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learn.steadfast
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I'm working on semiconductor physics and cyclotrons.
There's an article that I am trying to understand, but am having difficulties matching with experiment. I asked a question on an electronics site, but apparently the physics is too advanced.
The relativity article which sparked my interest is this one:
https://arxiv.org/pdf/1701.07067.pdf
What I'm having trouble understanding is section III and equations 24-30.
I get that in classical physics $$E=0.5 m v^2 $$ The classic formula makes a parabola with constant curvature if graphed as E vs m*v or k (AKA momentum is proportional to 'k' in semiconductor wave packet physics instead of being called 'p')
I realize that if the classical kinetic energy formula's second derivative is taken with respect to momentum (m*v), in one dimension, it yields inverse mass; eg: That's the same as taking the derivative of the equation with respect to velocity, twice; and dividing by mass twice. eg: $$ { 0.5 \cdot m \cdot 2 \cdot v^0 \over { m^2 }} = m^{-1} $$
In equation 30 of the arxiv article, the authors do something similar but instead of taking the derivative twice in one dimension; they take a partial derivative in two dimensions? I'm not sure why they are doing that. It's been too long since I've had physics. Can anyone explain why they conclude that mass is not a scalar? Is there a directional dependence of mass when traversing a circular path at constant energy that doesn't end up being constant? (Constant spherical energy means an idealized electron, neglecting Heisenburg, is essentially moving with constant momentum along the surface of a sphere. In general, it's typically a circular orbit of arbitrary radius. )
I know that in relativistic physics, the shape of an energy vs. momentum curve for a particle is a hyperbola. That means the curvature is not constant, and at high energies the relationship becomes linear. eg: see Figure 1 in the arxiv article. So, I assume that 'mass not being scalar' might be related to the fact that the second derivative of the curve goes to zero as energy gets large; hence "effective mass" goes toward infinity. Could someone verify my assumption?
If I am correct, then there has to be a way to convert cyclotron masses based on the Dirac equation/relativistic physics into a mass compatible with F=ma by knowing the E vs. k curve, and doing something like taking the second derivative of it. Would a second derivative (in one dimension) not give a mass which would allow F=m*a to approximate the force measured when accelerating an object a very small amount?
The electrical engineers I asked on another other site have no answers for me about converting the masses between a relativistic framework and a classical one. The graph at the bottom of the following thread shows a red line, which is from a cyclotron. The grey line is from an experimenter, called Barber, who's giving a mass that I assume is supposed to work with classical physics models. eg: F=m*a. The two don't agree, and I think it has to do with differences in classical and relativistic ideas of mass.
https://electronics.stackexchange.c...-mass-calculations-band-edge-vs-dos-vs-cyclot
There's an article that I am trying to understand, but am having difficulties matching with experiment. I asked a question on an electronics site, but apparently the physics is too advanced.
The relativity article which sparked my interest is this one:
https://arxiv.org/pdf/1701.07067.pdf
What I'm having trouble understanding is section III and equations 24-30.
I get that in classical physics $$E=0.5 m v^2 $$ The classic formula makes a parabola with constant curvature if graphed as E vs m*v or k (AKA momentum is proportional to 'k' in semiconductor wave packet physics instead of being called 'p')
I realize that if the classical kinetic energy formula's second derivative is taken with respect to momentum (m*v), in one dimension, it yields inverse mass; eg: That's the same as taking the derivative of the equation with respect to velocity, twice; and dividing by mass twice. eg: $$ { 0.5 \cdot m \cdot 2 \cdot v^0 \over { m^2 }} = m^{-1} $$
In equation 30 of the arxiv article, the authors do something similar but instead of taking the derivative twice in one dimension; they take a partial derivative in two dimensions? I'm not sure why they are doing that. It's been too long since I've had physics. Can anyone explain why they conclude that mass is not a scalar? Is there a directional dependence of mass when traversing a circular path at constant energy that doesn't end up being constant? (Constant spherical energy means an idealized electron, neglecting Heisenburg, is essentially moving with constant momentum along the surface of a sphere. In general, it's typically a circular orbit of arbitrary radius. )
I know that in relativistic physics, the shape of an energy vs. momentum curve for a particle is a hyperbola. That means the curvature is not constant, and at high energies the relationship becomes linear. eg: see Figure 1 in the arxiv article. So, I assume that 'mass not being scalar' might be related to the fact that the second derivative of the curve goes to zero as energy gets large; hence "effective mass" goes toward infinity. Could someone verify my assumption?
If I am correct, then there has to be a way to convert cyclotron masses based on the Dirac equation/relativistic physics into a mass compatible with F=ma by knowing the E vs. k curve, and doing something like taking the second derivative of it. Would a second derivative (in one dimension) not give a mass which would allow F=m*a to approximate the force measured when accelerating an object a very small amount?
The electrical engineers I asked on another other site have no answers for me about converting the masses between a relativistic framework and a classical one. The graph at the bottom of the following thread shows a red line, which is from a cyclotron. The grey line is from an experimenter, called Barber, who's giving a mass that I assume is supposed to work with classical physics models. eg: F=m*a. The two don't agree, and I think it has to do with differences in classical and relativistic ideas of mass.
https://electronics.stackexchange.c...-mass-calculations-band-edge-vs-dos-vs-cyclot
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