I have read that the incompressible Navier Stokes equation is preserved by the scaling

$$x',y',z'=\lambda x, \lambda y, \lambda z$$ $$t'=\lambda^2 t$$ $$u'=(1/\lambda) u$$

As I understand it, fluid energy is given by

$$\int u^2 dv$$

I am trying to understand what is meant by the claim that the fluid energy is invariant under the scaling $\frac{1}{\lambda^{d/2}}u(\frac{x}{\lambda},\frac{t}{\lambda^2})$ where $d$ is the dimension of space? Could someone explain/derive this scaling invariance of the energy

  • $\begingroup$ Your question is why the integral is invariant under this rescaling? Did you try to see what happens when you perform a rescaling of the coordinates? $\endgroup$
    – haelewiin
    Commented Aug 15, 2020 at 12:30
  • $\begingroup$ Yes, it worked. the factor comes out with $d=3$ when coordinates are rescaled, thanks $\endgroup$
    – cmdfrills
    Commented Aug 30, 2020 at 23:10


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.