# Why does a transformation to a rotating reference frame NOT break temporal scale invariance?

Naively, I thought that transforming a scale invariant equation (such as the Navier-Stokes equations for example) to a rotating reference frame (for example the rotating earth) would break the temporal scaling symmetry due to the introduction of the period of rotation as an external time scale.

Now I was told that such a transformation to a rotating reference frame does not break the temporal scale invariance, its form would just look different (more complicated ?) in such a case. I dont understand why this is and what it means exactly, as explained in the first paragraph it seems counterintuitive to me ...

My naive idea to verify that the transformation to a rotating reference frame does indeed not spoil the temporal scaling symmetry would be to show that the corresponding transformations commute. I mean I would start with

1) writing down the analogue of the equation

(1) $\qquad \exp(-ia^{\mu}\hat{P}_{\mu})\phi(x) = \phi(x+\delta x)$

for a translation in space(time) for both transformations,

2) Taylor expand the r.h.s of (1),

3) insert the coordinate transformations that correspond to the scaling transformations and the transformation to a rotating reference frame,

4) and approximate the exponential on the l.h.s of (1) to first order.

5) Comparing the l.h.s with the r.h.s should then provide the differential operators for the two transformations such that the commutator can be calculated.

My question now is:

Is testing if the two transformations commute the right procedure to show that a transformation to a rotating reference frame does not break temporal scale invariance? If this is not the case how can it be shown otherwise (mathematically or intuitively)?

(In addition to "direct help" I would appreciate being pointed to some literature that deals with this issue too)

• Yes, commutation relations would do just fine. Let $T$ be a temporal scale transform, and let $R$ be a trasformation to a rotating reference frame. Then what you want to show is that if my equation is initially invariant under $T$, that subsequently after $R$, it is still invariant under $T$. Similarly, if I change frames, and apply $T$, do I get the same thing if I apply $T$, and change frames. – kηives Aug 26 '12 at 23:11
• What happens when you carry out the procedure you've described? – kleingordon Aug 27 '12 at 0:43
• @kleingordon I did not yet try it since I was not sure if the procedure is applicable in this case too. But now I am encouraged to give it a try. – Dilaton Aug 27 '12 at 8:52

## 1 Answer

My answer will focus just on the mathematical parts pertaining to partial differential equations.

Scale invariance is the fact that some partial differential equations stay the same if you appropriately scale the variables.

For example the heat equation (where $\boldsymbol{x}$ is the position vector in 1, 2 or 3D, doesn't matter) $$\partial_t u(\boldsymbol{x},t) = \Delta u(\boldsymbol{x},t),$$ is scale invariant in the following way (I basically rip off an article of Terence Tao). If you have a solution $u$ to the heat equation, then $u^{(\lambda)}$, defined as $$u^{(\lambda)}(\lambda \boldsymbol{x}, \lambda^2 t) := u(\boldsymbol{x}, t),$$ is also a solution, because $$\lambda^2 u^{(\lambda)} (\lambda \boldsymbol{x}, \lambda^2 t) = \lambda^2 \Delta u^{(\lambda)} (\lambda \boldsymbol{x}, \lambda^2 t),$$ which is nothing else but $$\lambda^2 \partial_t u(\boldsymbol{x},t) = \lambda^2\Delta u(\boldsymbol{x},t),$$ where we can simplify the $\lambda^2$ factors to get our original equation.

The case of the wave equation reveals a different scaling. We have $$\partial_t^2 u(\boldsymbol{x},t) = \Delta u(\boldsymbol{x},t),$$ where, if we have a solution $u$, then the scaling which gives us other solutions is $$u^{(\lambda)}(\lambda \boldsymbol{x},\lambda t) := u(\boldsymbol{x},t).$$

Finally, the Navier-Stokes equation, without external forces and with all physical constants set to one, $$\partial_t \boldsymbol{u}(\boldsymbol{x},t) + (\boldsymbol{u}(\boldsymbol{x},t)\cdot\nabla)\boldsymbol{u}(\boldsymbol{x},t) +\nabla p(\boldsymbol{x},t) -\Delta\boldsymbol{u}(\boldsymbol{x},t)=0,\\ \nabla \cdot \boldsymbol{u} = 0,$$ scales as $$\lambda \boldsymbol{u}^{(\lambda)}(\lambda \boldsymbol{x},\lambda^2 t) := \boldsymbol{u}(\boldsymbol{x},t),\\ \lambda^2 p(\lambda \boldsymbol{x}, \lambda^2 t) := p(\boldsymbol{x},t).$$ Note that here the functions themselves need to be amplified as well (not only the space and time variables). This is important for the reasons given in Tao's article, but that's another topic.

