# Tag Info

0

Ok I think I solved the problem. So to divide the density FFT by k^2 i need actual values of k in k-space for my system. FFTW orders the result of transformation in so called "in-order" output, that means in first quadrant of FFT the first pixel corresponds to DC frequency and next to k/L frequency (k is from 0 to N-1) where L is length of whole system. ...

0

Substitute $$\varphi(x,y) = \int dx dy \phi(k_x, k_y) e^{i k_x x + i k_y y},~~\varrho(x,y) = \int dx dy \rho(k_x, k_y) e^{i k_x x + i k_y y}$$ in the first equation and this should immediately give the equation you desire. Also, just for potential future purposes, note $$\int dx e^{i k x } = 2\pi \delta (x)$$

-1

The speed of light is currently defined as "exactly 299,792,458 metres per second". Even if the measurements of the speed of light are not exact, or even if the speed of light changes, it is still defined as "exactly 299,792,458 metres per second". This adopted value for teh speed of light was specified at the General Conference of Weights And Measures on ...

2

Andrew the problem is that the speed of light is not an exact number. You give it as $2.99792458\times 10^{8}$ m/s; but suppose I instead tell you that it is really $1.8026175\times 10^{12}$ furlongs per fortnight. Neither of these numbers is any more correct than the other, but we have an accepted and defined system of units that we work in. Maxwell's ...

5

I reject the notion that $\epsilon_0=4\pi\times10^{-7}\,{\rm F/m}$ and $\mu_0\sim9\times10^{-12}\,{\rm H/m}$, they are precisely $\epsilon_0=1$ and $\mu_0=c^{-2}$. This obviously takes care of any issue with defining any unit with $\pi$ in it because, $$\sqrt{\frac{1}{\epsilon_0\mu_0}}=\sqrt{\frac{1}{1\cdot c^{-2}}}=\sqrt{c^2}=c$$ We use the MKSA system ...

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