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--- equations:continuity_equation [2018/04/19 09:25]
jakobadmin [Concrete]
+++ equations:continuity_equation [2020/03/03 10:38] (current)
128.179.254.165
@@ Line 1: / Line 1: @@
-<WRAP lag> $\color{blue}{\frac{\partial \rho}{\partial t}} +  \color{magenta}{\rho \vec \nabla  \vec v} = \color{red}{\sigma} $</WRAP>
+<WRAP lag> $\color{blue}{\frac{\partial \rho}{\partial t}}  = \color{red}{\sigma} - \color{magenta}{\rho \vec{\nabla} \cdot \vec{v}} $</WRAP>
 ====== Continuity Equation ======
@@ Line 5: / Line 5: @@
 <tabbox Intuitive>
-The continuity equation states that the $\color{red}{\text{total amount of a quantity (like water) that is produced (or destroyed) inside some volume}}$ is proportional to the $\color{blue}{\text{change of the quantity}}$ plus the $\color{magenta}{\text{total amount that flows in minus the amount that flows out of the volume}}$.
+The continuity equation states that the total $\color{blue}{\text{change of some quantity}}$ is equal to the $\color{red}{\text{amount that gets produced}}$ minus the amount that $\color{magenta}{\text{flows out of the volume}}$.
-Or formulated differently, the total $\color{blue}{\text{change of some quantity}}$ is equal to the $\color{red}{\text{amount that gets produced}}$ plus the amount that $\color{magenta}{\text{flows in minus the amount that flows out of the volume}}$.
 [{{ :equations:venturi.gif?nolink |Image by Thierry Dugnolle}}]
@@ Line 15: / Line 13: @@
 Whenever a system possesses some symmetry we know from [[theorems:noethers_theorems|Noether's theorem]] that some corresponding quantity is conserved. Using Noether's theorem, we can then also derive the corresponding continuity equation that describes how the conserved quantity flows through the system.
+----
+**Recommended further reading**
+  * [[https://www.wired.com/2016/08/perfection-continuity-equation-key-foundations-reality/|The Perfection of the Continuity Equation, Key to the Foundations of Reality]] by WIRED magazine
 <tabbox Concrete>
@@ Line 37: / Line 39: @@
 $$ \nabla \cdot  J  + \frac { \partial ( \nabla \cdot  D  ) } { \partial t } = 0.
 $$
-Finally, we use another [[equations:maxwell_equations|Maxwell equation]], namely [[equations:yang_mills_equations:gauss_law|Gauss law]],
+Finally, we use another [[equations:maxwell_equations|Maxwell equation]], namely [[formulas:gauss_law|Gauss law]],
 $$\nabla \cdot  D  = \rho
  $$
@@ Line 48: / Line 50: @@
 -->Derivation of the continuity equation in quantum mechanics#
+The probability density is quantum mechanics is $\rho = \Psi^* \Psi$. The partial derivative of $\rho$ with respect to time is therefore
+$$ \frac { \partial \rho } { \partial t } = \frac { \partial | \Psi | ^ { 2} } { \partial t } = \frac { \partial } { \partial t } ( \Psi ^ { * } \Psi ) = \Psi ^ { * } \frac { \partial \Psi } { \partial t } + \Psi \frac { \partial \Psi ^ { * } } { \partial t } . $$
+Next, we consider the [[equations:schroedinger_equation|Schrödinger equation]]
+$$ - \frac { \hbar ^ { 2} } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi ^ { * } = - i \hbar \frac { \partial \Psi ^ { * } } { \partial t }.$$
+Taking the complex conjugate of it yields
+$$- \frac { \hbar ^ { 2} } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi ^ { * } = - i \hbar \frac { \partial \Psi ^ { * } } { \partial t }. $$
+Multiplying the Schrödinger equation with $\Psi^\star$ and the complex conjugated Schrödinger equation with $\Psi$ yields the two equations
