0
$\begingroup$

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

Improve this question
$\endgroup$
2
  • $\begingroup$ 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
    Aug 15, 2020 at 12:30
  • $\begingroup$ Yes, it worked. the factor comes out with $d=3$ when coordinates are rescaled, thanks $\endgroup$
    – cmdfrills
    Aug 30, 2020 at 23:10

0

Reset to default

Your Answer

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