Monthly Archive: April 2016

Last column introduced the underlying formalism for describing waves in plasmas.  The key entities in the analysis are the dispersion tensor $$\overleftrightarrow{D}(\vec n, \omega)$$ and the dielectric tensor $$\overleftrightarrow{K}$$.  The latter quantity connects the behavior of the plasma to the underlying particle motions via the conductivity tensor $$\overleftrightarrow{\sigma}$$.  Setting the determinant of the dispersion tensor to zero results in the conditions for wave propagation; that is to say the dispersion relation that connects wavelength and frequency response.

The overall approach is to find a tractable, linear set of equations that related the state variables to each other.  Since one condition for a plasma is the maintenance of neutrality, at least at fairly large scales, the assumption will be that the state variables consist of a constant piece at zeroth order plus a first-order piece that acts as perturbations:  $$\vec B = \vec B^{(0)} + \vec B^{(1)} + \ldots$$.  The perturbation terms (e.g. $$\vec B^{(1)}$$) will then carry all of the action.

This installment looks at the most basic perturbation scenario available – waves in an unmagnetized cold plasma (see Section 4.3 of Gurnett and Bhattacharjee).  The technical conditions the plasma must satisfy to be classified this way are that its zeroth-order magnetic field $$\vec B^{(0)} = 0$$ (unmagnetized) and that its zeroth order particle velocities $$\vec v_s^{(0)}$$ and electric field $$\vec E^{(0)}$$ are also zero.

The particle density is also expanded perturbatively as

\[ n_s = n_s^{(0)} + n_s^{(1)} + \ldots \]

subject to the quasi-neutrality condition

\[ \sum_s e_s n_s^{(0)} = 0 \; .\]

The index $$s$$ specifies the species of the plasma: $$s=e$$ for electrons, $$s=i$$ for ions or $$s=h$$ for hydrogen, $$s=he$$ for helium, and so on.

As discussed last week, the approach is to start with the individual particle motions and derive a chain of results that link these with the conductivity tensor, then the dielectric tensor, then the dispersion tensor, and finally with the latter’s determinant.

For an unmagnetized cold plasma, the particle equations of motion (Lorenz force law) are:

\[ m_s \frac{d \vec v_s^{(1)}}{dt} = e_s \left[ \vec E^{(1)} + \vec v^{(1)} \times \vec B^{(1)} \right] \]

and

\[ \vec J = \sum_s e_s n_s \vec v_s = \sum_s e_s (n_s^{(0)} + n_s^{(1)}) \vec v_s^{(1)} \; . \]

Embracing the usual spirit of perturbative analysis, we throw away anything involving products of any first-order terms.  The resulting equations simplify considerably to become:

\[ m_s \frac{d \vec v_s^{(1)}}{dt} = e_s \vec E^{(1)} \]

and

\[ \vec J = \sum_s e_s n_s^{(0)} \vec v_s^{(1)} \; .\]

The Fourier transform of the differential equations yields:

\[ i \omega m_s \vec v_s^{(1)} = e_s \vec E^{(1)} \; .\]

Eliminating $$\vec v_s^{(1)}$$ from $$\vec J$$ gives

\[ \vec J = \sum_s e_s n_s^{(0)} i \frac{e_s}{\omega m_s} \vec E^{(1)} \; , \]

from which one immediately reads off the conductivity tensor as

\[ \overleftrightarrow{\sigma} = \sum_s \frac{e_s^2 n_s^{(0)} }{-i \omega m_s} \overleftrightarrow{1} \; . \]

The dielectric tensor follows immediately as

\[ \overleftrightarrow{K} = \overleftrightarrow{1}\left[1 – \sum_s \frac{e_s^2 n_s}{\omega^2 \epsilon_0 m_s} \right] \; .\]

Note that the perturbation order notation has been dropped for notational convenience.

The plasma frequency for an individual species $$\omega_{ps}$$ is a fundamental quantity that is defined as

\[ \omega_{ps}^2 = \frac{e_s^2 n_s}{\epsilon_0 m_s} \; . \]

The total plasma frequency of then defined as

\[ \omega_p^2 = \sum_s \omega_{ps}^2 \; .\]

Now, as there is no other physical direction in the problem, the direction of propagation can be taken, without loss of generality, to be along the $$z$$-axis. Thus the index of refraction vector is

\[ \vec n = \left[ 0, 0, n \right]^{T} \; .\]

