Template:Energy contour and preliminaries - Revision history

Meylan at 08:59, 11 March 2009

2009-03-11T08:59:55Z

← Older revision		Revision as of 08:59, 11 March 2009
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	The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.		The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

	[[Image:energy_schematic.jpg\|thumb\|right\|410px\|A diagram depicting the area <math>\~~mathcal{U}~~</math> which is bounded by the rectangle <math>\~~mathcal{S}~~</math>. The rectangle <math>\~~mathcal{S}~~</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]		[[Image:energy_schematic.jpg\|thumb\|right\|410px\|A diagram depicting the area <math>\Omega</math> which is bounded by the rectangle <math>\partial\Omega</math>. The rectangle <math>\partial\Omega</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
	<center><math>		<center><math>
	{ \~~int~~\~~int_\mathcal{U}~~\left(\phi^\nabla^2\phi - \phi\nabla^2\phi^ \right)\mathrm{d}x\mathrm{d}z		{ \iint_\Omega\left(\phi^\nabla^2\phi - \phi\nabla^2\phi^ \right)\mathrm{d}x\mathrm{d}z
	= \int_\~~mathcal{S~~}\left(\phi^\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^}{\partial n} \right)\mathrm{d}l },		= \int_{\partial\Omega}\left(\phi^\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^}{\partial n} \right)\mathrm{d}l },
	</math></center>		</math></center>
	where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.		where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.

Adi Kurniawan at 07:31, 24 February 2009

2009-02-24T07:31:38Z

← Older revision		Revision as of 07:31, 24 February 2009
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	The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.		The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

	[[Image:energy_schematic.jpg\|thumb\|right\|410px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]		[[Image:energy_schematic.jpg\|thumb\|right\|410px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>. The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]
	~~{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.~~
	The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives

Adi Kurniawan at 07:30, 24 February 2009

2009-02-24T07:30:47Z

← Older revision		Revision as of 07:30, 24 February 2009
Line 3:		Line 3:
	The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.		The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

	[[Image:energy_schematic.jpg\|thumb\|right\|~~500px~~\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]		[[Image:energy_schematic.jpg\|thumb\|right\|410px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
	{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.		{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
	The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]		The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Adi Kurniawan at 07:30, 24 February 2009

2009-02-24T07:30:33Z

← Older revision		Revision as of 07:30, 24 February 2009
Line 3:		Line 3:
	The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.		The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

	[[Image:energy_schematic.jpg\|thumb\|right\|~~400px~~\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]		[[Image:energy_schematic.jpg\|thumb\|right\|500px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
	{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.		{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
	The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]		The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Adi Kurniawan at 07:30, 24 February 2009

2009-02-24T07:30:15Z

← Older revision		Revision as of 07:30, 24 February 2009
Line 3:		Line 3:
	The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.		The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

	[[Image:energy_schematic.jpg\|thumb\|right\|~~300px~~\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]		[[Image:energy_schematic.jpg\|thumb\|right\|400px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
	{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.		{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
	The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]		The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Adi Kurniawan at 07:29, 24 February 2009

2009-02-24T07:29:52Z

← Older revision		Revision as of 07:29, 24 February 2009
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	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
	<center><math>		<center><math>
	{ \int\int_\mathcal{U}\left(\phi^\nabla^2\phi - \phi\nabla^2\phi^ \right)~~dxdz~~		{ \int\int_\mathcal{U}\left(\phi^\nabla^2\phi - \phi\nabla^2\phi^ \right)\mathrm{d}x\mathrm{d}z
	= \int_\mathcal{S}\left(\phi^\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^}{\partial n} \right)dl },		= \int_\mathcal{S}\left(\phi^\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^}{\partial n} \right)\mathrm{d}l },
	</math></center>		</math></center>
	where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.		where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
	As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to		As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
	<center><math>		<center><math>
	\Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} dl = 0,		\Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}l = 0,
	</math></center>		</math></center>

Meylan at 09:50, 9 February 2009

2009-02-09T09:50:54Z

← Older revision		Revision as of 09:50, 9 February 2009
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	Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions energy balance.		Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions energy balance.
	The energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.		The energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
	The domain of integration is shown in the figure on the right.		The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

	[[Image:energy_schematic.jpg\|thumb\|right\|300px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]		[[Image:energy_schematic.jpg\|thumb\|right\|300px\|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]

Meylan at 00:34, 28 October 2008

2008-10-28T00:34:42Z

← Older revision		Revision as of 00:34, 28 October 2008
Line 9:		Line 9:
	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
	<center><math>		<center><math>
	{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz		{ \int\int_\mathcal{U}\left(\phi^\nabla^2\phi - \phi\nabla^2\phi^ \right)dxdz
	= \int_\mathcal{S}(\phi\frac{\partial\phi^}{\partial n} - \phi^\frac{\partial\phi}{\partial n})dl },		= \int_\mathcal{S}\left(\phi^\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^}{\partial n} \right)dl },
	</math></center>		</math></center>
	where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.		where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
	As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to		As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
	<center><math>		<center><math>
	\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,		\Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} dl = 0,
	</math></center>		</math></center>

Meylan at 00:13, 28 October 2008

2008-10-28T00:13:28Z

← Older revision		Revision as of 00:13, 28 October 2008
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	~~== Energy Balance ==~~

	Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions energy balance.		Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions energy balance.
	The energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.		The energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.

Meylan at 00:10, 28 October 2008

2008-10-28T00:10:27Z

New page

== Energy Balance ==

Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions energy balance.
The energy balance equation is derived by applying Green's theorem to <math>\phi</math> and its conjugate.
The domain of integration is shown in the figure on the right.

[[Image:energy_schematic.jpg|thumb|right|300px|A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]
{A diagram depicting the area <math>\mathcal{U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.
The rectangle <math>\mathcal{S}</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]

Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
<center><math>
{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)dxdz
= \int_\mathcal{S}(\phi\frac{\partial\phi^*}{\partial n} - \phi^*\frac{\partial\phi}{\partial n})dl },
</math></center>
where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
<center><math>
\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,
</math></center>