# Restriction on vector fields

The 2D vector field (x,-y) does not transform like a vector under rotation(Arfken Vol. 1)! Does this mean we cannot have such a vector field physically?

-
What do you mean it doesn't transform like a vector? A vector transforms like a vector by definition. Details please. – Michael Brown Feb 22 '13 at 6:53
$V'_x=V_x cos \phi+V_y sin \phi$ , $V'_y=-V_x sin \phi+V_y cos \phi$ according to this the transformd vector is $(x cos \phi-y sin \phi, -x sin \phi-y cos \phi)$ which is not $(x',-y')$ as it should be – richard Feb 22 '13 at 6:57
In addition to rotating the vectors you also have to rotate the coordinates: $V'(\vec{r}')= \hat{R} V(\vec{r})$, where $\vec{r}' = R \vec{r}$. – Michael Brown Feb 22 '13 at 7:05
yes if you do that you will not get $(x',-y')$ – richard Feb 22 '13 at 7:12
That's okay. It's not a rotationally invariant vector field. There is no reason in general that you should get back the same thing. – Michael Brown Feb 22 '13 at 7:25

I'm going to address the important concepts at play here in three dimensions.

The issue here is to get straight the distinction between any function $\mathbf v:\mathbb R^3\to\mathbb R^3$, which we'll call a vector field, and an object that in addition to being a vector field in this sense, transforms in some prescribed way. To mathematically describe/formalize the latter sort of object, let me first introduce the following definition:

For any vector field $\mathbf v$, and for any rotation (special orthogonal transformation) $R$, we define a rotated vector field $\mathbf v^R$ by $$\mathbf v^R(\mathbf x) = R \mathbf v(R^{-1} \mathbf x)$$ This definition is precisely the mathematical formalization of what most people think of when they think of rotating a vector field.

We now define a 3-vector field as a function $\mathbf f(\mathbf v_1, \dots, \mathbf v_n):\mathbb R^3\to\mathbb R^3$ depending on a finite number of vector fields $\mathbf v_i$ such that for any rotation $R$, $$\mathbf f(\mathbf v_1^R, \dots, \mathbf v_n^R) = \mathbf f(\mathbf v_1, \dots, \mathbf v_n)^R$$

The idea here is that a 3-vector field is a vector field constructed out of a bunch of other vector fields such that if each of the constituent vector fields out of which it is built are rotated, then that is equivalent to just rotating the 3-vector field itself.

As an example, consider the function $\mathbf f(\mathbf{v})$ defined by $$\mathbf f(\mathbf{v}) = \nabla\times\mathbf v$$ In other words, $\mathbf f(\mathbf{v})$ is just the curl vector field. Is it a 3-vector field according to the definition above? Let's check: \begin{align} f(\mathbf{v}^R)(\mathbf x) &= \nabla\times \mathbf v^R(\mathbf x) \\ &=\epsilon_{ijk}\partial_j(R\mathbf v(R^{-1}\mathbf x))_k \mathbf e_i \\ &= \epsilon_{ijk}\partial_j(R_{kn} v_n(R^{-1}\mathbf x)) \mathbf e_i \\ &= \epsilon_{ijk}R_{kn}\partial_j(v_n(R^{-1}\mathbf x)) \mathbf e_i \\ &=\epsilon_{ijk}R_{kn}(\partial_m v_n)(R^{-1}\mathbf x)\partial_j(R^{-1}\mathbf x)_m \mathbf e_i\\ &= \epsilon_{ijk}R_{kn}(\partial_m v_n)(R^{-1}\mathbf x)\partial_j(R^{-1}_{m\ell} x_\ell) \mathbf e_i\\ &= \epsilon_{ijk}R_{kn}(\partial_m v_n)(R^{-1}\mathbf x) R_{jm}\mathbf e_i \\ &= R_{i\ell}\epsilon_{\ell m n}(\partial_m v_n)(R^{-1}\mathbf x)\mathbf e_i \\ &= R(\nabla\times \mathbf v)(R^{-1}\mathbf x) \\ &= f(\mathbf v)^R(\mathbf x) \end{align} or in summary $$\mathbf f(\mathbf v^R) = \mathbf f(\mathbf v)^R$$ so, indeed, the curl of a vector field is a 3-vector field! Now, let's return to your example. Let's take a 3D analog and define a function $\mathbf f(\mathbf v)$ by $$\mathbf f(\mathbf v)(\mathbf x) = (v_1(\mathbf x), -v_2(\mathbf x), v_3(\mathbf x))$$ Is this a 3-vector field? Namely, if we invert only one of the coordinates, then is the resulting vector field a 3-vector field according to our definition? I claim that no, it is not. To see this, consider the rotation $R$ that leaves the $z$ axis fixed, but that rotates in the $x$-$y$ plane by $\pi/2$ clockwise. I'll leave it to you to check that $$\mathbf f(\mathbf v^R)(x,y,z) = (-v_2(y,-x,z),-v_1(y, -x, z), v_3(y, -x, z))$$ while $$\mathbf f(\mathbf v)^R(x,y,z) = (v_2(y,-x,z),v_1(y, -x, z), v_3(y, -x, z))$$ so that for this rotation $R$, $$\mathbf f(\mathbf v_R)\neq \mathbf f(\mathbf v)^R$$ and therefore this particular $\mathbf f(\mathbf v)$ is not a 3-vector!

I think this is rigorously the meaning of what Arfken is trying to say.