The second and third examples are easy to check out for yourself: each derivative "generates" an appropriate power of $\lambda$, and then you just have to verify that each term indeed ends up with the same power of $\lambda$ in the end. In the Navier-Stokes example, all terms end up with a $\lambda^3$.

Now, why does a rotating reference frame NOT break this scale invariance? Say we have a reference frame $\mathfrak{R}$ with coordinates $(\boldsymbol{x},t)$. We shall define a rotating reference frame $\mathfrak{R}'$ rotating about the $z$-axis at a constant angular velocity $\Omega$. The new coordinates are given in terms of the old ones by $$x' = \cos{(\Omega t)} x + \sin{(\Omega t)} y\\ y' = -\sin{(\Omega t)} x + \cos{(\Omega t)} y\\ z' = z \\ t' = t$$

The partial derivatives transform like so, using the chain rule of differentiation, $$\partial_t u(x'(x,y,t),y'(x,y,t),z',t') = \partial_{x'} u(x',y',z',t') \cdot \frac{\partial x'}{\partial t} + \partial_{y'} u(x',y',z',t') \cdot \frac{\partial y'}{\partial t} + \partial_{t'} u(x',y',z',t') \cdot \frac{\partial t'}{\partial t}\\ = \Omega y' \partial_{x'} u(x',y',z',t') -\Omega x' \partial_{y'}u(x',y',z',t') + \partial_{t'} u(x',y',z',t'),$$ and so on. In the case of the heat equation in 2D, we get $$\partial_{t'} u(x',y',t') + \Omega y' \partial_{x'} u(x',y',t') -\Omega x' \partial_{y'}u(x',y',t') = \Delta' u(x',y',t').$$ The Laplacian is indeed invariant under rotation. Applying the appropriate scaling yields $$\lambda^2 \partial_{t'} u^{(\lambda)}(\lambda \boldsymbol{x}',\lambda^2 t') + \Omega (\lambda y') (\lambda \partial_{x'}) u^{(\lambda)}(\lambda \boldsymbol{x}',\lambda^2 t') - \ldots$$ and so on, and you get the $\lambda^2$ factor in front of each term. The equation has new terms due to the rotating reference frame, but in fact, it is still scale-invariant.

I'm not sure I'm explaining this quite right, but I'll have a go anyway, please, anyone, feel free to correct/complete my explanation. The transformation from $\mathfrak{R}$ to $\mathfrak{R}'$ doesn't break scale-invariance because it preserves the structure of each of the terms appearing in the equation in terms of counting the powers of $\lambda$. For example, $\partial_{t'} u$ will generate a $\lambda^2$. This is also the case for $\Omega y' \partial_{x'} u$, because $\Omega$ is just a number and $y'$ gets expanded to $\lambda y'$ and $\partial_{x'}$ "throws out" the necessary $\lambda$ to get to $\lambda^2$, etc.

The non-linear term in the Navier-Stokes equation gets quite ugly, but in the end, it's just jumbled-up partial derivatives in the space dimensions, but they always remain the "same" with respect to the power-counting.

I hope this helps!

• Thanks Christoph for this nice answer. I guess that this checking of the structure of each term after the rotation as you have shown by these examples should work for every equation which is scale invariant, such that one can say that rotation and scaling commute generally? – Dilaton Apr 17 '13 at 13:39
• Hi, I cannot help you regarding the commutation approach (isn't that quantum mechanics stuff? scale invariance exists in classical physics too), I've no idea how to treat that. As far as I can tell, it's just a question of making sure the change of reference frame doesn't induce weird terms. – Christoph B. Apr 17 '13 at 14:34
• It is ok, I just thought there should be a general way of profing that transformaton to a rotating reference frame does not break scale invariance once and for all, such that one does not need to check each term in the equation each time. Since I am specifically interested in the NS equations, your nice answer is good enough for me now :) – Dilaton Apr 18 '13 at 15:26