I'm trying to reproduce the work in computing turbulent energy spectrum from isotropic turbulence flow field in a box but I have a hard time understanding James' answer about the binning part.

Would you be so kind as to try to explain it to me ? Many thanks !


1 Answer 1


After a shallow read, I guess that the main point is that you have a limit.

When you expand your velocity ${\bf{u}}(x,y)$ (2D for simplicity) in a Fourier series you will have modes like: $$\exp{(ik_{xn}x)} \qquad\exp{(ik_{ym}y)}$$ Therefore you can write in discrete case with $N+1$ points: $${\bf u}(x,y) = \sum_{nm}{\hat{\bf{u}}_{nm}(k_{nx},k_{my},t)\exp{(ik_{xn}x)} \exp{(ik_{ym}y)}}\qquad n,m \in[-N/2,N/2]$$

Where $\hat{\bf{u}}_{nm}(k_{nx},k_{my},t)$ are the Fourier coefficients that describe $u$ in Fourier space given by wavenumbers $k_{xn}$ and $k_{ym}$ is: $$k_{xn} =\frac{2\pi n}{L}=k_0n \qquad k_{ym} = k_0m$$ The total energy $E_T(t)$ can be computed as: $$E_T(t) = \frac{k_0^2}{2}\sum_{n,m}{\hat{\bf{u}}_{nm}(k_{nx},k_{my},t)\cdot \hat{\bf{u}}^*_{nm}(k_{nx},k_{my},t)}$$

This energy can also be computed with the energy spectrum $E(k,t)$ with $k^2=k_{xn}^2+k_{ym}^2$

$$E_{T}(t) = \sum_{k}E(k,t)\Delta k$$ where $\Delta k$ must be defined with care and $E(k,t)$ is $$E(k,t) = \frac{1}{2}\sum_{k_{xn}^2+k_{ym}^2=k^2}{\hat{\bf{u}}_{nm}(k_{nx},k_{my},t)\cdot \hat{\bf{u}}^*_{nm}(k_{nx},k_{my},t)2\pi k }$$

The last sum is related with the discrete version of a line integral. It can be obtained from what he refers with "binning".

It is clear that values that do not meet with the restriction $k_{xn}^2+k_{ym}^2=k^2$ are discarded ( red zone in James' answer) and therefore you cannot reach values $k_{xN/2}$ or $k_{yN/2}$ and other points that are ourside the circle $k_{xn}^2+k_{ym}^2=k^2$. It can be also shown in the $3D$ case with the sphere: $k_{xn}^2+k_{ym}^2+k_{zl}^2=k^2$.

I am not an expert on this area... I would be grateful if someone sees necessary to correct me in aspects that are not correct.

  • $\begingroup$ Thank you for taking time to answer @HBR but are you sure that the red zone in James' answer matches the values that do not meet the restriction $k^2_{xn}+k^2_{ym}=k^2$ ? $\endgroup$
    – PandaMat
    Jul 31, 2017 at 7:09
  • $\begingroup$ Yes. I am sure. $\endgroup$
    – HBR
    Jul 31, 2017 at 7:11

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