The matrix $$(\vec n ^{\times})^2$$ simplifies to

\[ (\vec n ^{\times})^2 = \left[ \begin{array}{ccc} -n^2 & 0 & 0 \\ 0 & -n^2 & 0 \\ 0 & 0 & 0 \end{array} \right] \; . \]

The dispersion tensor is then

\[ \overleftrightarrow{D}(\vec n,\omega)= \left[ \begin{array}{ccc} 1 – \omega_p^2/\omega^2 -n^2 & 0 & 0 \\ 0 & 1 – \omega_p^2/\omega^2 -n^2 & 0 \\ 0 & 0 & 1 – \omega_p^2/\omega^2 \end{array} \right] \; . \]

Demanding that the determinant is zero gives the characteristic equation

\[ \left(1 – \frac{\omega_p^2}{\omega^2} – n^2 \right)^2 \left(1 – \frac{\omega_p^2}{\omega^2} \right) = 0 \; . \]

There are clearly two roots. The first, simple root, along the direction of propagation is called the longitudinal mode and has the dispersion relation

\[ \omega^2 = \omega_p^2 \; .\]

The second root, which is a double root, is associated with transverse mode and has the dispersion relation

\[ \omega^2 = \omega_p^2 + c^2 k^2 \; , \]

where $$c$$ is the speed of light and $$k$$ is the wave number.

Once the dispersion relation is obtained, the analysis of the resulting wave motion follows from looking at the phase and group velocities.

The dispersion relation for the longitudinal mode is independent of the wave number and the resulting motion, electrostatic in origin, is really an oscillation at the plasma frequency of the charge fluctuations. No wave actually propagates since the group velocity $$d \omega / d k$$ is zero.

The situation for the transverse mode is quite different. At frequencies large compared to the plasma frequency the wave behaves like a free-space wave with the dispersion relation $$\omega = \pm c k$$. As the frequency approaches the plasma frequency the nature of the propagation changes dramatically and propagation ceases entirely below the plasma frequency. The group velocity for the wave

\[ v_g = \frac{d \omega}{dk} = c \sqrt{1 – \frac{\omega_p^2}{\omega^2} } \]

clearly shows the cutoff as $$\omega \rightarrow \omega_p$$. It is interesting to note that while the group velocity is always lees than $$c$$, the phase velocity

\[ v_p = \frac{\omega}{k} = \frac{c}{\sqrt{1 – \frac{\omega_p^2}{\omega^2} }} \]

is always greater than $$c$$.

Next column will look at what happens when the assumption of $$\vec B = 0 $$ is relaxed.

This week is the start of a several columns devoted to studying wave propagation in plasmas. The wave behavior in a plasma are very rich due to the collective behavior of the plasma that results from its desire to keep charge neutrality. The presentation here is strongly influenced by and closely follows the approach presented in Introduction to Plasma Physics with Space and Laboratory Applications by Gurnett and Bhattacharjee.

Waves in a plasma represent the propagation over long distances of a complicated patterns of interaction between the charge particles that make up the plasma and the electromagnetic fields present. Thus the starting point is the Lorentz force law

\[ m_s \frac{d \vec v_s}{d t} = q_s (\vec E + \vec v_s \times \vec B ) \; , \]

where $$s$$ can be thought of as a multi-index labeling both particle and species, and
Maxwell’s equations.

The particular form of Maxwell’s equations – microscopic versus macroscopic – that is best to use is a bit of a thorny question. On one hand, the electrons and ions that comprise a plasma are free to move about over very large distances. On the other, they act together to maintain charge neutrality. Gurnett and Bhattacharjee seem to split the difference by analyzing both the microscopic set

\[ \begin{array}{ccc} \nabla \cdot \vec E & = & \rho/\epsilon_0 = (\rho_{free}+\rho_{pol})/\epsilon_0 \\ \nabla \cdot \vec B & = & 0 \\ \nabla \times \vec E & = & – \partial_t \vec B \\ \nabla \times \vec B & = & \mu_0 \vec J + \mu_0 \epsilon_0 \partial_t \vec E \end{array} \]

and the macroscopic set

\[ \begin{array}{ccc} \nabla \cdot \vec D & = & \rho_{free} \\ \nabla \cdot \vec B & = & 0 \\ \nabla \times \vec E & = & – \partial_t \vec B \\ \nabla \times \vec H & = & \vec J_{free} + \partial_t \vec D \end{array} \; \]

and then enforcing a sort of consistency relation between them. The central notion is that all of the charges are to be counted as bound charges so that the usual linear relation between the displacement vector, the electric field, and the polarization – $$\vec D = \epsilon_0 \vec E + \vec P$$ holds. Gurnett and Bhattacharjee are not particularly clear on how to treat the magnetic constitutive relation other than to say

This half-and-half approach seems to be related to the points raised in the discussion in Section 3.2 of Chen’s book, where he points out that the dependence of the magnetic moment of the individual particles is given by

\[ \mu = \frac{m v_{\perp}^2}{2 B} \; , \]

which makes the magnetization, $$\vec M$$, nonlinear in $$|\vec B|$$, thus complicating the purely-macroscopic approach.