+$$ \Psi \cdot \frac { \partial \Psi } { \partial t } = \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi ^ { * } } { 2m } \nabla ^ { 2} \Psi + U \Psi ^ { * } \Psi ]$$
+$$ \Psi \frac { \partial \Psi ^ { * } } { \partial t } = - \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi \Psi ^ { * } ] .$$
+Putting these two equations into our equation for $\frac { \partial \rho } { \partial t }$ from above yields
+$$ \frac { \partial \rho } { \partial t } = \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi ^ { * } } { 2m } \nabla ^ { 2} \Psi + U \Psi ^ { * } \Psi ] - \frac { 1} { i \hbar } [ - \frac { \hbar ^ { 2} \Psi } { 2m } \nabla ^ { 2} \Psi ^ { * } + U \Psi \Psi ^ { * } ]$$
+$$ = \frac { \hbar } { 2i m } [ \Psi \nabla ^ { 2} \Psi ^ { * } - \Psi ^ { * } \nabla ^ { 2} \Psi ] .$$
+The second puzzle piece that appears in the continuity equation is the current $j$ which in quantum mechanics is given by
+$$j =  \frac { \hbar } { 2m i } [ \Psi ^ { * } ( \nabla \Psi ) - \Psi ( \nabla \Psi ^ { * } )]$. $$
+Taking the divergence of it (since $\nabla j$ is what appears in the continuity equation)  yields
+$$ \nabla \cdot j = \nabla \cdot [ \frac { \hbar } { 2m i } ( \Phi ^ { * } ( \nabla \Phi ) - \Psi ( \nabla \Psi ^ { * } ) ) ]
+ $$
+$$ = - \frac { \hbar } { 2m i } [ \Psi ( \nabla ^ { 2} \Psi ^ { * } ) - \Psi ^ { * } ( \nabla ^ { 2} \Psi ) ]. $$
+This is exactly, except for the minus sign what we derived above for $\frac { \partial \rho } { \partial t }$ and therefore we can conclude
+$$ \frac { \partial \rho } { \partial t } = - \nabla \cdot j $$
+$$ \therefore \frac { \partial \rho } { \partial t } +  \nabla \cdot j =0. $$
+This is exactly the continuity equation that we wanted to derive.
 <--
+----
+**Integral Form of the continuity equation**
+By integrating the continuity equation over some volume $V$, we get
+$$ \int_V \frac { \partial \rho } { \partial t } +  \int_V \nabla \cdot j =0  $$
+For the second term, we can then use [[basic_tools:vector_calculus:gauss_theorem|Gauss divergence theorem]] that tells us that we can replace the volume integral over some divergence by a surface integral
+$$ \int_V \frac { \partial \rho } { \partial t } +  \int_S  j =0 , $$
+where $S$ denotes the surface of our volume $V$. This is the integral form of the continuity equation.
+  * The first term simply describes the total amount of the quantity, e.g. electric charge or mass, inside our volume.
+  * The second term describes the amount that flows into the surface minus the amount that flows out of the surface.
 <tabbox Abstract>
-<note tip>
+In relativistic theories, the charge density and the current live in one object called the current four-vector (or four-current) $j_\mu = (c \rho, \vec j),$ where $c$ denotes the speed of light and is inserted such that all components have the same dimensions. Using this definition, the continuity equation reads
-The motto in this section is: //the higher the level of abstraction, the better//.
-</note>
+$$ \partial_\mu j^\mu = 0. $$
+The continuity equation can also be written using the [[advanced_tools:differential_forms|3-form]] of charge density
+$$ d \gamma = 0$$
+where
+$$ \gamma  = - \frac { 1} { c } ( 1,d x ^ { 2} \wedge d x ^ { 3} + i _ { 2} d x ^ { 3} \wedge d x ^ { 1} + j _ { 3} d x ^ { 1} \wedge d x ^ { 2} ) \wedge d x ^ { 0} + \rho d x ^ { \prime } \wedge d x ^ { 2} \wedge d x ^ { 3}.$$
 <tabbox Why is it interesting?>

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