# Tag Info

## New answers tagged fourier-transform

1

Since $T_{\mu \nu}(x)$ is symmetric in $\mu$ and $\nu$, the RHS is also symmetric by default. If it is not manifestly symmetric (like in your example), you can always symmetrize it over the corresponding indices. So in your integral, change $k_\mu q_\nu \to \frac{k_\mu q_\nu+k_\nu q_\mu}{2}$. In order to find $T_{\mu \nu} (p)$ (where $p=$ momentum) from $T_{... 7 Your problem -- as far as I can see -- is a simple calculation error: while you are correct that$\sin(k_0 x) = \mathcal{I}(e^{ik_0 x})$, it does not follow that $$\mathcal{I}(e^{i(k_0-k) x}) = \sin(k_0 x)e^{ikx} \quad \text{(Wrong!)},$$ as you should quickly be able to see since the left-hand side should be real but the right hand side has an imaginary part.... 0 If you know how the B fields changes with position$r$, you could calculate the function$f(r)$. However, as you suggested, by Fourier transforming the function$f(B)$, you will obtain$F(\omega_B)$, where$\omega_B$is the "magnetic (angular) frequency". The Fourier convolution theorem still holds. So the only difference is that your result uses &... 1 I think this paper may answer your question, although I couldn't find the whole paper available anywhere, unless you have privileges to download from the American Journal of Physics: Monochromators as light stretchers. It showed that indeed the prism does hold on to light for a time period that is commensurate with the uncertainty relation that you wrote. ... 2 Note that$\int dp'\ p' e^{-ip'(x'-x'')} = 2\pi i\delta'(x'-x'')$, where$\delta'$is the distributional derivative of the delta function defined by $$\int dx \ f(x) \delta'(x-a) := -f'(a)$$ Note also that for a complex number$a$, we have that$a-a^* = 2\color{red}{i}\mathrm{Im}(a)$, not just$2\mathrm{Im}(a)$. Those two pieces of information, plus the ... 2 This is a basic property of Fourier transforms. Quoting from Cohen-Tannoudji (Photon and atoms, introduction to QED, beginning of section I.B.1): Since$\mathbf{E}(\mathbf{r}, t)$is real, it follows that $$\mathscr{E}^*(\mathbf{k}, t) = \mathscr{E}(-\mathbf{k}, t).$$ Note that$\mathbf{E}(\mathbf{r}, t) \leftrightarrow \mathscr{E}(\mathbf{k}, t)$are ... 3 While you can't use a spatial Fourier transform because of$l(x)$and$c(x)$not being constant, you can still use a Fourier transform in time to turn the PDE into an ODE. So using the Fourier transform $$\tilde{V}(x,\omega) \equiv \int_{-\infty}^\infty dt~V(x,t) e^{-i\omega t},$$ your equation transforms to $$\frac{\partial}{\partial x}\Big(\frac{1}{l(x)}\... 1 As Emilio Pisanty already said, you don't need to. It is actually possible to determine the reciprocal lattice for any lattice in an arbitrary number of dimensions: Let V be a n-dim. real vector space and let g\colon V\times V\to\mathbf{R} be a non-degenerate bilinear map (we don't need to assume that g is symmetric). If (a_1,\ldots,a_n) is a basis ... 1 You're integrating over all possible values of the momenta, so in this case the minus sign is not needed. -2 It helps to also recall the four-momentum invariant relation p_\mu p^\mu = m^2 which for the above case, m=0 gives, (p^0)^2 - \textbf{p}^2 = 0 where \textbf{p} is the spatial momentum. The zeroth component is energy/c, \hbar \omega/c and the momentum is given by ]textbf{p} = \hbar k. Plug in, to get the above result. 6 Well, the wave equation is$$ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} $$where c is just a constant, and has the interpretation of wave velocity (by simple dimensional analysis). Putting in u(t, x) = e^{i(\omega t−kx)}:$$ \frac{\partial^2}{\partial t^2} e^{i(\omega t−kx)} = c^2 \frac{\partial^2}{\partial x^2}e^{i(\... 0 You're integrating over$k$twice, which is a bad idea. If you introduce a second dummy variable$q$, then the first term becomes something like $$\int d^4x \int d^4k\int d^4q \left[ \tilde \phi(k)\tilde \phi(q)(-k^2-m^2+i\epsilon) e^{-ix\cdot (k+q)}\right]$$ Now the only thing which depends on$x$is that exponential. Noting that$\int d^4x \ e^{-ix\cdot ...

2

Your problem is in part due to the small dynamic range, the ratio between the maximum and minimum measurable light intensities, of your detector. This is a problem often encountered whenever one tries to "photograph" fringe patterns and you can usually tell if an image is real or simulated by looking to see if there are overexposed fringes or not. ...

2

Section 4.5 of paper "Vacuum Noise and Stress Induced by Uniform Acceleration" by Takagi uses the massive position space propagator (the Wightman function) to calculate the rate of excitation of a two-level detector which uniformly accelerated through Minkowski space (this then tells you the Unruh temperature). This calculation requires your ...

0

One interesting application of a 4f correlator is target identification. Let’s say that you want to identify a target in a non-controlled environment. The target can be as diverse as an enemy tank, a bacterium, or a security mark on a credit card. Your mask will be the Fourier transform of that target. You can then create that mask using a lithographic ...

5

The variable "$p$" in the definition of the wavefunction is not the same as the variable $p$ in the definition of the Fourier Transform. It's best if you call the first $p$ some other constant, say, $p_0$. This also makes sense physically, since it actually represents the expectation value of the momentum of the Gaussian wavepacket. (You should be ...

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