In any event, the Fourier transform is the usual tool used to tackle wave phenomena, since it provides a transform framework that handles both spatial and time variations in a convenient way. The Fourier transform replaces all time derivatives with multiplications by frequency and replaces all spatial derivatives with multiplication by a wave number:

\[ \partial_t \rightarrow – i \omega \]

\[ \nabla \cdot \rightarrow \vec k \cdot \]

\[ \nabla \times \rightarrow \vec k \times \]

Using this translations, the microscopic Maxwell equations become

\[ \begin{array}{ccc} i \hat k \cdot \vec E & = & \rho/\epsilon_0 \\ i \hat k \cdot \vec B & = & 0 \\ i \hat k \times \vec E & = & i \omega \vec B \\ i \hat k \times \vec B & = & \mu_0 \vec J – i \omega \mu_0 \epsilon_0 \vec E \end{array} \; ,\]

while the macroscopic equations become

\[ \begin{array}{ccc} i \hat k \cdot \vec E & = & 0 \\ i \hat k \cdot \vec B & = & 0 \\ i \hat k \times \vec E & = & i \omega \vec B \\ i \hat k \times \vec B & = & – i \omega \mu_0 \vec D \end{array} \; .\]

It is understood that the vector quantities in both set of equations are the Fourier transforms and, thus, depend on frequency and wave number and not time and position. Also note that the assumptions that a plasma has only bound charges and no magnetization results in no free density or current terms in the macroscopic set.

To close the set of equations, two constitutive relations need to be adopted. The first is the generalized Ohm’s law that relates the current to the electric field

\[ \vec J = \overleftrightarrow{\sigma} \cdot \vec E \; \]

through the conductivity tensor $$\overleftrightarrow{\sigma}$$.

The second one is a generalized dielectric relation between the displacement vector and the electric field

\[ \vec D = \epsilon_0 \overleftrightarrow{K} \cdot \vec E \]

through the dielectric tensor $$\overleftrightarrow{K}$$.

A formal connection between the conductivity and dielectric tensors can be made by equating the right-hand sides of Ampere’s law in microscopic and macroscopic form to yield

\[ \mu_0 \overleftrightarrow{\sigma} \cdot \vec E – i \omega \epsilon_0 \mu_0 \vec E = – i \omega \mu_0 \epsilon_0 \overleftrightarrow{K} \cdot \vec E \; .\]

A simple re-arrangement of terms results in the dielectric tensor express as

\[ \overleftrightarrow{K} = \overleftrightarrow{1} – \frac{ \overleftrightarrow{\sigma} }{i \omega \epsilon_0 } \; .\]

With the connection of the conductivity and dielectric tensors made, the microscopic equation now take a back seat to the macroscopic ones. The wave equation results when substituting Ampere’s equation into Faraday’s. Doing so yields

\[ \vec k (\vec k \times \vec E) + \frac{\omega^2}{c^2} \overleftrightarrow{K} \cdot \vec E = 0 \; .\]

Defining an index of refraction vector as

\[ \vec n = \frac{c}{\omega} \vec k\]

and using the fundamental relation

\[ \mu_0 \epsilon_0 = \frac{1}{c^2} \]

gives the form

\[ \vec n \times (\vec n \times \vec E) + \overleftrightarrow{K} \cdot \vec E = 0 \; .\]

The only way for this equation to be satisfied with a non-trivial electric field is for the operator that acts on it to be singular. If the equation can be written as

\[ \overleftrightarrow{D}(\vec n, \omega) \cdot \vec E = \vec 0 \]

then the determinant of $$\overleftrightarrow{D}(\vec n, \omega)$$ must be zero.

Unfortunately, $$\overleftrightarrow{D}(\vec n, \omega)$$ hides its structure with the complicated division into two pieces, one of which involves a double cross-product. Fortunately, there are convenient tools from mechanics that can tame the first term and join that which is split.

