When I encounter the parity transformation in physics,I feel often it's not really clear what we are doing and I want to understand it in a more rigorous way, can you help me with that?(I want to see it from the point of view of tensors.)I always hear that the electric field changes it's sign, often denoted as


I assume this means the following: there is a vector field $\vec{E}(\vec{r})$ in a euclidian vector-space, $\vec{r}$ is the position vector (both are independent of a coordiante system-> invariant tensors) and we have chosen a cartesian coordinate system, where the {x,y,z} coordinates are measured to be increasing in a certain direction, which gives rise to a local (covariant) basis in every point in space according to

$$\vec{e}_i(x,y,z)=\frac{\partial}{\partial x^i}\vec{r}(x,y,z)\tag{2}$$

(This equation will of course also work in curvilienar coordinate systems, then the basis will change from one point to the other, whereas in affine coordinates like the cartesian they will be constant)

Now the parity transformation is done by what is usually denoted as (x,y,z) -> (-x,-y,-z). I think this means that we now create a second coordinate system, where the {x',y',z'} coordinates are measured as increasing in exactly the opposite direction compared to the unprimed system.This will cause the basis vectors (who are covariant vectors) of the primed system to point in the opposite direction in every point.Now we can decompose the invariant E-field vectors in every point with respect to either the unprimed or the primed basis and compare their contravariant components. And since for every vector $\vec{V}$ we have

$$\vec{V}=V^i \vec{e}_i=V'^i \vec{e}'_i\tag{3}$$

this means that the components have the relationship


(the co- and contravariant tensors transform the same way in this case)

My first question would be if this is what physicists mean when they do a parity transformation and say that the E-field changes it's sign? (I think in this case it would be true for any configuration of charges, but let's consider a point charge at the origin).

But then if we didn't start with a cartesian system but with spherical coordinates, and perform a parity transformation, eq. (4) is still supposed to hold (at least if I have understood correctly what wikipedia and other sources mean when they speak about the sign flip, so please correct me if I'm wrong), which implies, because of how (2) relates the coordinate lines to the basis vectors, that in the primed system, the radius is now measured to be increasing towards the origin, which is not true.(Or is it?)

My second question is: how do the coordinate lines of the spherical system look like (how do the primed axes compare to the unprimed?), after the parity transformation, and how can one say that the E-field changes its sign under parity transformation, without having to measure the radius as increasing towards the origin? I mean that r is zero at the origin but becomes negative as we go outward, because this is required to make the $\vec{e}_r$ vectors point to the origin according to (2).

(For a point charge, the only nonzero $E^i$ will be the componets of the $\vec{e}_r$ basis vectors so only the orientation of the radius coordinate matters, doesn't it?)

In the literature I only encounter the case that the angles of the spherical coordinate system are transformed under parity and not the radius and imo this would not result in a sign flip of the E-field.


1 Answer 1


It turns out that a vector field $\vec F(x, y, z)$ is best thought of as a local directional derivative which we'd write $\mathsf F= \vec F \cdot \nabla$, taking scalar fields $f(x,y,z)$ and producing new scalar fields $g(x,y,z)$ from them. When we start from this place of$$g(x,y,z) = \mathsf F[f](x,y,z) = F_x\frac{\partial f}{\partial x} + F_y\frac{\partial f}{\partial y} + F_z\frac{\partial f}{\partial z}$$ for scalar fields $F_{x,y,z}(x, y, z)$, and we define that $\hat g(x, y, z) = g(-x, -y, -z)$ and $\hat f(x, y, z) = f(-x, -y, -z)$, as the parity transform asks us to perform, we will find that the chain rule tells us that $\partial_x \hat f = -\partial_x f,$ and we get one big minus sign on the right hand side. We can correct for this if we define that $\hat F_{\alpha}(x, y, z) = -F_\alpha(-x, -y, -z).$ So there is both a minus sign "inside" and "outside" the parentheses; the inner one comes from the fact that $\vec F$ needs to be using parity-transformed inputs and the outer one comes from the fact that this reflection actually changed the direction of $\vec F$ itself.

