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!
This post imported from StackExchange Physics at 2014-03-12 15:46 (UCT), posted by SE-user Christoph B.
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