The first step is to recognize that a vector cross product can be represented in terms of matrix multiplication

\[ \vec n \times \doteq \left[ \begin{array}{ccc} 0 & -n_z & n_y \\n_z & 0 & -n_x \\ -n_y & n_x & 0 \end{array} \right] \; ,\]

which is often denoted as

\[ \vec n \times \equiv \vec n^{\times} \; .\]

This step is very convenient since it puts the cross product into matrix form and it rids the need for the parentheses. The double product immediately results from a second application of the matrix $$\vec n^{\times}$$ giving

\[ \vec n \times (\hat n \times \vec E) \doteq \left[ \begin{array}{ccc} -n_y^2 – n_z^2 & n_x n_y & n_x n_z \\n_x n_y & -n_x^2 – n_z^2 & n_y n_z \\n_x n_z & n_y n_z & -n_x^2 -n_y^2 \end{array} \right] \; .\]

Of course, this same result falls out from the expansion of $$\vec n \times (\vec n \times \vec E) = \vec n (\vec n \cdot \vec E) – \vec E (\vec n \cdot \vec n) $$ using BAC-CAB.

The operator $$\overleftrightarrow{D}(\vec n, \omega)$$, which I will call the dispersion tensor, is then recognized as the sum

\[ \overleftrightarrow{D}(\vec n, \omega) = (\vec n^{\times})^2 + \overleftrightarrow{K} = (\vec n^{\times})^2 + \overleftrightarrow{1} – \frac{\overleftrightarrow{\sigma}}{i \omega \epsilon_0} \; .\]

Much of the relevant behavior of the waves come from mining relationships that come from the requirement that the determinant of the dispersion tensor is zero. This is the coupled-equation form of the requirement that the wave operator $$\nabla^2 – c^{-2} \partial_t^2$$ is zero when operating on the a valid waveform (e.g. $$f(x-ct) + g(x+ct)$$).

The strategy for finding the dispersion tensor is as follows. First compute the conductivity tensor using some model of the plasma. The possible ways of tackling this seem to be single particle motion from the Lorentz law, magnetohydrodynamic methods, and kinetic theory. Next get the dielectric tensor by plugging the conductivity tensor into the relation above. Finally, add this piece to $$({\vec n^{\times}})^2$$.

Next week I’ll examine the implementation of this strategy for a cold plasma.

This week’s column is relatively short but sweet. It covers the general operations that can be carried out on the Gaussian family of integrals. Since these integrals are relatively easy to compute and they find applications in numerous fields they are a cornerstone of many physical disciplines and appear in many models of basic systems.

The prototype integral upon which the whole family is built is the traditional

\[ I = \int_{-\infty}^{\infty} dx \; e^{-a x^2} \; .\]

Although the derivation of this integral is well known, I’ll cover the computation for the sake of completeness.

Solving for the value of $$I$$ is much easier when one squares both sides to get

\[ I^2 = \int_{-\infty}^{\infty} dx \, e^{-a x^2} \int_{-\infty}^{\infty} dy \, e^{-a y^2} \; . \]

The next step is to put the repeated integral together as a double integral and then to change coordinates from Cartesian to plane polar. Following this program, one soon arrives at

\[ I^2 = \int_{0}^{\infty} \int_{0}^{2\pi}\, dr \, d\theta \, r e^{-ar^2} \; .\]

The $$\theta$$ integral is easily done leaving

\[ I^2 = 2 \pi \int_{0}^{\infty} \, dr \, r e^{-a r^2} \; .\]

Letting $$s = a r^2$$ gives $$ds = 2\, a\, r\, dr$$ with limits $$s(r=0) = 0$$ and $$s(r=\infty) = \infty$$. Plugging these in yields

\[ I^2 = \frac{\pi}{a} \int_{0}^{\infty} ds \, e^{-s} \; ,\]

which is a standard integral that can be immediate solved to give

\[ I^2 = \frac{\pi}{a} \left. e^{-s} \right|_{0}^{\infty} = \frac{\pi}{a} \]

or, more simply

\[ I = \int_{-\infty}^{\infty} \, dx \, e^{-a x^2} = \sqrt{ \frac{\pi}{a} } \; .\]

This can be generalized without much more work to the more fearsome-looking (but actually not much harder) integral family

\[ I_n = \int_{-\infty}^{\infty} \, dx \, x^n e^{-a x^2 + b x } \; n = 0, 1, 2, \ldots \; .\]

The first step is to evaluate $$I_0$$, which is the only true integral that will have to be done. The integral

\[ I_0 = \int_{-\infty}^{\infty} \, dx \, e^{-a x^2 + b x} \]

is tackled by completing the square:

\[ \int_{-\infty}^{\infty}\, dx e^{-a x^2 + b x} = \int_{-\infty}^{\infty} dx \; e^{-(\sqrt{a} x – \frac{b}{2\sqrt{a}})^2} e^{b^2/4a} \; . \]

