPLE

An unchanged arrow can have changing components

Can a vector have a nonzero component derivative while the vector itself stays fixed?

Builds onA vector and its basis Polar metric and coordinate lengths

See the idea

Carry a little radial-and-angular frame around the flat plane. Keep a unit arrow pointing east in the fixed Cartesian picture. The frame rotates beneath it, so its two component readings change. A derivative of the whole arrow must include both changes: the component readings and the basis vectors that give those readings meaning.

Work it out

1. 01
   Make the moving rulers explicit

   The hats mark unit vectors. The angle $\theta$ is in radians, and these basis vectors are written in a fixed Cartesian frame. They are perpendicular and have length one.

   $$\hat e_r=(\cos\theta,\sin\theta),\qquad\hat e_\theta=(-\sin\theta,\cos\theta).$$

   Why this step works The radial unit vector points outward; rotating it by a right angle gives the angular unit vector.

2. 02
   Describe the fixed eastward arrow

   Take $V=(1,0)$ in the Cartesian frame. Dot products with the two unit rulers give its moving-frame components. At $\theta=\pi/2$, east is the negative angular direction.

   $$V=\cos\theta\,\hat e_r-\sin\theta\,\hat e_\theta.$$

   Why this step works In an orthonormal frame, each component is the dot product with its unit basis vector.

3. 03
   Differentiate the basis as well as the numbers

   Direct differentiation gives $d\hat e_r/d\theta=\hat e_\theta$ and $d\hat e_\theta/d\theta=-\hat e_r$. The product rule now contains four terms.

   $$\frac{dV}{d\theta}=\underbrace{-\sin\theta\,\hat e_r-\cos\theta\,\hat e_\theta}_{\text{component change}}+\underbrace{\cos\theta\,\hat e_\theta+\sin\theta\,\hat e_r}_{\text{basis change}}=0.$$

   Why this step works The apparent component change cancels exactly against the change of the moving rulers.

4. 04
   Recognize the correction a connection must supply

   For any vector with unit-frame components $V^{\hat r}$ and $V^{\hat\theta}$, the corrected derivative includes the same basis terms. Our eastward arrow makes both corrected components zero.

   $$\left(\nabla_{\partial_\theta}V\right)^{\hat r}=\partial_\theta V^{\hat r}-V^{\hat\theta},\qquad\left(\nabla_{\partial_\theta}V\right)^{\hat\theta}=\partial_\theta V^{\hat\theta}+V^{\hat r}.$$

   Why this step works Collect the product-rule terms along each unit basis vector. This is the flat-plane connection written in a moving frame.

Go deeper

These are orthonormal-frame components, not polar coordinate components. The coordinate basis satisfies $\partial_r=\hat e_r$ and $\partial_\theta=r\hat e_\theta$ for $r>0$, so $V^r=V^{\hat r}$ but $V^\theta=V^{\hat\theta}/r$. This scale factor matters when comparing the formulas with Christoffel symbols. The calculation above is an exact product-rule proof in the Euclidean plane. On a general manifold, a connection supplies the comparison between neighboring tangent spaces; an ambient Cartesian frame need not exist. A nonzero connection coefficient can describe changing rulers in a completely flat geometry. Curvature asks whether consistent local comparisons fail to agree around a loop.

Further readingDavid Tong · Metrics, connections, and transport, §§3.1–3.3

A STATE YOU CAN CHECK

At an angle of 45°, the polar unit basis has rotated, but the Cartesian vector still points east.

Polar components

$(V^{\hat r},V^{\hat\theta})=(0.7071,-0.7071)$

Cartesian vector

$(V^x,V^y)=(1,0)$

The worked steps explain these measurements. Interactive controls appear when available.

Open the reference diagram

Test the idea

FIRST, PREDICT

The eastward arrow is examined at $\theta=\pi/2$. Which description is correct?

