Timeline for Finding turbulent velocity Fourier mode amplitudes from kinetic energy spectrum
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2015 at 17:30 | comment | added | honeste_vivere | I am not entirely sure what you are asking but I think the answer is yes. I just used subscripts for $x$ and $k$ as a matter of choice, but could have just as easily summed over $k$. | |
| Feb 5, 2015 at 13:57 | comment | added | Dipole | Sorry I mean dual as in the fourier transform pair $E(k)$ and $E(x)$. I get that $E(k_i)$ are the Fourier amplitudes of the function $E(x)$, but what I want are the Fourier amplitudes $\hat{u}(k)$ of the function $u(x)$, given knowledge of $E(k)$. You seem to be suggesting that $\hat{u}(k) = E(k)$? | |
| Feb 4, 2015 at 21:24 | comment | added | honeste_vivere | I am not sure what you mean by "dual" here, but I am only showing the relationships between Fourier transforms and illustrating that the amplitudes are independent of the exponent term. Thus, you already have the amplitudes in the form of $E(\kappa)$. | |
| Feb 4, 2015 at 16:50 | comment | added | Dipole | I'm not sure I'm following this- you are saying $E(x)$ is the dual of $E(k)$ and that therefore implies that since $u(x)$ is the dual of $u(k)$ then $u(k_i) = E(k_i)$? | |
| Feb 3, 2015 at 21:48 | history | answered | honeste_vivere | CC BY-SA 3.0 |