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primetimeblues
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For those unaware, geometric algebra is a mathematical language that generalizes and simplifies a lot of the tools physicists work with (vectors, complex numbers, tensors, etc). I found a neat example in classical mechanics to illustrate its power. Simply, a charge in a constant magnetic field. A bit of (very simplified) background though:
Geometric algebra has vector-like objects that have the property that they square to -1. This let's complex rotations operate directly on vectors.
The magnetic field is one of these vector-like objects. Since this is a 2-D example, we'll be defining the magnetic field vector as [tex] \vec{B} = iB [/tex]
The curl in geometric algebra is a different kind of product. In the case of the Lorentz force law with no electric fields, it happens to turn into a regular product.
[tex] \vec{F} = q\vec{v} \vec{B}[/tex]
By Newton's 2nd law:
[tex] m\frac{d\vec{v}}{dt} = q\vec{v} \vec{B}[/tex]
using separation of variables,
[tex] \frac{1}{\vec{v}}d\vec{v} = \frac{q}{m}\vec{B}dt[/tex]
and integrating:
[tex] \ln(\vec{v}) = \frac{q}{m}\vec{B}t+C[/tex]
now, solving for velocity
[tex]\vec{v}=Ce^{(q\vec{B}/m)t}[/tex]
qB/m is the well known angular velocity of this system, so using our original definition of the magnetic field, and substituting in omega:
[tex]\vec{v}=Ce^{i\omega t}[/tex]
Setting t=0, we learn that our constant is our initial velocity.
[tex]\vec{v}=\vec{v}_0e^{i\omega t}[/tex]
Thus we have a rotating velocity, with some initial direction, exactly like we would expect. We could integrate again to get position as a function of time, but I think the general method is clear enough. What we would get is a rotating position, with 90 degrees relative to velocity.
This is a nice example of how geometric algebra can simplify cross products, and combines the mathematical language of complex numbers and vectors, thus allowing us to directly solve this problem without resorting to components.
Geometric algebra has vector-like objects that have the property that they square to -1. This let's complex rotations operate directly on vectors.
The magnetic field is one of these vector-like objects. Since this is a 2-D example, we'll be defining the magnetic field vector as [tex] \vec{B} = iB [/tex]
The curl in geometric algebra is a different kind of product. In the case of the Lorentz force law with no electric fields, it happens to turn into a regular product.
[tex] \vec{F} = q\vec{v} \vec{B}[/tex]
By Newton's 2nd law:
[tex] m\frac{d\vec{v}}{dt} = q\vec{v} \vec{B}[/tex]
using separation of variables,
[tex] \frac{1}{\vec{v}}d\vec{v} = \frac{q}{m}\vec{B}dt[/tex]
and integrating:
[tex] \ln(\vec{v}) = \frac{q}{m}\vec{B}t+C[/tex]
now, solving for velocity
[tex]\vec{v}=Ce^{(q\vec{B}/m)t}[/tex]
qB/m is the well known angular velocity of this system, so using our original definition of the magnetic field, and substituting in omega:
[tex]\vec{v}=Ce^{i\omega t}[/tex]
Setting t=0, we learn that our constant is our initial velocity.
[tex]\vec{v}=\vec{v}_0e^{i\omega t}[/tex]
Thus we have a rotating velocity, with some initial direction, exactly like we would expect. We could integrate again to get position as a function of time, but I think the general method is clear enough. What we would get is a rotating position, with 90 degrees relative to velocity.
This is a nice example of how geometric algebra can simplify cross products, and combines the mathematical language of complex numbers and vectors, thus allowing us to directly solve this problem without resorting to components.