The last term is a constant that can be brought out of the integral. The remaining piece is best computed by first defining

\[ y = \sqrt{a} x – \frac{b}{2\sqrt{a}} \; \]

and

\[ dy = \sqrt{a} dx \; .\]

The limits of integration stay the same the new form of the integral is

\[ I_0 = e^{b^2/4a} \frac{1}{\sqrt{a}} \int_{-\infty}^{\infty} \, dy e^{-y^2} = \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; . \]

This result is so important it deserves its own box

All the other integrals come from the clever trick that says that powers of $$x$$ can be obtained from $$I_0$$ by differentiating with respect to either $$b$$ (all powers) or $$a$$ (even powers only).

For example, by definition

\[ I_1 = \int_{-\infty}^{\infty} \, dx \, x e^{-a x^2 + bx} \; ,\]

but, by construction

\[ \frac{\partial}{\partial b} I_0 = \frac{\partial}{\partial b} \int_{-\infty}^{\infty} \, dx \, e^{-a x^2 + bx} = \int_{-\infty}^{\infty} \, dx \, \frac{\partial}{\partial b} e^{-a x^2 + bx} \\ = \int_{-\infty}^{\infty} \, dx \, x e^{-a x^2 + bx} \; .\]

So

\[ I_1 = \frac{\partial}{\partial b} I_0 = \frac{\partial}{\partial b} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \]

and the computation of this integral is done by performing a differentiation – a much easier task – to get

\[ I_1 = \frac{b}{4a} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; . \]

This generalizes immediately to

\[ I_n = \frac{\partial^n}{\partial b^n} I_0 \; .\]

Sometimes the integrals we wish to compute have $$b=0$$. In this case, which is fairly common, we can get the desired results almost as easily.

For $$n$$ odd, the integrand will be a product of an even function $$\exp({-ax^2})$$ and an odd function $$x^n$$ over an even domain. The result is exactly zero.

The $$n$$ even case comes from a similar differentiation with respect to $$a$$. For example,

\[ I_2 = -\frac{\partial}{\partial a} I_0 = \int_{-\infty}^{\infty} \, dx \, x^2 e^{-a x^2} \; .\]

Evaluating the derivative gives

\[ I_2 = -\frac{\partial}{\partial a} \sqrt{\frac{\pi}{a}} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}} \; .\]

There are two consistency checks that are worth noting.

First, the recipe involving $$b$$ can still be used in the $$b = 0$$ case by taking the limit as $$b$$ approaches zero as the last step. In this limit,

\[ \lim_{b\rightarrow 0} I_1 = \lim_{b\rightarrow 0} \frac{b}{4a} \sqrt{\pi}{a} e^{b^2/4a} = 0 \; , \]

which checks with the partity argument given above.

Second, one would like to see that

\[ -\frac{\partial}{\partial a} I_0 = \frac{\partial^2}{\partial b^2} I_0 \]

for the computation of $$I_2$$. The left-hand side gives

\[ -\frac{\partial}{\partial a} I_0 = \frac{1}{2} \sqrt{\frac{\pi}{a^3}} e^{b^2/4a} + \frac{b^2}{4 a^2} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; .\]

Happily, the right-hand side gives the same:

\[ \frac{\partial^2}{\partial b^2} I_0 = \frac{\partial}{\partial b} \frac{b}{4a} \sqrt{\frac{\pi}{a}} e^{b^2/4a} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}} e^{b^2/4a} + \frac{b^2}{4 a^2} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; .\]

In closing, a note on strategy. It seems that even in the case where $$b \neq 0$$ it is easier to differentiate with respect to $$a$$ for even powers.

Most students find the study of the motion of a rigid body a difficult undertaking. There are many reasons for this and it is hard to rank which one is the greatest stumbling block. Regardless of its rank, fleshing out the relationship between the angular velocity in the body-frame and the angular in the inertial frame is certainly near the top.

First off, the very concept of angular velocity being represented by a vector is foreign concept and difficult to reconcile with the idea that finite rotations are represented by matrices that do not commute. Second, we usually want to express the angular velocity in terms of time derivatives of certain Euler angles and each of these is most naturally expressed in a frame different from all the others.

This short column is intended to give a step-by-step calculation that addresses some of the difficulties encountered in this second point. While the results are not new, the presentation is fairly novel in the pedagogy, which is clean and straightforward.

