Commit 0cba1d18 authored by Tianbai Xiao's avatar Tianbai Xiao
Browse files

Fix physics.rst


Former-commit-id: 7d1e77c0
parent 6a97b0a1
......@@ -7,15 +7,15 @@ Authors
KiT-RT Software
*****************
+--------------------+------------------------+
|Main contributors | - Jannick Wolters |
| | - Jonas Kusch |
| | - Steffen Schotthöfer |
| | - Tianbai Xiao |
| | - Pia Stammer |
+--------------------+------------------------+
|Other contributors | |
+--------------------+------------------------+
+--------------------+----------------------------------------+
|Main contributors | - Jannick Wolters |
| | - Jonas Kusch |
| | - Steffen Schotthöfer |
| | - Tianbai Xiao (tianbai.xiao@kit.edu) |
| | - Pia Stammer |
+--------------------+----------------------------------------+
|Other contributors | |
+--------------------+----------------------------------------+
*********************
KiT-RT Dokumentation
......
......@@ -39,9 +39,9 @@ The software is being developed by members of the group `CSMM <https://www.scc.k
If you are interested in using KiT-RT or are trying to figure out how to use it, please feel free to get in touch with `us <authors.html>`_.
Do open an issue or pull request if you have questions, suggestions or solutions.
--------
Table of contents
--------
------------------------
.. toctree::
:maxdepth: 1
......
......@@ -11,13 +11,13 @@ Down to the finest scale of a many-particle system, the Newton’s second law de
.. math::
F = m a
F = m a,
which leads
.. math::
\frac{d x}{dt} = v, \ \frac{d v}{dt} = \frac{F}{m}
\frac{d x}{dt} = v, \ \frac{d v}{dt} = \frac{F}{m}.
An intuitive numerical solution algorithm is to get the numerous particles on board and track the trajectories of them.
A typical example is the molecular dynamics (MD) method.
......@@ -32,11 +32,11 @@ Many realizations must be simulated successively to average the solutions and re
An alternative strategy can be made from ensemble averaging, where the
coarse-grained modeling is used to provide a bottom-up view.
At the mean free path and collision time scale of particles. Such dynamics can be described with kinetic theory.
The Boltzmann equation can be formulated via an operator splitting approach.
The Boltzmann equation can be formulated via an operator splitting approach, i.e.
.. math::
\partial_{t} f(v)+v \cdot \nabla_{x} f(v)=\int_{\mathcal R^3} \int_{\mathcal S^2} k\left(v, v^{\prime}\right) \left(f\left(v^{\prime}\right)f\left(v_*^{\prime}\right)-f(v)f(v_*)\right) d\Omega d v_*
\partial_{t} f(v)+v \cdot \nabla_{x} f(v)=\int_{\mathcal R^3} \int_{\mathcal S^2} k\left(v, v^{\prime}\right) \left(f\left(v^{\prime}\right)f\left(v_*^{\prime}\right)-f(v)f(v_*)\right) d\Omega d v_*,
where the left and right hand sides model particle transports and collisions correspondingly.
The distribution function :math:`f` is the probability of finding a particle with certain location, and :math:`\{v, v_*\}` denotes the velocities of two classes of colliding particles.
......@@ -44,18 +44,18 @@ The collision kernel :math:`k` models the strength of collisions at different ve
Different collision models can be inserted into the Boltzmann equation.
In the KiT-RT solver, we are interested in the linear Boltzmann equation, where the particles don't interact with one another but scatter with the background material.
Therefore, the Boltzmann can be simplified as the linear equation with respect to :math:`f`.
Therefore, the Boltzmann can be simplified as the linear equation with respect to :math:`f`
.. math::
\partial_{t} f(v)+v \cdot \nabla_{x} f(v)=\int k\left(v, v^{\prime}\right)\left(f\left(v^{\prime}\right)-f(v)\right) d v^{\prime}-\tau f(v)
\partial_{t} f(v)+v \cdot \nabla_{x} f(v)=\int k\left(v, v^{\prime}\right)\left(f\left(v^{\prime}\right)-f(v)\right) d v^{\prime}-\tau f(v).
For convenience, it is often reformulated with polar coordinates :math:`\{r, \phi, \theta \}`.
For convenience, it is often reformulated with polar coordinates :math:`\{r, \phi, \theta \}`,
.. math::
&\left[\frac{1}{v(E)} \partial_{t} +\Omega \cdot \nabla+\Sigma_t (r, E, t)\right] \psi(r, \Omega, E, t) \\
&=\int_{0}^{\infty} d E^{\prime} \int_{\mathcal R^2} d \Omega^{\prime} \Sigma_{s}\left(r, \Omega^{\prime} \bullet \Omega, E^{\prime} \rightarrow E\right) \psi\left(r, \Omega^{\prime}, E^{\prime}, t\right) + Q(r, \Omega, E, t)
&=\int_{0}^{\infty} d E^{\prime} \int_{\mathcal R^2} d \Omega^{\prime} \Sigma_{s}\left(r, \Omega^{\prime} \bullet \Omega, E^{\prime} \rightarrow E\right) \psi\left(r, \Omega^{\prime}, E^{\prime}, t\right) + Q(r, \Omega, E, t).