Its components are $(0,-1)$, proving that the vector field is turning.Its unit-frame components are $(0,-1)$ and its geometric derivative is zero.Its components are $(1,0)$ because the arrow stayed fixed.

Compare the reasoning

Its components are $(0,-1)$, proving that the vector field is turning.

A changing component list can come entirely from changing rulers. Include the basis derivative before concluding that the field changes.

Its unit-frame components are $(0,-1)$ and its geometric derivative is zero.

Exactly. At that position $\hat e_\theta$ points west, so the unchanged arrow has angular component $-1$. The changing basis cancels its component derivative.

Its components are $(1,0)$ because the arrow stayed fixed.

Those are its Cartesian components. The local radial-and-angular frame has rotated.

A hint

Evaluate both moving basis vectors at $\theta=\pi/2$, then express east using them.

NOW CHANGE THE EXAMPLE

At $r=3\,\mathrm m$ and $\theta=\pi/3$, a particle’s Cartesian velocity is $(0,6)\,\mathrm{m\,s^{-1}}$. What is the coordinate rate $d\theta/dt$?

A hint

Project onto $\hat e_\theta$ to obtain the physical angular velocity component, then divide by $r$.

Work through the solution

The unit-frame component is $V^{\hat\theta}=6\cos(\pi/3)=3\,\mathrm{m\,s^{-1}}$. The coordinate component is $V^\theta=V^{\hat\theta}/r=1\,\mathrm{s^{-1}}$, or one radian per second. Mixing these two components would lose a factor of $r$.

The covariant derivative combines changing components with the comparison of local bases so that an unchanged vector has zero derivative.

6.3 Why covectors acquire a minus sign#

A covector $\omega$ assigns a scalar to a vector $V$. In components,

A scalar has no moving basis indices to correct, so $\nabla_\mu f=\partial_\mu f$. We also require a derivative to obey the product rule and respect contractions:

Substitute the vector derivative. The final term contributes $+\omega_\nu\Gamma^\nu{}_{\mu\lambda}V^\lambda$. There is no corresponding connection term in the ordinary derivative of the scalar on the left. The covector derivative must cancel it, for every possible vector. Thus

The opposite signs keep differentiation of the scalar pairing consistent: the vector correction and covector correction cancel.

For a general tensor, every upper index gets a plus correction and every lower index gets a minus correction. For example,

Each correction replaces exactly one index with a summed index. Every term retains the same free indices $\mu,\alpha,\beta,\gamma$. Checking those free indices helps catch an incorrectly placed index.

Respecting contractions means, for example,

The two connection corrections cancel after a dummy-index relabeling. However, raising an index is an additional operation involving the metric. Commuting differentiation with raising and lowering requires metric compatibility, $\nabla g=0$, which we will impose and examine in §7.2. Compatibility with vector–covector contraction and compatibility with the metric are related ideas, but are not identical assumptions.

6.4 Taking a second derivative of a scalar#

For a scalar field $f$,

After one differentiation, the result is the covector $df$, with components $\partial_\nu f$. Its derivative therefore needs the covector correction:

This is the covariant Hessian. The metric connection derived in §5.5 has $\Gamma^\lambda{}_{\mu\nu}=\Gamma^\lambda{}_{\nu\mu}$, so this Hessian is symmetric in $\mu,\nu$. Section 7.2 will explain the name torsion-free for that symmetry in a coordinate basis. Ordinary second partial derivatives are symmetric too, but generally lack the correction needed to transform as a tensor.

There is one useful special case. At a critical point, where $\partial_\lambda f=0$, the correction vanishes. The Hessian computed there with ordinary second derivatives has a coordinate-independent meaning as a bilinear form. This is why classifying a stationary point as a maximum, minimum, or saddle can be done intrinsically despite using coordinate derivatives.

A related distinction: $df$, with components $\partial_\mu f$, is a covector defined without a metric. The gradient vector is

and does require a metric. In Euclidean space these objects are often merged into one mental image. For example, in Cartesian Minkowski coordinates, $(\operatorname{grad}f)^0=-\partial_0f$. The covector and gradient vector therefore need different component lists even in this simple frame.

