Target Diamond Structural Analysis - Revision history

Tnemiger: Fixing various minor problems

2009-05-14T21:12:43Z

Fixing various minor problems

Show changes

Tnemiger at 22:30, 11 May 2009

2009-05-11T22:30:54Z

Tnemiger at 20:45, 30 April 2009

2009-04-30T20:45:31Z

Tnemiger: /* Ideal Amplitude Calculation */

2009-04-30T20:37:39Z

Ideal Amplitude Calculation

Tnemiger: /* Interference */

2009-04-30T20:32:02Z

Interference

Tnemiger: /* Interference */

2009-04-30T20:31:06Z

Interference

Tnemiger: /* Interference */

2009-04-30T20:28:40Z

Interference

Tnemiger: /* Interference */

2009-04-30T20:24:27Z

Interference

Tnemiger: /* Interference */

2009-04-30T20:22:12Z

Interference

Tnemiger: /* Interference */

2009-04-30T20:19:56Z

Interference

@@ Line 1: / Line 1: @@
 __TOC__
-== The Target Diamond ==
+== Probing the Target Diamond's Structure ==
 [[Image:01 Apparatus with Red Laser.png|thumb|The Michelson interferometer to be used. The diamond is not to scale.]]

@@ Line 37: / Line 37: @@
 == Ideal Amplitude Calculation ==
-[[Image:03 Apparatus for Beam 00.png|thumb|The zeroth wave.]]
+[[Image:03 Apparatus for Beam 00.png|thumb|The zeroth wave path. Although some light passes through the splitter, it is lost, cannot affect the experiment, and therefore is not illustrated.]]
 Because all three waves are reflections of the same original wave, they all have the same wavelength. However, the processes of reflection and transmission will modify the ''amplitude'' of each wave. By removing the diamond and reflecting the laser solely off of the mirror, we will be able to calculate the amplitude of the initial light after it has reflected off the mirror and beam splitter once and been transmitted through the splitter once.
@@ Line 49: / Line 49: @@
 == Ideal Thickness Calculation ==
-[[Image:04 Appa for Beam 01.png|thumb|The first wave reflection.]]
+[[Image:04 Appa for Beam 01.png|thumb|The first wave path. Although some light reflects off the splitter, it is lost, cannot affect the experiment, and therefore is not illustrated.]]
 To find the thickness of the diamond, we ideally only need the first two waves. To remove the third wave, which reflects from the mirror, we can simply remove the mirror.
@@ Line 72: / Line 72: @@
 <math>A^2 _{012} = C^2 _0 A^2 + C^2 _1 A^2 + C^2 _2 A^2 + 2 C _1 C _2 A^2 \cos ( d _2 - d _1 ) + 2 C _0 C _1 A^2 \cos ( - d _1 ) + 2 C _0 C _2 A^2 \cos ( - d _2 ) </math>
 Although this equation looks very complicated, we know that
@@ Line 101: / Line 103: @@
 == Compensating for Internal Reflection ==
-[[Image:02 Internal Reflection with Labels.png|thumb|Internal reflection, where C is a coefficient to be multiplied by the amplitude. The angles have been exaggerated from zero degrees.]]
+[[Image:02 Internal Reflection with Labels.png|thumb|Internal reflection of the diamond, where C is a coefficient to be multiplied by the incoming amplitude. The angles have been exaggerated from zero degrees. The diagram also shows that all error comes from about two percent of the amplitude.]]
 Realistically, the laser will not miraculously split in two upon reaching the diamond, creating one wave that reflects back and a second that reflects off of the back of the diamond and then passes perfectly through the front. Internal reflection will occur; we must calculate how much there will be and whether or not we must compensate for it.

← Older revision		Revision as of 20:37, 30 April 2009
Line 36:		Line 36:

	== Ideal Amplitude Calculation ==		== Ideal Amplitude Calculation ==
		+
		+	[[Image:03 Apparatus for Beam 00.png\|thumb\|The zeroth wave.]]