The particle distribution :math:`\psi(r, \Omega, E, t)` here is often named as angular flux, :math:`\{\Sigma_s, \Sigma_t \}` are the scattering and total cross sections correspondingly, and :math:`Q` denotes a source term.
......@@ -104,7 +104,7 @@ Leaving out the superscript :math:`H_2O`, the CSD equation can be simplified as
&= \int_{\mathbb{S}^2}\rho(x)\Sigma_s(E,\Omega\cdot\Omega')\psi(E,x,\Omega')d\Omega'.
Now, we bring this system in a form which resembles the standard Boltzmann equation.
Multiplying :ref:`CSD2` with :math:`S(E)` gives
Multiplying :eq:`CSD2` with :math:`S(E)` gives
.. math::
:label: CSD3
......@@ -148,20 +148,20 @@ And by rearranging the terms, we finally get
\partial_{ E}\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega) = \partial_{\widetilde{E}}\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega)\frac{1}{S(E(\widetilde E))},
since :math:`S(E(\widetilde E))` is nonzero.
Therefore, substituting :math:`\widetilde E` in :ref:`CSD4` gives
Therefore, substituting :math:`\widetilde E` in :eq:`CSD4` gives
.. math::
:label: CSD5
&-\partial_{\widetilde E}\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega)+\Omega\cdot\nabla_x \frac{\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega)}{\rho}+\widetilde\Sigma_t(\widetilde E)\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega)
&= \int_{\mathbb{S}^2}\widetilde\Sigma_s(\widetilde E,\Omega\cdot\Omega')\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega')d\Omega'.
& -\partial_{\widetilde E}\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega)+\Omega\cdot\nabla_x \frac{\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega)}{\rho}+\widetilde\Sigma_t(\widetilde E)\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega) \\
& = \int_{\mathbb{S}^2}\widetilde\Sigma_s(\widetilde E,\Omega\cdot\Omega')\widetilde{\widehat{\psi}}(\widetilde E,x,\Omega')d\Omega'.
Here, we define :math:`\widetilde\Sigma_{t}(\widetilde E):=\Sigma_t(E(\widetilde E))` and :math:`\widetilde\Sigma_{s}(\widetilde E,\Omega\cdot\Omega'):=\Sigma_s(E(\widetilde E),\Omega\cdot\Omega')`. Finally, to obtain a positive sign in front of the energy derivative, we transform to
.. math::
\bar{E}(\widetilde{E}) = \widetilde{E}_{\text{max}}-\widetilde{E}.
Then, with $\bar{\psi}(\bar{E},x,\Omega):=\widetilde{\widehat{\psi}}(\widetilde{E}(\bar{E}),x,\Omega)$ and $\bar\Sigma_{t}(\bar E):=\widetilde{\Sigma}_t(\widetilde{E}(\bar{E}))$ as well as $\bar\Sigma_{s}(\bar E,\Omega\cdot\Omega'):=\widetilde{\Sigma}_s(\widetilde{E}(\bar{E}),\Omega\cdot\Omega')$ equation \eqref{eq:CSD4} becomes
Then, with :math:`\bar{\psi}(\bar{E},x,\Omega):=\widetilde{\widehat{\psi}}(\widetilde{E}(\bar{E}),x,\Omega)`, :math:`\bar\Sigma_{t}(\bar E):=\widetilde{\Sigma}_t(\widetilde{E}(\bar{E}))` as well as :math:`\bar\Sigma_{s}(\bar E,\Omega\cdot\Omega'):=\widetilde{\Sigma}_s(\widetilde{E}(\bar{E}),\Omega\cdot\Omega')` equation \eqref{eq:CSD4} becomes
.. math::
:label: CSD6
......@@ -176,7 +176,7 @@ Dropping the bar notation and treating :math:`\bar E` as a pseudo-time :math:`t`
\partial_{t}\psi(t,x,\Omega)+&\Omega\cdot\nabla_x \frac{\psi(t,x,\Omega)}{\rho}+\Sigma_t(t)\psi(t,x,\Omega) = \int_{\mathbb{S}^2}\Sigma_s(t,\Omega\cdot\Omega')\psi(t,x,\Omega')d\Omega'\\
&\psi(t=0,x,\Omega) = S(E_{\text{max}})\rho(x)\psi(E_{\text{max}},x,\Omega).
We are interested in computing the dose, which (when again using the original energy $E$ and angular flux $\psi$) reads
We are interested in computing the dose, which (when again using the original energy :math:`E` and angular flux :math:`\psi`) reads
.. math::
D(x) = \int_0^{\infty} \int_{\mathbb{S}^2} S(E)\psi(E,x,\Omega)\,d\Omega dE = \int_0^{\infty} \int_{\mathbb{S}^2} \frac{1}{\rho(x)}\widehat\psi(E,x,\Omega)\,d\Omega dE.
......@@ -184,7 +184,6 @@ We are interested in computing the dose, which (when again using the original en
So let us check how we can compute the dose from our solution :math:`\bar \psi(\bar E,x,\Omega)`. For this, let us substitute
.. math::
:label: BarE
\bar E(E) = \tilde{E}(E_{max}) - \int_0^E \frac{1}{S(E')}dE'.
......
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