# Rotation of scalar

I came across this problem in Intro to ED by Griffiths: "In two dimensions, show that the divergence transforms as a scalar under rotation". Now, I was able to prove that this statement is true, but something bothered me. Intuitively, I know that a scalar does not change under rotation, but how can this be showed rigorously? We cannot have the rotation matrix operate on a scalar, because that is not defined. Is it from this fact that it is not defined that we say a scalar is not changed by 2D-rotation? Or does it have to due with the magnitude of a vector not changing under rotation?

I know this is more of a math question and I have already asked this on math.stackexchange, but it seems to have been overlooked.

• Cross-posted to Mathematics: math.stackexchange.com/q/2903119/166791 – rob Sep 3 '18 at 2:52
• This should help physics.stackexchange.com/questions/155878/… – InertialObserver Sep 3 '18 at 3:59
• This is Problem 1.17 in $\:\boldsymbol{\S}\:$ 1.2.4 The Divergence in both 3rd and 4th Edition. In the footnote (4) under the $\:\boldsymbol{\S}\:$ 1.1.5 How Vectors Transform : A scalar does not change when you change coordinates. In particular, the components of a vector are not scalars, but the magnitude is. So, in the textbook the term scalar refers to the invariant scalar quantities not to any scalar. – Frobenius Sep 3 '18 at 6:46
• The divergence of a vector function is a scalar function, the output of the application of the known differential operator $\:\boldsymbol{\nabla\cdot}\:$. You are confused. We DO NOT rotate the divergence, we DO rotate the vector function and the problem is to prove that the divergence is an invariant scalar under rotations of this vector function. – Frobenius Sep 3 '18 at 13:03
• Interesting question. I think it was Grassmann that thought of vectors as how they behaved under rotations. It's also useful as physics textbooks, in my experience, aren't clear enough by quite what they mean by this. – Mozibur Ullah Sep 3 '18 at 22:28

I actually believe that it can be proven.

Proof: Let $R$ be an element of $SO(n)$. So, in 3D this is just the usual rotation operator. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, we may write $q = q e^1_ie^1_j \delta^{ij}$. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q.$$

Hence, we have used the fact that the kronecker delta transforms as a scalar under rotations to show that if $q$ is a real number then it transforms as a scalar under rotations.

• +1: great answer. – Mozibur Ullah Sep 3 '18 at 22:29

I know the other answers are quite good, just want to point out two things:

What we have is that a scalar is precisely "something invariant under rotations". Otherwise, how do you tell if something is a scalar or not?

Okay, they tell you, but suppose you're not seeing it written, you have to decide yourself. Is pressure a vector or not? Before you choose the notation, how do you decide if something must have an arrow upside? Just rotate your reference frame and check how it changes.

IF the coordinates change, it is a higher order tensor (Velocity, dielectric tensor, whatever). On the other hand, if it keeps its value (pressure, temperature, volume...) then it's a scalar.

Then, you say that

We cannot have the rotation matrix operate on a scalar

We can, but that matrix is $1\times1$, for obvious reasons (it must act on a number) and its value is 1, because the numebr "1" is the only $1\times 1$ operator that preserves the norm. So, trivially

$$x'=Rx=1\cdot x$$

I think it can't be showed rigorously, because a scalar is defined to be invariant under rotation. See https://en.wikipedia.org/wiki/Scalar_(physics) or chapter 1 of "Classical Dynamics of Particles and Systems" by Marion, Thornton.

• Yes, in fact, there's a proof in the book I mentioned above. – Alberto Navarro Sep 3 '18 at 20:41