For sake of concreteness, this presentation will focus on the traditional Euler 3-1-3 sequence used in celestial mechanics to orient the plane of an orbit. This selection has two advantages: it is commonly found in textbooks (e.g. Goldstein) and it involves only two transformation primitives. Its one drawback is that modern attitude control typically uses a different sequence. This latter point may be address in a companion entry at a future date.

We will use the transformation primitives

\[ {\mathcal T}_x(\alpha) = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right] \; \]

and

\[ {\mathcal T}_z(\alpha) = \left[ \begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right] \; . \]

The classical Euler 3-1-3 sequence gives a transformation matrix from inertial to body frame expressed as

\[ {\mathcal A} = {\mathcal T}_z(\psi) {\mathcal T}_x(\theta) {\mathcal T}_z(\phi) \; . \]

Before finding the angular velocity in the body frame, an explicit form of the attitude, $${\mathcal A}$$, is needed as it serves two purposes. It is the operator that maps body rates in the inertial frame to the body frame. It is also the device by which we get the intermediate frame orientations during the transformation. These intermediate frames provide the most straightforward route for expressing the Euler angle rates as inertial angular velocity vectors.

Start with a a known inertial frame here denoted as $$\left\{\hat x_{GCI}, \hat y_{GCI}, \hat z_{GCI} \right\}$$

and begin by applying the operator $${\mathcal T}_z(\phi)$$ corresponding to a tranformation into a frame rotated by an angle $$\phi$$ about $$\hat z_{GCI}$$.

The intermediate transformation matrix

\[ {\mathcal A}^{(1)} = \left[ \begin{array}{ccc} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{array}\right] \; \]

allows us to read off the basis vectors in the intermediate frame from the rows of $${\mathcal A}^{(1)}$$:

\[ \hat x^{(1)} = \left[ \begin{array}{c} \cos \phi \\ \sin \phi \\ 0 \end{array} \right]_{GCI} \; ,\]

\[ \hat y^{(1)} = \left[ \begin{array}{c} -\sin \phi \\ \cos \phi \\ 0 \end{array} \right]_{GCI} \; ,\]

and

\[ \hat z^{(1)} = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]_{GCI} \; .\]

The subscript $$GCI$$ reminds us that these vectors are expressed in the inertial frame even though they point along the axes of the intermediate frame.

Next apply the operator $${\mathcal T}_x(\theta)$$ corresponding to a tranformation into a frame rotated by an angle $$\theta$$ about $$\hat x ^{(1)}$$.

The corresponding intermediate transformation matrix

\[ {\mathcal A}^{(2)} = \left[ \begin{array}{ccc} \cos \phi & \sin \phi & 0 \\ – \sin \phi \cos \theta & \cos \phi \cos \theta & \sin \theta \\ \sin \phi \sin \theta & -\cos \phi \sin \theta & \cos \theta \end{array}\right] \; \]

and the basis vectors in this frame are

\[ \hat x^{(2)} = \left[ \begin{array}{c} \cos \phi \\ \sin \phi \\ 0 \end{array} \right]_{GCI} \; ,\]

\[ \hat y^{(2)} = \left[ \begin{array}{c} -\sin \phi \cos \theta \\ \cos \phi \cos \theta \\ \sin \theta \end{array} \right]_{GCI} \; ,\]

and

\[ \hat z^{(2)} = \left[ \begin{array}{c} \sin \phi \cos \theta \\ -\cos \phi \sin \theta \\ \cos \theta \end{array} \right]_{GCI} \; .\]

The last step is to appluy $${\mathcal T}_z(\psi)$$ corresponding to a transformation into a frame rotated by an angle $$\psi$$ about $$\hat z^{(2)}$$.

The final transformation matrix is one of the ingredients we were seeking as is given by

\[ {\mathcal A} = \left[ \begin{array}{ccc} \cos \phi \cos \psi – \sin \phi \sin \psi \cos \theta & \cos \phi \sin \psi \cos \theta + \sin \phi \cos \psi & \sin \psi \sin \theta \\ -\sin \phi \cos \psi \cos \theta – \cos \phi \sin \psi & \cos \phi \cos \psi \cos \theta – \sin \phi \sin \psi & \cos \psi \sin \theta \\ \sin \phi \sin \theta & -\cos \phi \sin \theta & \cos \theta \end{array}\right] \; . \]

Each of the Euler angles has an associated rate $$\left\{\dot \phi, \dot \theta, \dot \psi\right\}$$. The angular velocity in the body frame is related to these rates through

\[ \vec \omega = \vec \omega_{\phi} + \vec \omega_{\theta} + \vec \omega_{\psi} = \dot \phi \hat z_{GCI} + \dot \theta \hat x^{(1)} + \dot \psi \hat z^{(2)} \; .\]

The utility of pulling out the intermediate basis vectors should now be clear.

