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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 15:26, 2 July 2009
, 15:26, 2 July 2009no edit summary
== Determining Angle for First Diffraction Minimum ==
We start off with Maxwell's Equation in the Lorentz gauge:
<math>\square^2A^\mu(\mathbf{x},t) = \square^2A^\mu (x)=(-\mu_1 j^\mu (x))</math>
Where:
<math>A^\mu = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part t^2}</math>
<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math>
Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math>
Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')
<math>\square^2_xG(x,x')=\delta^4(x-x')</math>
Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4x e^{iqx} G(x,0)</math>
Then <math> G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqx} \tilde{G}(x,0)</math>
In free space, translational symmetry implies:<br>
<math>G(x-x',0)=G(x,x')</math>
∴<math> G(x,x')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(x-x')} \tilde{G} (q)</math><br>
We start off with Maxwell's Equation in the Lorentz gauge:
<math>\square^2A^\mu(\mathbf{x},t) = \square^2A^\mu (x)=(-\mu_1 j^\mu (x))</math>
Where:
<math>A^\mu = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part t^2}</math>
<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math>
Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math>
Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')
<math>\square^2_xG(x,x')=\delta^4(x-x')</math>
Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4x e^{iqx} G(x,0)</math>
Then <math> G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqx} \tilde{G}(x,0)</math>
In free space, translational symmetry implies:<br>
<math>G(x-x',0)=G(x,x')</math>
∴<math> G(x,x')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(x-x')} \tilde{G} (q)</math><br>