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Construction of a Tabletop Michelson Interferometer
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Revision as of 17:55, 2 July 2009
361 bytes added
,
17:55, 2 July 2009
→Determining Angle for First Diffraction Minimum
Line 37:
Line 37:
Let <math>\mathbf{F}=A(x)\mathbf{\nabla}G_1(x,x')-G_1(x,x')\mathbf{\nabla}A(x)</math><br><br>
Let <math>\mathbf{F}=A(x)\mathbf{\nabla}G_1(x,x')-G_1(x,x')\mathbf{\nabla}A(x)</math><br><br>
<math>\int \mathbf{\nabla} \cdot \mathbf{F}d^4x= \int cdt \int d^3x[\mathbf{\nabla}A \cdot \mathbf{\nabla}G+A\nabla^2G_1-\mathbf{\nabla}G \cdot \mathbf{\nabla}A -G_1\nabla^2A]</math><br><br>
<math>\int \mathbf{\nabla} \cdot \mathbf{F}d^4x= \int cdt \int d^3x[\mathbf{\nabla}A \cdot \mathbf{\nabla}G+A\nabla^2G_1-\mathbf{\nabla}G \cdot \mathbf{\nabla}A -G_1\nabla^2A]</math><br><br>
−
But <math>\nabla^2G_1(x,x')=\delta^4(x-x')+\frac{1}{c^2}\frac{\part^2}{\part t^2}
+
But <math>\nabla^2G_1(x,x')=\delta^4(x-x')+\frac{1}{c^2}\frac{\part^2}{\part t^2}
G_1(x,x')</math><br><br>
−
G_1
(x,x')</math><br><br>
+
<math>\nabla^2A(x)=\mu j(x)+\frac{1}{c^2}\frac{\part^2}{\part t^2}A
(x
)</math>
,
let <math>j(x)=0 \quad</math><br><br>
+
<math>\int \nabla \cdot \mathbf{F} d^4x=A(
x')
+\frac{1}{c^2}\int d^4x\left[A\frac{\part^2}{\part t^2}G_1 - G_1\frac{\part^2}{\part t^2}A\right]
</math><br><br>
+
The last term vanishes if G<sub>1</sub>(x,x')and A(x) fall off sufficiently fast at t
Pe1505686
196
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