Carrying out the multiplication gives

\[ \vec \omega = \left[ \begin{array}{c} \dot \theta \cos \phi + \dot \psi \sin \phi \sin \theta \\ \dot \theta \sin \phi – \dot \psi \cos \phi \sin \theta \\ \dot \phi + \psi \cos \theta \end{array} \right]_{GCI} \; ,\]

which is the accepted answer (e.g. problem 19 in Chapter 4 of Goldstein).

The final step is to express this vector’s components in the body frame. This is done by multiplying the column array of inertial components by $${\mathcal A}$$ to get

\[ \vec \omega = \left[ \begin{array}{c} \dot \phi \sin \psi \sin \theta + \dot \theta \cos \psi \\ \dot \phi \cos \psi \sin \theta – \dot \theta \sin \psi \\ \dot \phi \cos \theta + \dot \psi \end{array} \right]_{Body} \; ,\]

which, again, can be favorably compared with Section 4-9 of Golstein (page 179 in the second edition).

The method outlined above is the most straightforward and pedagogically sound one I know. Admittedly, the multiplications and trigonometric simplifications are tedious but with a modern computer algebra system, such as wxMaxima, it can be readily done.

The world of particle physics abounds with a lot of convenient but specialized jargon that takes some getting used too. This is particularly true in the context of natural units where the speed of light $$c$$ is a fundamnetal yark stick for measuring velocity and mass/energy and where electric charge is measured in units of the charge of an electron. The use of natural units traces back to the roots of the discipline in the studies into electromagnetism. Those practiced in the art specfy energies and masses in units of electron-volts ($$eV$$) and electron-volts per $$c^2$$.

