Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Classical mass vs. relativistic mass for electronic analogy

  1. Aug 8, 2017 #1
    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:

    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.

    Last edited: Aug 9, 2017
  2. jcsd
  3. Aug 9, 2017 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The idea is known as "quasi particles" and goes back to L. D. Landau. Often you can describe the excitation spectrum of a many-body system in terms of "quasi-particles". Formulated in terms of QFT ("2nd quantization") you can evaluate the (retarded) two-point function of some appropriate quantum field. If its imaginary part has narrow-peaked structures, i.e., if the imaginary part of the self-energy is small around a pole of the two-point function, you have defined a quasi-particle. Then you can define an energy-momentum relation ##\omega=\omega(\vec{k})## by solving
    $$\mathrm{Re} G^{-1} (\omega,\vec{k})=0.$$
    Then the group velocity is
    $$v_{\text{g}}=\nabla_{\vec{k}} \omega(\vec{k}).$$
    It's of course most simply illustrated on the case of a free particle. It's retared propagator is given by
    $$G(\omega,\vec{k})=\frac{1}{\omega-\vec{k}^2/(2m)+\mathrm{i} 0^+}.$$
    Of course, here you get a real particle, i.e., the spectral function
    $$\rho(\omega,\vec{k})=-2 \mathrm{im} G=2 \pi \delta(\omega-\vec{k}^2/2m),$$
    and the dispersion relation, defined by the pole is precisely
    and the group velocity is
    $$\vec{v}_{\text{g}}=\vec{\nabla}_{\vec{k}} \omega(\vec{k})=\frac{\vec{k}}{m}. \qquad (1)$$
    Now in the paper they define the inverse mass tensor by
    $$v_{\text{g}i}=(\mathcal{M}^{-1})_{ij} k_j.$$
    comparing with (1) you get
    $$(\mathcal{M}^{-1})_{ij}=\frac{1}{m} \delta_{ij}.$$
    In the general case this matrix can be an arbitrary symmetric matrix (if the problem has no symmetries).

    In the paper they consider the often realized case of an isotropic dispersion relation, i.e.,
    $$\omega=\omega(k) \quad k=|\vec{k}|.$$
    $$v_{\text{g}i}=\frac{\partial \omega}{\partial k_i}=\frac{k_i}{k} \omega'(k) \stackrel{\text{!}}{=} (\mathcal{M}^{-1})_{ij} k_j,$$
    from which you get
    $$(\mathcal{M}^{-1})_{ij}=\frac{\omega'(k)}{k} \delta_{ij}=\frac{1}{m^*} \delta_{ij},$$
    which defines the mass ##m^*## of the quasi-particle.
  4. Aug 9, 2017 #3
    Thanks for your help.
    I am humbly struggling to keep up with all the math notation. :)

    So, after staring at (1) in your post and the inverse mass tensor for a while; What I am seeing is that for typical cases in Cartesian coordinates where the domains of i and j are the same 1,2,3 : $$ \mathcal{M}^{-1} = {1 \over m} \left[ {\begin{array}{xyz} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}} \right] = { 1 \over m} \mathit{I}_3 $$
    It's a symmetric matrix because of the Kronecker delta, and I'm not sure what you meant by "(if the problem has no symmetries)"

    I'd like to concentrate on the isotropic case, unless that's the source of confusion.

    What you seem to be showing is that taking the partial derivative of energy with respect to momentum in three dimensions, is the same as a unit vector pointing in the direction of motion; multiplied by a derivative of a scalar energy with respect to momentum. That makes perfect sense to me. I am not sure what the ! over an equals sign means, though ; This is the first time I've seen that math symbol. However, the right side of your relationship is showing an inverse mass times a momentum. I would also expect that product to have units of velocity.

    Is there any significant difference between m and m* in your equations? eg: Is the difference solely the scalar mass of one free particle (m) vs the scalar mass of a quasi-particle (m*)?

    If I'm following you correctly, you've discussed the background of arxiv's equations (23) to (29); but not (30) yet.
    Last edited: Aug 9, 2017
  5. Aug 9, 2017 #4
    I think I might see part of whats been confusing me. In equation 30, if I write differentiation with respect to x,y,z in place of indicies 1,2,3 for I and j; the partial derivatives make a matrix/tensor with the following elements: I am assuming that i is rows, and j is columns. The subscripts in this matrix mean partial differentiation with respect to momentum along a given axis.

    $$ \mathit{M^*}^{-1} = \left[ { \begin{array}{ccc} \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{array} } \right] $$

    The diagonal of the matrix, then, is the classical second derivative of scalar energy with respect to momentum in three dimensions. That's when both derivatives are taken along the same dimension. What do the "off" diagonal elements represent? And what does the arxiv article mean by mass "not being a scalar" in the text immediately following equation (30) ?
  6. Aug 10, 2017 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    (30) is what one calls usually the inverse-mass matrix, and indeed it's a tensor, which doesn't reduce to a multiple of the identity matrix, even in the case that you have
    $$\epsilon(\vec{k})=\epsilon(k), \quad k=|\vec{k}|.$$
    Then you have
    $$\frac{\partial \epsilon}{\partial k_i}=\frac{\partial k}{\partial k_i} \epsilon'(k)=\frac{k_i}{k} \epsilon'(k)$$
    $$\frac{\partial^2 \epsilon}{\partial k_i \partial k_j} = \frac{k_i k_j}{k^2} \epsilon''(k) + \frac{k^2 \delta_{ij}-k_i k_j}{k^3} \epsilon'(k).$$
  7. Aug 10, 2017 #6
    hmm.... OK. I'm going to assume that "not being scalar" means not reducible to a scalar and the identity matrix.
    Last night, however, I worked the problem that you are showing in one dimension and produced some interesting results.