In spherical coordinates (I will take $0\le \theta < 2\pi$ as azimuthal while $0\le \varphi\le\pi$ comes down from the North Pole) both of these ideas get a little complicated; we instead have $$g(r,\theta,\varphi) = \mathsf F[f](r, \theta, \varphi) = F_r \frac{\partial f}{\partial r} + F_\theta \frac1{r\sin\varphi} \frac{\partial f}{\partial \theta} + F_\varphi\frac1{r}\frac{\partial f}{\partial\varphi}.$$Our parity transform needs to map latitudes to opposite latitudes $\varphi\mapsto\pi-\varphi$ and needs to also rotate a point 180 degrees about the globe, $\theta\mapsto\theta + \pi.$ Under this mapping actually $\sin\varphi\mapsto\sin\varphi,$ so there's a sort of "double negative" in this $\sin$ term and we therefore find that: $$\begin{array}{ll} F_r(r,\theta,\varphi) &\mapsto~ {+F_r(r, \theta + \pi, \pi - \varphi)},\\ F_\theta(r,\theta,\varphi) &\mapsto~ {+F_\theta(r, \theta + \pi, \pi - \varphi)},\\ F_\varphi(r,\theta,\varphi) &\mapsto~ {-F_\varphi(r, \theta + \pi, \pi - \varphi)}.\end{array}$$Another way to interpret this is to just think about the unit vectors at a given point in space; remember that when we're using curvilinear coordinates these unit vectors vary over space itself, so unlike $\hat x$, the unit vector in the $x$-direction which is the same at all points in a Cartesian space, the unit vector $\hat r$ changes to point away from the origin at all points in space. In fact one can work out that these unit vectors are:$$\begin{array}{ll} \hat r &=~ {+\hat z}~\cos\varphi + \hat x~ \sin\varphi \cos\theta + \hat y\sin\varphi\sin\theta\\ \hat \theta &=~ {-\hat x}~\sin\theta + \hat y~ \cos\theta\\ \hat \varphi &=~ {-\hat z}~\sin\varphi + \hat x~ \cos\varphi \cos\theta + \hat y~\cos\varphi\sin\theta \end{array}$$

When we reflect space our starting $\hat r$ gets mapped to the reflected $\hat r$; they both are unit vectors pointing away from the origin. Similarly our starting $\hat \theta$ gets mapped to the reflected $\hat\theta$; they both are unit vectors pointing "West". But when we reflect $\hat \varphi$, which points "South" everywhere on the sphere, we get that the reflected vector points "North", which is opposite of its new local-unit-vector. So that component, and only that component, must pick up the minus sign.

  • $\begingroup$ thanks for the reply but i still havn't got it. First i don't see the point in this scalar representation of the vector field, could you please explain why you do this? Isn't the E-vectorfield a sensible thing to work with? Then you write that φ↦π−φ and θ↦θ+π. I see that this will bring me from one point in euclidian space to the reflected one through the origin and I have seen this change of parameters before, but i dont get why this will result in the E-field getting a minus sign, since now we are still measuring the radius to be increasing away from the origin and with (2) this means ... $\endgroup$
    – curio
    Commented Sep 1, 2017 at 18:05
  • $\begingroup$ ... that the covariant radius basis vector of our new coordinate system will at any point in space point into the same direction as the radius basis vector of the unprimed system and so the decomposition of the E-field will in both systems yield the same value since E is always parallel to the radius basis vector. Which implies that the E-field is invariant under parity which is not what seems to be the consensus? $\endgroup$
    – curio
    Commented Sep 1, 2017 at 18:10
  • $\begingroup$ is it even true to think of the parity transform as a transformation of my coordinate system? in that case we will have different coordinate lines and basis vectors. How do they look? And are we comparing the contravariant components of the E-field before and after the parity transformation when we say that the E-field gets a minus sign? $\endgroup$
    – curio
    Commented Sep 1, 2017 at 18:14
  • $\begingroup$ (a) The best reason that this is the best way to think about vector fields is that it makes sense in curved spaces: that is maybe not what you want. Here's an intuitive idea: two close points in space $(x,y,z)$ and $(x+\delta x,y+\delta y,z+\delta z)$ should define The Very Concept of Vector at that point, via this displacement vector $[\delta x;\delta y;\delta z]$; our definition of vector is "anything that transforms like this vector." And that maybe helps also to see why vectors must invert $\vec E\mapsto -\vec E$ under the 3D parity transform. $\endgroup$
    – CR Drost
    Commented Sep 1, 2017 at 19:18
  • $\begingroup$ (b) Now there are many $\vec E$ fields and not all of them are going to be invariant under the parity transform; I think you're thinking specifically of the electric field of a point charge at the origin. But put that point charge anywhere else and you will not have parity symmetry. Heck, imagine that you're inside of a huge capacitor $\vec E = E_0~\hat z$ and this E-field inverts under the parity transform, too. $\endgroup$
    – CR Drost
    Commented Sep 1, 2017 at 19:21

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