To put things in perspective, note that room temperature $$T \approx 300 k$$ corresponds to an energy content of about $$1/40 eV$$. In contrast, the energy content associated with the mass of an electron is approximately $$511,000 eV$$ or $$511 KeV$$. The key for determining when particle motion can be treated classically versus relativistically is the ratio of kinetic energy to the rest mass energy. At the low energies representative of routine atomic processes ($$< 100 eV$$), the kinetic energy is very small compared with the rest energy and classical methods work well. On the other side are ultra-relativistic energies (such is seen in the LHC) where the kinetic energy is so much larger than the rest energy that the latter can be ignored. The grey area is that stretch inbetween low and ultra-high energy where the ratio ranges from 0.1 to 10 or so. The aim of this note is to look a little at this inbetween zone where the classical description breaks down and gives way to the relativistic one. To set the notation for the relativistic descripion, note that the interval is \[ c^2 dt^2 - dx^2 - dy^2 - dz^2 = c^2 d\tau^2 \] and that coordinate velocities are denoted as $$v^x \equiv \frac{dx}{dt} $$ and so on. Dividing out $$c dt$$ from all terms in the interval gives \[ c^2 dt^2 \left[ 1 - \frac{\vec v^2}{c^2} \right] = c^2 d\tau^2 \; ,\] from which the traditional, relativistic $$\gamma$$ is defined as: \[ \left(\frac{dt}{d\tau}\right)^2 = \frac{1}{\left[1-\vec v^2/c^2\right]} \equiv \gamma^2 \; .\] If a differential movement in spacetime is denoted by \[ dq^{\mu} = \left[\begin{array}{cccc} c dt & dx & dy & dz \end{array} \right] \] and the Minkowski metric \[ \eta_{\mu\nu} = diag(1,-1,-1,-1) \] is used, then the four-velocity is given by \[ u^{\mu} \equiv \frac{d q^{\mu}}{d \tau} = \left[\begin{array}{cccc} c \frac{dt}{d\tau} & \frac{dx}{d\tau} & \frac{dy}{d\tau} & \frac{dz}{d\tau} \end{array} \right] \; .\] The chain rule proves particularly helpful, allowing the coordinate time $$t$$ to be used in place of the proper time $$\tau$$:0 \[ u^{\mu} = \frac{dt}{d\tau}\left[\begin{array}{cccc} c & \frac{dx}{d\tau} \frac{d\tau}{dt}& \frac{dy}{d\tau}\frac{d\tau}{dt} & \frac{dz}{d\tau}\frac{d\tau}{dt} \end{array} \right] \; .\] Factoring out the $$\gamma$$ gives a compact form for the four-velocity of \[ u^{\mu} = \gamma \left[\begin{array}{cc}c & \vec v \end{array} \right] \; .\] The four-momentum is defined as \[ p^{\mu} = m u^{\mu} = m \gamma \left[\begin{array}{cc}c & \vec v \end{array} \right] \; . \] The normalization of the four-momentum is of particular physical interest. It is defined as: \[ p^{\mu} p_{\mu} = m^2 \gamma^2 \left[\begin{array}{cc}c & \vec v \end{array} \right] \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right] \left[\begin{array}{c}c \\ \vec v \end{array} \right] \; ,\] which simplifies to \[ p^{\mu}p_{\mu} = m^2 \gamma^2 (c^2 - \vec v^2) = m^2 \gamma^2 c^2 \left[1-\vec v^2/c^2\right] = m^2 c^2 \; . \] The zeroth component, \[ p^0 = m c \gamma = \frac{m c}{\sqrt{1-\frac{\vec v^2}{c^2} }} \; ,\] is related to the total energy by the equation \[ E = c p^0 = mc^2 \gamma = \frac{RE}{\sqrt{1 - \frac{\vec v^2}{c^2}}} \; .\] This last relation is cornerstone for looking at the cross-over from classical to relativistic descriptions. The approach will be to compute the coordinate velocity both ways and to compare the relative error made in the classical expression. Two different scenarios will be explored. The first is the usual one where we know the kinetic energy of the particle, having accelerated it through fixed potential difference. The second, and far less common scenario, is where we know the relativistic momentum of the particle after a collision. Scenario 1: Measure the kinetic energy By its thermodynamic definition, energy is additive and the total energy of a particle is given by the sum \[ E = K + RE \] of its kinetic energy $$K$$ and rest energy $$RE = m c^2$$. Recasting the definition of energy given above in terms rest energy gives \[ E = \frac{RE}{\sqrt{1 - \frac{\vec v^2}{c^2}}} \; .\] Equating the two expressions gives \[ \frac{RE}{\sqrt{1 - \frac{\vec v^2}{c^2}}} = K + Re \; ,\] which can be simplified to \[ K = (\gamma - 1) RE \; .\] Next define the ratio of kinetic and rest energies as \[ b = \frac{K}{RE} \; .\] Using this definition leads to the identity \[ b + 1 = \gamma \] and the subsequent expression for the coordinate velocity \[ v = c \sqrt{ 1 + \frac{1}{(1+b)^2} } \; \] For a $$K = 30 \; KeV$$ electron, with rest energy $$RE = 510.839 \; KeV$$, the resulting coordinate velocity (really speed) is \[ v = 98458.806 \; km/s \; .\] The classical method, expressed via the usual relation $$K = 1/2 m v^2$$, yields \[ v_{classical} = \sqrt{\frac{2 K}{m_e}} \] or, numerically, \[ v_{classical} = 102743.45 \; km/s \; ,\] corresponding to a relative error of $$\epsilon_1 = 0.0435171$$. While not a gross error, using the classical expression results in a non-negligible difference. Scenario 2: Measure the relativistic momentum If instead of knowing the kinetic energy, suppose we know the spatial portion of the relativistic four-momentum $${\mathcal P}^2 = \vec p^2 c^2$$. The normaliztion of the four-momentum means that \[ E^2 = m^2 c^2 + p^2 c^2 \equiv RE^2 + {\mathcal P}^2 \; .\] Defining the ratio \[ d = \frac{{\mathcal P}}{RE} \] and factoring and taking the square root gives \[ E = RE \sqrt{1 + d^2} \; .\] Combining with the formula for $$E$$ in terms of $$\gamma$$ allows us to determine \[ v = c\sqrt{1-\frac{1}{1+d^2}} = c \sqrt{\frac{d^2}{1+d^2}} \; .\] If we suppose that the energy associated with $${\mathcal P} = 30 KeV$$ then the resulting speed is \[ v = 17575.59 \; km/s \] whereas the classical expression is \[ v_{classical} = 17598.29 \; km/s \; .\] The relative error is $$\epsilon_2 = 0.00192$$; substantially smaller than the case reflecting the fact that the speed is so much smaller (almost a factor of 6 smaller). It is interesting to note that, in both cases, the classical expression overestimates the speed. This is consistent with the notion that the classical theory has no fundamental speed limit like relativity but it this is only a conjecture at this point.