6.5 Divergence and physical volume#

The divergence contracts the derivative index with the vector index. It extends the net-outflow calculation from Chapter 0 to general coordinates:

Use the Christoffel formula obtained from the free-particle action in §5.5. We will derive it again from geometric requirements in §7.3. In its trace $\Gamma^\mu{}_{\mu\nu}$, the first and third metric-derivative terms cancel after relabeling their summed indices. The remaining term gives

Here $g$ is the metric determinant, which is negative for our spacetime signature. The logarithmic derivative is shorthand for $(1/\sqrt{-g})\partial_\nu\sqrt{-g}$; equivalently one can take the logarithm of the ratio to a fixed reference with the same units.

To see the matrix identity behind this step, perturb an invertible matrix $A$ by a small amount $\epsilon B$:

Why does the trace appear? In the determinant of $I+\epsilon C$, a first-order contribution chooses one diagonal perturbation and leaves every other diagonal entry equal to one. Their sum is $\epsilon\sum_i C^i{}_i$. Terms involving off-diagonal entries require at least two perturbations. Thus the fractional change of $\det A$ is $\epsilon\operatorname{tr}(A^{-1}B)$ to first order. Taking its square root supplies the factor $1/2$ used above.

Combining the two terms by the product rule gives

Why should the determinant appear? A coordinate box of side lengths $dx^\mu$ represents physical four-volume $\sqrt{-g}\,d^4x$. A flow can have changing coordinate components merely because the coordinate boxes expand or shrink. Divergence measures net outflow per physical volume, so it must include that change in the measuring boxes.

In an $n$-dimensional Riemannian space, replace $\sqrt{-g}$ by $\sqrt{g}$; the general expression uses $\sqrt{|g|}$. The idea is the same.

We can now take the divergence of a gradient. In spacetime this combination is called the wave operator, written $\Box$ (read “box”):

In Cartesian Minkowski coordinates, the metric and its determinant are constant. The inverse metric supplies one minus sign; $x^0=ct$ gives $\partial_0=c^{-1}\partial_t$. Therefore

In the simplest scalar wave model, the field obeys the wave equation $\Box f=0$. As in Chapter 2, a scalar field assigns one coordinate-independent value to each event. We can see why this equation describes waves by constructing a solution. Choose a smooth profile $F$ and set $f(t,x)=F(x-ct)$, with no $y$ or $z$ dependence. The profile moves to the right at speed $c$: after time $\Delta t$, the same value of its argument occurs a distance $c\Delta t$ farther right. Differentiating gives

The primes here mean derivatives of $F$ with respect to its one argument. We have verified a traveling-wave solution. The curved-spacetime formula above includes the metric and volume factors needed to state the same scalar-field law in other geometries. Other kinds of field can have different equations; this is the scalar example.

A tensor with two upper indices has a second basis index to correct. Contracting one with the derivative index does not remove the other. Thus

The volume factor accounts for the contracted index $\mu$; the final connection term accounts for the remaining index $\nu$. We will need both terms when differentiating the stress-energy tensor in Chapter 11.

6.6 When the order of two flows matters#

A vector field can tell a point how to move: at every instant, follow the arrow at the point’s current location. The resulting motion is called the field’s flow. A velocity field uses time as its parameter; a mathematical flow can use another curve parameter.

Consider two instructions on a flat plane. The field $X$ moves a point horizontally at unit rate. The field $Y$ moves it vertically at rate $x$, so its arrows get longer farther to the right. Follow each instruction for a small parameter step $h$. The coordinates $x,y$ and the parameter $h$ are dimensionless in this example.

Starting from $(1,0)$, does X followed by Y reach the same place as Y followed by X? The Lie bracket—pronounced “lee”—measures the leading difference between these orders. First watch the two routes.

