# Tag Info

0

Since in GR equations can get very complicated very quickly it's common practice to simplify then by using a system of units where pretty much every constant is set to unity. These are generally referred to as geometrised units. The trouble is that people tend to get a bit causal with the units. For example since $E = mc^2$ and in geometrised units $c = 1$ ...

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 $$\int dx e^{i k x } = 2\pi \delta (x)$$

5

The most general form of Maxwell's equations are (setting $\mu_0 = \varepsilon_0 = 1$) \begin{align} \vec{\nabla} \cdot \vec{B} &= 0 \\ \vec{\nabla} \times \vec{E} &= - \frac{ \partial \vec{B} }{ \partial t} \\ \vec{\nabla} \cdot \vec{E} &= \rho \\ \vec{\nabla} \times \vec{B} &= \vec{J} + \frac{ \partial \vec{E} }{ \partial t} \end{align} ...

1

Maxwell equations are not "derived" from this tensor. 1) This is a definition of $E_x$ via $\partial A_x/\partial t -(\nabla\phi)_x$. The other equations are derived from the corresponding action or postulated, if you like. In the former case you can use the equations for $A_{\mu}$ if you have them from the action.

-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 ...

0

The British thermal unit (BTU or Btu) is a traditional unit of energy equal to about 1055 joules (ref). A propane torch rated at 200,000BTU will contain this amount of energy in total. The rate at which it supplies this energy is down to how you. In a propane torch, you can control the rate at which the gas is burned. Or you can burn it all at once ...

1

For the sake of doing a strict units analysis in data conversions, to confirm that the code converts correctly; for that, each variable must be documented with the correct units. Then you're straight outa luck (or some cruder version of SOL). This unsigned eight bit integer contains a value that represents a temperature in a non-standard unit. The value ...

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