	Because all three waves are reflections of the same original wave, they all have the same wavelength. However, the processes of reflection and transmission will modify the ''amplitude'' of each wave. By removing the diamond and reflecting the laser solely off of the mirror, we will be able to calculate the amplitude of the initial light after it has reflected off the mirror and beam splitter once and been transmitted through the splitter once.		Because all three waves are reflections of the same original wave, they all have the same wavelength. However, the processes of reflection and transmission will modify the ''amplitude'' of each wave. By removing the diamond and reflecting the laser solely off of the mirror, we will be able to calculate the amplitude of the initial light after it has reflected off the mirror and beam splitter once and been transmitted through the splitter once.

@@ Line 222: / Line 222: @@
 This is a complicated equation, but it can be solved for G.
-<math>G(\mathbf{k},\omega) = \frac{1}{4\pi^2}\frac{1}{c^2k^2-\omega^2}</math>
+<math>G(\mathbf{k},\omega) = \frac{1}{4\pi^2}\left(\frac{1}{c^2k^2-\omega^2}\right)</math>
 Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

@@ Line 220: / Line 220: @@
 <math> \frac{d^2}{dt^2} \left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right)-c^2\frac{d^2}{dx^2}\left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right) = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>
-This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into
+This is a complicated equation, but it can be solved for G.
 <math>\int{dt_i} \int{f(x_i,t_i)g(\Delta x, \Delta t) dx_i dy_i}</math>

@@ Line 206: / Line 206: @@
 This is not an easy equation to solve without using a Fourier transform. Therefore, we'll do just that, with G as the transformed function.
-<math>G = \frac{1}{4\pi^2} \int{g(\Delta x,\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}d^3x dt}</math>
+<math>G(\mathbf{k},\omega) = \frac{1}{4\pi^2} \int{g(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}d^3x dt}</math>
 Therefore,
-<math>g = \frac{1}{4\pi^2} \int{G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta x}e^{i\omega \Delta t}d^3k d\omega}</math>
+<math>g(\Delta \mathbf{x},\Delta t) = \frac{1}{4\pi^2} \int{G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta x}e^{i\omega \Delta t}d^3k d\omega}</math>
 We will then need to plug this function into the earlier equation. To make this easier, we know that the four-dimensional delta function on the right-hand side can be simplified.
@@ Line 218: / Line 218: @@
 Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.
-<math> \frac{d^2}{dt^2} \left(\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right)-c^2\frac{d^2}{dx^2}\left(\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right) = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>
+<math> \frac{d^2}{dt^2} \left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right)-c^2\frac{d^2}{dx^2}\left(\frac{1}{4\pi^2} G(\mathbf{k},\omega)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right) = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>
 This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

@@ Line 214: / Line 214: @@
 We will then need to plug this function into the earlier equation. To make this easier, we know that the four-dimensional delta function on the right-hand side can be simplified.
-<math>\delta = \frac{1}{4\pi^2} \int{e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}d^3k d\omega}</math>
+<math>\delta^4(\Delta \mathbf{x},\Delta t) = \frac{1}{16\pi^4} \int{e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}d^3k d\omega}</math>
 Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.
-<math> \frac{d^2 \frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}}{dt^2}-c^2\frac{d^2\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}}{dx^2} = \frac{1}{4\pi^2} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>
+<math> \frac{d^2 \frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}}{dt^2}-c^2\frac{d^2\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}}{dx^2} = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>
 This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

@@ Line 198: / Line 198: @@
 Calculation of g proves more challenging. We know the wave equation
-<math>\frac{d^2y}{dt^2} - \frac{c^2d^2y}{dx^2} = 0</math>
+<math>\frac{d^2y}{dt^2} - c^2\frac{d^2y}{dx^2} = 0</math>
 However, this is for a uniform, sourceless wave. Like most generalizations, this is an unrealistic situation in the real world. What we need is a function that can generate a brief pulse. This sounds like a delta function.
-<math>\frac{d^2g}{dt^2} - \frac{c^2d^2g}{dx^2} = \delta(\Delta \mathbf{x})\delta(\Delta t)</math>
+<math>\frac{d^2g}{dt^2} - c^2\frac{d^2g}{dx^2} = \delta^3(\Delta \mathbf{x})\delta(\Delta t)</math>
 This is not an easy equation to solve without using a Fourier transform. Therefore, we'll do just that, with G as the transformed function.