GEOMETRY LAB / 04

Try the other order.

Move right into a stronger upward flow. Now reverse the order of those instructions. The two dots start together; watch where they finish.

A · X then YB · Y then X

Faint arrows show the Y field. Farther right, the same flow step carries you farther upward.

Final vertical gap · h²0.3600

Gap divided by h²1.0000

Shrink h. The gap shrinks as h², but the gap divided by h² stays equal to one. This is the upward component of the Lie bracket.

Why the endpoints differ

Both dots start at $P=(1,0)$. X changes x at unit rate, leaving y fixed. Y changes y at rate x, leaving x fixed. Follow each field for the same parameter step $h$.

For A, X first takes the point to $(1+h,0)$. The Y step then raises it by $h(1+h)$. For B, the Y step happens while x is still one, so it raises the point only by $h$. X then changes only its horizontal position. The final vertical gap is

$$h(1+h)-h=h^2$$

The coordinates and h are dimensionless. This is a mathematical flow parameter, not a physical clock. The circles marked A and B indicate the final endpoints even during playback; filled dots track progress. Both paths use exact flow solutions. The faint field arrows use one shared scale.

The plane is flat. The gap records the fact that Y’s instruction changes with position. It is a difference between endpoints of two flows; the parallel-transport experiment compares directions after a vector follows a specified path.

For X then Y, the horizontal step first changes $x$ to $1+h$. The vertical flow is now stronger, and its step raises $y$ by $h(1+h)$. For Y then X, the vertical step happens while $x$ is still one, so it raises $y$ only by $h$. The endpoints are

Their difference is $(0,h^2)$. A smaller step makes the gap smaller, but dividing the gap by $h^2$ always gives the same upward vector $(0,1)$. This remaining vector is $[X,Y]$ for these fields.

Why does the square of the step appear? The first move changes where the second instruction is sampled. Over a small X step, the component $Y^i$ changes by $hX^j\partial_jY^i$ to first order. Following that changed field for another step $h$ contributes $h^2X^j\partial_jY^i$. Reversing the order gives $h^2Y^j\partial_jX^i$ instead. The other second-order terms occur in both routes and cancel. Thus, for smooth fields, the leading endpoint difference is $h^2[X,Y]$, where

The $i$ index labels the component of the resulting vector; $j$ is summed over the coordinate directions. In our example, $X=(1,0)$ and $Y=(0,x)$. Only $\partial_xY^y=1$ contributes, so $[X,Y]=(0,1)=\partial_y$.

We can also recognize this vector by its action on a scalar function. Recall from §4.3 that $X[f]$ means the directional derivative of $f$ along $X$. The ordinary product rule gives

The mixed second derivatives of the smooth function $f$ cancel. In this operator language, the bracket is a commutator: one composition minus the reversed composition. A bracket of zero means the flows agree through this leading comparison; constant coordinate directions such as $\partial_x$ and $\partial_y$ have zero bracket.

The plane in the experiment has remained flat throughout. The endpoint gap comes from position-dependent instructions. In §7.2, subtracting this effect will be essential to defining torsion without confusing it with the behavior of the chosen vector fields.

Check the distinction in a polar frame

On a regular polar patch with $r>0$, the coordinate fields $\partial_r$ and $\partial_\theta$ commute. Their components in their own coordinate basis are constant. The unit angular direction is instead $\hat e_\theta=r^{-1}\partial_\theta$, while $\hat e_r=\partial_r$. Its factor $1/r$ changes with position:

$$[\hat e_r,\hat e_\theta] =[\partial_r,r^{-1}\partial_\theta] =-r^{-2}\partial_\theta =-\frac1r\hat e_\theta.$$

The nonzero bracket belongs to this choice of unit frame on an ordinary flat plane. It does not indicate curvature or torsion. This is why a formula stated for a coordinate basis cannot always be copied unchanged into a moving unit frame.