Physical Interpretations

You might ask why the definition of a 3-vector field given above is useful in physics. Well here's the main idea. Let's say that we measure a certain physical vector field using some apparatus, like the velocity field $\mathbf v$ on the surface of a lake for instance. Then given this velocity field, one could compute the curl of this field $\mathrm{curl}(\mathbf v)$. Now suppose that we were to rotate our measuring apparatus, then we would measure the rotated velocity field $\mathbf v^R$. We could now calculate the curl $\mathrm{curl}(\mathbf v^R)$ of this rotated vector field, but because, as we showed above, the curl of a vector field is a 3-vector field, we could just as easily compute the rotated curl $\mathrm{curl} (\mathbf v)^R$, and we would get the same answer. So in some sense, a 3-vector can be interpreted as a computed physical quantity which rotates in the same way as measured physical quantities.

-
I've never understood why people leave out translations when defining a 3-vector field. – Larry Harson Feb 24 '13 at 1:08
@LarryHarson Hmmm yeah interesting observation. I've never really thought about that. I see no reason why we couldn't include translations in the definition in principle. Except I suppose that if we did, then the curl, for example, would not be a 3-vector field since the derivative would kill the translation. Actually I think that might be a significant point. I'll think about it and see if I come to any other conclusions. – joshphysics Feb 24 '13 at 1:24
@joshphysics I think this is good. Seperating the transformations of direction and coordinates is confusing. – Michael Brown Feb 28 '13 at 7:32
@LarryHarson I think translations are treated seperately is that it is trivial to do so and doesn't add anything: to translate a vector field you just translate the arguments. There is nothing specifically "vectorial" about that. The Euclidean group (of translations + rotations) is a semi-direct product of the translation and rotation groups: $E_3 \sim \mathbb{R}_{3}\rtimes\mathrm{SO}\left(3\right)$. So once you understand the representations of the rotation group $\mathrm{SO}\left(3\right)$ you immediately get reps of the Euclidean group for free. This is Wigner's "little group" idea. – Michael Brown Feb 28 '13 at 7:45

Transforming the positions and rotating the vector field are two different things.

Let $p$ be a point--for instance, $p = x e_1 + y e_2$, with $e_1, e_2$ being the usual Cartesian basis vectors. We can define a transformation $f(p) = p' = x' e_1 + y' e_2$.

Now then, let $A(p)$ be a vector field--in our case, $A(p) = x e_1 - y e_2$. As a vector field, $A$ can be considered as the derivative of some curve through the point $p$. Let $c(\tau,p)$ be such a curve. Then $A = \partial c/\partial \tau$ for some curve $c$. There are necessarily different curves through each point $p$, but such curves always exist.

It is this construction that is crucial to understanding the transformation properties of vector fields. Let us now consider $c' = f \circ c$, which is a family of curves in the primed space. The chain rule then tells us that

$$\frac{\partial c'}{\partial \tau} \Big|_{p'} = \frac{\partial c}{\partial \tau} \cdot \nabla f \Big|_p$$

The object $a \cdot \nabla f$ for any vector $a$ is special--we call it the "differential" of $f$, or the "Jacobian". Denote this as $\underline f(a)$, and we get

$$\frac{\partial c'}{\partial \tau} \Bigg|_{p'} = \underline f \left[\frac{\partial c}{\partial \tau} \right]_{p}$$

Or, more concisely,

$$A'(p') \equiv \underline f(A[p])$$

Linear transformations are special--they obey $f = \underline f$. This means that $A'$ is necessarily just the rotation of $A$, but evaluated at the rotated point also.

Let's work out what this means for your vector field. First work out the rotation:

$$\underline f(e_1) = e_1 \cos \theta + e_2 \sin \theta \\ \underline f(e_2) = e_2 \cos \theta - e_1 \sin \theta$$

Now work through the vector field: \begin{align*} A(p) &= x e_1 - y e_2 \\ \underline f(A[p]) &= (x \cos \theta + y \sin \theta) e_1 + (x \sin \theta - y \cos \theta) e_2 \end{align*}

We need to convert this into the primed coordinates. See that

$$x = x' \cos \theta + y' \sin \theta \\ y = y' \cos \theta - x' \sin \theta$$

Let's look at the first component.

$$x \cos \theta + y \sin \theta = x' \cos^2 \theta +2 y' \sin \theta \cos \theta - x' \sin^2 \theta = x' \cos 2\theta + y' \sin 2\theta$$

Similar logic applies for the other component, yielding

$$A'(p') = (x' \cos 2 \theta + y' \sin 2 \theta) e_1 + (x' \sin 2\theta - y' \cos 2\theta)e_2$$

You may be skeptical that this is the correct transformation of the vector field, but I assure you it is. To check, pick the point $p = e_1$ and a vector $v = e_2$. We know that $A(e_1) = e_1$, so that $v \cdot A = 0$. If we transform both vectors according to $\underline f$, we should get $\underline f(v) \cdot \underline f(A) = 0$ as well--after all, a rotation should not change angles, should not change orthogonality.

Of course, it would be onerous to find the coordinates of $p = e_1$ in the primed frame. It's much easier to see that

$$A'(p') = e_1 \cos \theta + e_2 \sin \theta$$

Similarly, $v' = \underline f(v) = e_2 \cos \theta - e_1 \sin \theta$. This does indeed satisfy that $v' \cdot A' = 0$, as required.

In short, you cannot merely rotate the vector itself. There are two rotations involved: one of the underlying positions, and then one of the vector field. You should not expect that $A' = x' e_1 - y' e_2$. This is not founded in the transformation law. Perhaps this is what Arfken meant--that you can't be naive, expecting that even a simple rotation will preserve components of vectors and their relations to positions. Once you derive the transformation law for vectors, though, it becomes a bit meaningless to say something is or is not a vector field. It's easy enough to impose this transformation law. Perhaps Arfken meant to say that, given $A$ and $A'$ as prescribed in your question, it's clear that is inconsistent with the vector field transformation law. Really, though, this strikes me as quite unclear.

-