    Let me define a variable name convention, to be consistent; It can be changed to holes or electrons by inspection after working equations for a generic carrier. 'c' I also want to keep the variable name prefixes the same as used in special relativity so it's easy to see how the equations translate back to the basic Einstein relationships. Therefore, I'm going to use subscripts to denote the particular semiconductor values. E or epsilon will just represent total energy as usual, C is analogous to the capping speed of light, but it's subscript, sc, will stand for speed of carrier. E_mc0 will represent the Energy of the mass of a carrier with no momentum. eg: rest mass. m_c0 represents the semi-relativity scalar mass of a carrier with no momentum. etc.

    I expect that:

    $$ \epsilon( \vec{k} ) = \epsilon ( k ), k = | { \vec{k} | } $$

    From the general Einstein energy and mass relationship; we know the following (by direct analogy):
    $$ \epsilon = m_c^{*} C_{sc}^2 = \pm \sqrt { E_{mc0}^2 + (P_c \cdot C_{sc})^2 } = \pm \sqrt { E_{mc0}^2 + C_{sc}^2 \cdot ( \hbar k )^2 } $$

    In the ARXIV article, eqn (16) with (17) to (19) defines the same correspondence. The rest energy is half the band-gap of the semiconductor AKA E_mc0.

    Basic Cyclotron resonance theory is described at these links:
    http://physics.usask.ca/~alex/phys470/week1.pdf http://large.stanford.edu/courses/2007/ap273/hellstrom1
    I came to the conclusion that I can use equation (21) in the ARXIV article as a starting point to get the same result as the general Einstein relationship. Equation (21) is:
    $$ \epsilon
    = \pm \sqrt { E_{mc0}^2 + 2 E_{mc0} \left( {{ \hbar^2 k_{ct}^2} \over {2 m_{ct0}} } + { { \hbar^2 k_{cl}^2 } \over {2 m_{cl0}}} \right) } $$
    ARXIV cites Arnov and Pikus to transform the directional masses into an isotropic one. http://www.jetp.ac.ru/cgi-bin/dn/e_024_02_0339.pdf
    By inspection, I think the original ARXIV eqn. (21) has to transform into the following, but I don't know all the steps or implications.
    $$ \epsilon
    =\pm \sqrt { E_{mc0}^2 + {{ 2 E_{mc0} \cdot ( \hbar k )^2 } \over { 2 m_{c0}^{*}} } }
    = \pm \sqrt { E_{mc0}^2 + C_{sc}^2 \cdot ( \hbar k )^2 }

    So, I'm going to use the final step of the transform, to see if I can change the (m*) of a carrier into an F=Ma type mass (M_c) .
    The first derivative is of the transform is:
    $$ { { d \epsilon } \over { d ( \hbar k ) } }
    = { 1 \over 2 } { { C_{sc}^2 \cdot 2( \hbar k ) } \over { \epsilon } }
    = { C_{sc}^2 ( \hbar k \cdot \epsilon^{-1} ) } $$
    The second derivative is:
    $$ { { d^2 \epsilon } \over { d ( \hbar k )^2 } }
    = C_{sc}^2 \left({ 1 \cdot \epsilon^{-1} + \hbar k \cdot (-1) \epsilon^{-2} } { { d \epsilon } \over { d ( \hbar k ) } } \right )
    = C_{sc}^2 \left({ \epsilon^{-1} - C_{sc}^2 ( \hbar k )^2 \epsilon^{-3} } \right)
    = C_{sc}^2 \left( { { \epsilon^{2} - C_{sc}^2 ( \hbar k )^2 } \over \epsilon^{3} } \right)
    = { C_{sc}^2 \over \epsilon^3 } \left( { E_{mc0}^2 + C_{sc}^2 \cdot ( \hbar k )^2 - C_{sc}^2 \cdot ( \hbar k )^2 } \right) ={ { C_{sc}^2 \over \epsilon } { E_{mc0}^2 \over \epsilon ^2 } }

    The result ought to be 1 over Mass as a scalar.
    Given that total energy divided by a square C is the same as relativistic mass, the formula can algebraically reworked as:
    $$ M_c^{*} = m_c^{*} \cdot \left ( { \epsilon \over E_{mc0} } \right )^2 $$

    Since the total energy is always greater or equal to the rest energy, the Mass of the carrier is always larger than it's quasi-relativistic equivalent mass. I have a graph at the bottom of another thread, which I am updating as I work the problem here. I converted from red m_c* to orange M_c using a carrier mass at energy of kt/2 from the band edge. The conversion I derived in this post is not quite sufficient to match the original data; although it might be that I've chosen the wrong energy offset from the band-edge to represent the carrier who's mass most affects the cyclotron measurement.

    Last edited: Aug 10, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted