text:
Best Classic Heavy Metal Bands
1 Judas Priest Judas Priest are a British heavy metal band that formed in Birmingham, England, in 1969. They are often referred to as one of the greatest metal bands of all time, and are even commonly called "The Metal Gods", after one of the songs on their 1980 album "British Steel". ...read more.
2 Iron Maiden Iron Maiden are an English heavy metal band formed in Leyton, East London, in 1975 by bassist and primary songwriter Steve Harris. The most critically acclaimed period for the band was from 1983-1989. With vocalist Bruce Dickinson, bassist Steve Harris, lead guitarist Dave Murray, rhythm guitarist Adrian ...read more.
3 Black Sabbath Black Sabbath were a Heavy Metal band formed in Birmingham, England in 1968 by guitarist and songwriter Tony Iommi, singer John "Ozzy" Osbourne, bassist and main lyricist Geezer Butler and drummer Bill Ward. ...read more.
Hands Down, #1, Greatest Meatal Band of ALL TIME! Ozzy Osborne's snotty powerful voice, + best guitar riff master ever-Tony Iommi, = equals only the half of it! One of the powerful rhythm sections with drummer Bill Ward and Bassist and lyrics Geezer Butler, Well, Let's just say - I'm sure God was a little bit nervous about these guys joining forces! - Triklabel
4 Metallica Metallica is an American Thrash Metal band formed in 1981 . The original lineup was James Hetfield (Vocalist and Rhythm guitarist), Dave Mustaine (Lead Guitar), Lars Ulrich (Drums), Ron McGovney (Bassist). The group came into mainstream from their Self-Titled album and the hit single "Enter Sandman". ...read more.
This is thrash and not classic heavy metal. - Metal_Treasure
5 Megadeth Megadeth is an American thrash metal band from Los Angeles, California. Megadeth was formed in 1983 by guitarist Dave Mustaine and bassist David Ellefson, shortly after Mustaine was fired from Metallica. They have released 15 studio albums to date, and have gone through many lineup changes, with Dave ...read more.
thrash - SoldierOfFortune
6 Pantera Pantera was an American Heavy Metal band formed in 1981 by guitarist Dimebag Darrell and drummer Vinnie Paul. The band started out as a Glam Metal act in the 1980s with little success. After discarding their original lead vocalist and enlisting Phil Anselmo, the band signed a major record deal with ...read more.
I mean NWAHM - SoldierOfFortune
7 Dio Dio was an American heavy metal band formed in 1982 and led by vocalist Ronnie James Dio, after he left Black Sabbath with intentions to form a new band with fellow former Black Sabbath member, drummer Vinny Appice.
8 Deep Purple Deep Purple are an English rock band formed in Hertford in 1968. They are considered to be among the pioneers of heavy metal and modern hard rock, although their musical approach changed over the years. Originally formed as a progressive rock band, the band shifted to a heavier sound in 1970. Deep Purple, ...read more.
9 Motorhead Motörhead were an English heavy metal band formed in June 1975 by bassist, singer, and songwriter Ian Fraser "Lemmy" Kilmister, who was the sole constant member; guitarist and songwriter Larry Wallis; and drummer Lucas Fox. The band's harsh vocals and fast paced tempos helped inspire speed metal and ...read more.
10 Rainbow Rainbow are a British rock band led by guitarist Ritchie Blackmore from 1975 to 1984 and 1993 to 1997. They were originally established with Ronnie James Dio as the lead singer and lyricist. Rainbow were ranked No. 90 on VH1's 100 Greatest Artists of Hard Rock. The band has sold over 28 million albums ...read more.
11 Slayer Slayer is an American thrash metal band from Huntington Park, California, formed in 1981 by guitarists Jeff Hanneman and Kerry King. They rose to fame with their 1986 album Reign in Blood, and is credited as one of the big four of thrash metal bands, the others being Metallica, Megadeth, and Anthrax. ...read more.
Thrash/speed - CaptDeadeye
12 Manowar Manowar is an American heavy metal band from New York City, New York. Formed in 1980, the group is known for lyrics based on fantasy and mythology.
13 Diamond Head Diamond Head are an English heavy metal band formed in 1976 in Stourbridge, England. The band is recognised as one of the leading members of the new wave of British heavy metal and is acknowledged by thrash metal bands such as Metallica and Megadeth as an important early influence.
14 Mercyful Fate Mercyful Fate was a Danish heavy metal band from Copenhagen, formed in 1981 by vocalist King Diamond and guitarist Hank Shermann.
15 Accept Accept is a German heavy metal band from the town of Solingen, originally assembled by former vocalist Udo Dirkschneider, guitarist Wolf Hoffmann and bassist Peter Baltes.
16 Queensryche Queensrÿche is an American progressive heavy metal band formed in 1982 in Bellevue, Washington, out of the local band the Mob. They released their self titled EP in 1984, but didn't gain notoriety until the release of their critically acclaimed album Operation: Mindcrime in 1988. As of now, they ...read more.
17 Uriah Heep Uriah Heep are an English rock band formed in London in 1969 that was one of the top rock bands in the early 1970s. Twelve of the band's albums have made it to the UK Albums Chart (Return to Fantasy reached No. 7 in 1975). Uriah Heep's distinctive features include a massive keyboard sound, strong vocal ...read more.
Not a metal band.
Basically not a metal band but in the early 70s they released several metal songs - it was important because in the early 70s every metal song was something significant. - Metal_Treasure
18 Helloween Helloween is a German power metal band founded in 1984 in Hamburg, Northern Germany. The band is a pioneering force in the power metal genre, and their second and third studio albums, Keeper of the Seven Keys: Part I and Part II, are considered masterpieces of the genre.
19 Savatage Savatage is an American heavy metal band founded by the brothers Jon and Criss Oliva in 1978 at Astro Skate in Tarpon Springs, Florida.
20 Thin Lizzy Thin Lizzy are an Irish rock band formed in Dublin in 1969. Their leader Phil Lynott was composer or co-composer of almost all of the band's songs, and the first black Irishman to achieve commercial success in the field of rock music. Thin Lizzy's most successful songs, "Whiskey in the Jar", "Jailbreak" ...read more.
21 Anthrax Anthrax is an American thrash metal band from New York City, formed in 1981 by guitarist Scott Ian and bassist Dan Lilker.
22 Primal Fear Primal Fear are a German metal band formed in 1997 by Ralf Scheepers and Mat Sinner. They play heavy metal, power metal and speed metal. Primal Fear remind of Judas Priest but Primal Fear are faster and heavier. They use twin (dual) lead guitars. Ralf Scheepers voice reminds of Rob Halford.
"Classic Heavy Metal" can be a very confusing term, moreover the author didn't say what he meant. But I guess it's traditional / classic heavy metal (which excludes other subgenres such as thrash, prog, death, etc.)
Now, Primal Fear are a modern band but they play heavy metal of Judas Priest type. Pure metal. And they are very good. They also have speed and power metal songs. - Metal_Treasure
23 Led Zeppelin Led Zeppelin was an English rock band formed in London in 1968. The group consisted of Robert Plant (Vocal), Jimmy Page (Guitar), John Paul Jones (Bass, Keyboard) and John Bonham (Drums). The band's heavy, guitar-driven sound, rooted in blues and psychedelia on their early albums, has earned them recognition ...read more.
They have never been a metal band. - Metal_Treasure
24 Blue Cheer Blue Cheer was an American rock band that initially performed and recorded in the late 1960s and early 1970s and was sporadically active until 2009. Based in San Francisco, Blue Cheer played in a psychedelic blues rock style, and are also credited as being some of the earliest pioneers of heavy metal, ...read more.
This isn't metal. - Metal_Treasure
25 Blue Oyster Cult Blue Öyster Cult is an American rock band from Long Island, New York, whose most successful work includes the hard rock and heavy metal songs "(Don't Fear) The Reaper", "Godzilla" and "Burnin' for You".
Not a metal band. - Metal_Treasure
26 Nuclear Assault Nuclear Assault is an American thrash metal band from New York City, formed after bassist Dan Lilker left Anthrax in 1984. After releasing five full-length albums and touring relentlessly throughout the 1980s and early 1990s, Nuclear Assault split up in 1995. They briefly reunited in 1997 and again ...read more.
27 Annihilator Annihilator is a Canadian thrash metal band founded in 1984 by vocalist, guitarist, bassist, songwriter, engineer, producer, mixer and mastering engineer Jeff Waters and former vocalist John Bates.
28 Anal C*** Anal C*** was an American grindcore band founded by vocalist Seth Putnam in 1988. The band is known for their highly offensive lyrics which included themes of misogyny, homophobia, racism, Nazism, antisemitism, and schadenfreude (ridicule of the unfortunate). Their music has been received negatively ...read more.
It's under question whether they are metal at all. Metal Archives don't consider them metal because they are mostly extreme punk.
Whatever it is, I am sure their genre is NOT classic/traditional heavy metal and they shouldn't be on this list. - Metal_Treasure
29 Bolt Thrower Bolt Thrower are a British death metal band from Coventry, England. They formed in 1986 and released their first album with Vinyl Solution in 1988. The band then shifted to a new record label, Earache Records, soon becoming one of the best selling bands on that label.
This is a good band but they don't play classic heavy metal - Metal_Treasure
30 Ratt Ratt is an American heavy metal band that had significant commercial success in the 1980s, with their albums having been certified as gold, platinum, and multi-platinum by the RIAA.
31 Dream Theater Dream Theater is an American progressive metal band from Boston, Massachusetts. The band was formed in 1985 under the name of "Majesty", only to change its name to Dream Theater later. ...read more.
32 Iced Earth Iced Earth is an American heavy metal band from Tampa, Florida. It was formed in 1985 under the name Purgatory by guitarist and main songwriter Jon Schaffer and original drummer Greg Seymour.
33 Saxon Saxon are an English heavy metal band formed in 1976, in South Yorkshire. As one of the leaders of the new wave of British heavy metal, they had eight UK Top 40 albums in the 1980s including four UK Top 10 albums and two Top 5 albums.
34 Machine Head Machine Head is an American heavy metal band from Oakland, California. Formed on October 12, 1991, the group was founded by vocalist and guitarist Robb Flynn and bassist Adam Duce. They have received great critical success especially for their album "The Blackening" which is considered a modern metal ...read more.
Greatest, hardest-working classic metal band out there! In the past 20 years they have not compromised their dedication to bringing an aggressive, thundering, guitar assault that's a delight to the ears! Even if they are BLEEDING!
35 KISS Kiss is an American hard rock band formed in New York City in January 1973 by Paul Stanley, Gene Simmons and the band's original manager, Lew Linet.
Best Classic Heavy Metal Bands (Without Thrash Metal) Best Technical Heavy Metal / Post-hardcore Bands Best Heavy Metal Bands Top 10 New Wave of American Heavy Metal Bands Best Debut Albums From Heavy Metal Bands
Top Remixes (6)
2. Deep Purple
zxm
3. Motorhead
jared5


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Louisiana Katrina/Rita Relife
SecureTN
Couldn't figure out the best place to post this... I'm curious if anyone else worked in Louisiana after Hurricane Katrina and Rita, and if so... what part.
I worked in Slidell, Kenner, Butte (sp?), Choputulas (sp?) and New Orleans for Securitas... I (we) drove down shortly after Thanksgiving.
Also, are there any companies with Officers still "deployed" down south?
"What if this is as good as it gets?" ~ Melvin Udall
I was down there Oct 9th 2005- Nov. 14th 2005 with The Wackenhut Corp.
Stayed in Kenner, worked in Plaquemines Parish.
"Alright guys listen up, ya'll have probably heard this before, Jackson vs. Securiplex corporation; I am a private security officer, I have no State or governmental authority. I stand as an ordinary citizen. I have no right to; detain, interrogate or otherwise interfere with your personal property-... basically all that means is I'm a cop."-Officer Ernie
"The Curve" 1998
EMTGuard
Well, I live in Louisiana and work at Bayou Steel In LaPlace. The Sunday morning before Katrinia made landfall my supervisor went in to work. He called me and said, "If I'm still alive tommorrow I expect you to releive me as usual." I told him he was crazy to go down there with a (at that time) Cat 5 storm headed right at him but the plant had a policy of leaving a fire watch of about 6 people to ride out storms. That included Us, the Emergency Med Techs. The Melt Shop furnace bunker would be the place where the crew would shelter when things got bad.
That night the storm came ashore, I had evaced my home, a Fleetwood single wide moblie home, for the relative safety of the brick home of friends near Baton Rouge. By Monday afternoon the storm had come through and the winds had died down enough to drive. Problem was all the trees down. Most roads had only one lane open where some public works or fire dpeartment crew had come through and cut just enough to open part of a road. It took me a while to get home. When I pulled into my driveway I saw that a large section of a oak tree had split off and fallen toward my home. Luckily it did no major damage, falling on my porch and against the house and front door. With the help of a friend who owned a chainsaw we got the tree cleared and by monday evening I was back home but still without power.
Things were crazy but by late Tuesday morning I got a call on my cellphone from my supervisor who was still alive. "I don't care if you have to backstroke down the Mississippi River, just come relieve me." I used my Ham radio to contact the Office of Emergency Management in St John Parish, where the plant is, and was told that there were several roadblocks up between my home and work but since I had ligitimate business, and not sightseeing or looting, then I should have no problem getting through. I hit road blocks along I-10 twice and then once in LaPlace before gettign to work but my uniform and paperwork from the plant stating I was an essential employee got me through with hardly a word. Power was out through out LaPlace and moany homes and businesses were damaged but the city became a major staging point for police and releif agencies going into New Orleans.
I spent all of Tuesday night at work and by wednesday morning department heads were arriving at the plant to check the damage and get things up and operating agian. Several plant workers stopped by and staged in the parking lots with boats they would later use to rescue people trapped by flood waters in and around New Orleans. The Feds were holding up rescue efforts so it was groups of private citizens like the ones at our plant who banded together, found a way around roadblocks and went in to the area to help pull people out of the water.
By Wednesday moring my regular releif showed up. A hurricane had hit but we were pretty much intact and our schedule wasn't affected much since all of us EMTs lived outside of the main strike area. The regular security officers were scattered though and one SO would end up staying at the guard shack for 4 days straight eating meals brought in by department and plant heads. We had sandwiches, hamburgers, military MREs and other food galore in the guard shack after the hurricane. A folding cot was put up in the Scale house side of the guard shack so the guard could go there for a few hours of sleep with the door closed. While he slept the EMT/SO would watch the gate and handle the one working phone we had. Cellphones were iffy at best and our main switchboard was down. Luckily we had one direct line which bypassed the main switchboard. It was out main form of communication after the storm for about a week.
After being releived wednesday morning I got in my car, turned on my ham radio and signed on as mobile and headed home. My lady came up on the radio and asked how my night was and then told me to meet her at the courthouse in a neighboring Parish. What was she doing there? Had a tree, damaged by the storm, fallen on the house while I was at work? Before I could panic she told me that she was in the EOC there and had answered a request for hams with emergency training. While I had been at work she had driven over and had been manning the radios at that EOC all night. I drove there, met with the other people volunteering and ended up promissing to return for a night shift after I got some sleep. By wednesday eveing my lady and I were back at that EOC and she was assigned to the same radio desk from the night before while I was sent to a large shelter being set up at a Expo center a couple of miles down the road. The ham radios we used were the only communciations the shelter had with the local EOC, the State EOC in Baton Rouge and their Chapter HQ, also in Baton Rouge. By noon Thursday morning the shelter was full but the EOC had brought in working public safety radios and extra manpower and we were headed back home. By the time I left the shelter there were 1500 people there, most who had been driven straight from the floodwaters fo New Orleans to the Expo center in private vehicles and emergency vehicles.
By Friday evening I was back at the plant in LaPlace facing a 3 night weekend and employees were slowly returning to work as they were able.
During the hurricane emergency guards and EMTs were paid a Hurricane bonus of $100 a shift extra so some of us got pretty nice checks that following payday. Since then the policy has changed and now the plant closes and nobody is left at the plant for anything bigger than a cat 3 storm.
Attached is a photo of me taken the Thurday morning after Katrina in the shelter at Lamar Dixon Expo Center in Gonzales, La.
Last edited by EMTGuard; 10-26-2006, 06:08 PM.
LaPlace was our first point of contact when we arrived... I've actuallt debated moving to Louisiana... I liked the people and the area, minus the damage.
EMTGUARD, sometimes we do not know just how lucky we are, interesting story, thanks for sharing.
Quote me as saying I was mis-quoted.
WE went down and worked under a contract thru Shaw under a FEMA contract.
I spent three months down there from before November till February 4th. I then came back home as I had had enough. WE were shipped down by the company and worked in both New Orleans and Baton Rouge.
12 hour days 7 days a week even with $1200 a week gets a little old when your employer is always yanking your chain.
we worked at several hotels, 3 trailer distribution sites, and several trailer parks.
WE also worked at the only full service hospital still in operation.
I worked Home Depot's, Wal-mart, one of the main office's. I finally had enough with 15 hours a day, 7 days a week, standing 15 hours a day non-stop is not worth 1200 to me.

redpajama_set_name: RedPajamaCommonCrawl


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All posts tagged "Lee Grant"
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Manchester United register interest in signing teenage goalkeeping sensation
By Thomas Hein on October 12, 2018
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Manchester United have completed their third signing of the summer and it is a bit of a surprising one as Lee Grant has moved from Stoke City...
Goalkeeper who is closing in on a Manchester United move has previously earned praise from Mourinho
By Rohan Sengupta on July 4, 2018
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Manchester United are interested in signing Stoke City goalkeeper Lee Grant, according to The Mirror. The report claims that Jose Mourinho wants to sign cover for...


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text:
\section{Introduction}

Dirac formulated relativistic quantum mechanics in the late 1920's.\cite{Dirac}$\ $
Since field theory had not yet been developed, the relativistic aspect led to a number
of confusions related to currents since only total charge is conserved. With the
recognition of the appearance of antiparticles, and the development of field theory,
analogous to the transition from the Canonical to the Grand Canonical ensemble in
statistical mechanics, these issues were resolved. Nonetheless, the solution of the
equation itself, ignoring these deeper aspects, has proven valuable even in more
modern contexts.

For example, the MIT bag model of quark confinement,\cite{Bag} and the theories
that evolved from it (chiral bag, cloudy bag, etc.)\cite{AWT} depend on
solving the Dirac equation as a wave function for a particle in an effective potential.
Through boundary conditions and other approximations, the Dirac equation has
even been employed in this way in the study of nuclear structure.\cite{AWT2,GMSS}$\ $

In particle physics, the quark model has been employed, with great success,
to describe baryon and meson states and their structure.
Usually this is done in a {\it non}-relativistic model,
\cite{CorPot} where the potentials follow the patterns expected by
a Foldy-Wouthuysen reduction\cite{FW} of a simpler, theoretically motivated
potential in the Dirac equation.
In our case, however, we wanted to see how these structures and
solutions appear {\it without} the non-relativistic reduction approximations, by solving
the Dirac equation itself for the simpler potential with fewer adjustable parameters.

Going back somewhat in time, very soon after Dirac's initial formulation,
Darwin \cite{Darwin} found an analytic solution (!) for the radial wave
functions for hydrogen-like atoms in terms of confluent hypergeometric functions.
Since then, solutions of the Dirac equation have been important in atomic physics,
even to recent times.\cite{BetheSal,AuRogers}$\ $
For example, more complicated central potentials than $Z\alpha /r$, the fourth component
of a Lorentz four-vector, usually require finding a numerical solution.
Another example is a muonic atom, in which a muon is in an s-state about a heavy
nucleus having a realistic charge distribution.\cite{mu_xray}

Finding such numerical solutions involves solving coupled ordinary differential
equations for the upper and lower components of the Dirac wave function, \cite{BjD,AuRogers}
\begin{equation}
 \psi_{jlm}({\bf r}) = \left[ \begin{array}{c}
          \psi_a(r)\; {\cal{Y}}^l_{jm} \\
	   -i \psi_b(r) \; {\cal{Y}}^{l'}_{jm}
	   \end{array} \right]  = \frac{1}{r}
	   \left[ \begin{array}{c}
          i g(r)\; {\cal{Y}}^l_{jm} \\
	      -f(r) \; {\cal{Y}}^{l'}_{jm}
	   \end{array} \right]
	   \ ,
  \label{eq:psicolvec}
\end{equation}
where $l' = 2 j - l$ and the ${\cal{Y}}^l_{jm}$ are the spin-orbital angular momentum
wave functions,
\begin{equation}
	 {\cal{Y}}^{l'}_{jm}(\theta,\phi) = \sum_{m_s} <j m \;
	 	|\; l \;  {\textstyle \frac{1}{2}},\; m-m_s \; m_s>
	 	Y^l_m (\theta,\phi)\; \chi_{m_s} \ ,
\end{equation}
where $<j m \; |\; l \;  {\textstyle \frac{1}{2}},\; m-m_s \; m_s> $
is a Clebsch-Gordan coefficient and $\chi$ is a Pauli spinor.

We consider here the case of bound-state wave functions for a central potential with
energy eigenvalue $E$.
For example, for hydrogen-like atoms the potential is $V_v(r) = -Z\alpha/r$.
With appropriate boundary conditions, the radial wave functions $g(r)$ and $f(r)$,
which can be taken as real functions, are solutions of the following coupled first-order
ordinary differential equations (ODEs), \cite{BjD,BetheSal}
\begin{eqnarray}
	&g'(r)& + \;\frac{k}{r} g(r)\; - \;(E - V_v(r) + m)\; f(r) = 0 \ , \label{eq:Vv_geqn} \\
	&f'(r)& - \;\frac{k}{r} g(r)\; + \;(E - V_v(r) - m)\; g(r) = 0 \ . \label{eq:Vv_feqn}
\end{eqnarray}
Here the integer $k$ is determined by the angular momentum quantum numbers according to
\begin{eqnarray}
	k &=& -(l+1) {\rm \quad\quad if \ } j = l + {\textstyle \frac{1}{2}} \ , \nonumber \\
	k &=& l \quad\quad\ \ {\ \ \rm \quad\quad if \ } j = l - {\textstyle \frac{1}{2}} \ . \label{eq:kdef}
\end{eqnarray}

The reason for the subscript on the potential $V_v$ in those equations is to indicate that
it is the fourth component of a Lorentz four-vector, such as the Coulomb potential.
However, {\em our} motivation is not so much in atomic physics applications as it is
for treating mesons as $q \bar{q}$ states in a {\em relativistic} quark model.
The {\em non}-relativistic quark model often assumes the confining potential for
quarks to be the so-called ``Cornell potential,'' \cite{CorPot}
\begin{equation}
	V(r) = -\frac{\alpha_s}{r} + \kappa^2 r \ , \label{eq:CorPot}
\end{equation}
where $\alpha_s = g_{s}^{\ 2}/{4\pi}$, with $g_s$ being the (running) quark-gluon coupling
constant, and $\kappa^2$ (or $\sigma$) is the string tension.
It is the linear term in $V(r)$ that confines the quarks,
corresponding to the effective potential in the relativistic Bag Model.\cite{Bag}

For a relativistic quark model, the two pieces of that non-relativistic potential have
different Lorentz transformation properties.
The color Coulomb potential, $\alpha_s/r$, is the fourth component of a Lorentz vector,
while the confining linear potential transforms as a Lorentz scalar,
which we will write as $V_s(r)$.
Thus, in solving the relativistic radial Dirac equations, the two terms in the non-relativistic potential of Eq.\ (\ref{eq:CorPot}) enter the coupled ODE's differently.
The equations to be solved are
\begin{eqnarray}
	&g'(r)& + \;\frac{k}{r} g(r)\; - \;(E - V_v(r) + V_s(r) + m)\; f(r) = 0 \ ,
		\label{eq:VvVs_geqn} \\
	&f'(r)& - \;\frac{k}{r} f(r)\; + \;(E - V_v(r) - V_s(r) - m)\; g(r) = 0 \ .
		\label{eq:VvVs_feqn}
\end{eqnarray}
Note that in these equations  the sign for $V_v$ is opposite to that of the energy
eigenvalue $E$, the fourth component of the momentum four-vector, while
that for $V_s$ matches that for the fermion mass $m$, also a Lorentz scalar.

So, we were motivated to study solutions of Eqs.\ (\ref{eq:VvVs_geqn}) and (\ref{eq:VvVs_feqn}).
After submitting an earlier version of this paper to the archives,
we learned of related studies of these equations by Paris\cite{Paris} and
by Soares de Castro and Franklin\cite{JF}, who discussed analytic solutions.
We, however, have adopted a numerical approach and,
in this paper, we discuss how to go about solving these coupled ODE's numerically.
The presentation follows our numerical journey more or less historically.
Our hope in writing this paper is that, by reading it, students might avoid some
of the pitfalls we encountered and then overcame.

It turns out that solving for a linear scalar potential \cite{Critch,GMSS}
is quite a bit easier (numerically) than the relativistic hydrogen atom,
so we treat that first in Sec. II.
We then turn, in Sec. III, to the problems we had with in solving the radial equations
for a Coulomb-like potential for hydrogen-like atoms and how we dealt with them.
This is followed, in Sec. IV, by a discussion of the mixed problem having {\it both}
a scalar and a vector potential, as in Eqs.\ (\ref{eq:VvVs_geqn}) and (\ref{eq:VvVs_feqn}).

\section{Linear Scalar Potential}

For the case of massless quarks (an approximation we have made \cite{piExchg}
for the $u$ and $d$ quarks),
we want to solve the equations for a confining linear
scalar potential, \cite{Critch,GMSS}
\begin{equation}
   V_s(r) = \kappa^2 (r - r_0) \; , \label{eq:GMSSpotnl}
\end{equation}
where (with $\hbar = c = 1$), $\kappa$ has dimensions of fm$^{-1}$.
The negative offset, $-\kappa^2 r_0$, effectively gives these quarks a constituent mass.\cite{GMSS}$\ $
It also provides a rough representation of the effect of how the short-distance color
Coulomb interaction between quarks leads to quark confinement, albeit without the
correct Lorentz representation properties.

The case of massless quarks exhibits the role of relativity in a maximal way, meaning
that the lower component $f(x)$ is comparable in size to the upper component $g(x)$.
We will, later, choose an appropriate mass $m$ for the massive quark flavors, $s$, $c$, and $b$,
and solve for those wave functions.
In these cases, as the quark mass $m$ increases, $f(x)$ becomes smaller relative to $g(x)$,
reflecting the transition to a non-relativistic limit where $f(x)$ vanishes.
Also, as will be seen, with increasing mass $m$, the heavier quark radial wave functions
are progressively narrower than in the massless case.
Thus we will, in Sec.\ \ref{VsAndVv}, also include the effects of a color
Coulomb-like vector potential $V_v$ coming from one-gluon exchange.
This should prevent a too-rapid fall-off of the
quark-quark interaction energies as the quark masses increase \cite{AprilTalk},
but that is a matter for a separate paper.

It simplifies the equations if we convert the ODEs to dimensionless form,
dividing through by $\kappa$ and defining a dimensionless distance $x = \kappa r$:
\begin{eqnarray}
	&g'(x)& + \;\frac{k}{x} g(x)\; - \;(\tilde{E} + V_s(x) + \tilde{m})\; f(x) = 0 \ ,
		\label{eq:Vs_geqnx} \\
	&f'(x)& - \;\frac{k}{x} f(x)\; + \;(\tilde{E} - V_s(x) - \tilde{m})\; g(x) = 0 \ ,
		\label{eq:Vs_feqnx}
\end{eqnarray}
where $\tilde{E} = E/\kappa$, $V_s(x) = x-x_0$, and $\tilde{m} = m/\kappa$
are now also dimensionless.

Numerical integrations of these equations can be done by one or another of the
Runge-Kutta methods.\cite{RK}$\ $
To integrate these two first-order ODEs one needs to specify two boundary conditions (BCs)
at the place where one begins the integration.
However, as the radial wave functions eventually need to satisfy
a normalization condition,
\begin{equation}
	\int_{0}^{\infty} \; x^2 dx \;  [\psi_a^2(x) + \psi_b^2(x)] =
	\int_{0}^{\infty} \; dx \;  [g^2(x) + f^2(x)] = 1 \ , \label{eq:normzn}
\end{equation}
this will determine one of the BCs, for example, the scale of the wave functions at
asymptotically large distances.

\subsection{Shooting Outwards}

In our first attempt to solve these equations we thought to integrate them outward
from the origin
with the hope of adjusting the energy eigenvalue $\tilde{E}$ to assure
that both $g(x)$ and $f(x)$ fall off to zero asymptotically.
The first problem in doing this is the singularity in the equations at $x=0$.
This singularity, however, is easily avoided by starting instead at a small value
of $x = \epsilon$ away from the origin.

For starting values (BCs) we can take advantage of the expected
power-law behaviors of $\psi_a(x)$ and $\psi_b(x)$ near the origin for their
known orbital angular momenta $l$ and $l'$.
For example, if $j = l + \frac{1}{2}$, for which $k = -(l+1)$ and the lower
component's $l' = l+1$, one can choose
\begin{equation}
	g(\epsilon) = a_0 \;\epsilon^{l+1} \ , \quad
	f(\epsilon) = b_0 \;\epsilon^{l+2} \ , \label{eq:BC34_l+1/2}
\end{equation}
as motivated by the non-relativistic limit, where
$\psi_a \rightarrow \Psi_{\rm NR} \sim r^l$ at small $r$.
Substituting this $f(\epsilon)$ into the second ODE allows us to determine $b_0$:
\begin{equation}
	f(\epsilon) = -a_0 \; (\tilde{E} - V_s(\epsilon) - \tilde{m})\;
		\epsilon^{l+2}/(2 l + 3) \ . \label{eq:fBC4}
\end{equation}
The parameter $a_0$ at this point is arbitrary for now and will be fixed later
by the normalization condition, Eq.\ (\ref{eq:normzn}).

The problem with simply shooting outwards is what to choose for the energy $E$.
The ODE solver for almost all initial guesses for $E$ will have the
well-known problem that the calculated
$g(r)$ and $f(r)$ very soon blow up exponentially, either positively or negatively.
It might be possible to iteratively refine the guess for $E$ to eventually find
decaying solution, but we soon decided to try a different method.

\subsection{Shooting Inwards}

A better way of assuring decaying solutions is to start from an asymptotic distance,
where that behavior is built-in, and integrate the ODEs inward toward the origin.
At large distances the ODEs simplify, for this potential, to
\begin{eqnarray}
	&&g'(x) -  x f(x) = 0 \ , \nonumber \\
	&&f'(x) -  x g(x) = 0 \ ,
\end{eqnarray}
which have solutions at large $x = x_{\rm max}$,
\begin{eqnarray}
	&&g(x_{\rm max}) =  a_1 e^{-x_{\rm max}^2/2} \ , \nonumber \\
	&&f(x_{\rm max}) = -a_1 e^{-x_{\rm max}^2/2} \ . \label{eq:BC12}
\end{eqnarray}
Note that this asymptotic behavior is the same as that of the non-relativistic
simple harmonic oscillator wave functions.\cite{Critch}$\ $
These forms will be used as starting values (BCs) for the inward integration of the
full ODEs from $x_{\rm max}$.
Again, $a_1$ can be taken as arbitrary for now, to be later fixed by the normalization
condition, but we need to make an initial guess for the energy eigenvalue $\tilde{E}$.
The integration will go from $x_{\rm max}$ back to $x = \epsilon$, again to avoid
the singularity at $x=0$.

The problem with this shooting inwards method is much the same as that for
shooting outwards.
Unless one somehow is able to guess the exact value of $\tilde{E}$,
the solutions will blow up exponentially as one nears the
origin, and usually well before that.

\subsection{Shooting In and Out and Matching}

Thus, after experiencing these well-known problems with ODEs with eigenvalues,
we concluded that we had to combine the two methods, invoking a well-known
numerical method, ``shoot and match.''
In this case we shoot both outwards from $x = \epsilon$ to a point in the middle,
$x_{\rm match}$, and inwards from $x_{\rm max}$ back to $x_{\rm match}$.
The two values for $g_{\rm out}(x_{\rm match})$ and $g_{\rm in}(x_{\rm match})$
will generally differ for a given choice of $\tilde{E}$ and another parameter,
which we choose to be $a_0$.
That is likewise the case for $f_{\rm out}(x_{\rm match})$ and $f_{\rm in}(x_{\rm match})$.
What we therefore need is to find a way to {\it iteratively vary} $\tilde{E}$ and $a_0$
to reduce the two gaps, $\Delta g$ and $\Delta f$, to zero.
[The asymptotic parameter $a_1$ is chosen to be fixed, i.e., not to be varied,
but of a size that makes the initial determination of the gaps at $x_{\rm match}$
reasonably small; it will in the end be determined by the normalization condition,
Eq.\ (\ref{eq:normzn}).]

We need to define the gaps $\Delta g$, etc., at the match point.
Initially, we simply chose them in terms of their actual differences,
\begin{equation}
	\Delta g(\tilde{E},a_0) = g_{\rm out}(x_{\rm match}) - g_{\rm in}(x_{\rm match})
	\ , \label{eq:actualgaps}
\end{equation}
and likewise for $\Delta f$, $\Delta g'$, and $\Delta f'$.
Indeed, this often works for the problem we originally set out to do, namely, to calculate
wave functions for the (nearly) massless $u$ and $d$ quarks.
However, as the quark mass $\tilde{m}$ increases, the system becomes more and more
non-relativistic.
This means that the lower component wave function, $f(x)$, becomes relatively small
compared with $g(x)$.
Thus a better definition for the gaps would be to make them relative to their average
value, e.g.,
\begin{equation}
	\Delta g(\tilde{E},a_0) =
		\frac{2\;[\;g_{\rm out}(x_{\rm match}) - g_{\rm in}(x_{\rm match})\;]}
			{g_{\rm out}(x_{\rm match}) + g_{\rm in}(x_{\rm match})}
		\ , \label{eq:scaledgaps}
\end{equation}
and likewise for $\Delta f(\tilde{E},a_0)$.

\subsection{Zeroing the Gaps}

The two gaps can be driven to zero by using a generalized Newton-Raphson method.
First, recall how one derives the Newton-Raphson method for finding the zero
of a function $y(x)$.
Make an initial guess for the solution, call it $x_1$.
To find a better guess, $x_2$, expand in a Taylor's expansion,
\begin{equation}
	y(x_2) = y(x_1) + y'(x_1) (x_2 - x_1) + \cdots \ . \label{eq:Taylor}
\end{equation}
As we want the left-hand side to vanish, the next (and better) guess for the root is
\begin{equation}
	x_2 = x_1 - y(x_1)/y'(x_1) \ .  \label{eq:Newton}
\end{equation}
This procedure can be repeated until $y(x_n)$ is small enough to be considered zero.

The generalization of Eqs.\ (\ref{eq:Taylor}) and (\ref{eq:Newton}) for our problem is
to solve a two-by-two linear system for new values of the parameters,
$\tilde{E}_{\rm new}$ and $a_{0,{\rm new}}$,
from these equations:
\begin{eqnarray}
	0 &=& \Delta g +  \left(\frac{\partial\;\Delta g}{\partial\tilde{E}}\right)
	(\tilde{E}_{\rm new} - \tilde{E}_{\rm old}) +
	\left(\frac{\partial\;\Delta g}{\partial a_0}\right) (a_{0,{\rm new}} - a_{0,{\rm old}})
	\nonumber \\
	0 &=& \Delta f +  \left(\frac{\partial\;\Delta f}{\partial\tilde{E}}\right)
	(\tilde{E}_{\rm new} - \tilde{E}_{\rm old}) +
	\left(\frac{\partial\;\Delta f}{\partial a_0}\right) (a_{0,{\rm new}} - a_{0,{\rm old}})
	\ . \label{eq;delgf}
\end{eqnarray}

We will calculate the partial derivatives needed above numerically.
The procedure to find better values of the two parameters $\tilde{E}_{\rm new}$ and
$a_{0,{\rm new}}$ is to be iterated until the gaps are sufficiently small.

To solve this linear system in Eq.\ (\ref{eq;delgf}), it is convenient to define
a matrix
\begin{equation}
	M = \left[ \begin{array}{cccc}
	\frac{\partial\;\Delta g}{\partial\tilde{E}} & 	\frac{\partial\;\Delta g}{\partial a_0}\\
	\frac{\partial\;\Delta f}{\partial\tilde{E}} & 	\frac{\partial\;\Delta f}{\partial a_0}\\
	\end{array} \right] \ \label{eq:2x2partials}
\end{equation}
and a column vector containing the gaps and partials obtained using the old parameters
$\tilde{E}_{\rm old}$ and $a_{0,{\rm old}}$,
\begin{equation}
	C_{\rm old} = \left[ \begin{array}{c}
	\left(\frac{\partial\;\Delta g}{\partial\tilde{E}}\right) \tilde{E}_{\rm old}
		+ \left(\frac{\partial\;\Delta g}{\partial a_0}\right) a_{\rm old} - \Delta g_{\rm old} \\
	\left(\frac{\partial\;\Delta f}{\partial\tilde{E}}\right) \tilde{E}_{\rm old}
		+ \left(\frac{\partial\;\Delta f}{\partial a_0}\right) a_{\rm old} - \Delta f_{\rm old}
	\end{array} \right] \ . \label{eq:2x2oldC}
\end{equation}
Multiplying $C_{\rm old}$ by the inverse matrix $M^{-1}$ then provides us with a
column vector containing the improved parameters for the next iteration:
\begin{equation}
	P_{\rm new} = M^{-1} \; C_{\rm old}
		= [\tilde{E}_{\rm new}, a_{0,{\rm new}}]^T
		\ . \label{eq:2x2Pnew}
\end{equation}

It turns out that this iterative procedure works very well,
with the gaps in the {\it slopes} of $g(x)$ and $f(x)$ at the match point
{\it also} going to zero.

At this point we should remark that others might have done this calculation
but by zeroing the gaps in the logarithmic derivatives, such as$(dg(x)/dx)/g(x)$.
This has the advantage of removing the scale dependence of the functions $g$ and $f$,
but our choice of defining the gaps as relative, as in
Eq.\ (\ref{eq:scaledgaps}), is essentially equivalent, as
it also involves (semi-)local differences divided by the (semi-)local value.

\subsection{The $1s$ Radial Wave Functions and Some Programming Details \label{sec:Vs1s}}

We implemented the iterative numerical process discussed above
by developing Mathematica notebooks, \cite{MMa} but any reasonable programming language
could be used instead.
An example notebook is available from our group's web site.\cite{RRSwebpage}$\ $
(We especially invite our Canadian colleagues to convert this to a Maple program!)
Here,
as we describe the calculation of the ground state radial wave functions for
the GMSS potential of Eq.\ (\ref{eq:GMSSpotnl}), we will comment on a number of the programming issues that we encountered.

The first thing to be done is to set the value of $k$ in the equations to be solved.
It is an integer and depends upon the angular momentum quantum numbers, Eq.\ (\ref{eq:kdef}).
For the $1s$, $j=1/2$ ground state, $l = 0$, $l' = 1$, and thus $k = -1$.

As noted earlier, we work in units where $\hbar = c = 1$ \cite{hbarc} so that the
dimensions of Eqs.\ (\ref{eq:VvVs_geqn}) and (\ref{eq:VvVs_feqn}) are fm$^{-1}$.
For the cases discussed here and below, we have fixed the constants in the GMSS potential
at $\kappa^2 = 0.9 {\rm \ GeV/fm}$ (i.e., $\kappa = 2.14 {\rm\ fm^{-1}}$),
and $r_0 = 0.705$ fm.
Making the equations dimensionless, as mentioned after Eq.\ (\ref{eq:Vs_feqnx}),
this linear scalar potential simplifies to $\tilde{V}_s(x) = x - x_0$,
where $x_0 = \kappa r_0 = 1.506$.

For dimensionless distances we chose $\epsilon = 10^{-6}$, $x_{\rm match} = 1.0$, and
$x_{\rm max} = 6.0$.
We also define a small increment $\delta = 0.0001$, which will be used
when we calculate the partial derivatives in Eq.\ (\ref{eq:2x2partials}) numerically
and for testing when the gaps to be zeroed are small enough.

The two integrations, inwards and outwards, were done using Mathematica's {\tt NDSolve}
function, but one could use any standard Runge-Kutta procedure.\cite{RK}$\ $
To proceed, we need the four boundary conditions (BCs) from Eq.\ (\ref{eq:BC34_l+1/2})
and Eq.\ (\ref{eq:BC12}).
We have defined a subroutine, {\tt  shootinandout}, to do this.
This subroutine is, of course, a function of the parameters $\tilde{E}$ and $ a_0$.
It saves the results of the integrations as {\tt insoln} and {\tt outsoln}, respectively.

In this case, we already had a good idea \cite{GMSS} of what the energy eigenvalue $\tilde{E}$
is for the $1s$ ground state.
After some fiddling, we found an initial choice of parameters
\begin{equation}
	\tilde{E} = 0.82, \quad a_0 = 0.2, \quad a_1 = 1000.0 \label{eq:params1sVs}
\end{equation}
for which the resulting outwards and inwards integrations yielded the curves for
$g(x)$ and $f(x)$ shown in Fig.\ \ref{fig:initialVs1s}.
(As mentioned above, the asymptotic normalization $a_1$ is at this point arbitrary, to be fixed later by the normalization condition.)

It is useful at this point to define a function, {\tt calcmatchgaps}, that uses {\tt insoln}
and {\tt outsoln} to calculate and print out the values of $g(x)$ and $f(x)$
at the match point $x_{\rm match}$, along with their slopes and their gaps scaled as in
Eq.\ (\ref{eq:scaledgaps}).
These numbers provide guidance as to how to proceed when developing the Mathematica notebook.

Next, we need the four partial derivatives in the two-by-two matrix $M$ of
Eq.\ (\ref{eq:2x2partials}).
In pseudocode, the subroutine for calculating ${\partial\;\Delta g/\partial\tilde{E}}$ is

{\samepage
{\baselineskip=12 pt
\begin{verbatim}
   subroutine dggapbydE(tildeE,a0)
      delE = delta*tildeE
      outsoln1 = outward integration with parameters (tildeE + delE,a0)
                 from epsilon to xmatch
      g_out1 = g(x) at x_match from outsoln1
      insoln1 = inward integration with parameters (tildeE + delE,a0)
                from xmax to xmatch
      g_in1 = g(x) at x_match from insoln1
      ggap1 = 2.0*(g_out1 - g_in1)/(g_out1 + g_in1)
      dggapdE = (ggap1 - ggap)/delE
      return dggapdE
      end subroutine
\end{verbatim}
}}
\vspace{-12 pt}\noindent
Similar subroutines are also implemented for the three other partial derivatives.
We also found it useful, once these subroutines were in place, to define another subroutine,
{\tt gapsandpartials}, which calls {\tt shootinandout} for the present values of the
parameters $\tilde{E}$ and $a_0$, followed by calls of {\tt calcmatchgaps} and the
four subroutines for the partial derivatives.

At this point we are ready to solve the two-by-two linear system for improved values
of the parameters $\tilde{E}$ and $a_0$.
Manipulating arrays is a bit tricky in a programming language such as Fortran or C,
but in Mathematica one can simply define the matrix $M$ in Eq.\ ({\ref{eq:2x2partials}) as a list of lists and the column vector $C_{\rm old}$ is a simple list.
Mathematica also provides a built-in function to invert the matrix $M$,
so the evaluation of the right-hand-side of Eq.\ (\ref{eq:2x2Pnew}) is easy.
This procedure is encapsulated in a subroutine we called {\tt solvelinsys()}.

Before going on to the program for the iterative loop we need to decide when
the gaps are small enough to stop the iteration.
Thus we defined a test function which returns a boolean
{\tt True} if {\em any} of the {\it four} gaps, $\Delta g$ and $\Delta f$ as well as their slopes
$\Delta g'$, and $\Delta f'$, is greater than $\delta$.
Again, in pseudocode,

{\samepage
{\baselineskip=12 pt
\begin{verbatim}
   subroutine testgaps()
      boolean p1 = Abs[ggap] > delta
      boolean p2 = Abs[fgap] > delta
      boolean p3 = Abs[gpgap] > delta
      boolean p4 = Abs[fpgap] > delta
      boolean q = p1 OR p2 OR p3 OR p4
      return q
      end subroutine
\end{verbatim}
}
}

\vspace{-12 pt}\noindent
The iteration will stop when {\tt q = False}.

The coding for running the iteration from an initial choice of parameters
$\tilde{E}$ and $a_0$ is as follows:

{\samepage
{\baselineskip=12 pt
\begin{verbatim}
   {newtildeE,newa0} = {start_tildeE, start_a0}
   print starting parameters {newtildeE, newa0, a1}
   iter = 0
   q = True
   do loop
      iter = iter+1
      {oldtildeE,olda0} = {newtildeE,newa0}
      shootinandout(oldtildeE,olda0)
      gapsandpartials(oldtildeE,olda0)
      solvelinsys()   // returns improved values of newtildeE and newa0
      print iter and gaps
      print newtildeE and newa0
      if q
         continue to next iteration, i.e., go to the top of the do loop
      else
         stop and break out of the do loop
      end if
      end do loop
\end{verbatim}
}
}

\vspace{-12 pt}

For the starting parameters used above, $\tilde{E} = 0.82$ and $a_0 = 0.2$,
we find that our program closes the gaps  in four iterations.
The final parameter values are $\tilde{E} = 0.727102$ and $a_0 = 0.194709$.
That $\tilde{E}$ correponds, in more conventional units, to an energy of $E = 0.306$ GeV.
Figure 2 shows plots of the final, converged $g(x)$ and $f(x)$ after normalization.
Note the relatively large size of the lower component $f(x)$, showing the importance of
relativity in this case of massless quarks.

Figure 3 shows the corresponding plots for $\psi_a(x)$ and $-\psi_b(x)$.
As one ought expect for a ground state wave function,
the upper component $\psi_a(x)$ has no nodes and the $p$-wave lower component
$\psi_b(x)$ has one (at the origin).

We have plotted the $\psi$'s in Fig.\ 3 this way to compare with GMSS's Fig.\ 2.
The $\psi$'s here are for fitting the non-strange $q\bar{q}$ mesons and extend
out further than those of GMSS.
The reason for that is that GMSS calculated the $u$ and $d$ quark wave functions to fit the non-strange baryon spectrum, which is why they used a value of $r_0 = 0.57$.
Physically, the effective origin of the confining potential for mesons is
not as strongly localized as in the baryon case, where one more quark damps
fluctuations additionally.

\subsection{Some Excited States \label{VsExciteds}}

The procedure outlined in the previous sub-section can be applied to calculate the energy
eigenvalues and wave functions for excited states.
One expects that the first excited state is the $2s\;\frac{1}{2}$ state.
This state also has $l = 0$, $l' = 1$ and $k = -1$, so the BCs at $x = \epsilon$ are the same
as for the $1s$ case.
The upper component wave function $g(x)$ should now have a new node, i.e., cross the $x$-axis
somewhere between the origin and infinity.
Thus, if it is desirable to have $g(x)$ start off from the origin going positive,
one should choose the asymptotic normalization parameter $a_1$ to be negative.

In view of the asymptotic behavior here being similar to that of a simple
harmonic oscillator, we might expect that the energy eigenvalue of this state is
roughly twice that of the ground state.
Choosing the parameters needed for the initial integrations to be
\begin{equation}
	\tilde{E} = 2.1, \quad a_0 = 0.3, \quad a_1 = -4000.0 \ , \label{eq:params2sVs}
\end{equation}
the iterative loop again converges in four iterations.
The final parameters are $\tilde{E} = 1.91897$ (i.e., 0.809 GeV) and $a_0 = 0.378981$.
Plots of $\psi_a(x)$ and $-\psi_b(x)$ are displayed in Fig.\ 4.
For the $2s\;1/2$ state, $\psi_a(x)$ has one node and $\psi_b(x)$ has two.

Why are we counting nodes, anyway?  Because a state with more nodes has more energy.
Qualitatively, more nodes means more curvature, and that means a bigger contribution from
$\nabla^2$, which in turn means a bigger kinetic energy.

Besides the $2s\;\frac{1}{2}$ state, there are also two nearby $p$-wave excited states.
The $2p\;\frac{3}{2}$ state also has $j = l + \frac{1}{2}$, but now with $l = 1$, $l' = 2$, and thus $k = -2$.
The BCs near the origin are given by Eqs.\ (\ref{eq:BC34_l+1/2}) and (\ref{eq:fBC4}),
but otherwise the coding
is very similar to (basically can be copied from) the $1s\;\frac{1}{2}$ and
$2s\;\frac{1}{2}$ cases.
Starting this time with
\begin{equation}
	\tilde{E} = 2.1, \quad a_0 = 3.0, \quad a_1 = -4000.0 \ , \label{eq:params2p3by2Vs}
\end{equation}
the gaps close in four iterations, with final parameters $\tilde{E} = 2.23003$
(i.e., 0.940 GeV) and $a_0 = 1.57784$.
The $2p\;\frac{3}{2}$ wave functions $\psi_a(x)$ and $-\psi_b(x)$ are shown in Fig.\ 5.
In contrast to the $2s$ case, here {\it both} $\psi_a(x)$ and $\psi_b(x)$ have two nodes,
reflecting a higher energy eigenvalue.

The other $p$-wave state, $2p\;\frac{1}{2}$, is qualitatively different.
This is the first case for which $j = l - 1/2$, and $k$ is now a positive integer.
In a sense, this switches the roles of $g(x)$ and $f(x)$.
The boundary conditions for the outward integration when $j = l - \frac{1}{2}$ are
different from those of Eq.\ (\ref{eq:BC34_l+1/2}):
\begin{equation}
	f(\epsilon) = a_0 \epsilon^l \ , \quad
	g(\epsilon) = b_0 \epsilon^{l+1} \ . \label{eq:BC34_l-1/2}
\end{equation}
Substituting $g(\epsilon)$ into the first ODE allows us to determine this $b_0$, so
\begin{equation}
	g(\epsilon) = a_0 \; (\tilde{E} + V_s(\epsilon) + \tilde{m})
		\epsilon^{l+1}/(2 l + 1) \ . \label{eq:gBE4}
\end{equation}
That is, if $a_0 >0$, {\it both} $g(x)$ and $f(x)$ start out from the origin with
positive slopes.

Starting for this $2p\;\frac{1}{2}$ case with initial parameters
\begin{equation}
	\tilde{E} = 2.1, \quad a_0 = 3.0, \quad a_1 = -4000.0 \ , \label{eq:params2p1by2Vs}
\end{equation}
the iterative procedure converges on the fifth iteration, with final parameter values
$\tilde{E} = 2.37846$ (i.e., 1.002 GeV) and $a_0 = 0.171483$.
The $\psi$'s are plotted in Fig.\ 6.
Here also both component wave functions have two nodes.

We note in passing that the higher spin state of this pair has the lower energy,
contrary to the well-known case for the Coulomb potential. This is a common
feature for nuclear states (even ground states) and reflects the presence of an
effective scalar potential as found in many nuclear potential models.\cite{Walecka}

The energy difference between the $2p\;\frac{3}{2}$ and $2p\;\frac{1}{2}$ states,
62 MeV, is due to a spin-orbit interaction.
However, there is evidence in the meson spectrum that the spin-orbit interaction
is suppressed.\cite{Isgur}
Page, Goldman, and Ginocchio (PGG) \cite{PGG} claim this reflects a
cancellation between a scalar potential $V_s$ and a vector potential, $V_v$, having
the same linear slope at large distances.
We will return to this point in Sec.\ \ref{VsAndVv}.

\subsection{Solutions for Massive Quarks}

The cases for the linear scalar potential $V_s(r)$ of Eq.\ (\ref{eq:GMSSpotnl})
when the quarks are massive are computed straightforwardly using the program
described above.
For example, using masses appropriate \cite{Qqbar} for the $Q\bar{q}$ mesonic states,
where $q$ stands for a massless non-strange quark ($u$ and $d$) and $Q$ for the
massive strange ($s$), charmed ($c$), and bottom ($b$) quarks,
we find the following eigenenergies:

\begin{eqnarray}
	m_u = m_d = 0.0  \  {\rm GeV}\ , \quad & & E_u = E_d = 0.306 \  {\rm GeV} \ , \label{eq:m_ud} \\
	m_s = 0.3445  \  {\rm GeV} \ , \quad & & E_s = 0.486 \  {\rm GeV} \ , \label{eq:m_s}\\
	m_c = 1.803  \  {\rm GeV} \ , \quad & & E_c = 1.667 \  {\rm GeV} \ , \label{eq:m_c} \\
	m_b = 5.298  \  {\rm GeV} \ , \quad & & E_b = 5.007 \  {\rm GeV} \label{eq:m_b} \ ,
\end{eqnarray}
by requiring a match to the appropriately weighted average of the lowest
pseudoscalar and vector states.\cite{PDG}
Note that for the heaviest quarks the states are conventionally bound,
$E_Q < m_Q$, and confinement need not be invoked to understand the stability
of the state, in contrast to the situation for the light quarks.

Figure \ref{fig:upperWFs} compares the results for the $u$, $s$, $c$, and $b$ quarks
for the upper components, $\psi_a(x)$, and Fig.\ \ref{fig:lowerWFs} does the same for
the lower components, $-\psi_b(x)$.
Note that as the mass $m_q$ increases, the $Q\bar{q}$ system becomes
more and more non-relativistic,
i.e., the lower component wave function $\psi_b$ gets smaller, relative to the
upper component $\psi_a$.
Also, as $m_q$ gets larger, the wave functions are more and more concentrated
near the origin.

\section{Vector Potential -- Hydrogen-like Atoms \label{sec:hatom}}

In contrast with the case of massless quarks, an electron bound in a Coulomb potential,
as in the hydrogen atom, is at the opposite extreme, i.e., is
to a very good approximation a completely non-relativistic system.
The binding energy of the ground state, Ry = 13.6 eV, is very small compared to the
mass of the electron, $m_e =  0.511$ MeV.
(In this section we will use MeV instead of GeV.)
The Schr\"odinger equation for this problem predicts its energy levels basically correctly,
missing only the fine-structure splitting, a spin-orbit effect about $10^{-4}$ times
smaller than the binding of the $n = 2$ levels.\cite{Eisberg}.

As noted in the Introduction, the analytic solutions of the Dirac equation for the
hydrogen atom were found long ago, \cite{Darwin} and these solutions do give the
fine-structure splitting as a relativistic effect.
Nonetheless, it is an interesting numerical exercise to see if the program for the
linear scalar potential outlined above (or something like it) can be applied to the
hydrogen atom case, and with enough accuracy.
We want to solve Eqs.\ (\ref{eq:Vv_geqn}) and (\ref{eq:Vv_feqn}) with
$V_v(r) = -Z \alpha /r$.
We will set $Z=1$, as these solutions are somewhat more delicate.
Heavier, hydrogen-like atoms with $Z>1$ can be done in a similar manner.

Converting to dimensionless equations by dividing through by $m_e$ (instead of $\kappa$ as before),
\begin{eqnarray}
	&g'(x)& + \;\frac{k}{x} g(x)\; - \;(1 + \tilde{E} + Z\alpha /x)\; f(x) = 0 \ , \label{eq:Vv_geqnx} \\
	&f'(x)& - \;\frac{k}{x} f(x)\; - \;(1 - \tilde{E} - Z\alpha /x)\; g(x) = 0 \ , \label{eq:Vv_feqnx}
\end{eqnarray}
where now $x = m_e r$ and $\tilde{E} = E/m_e$.
The energy $E$ is better written as $E = m_e - B$, where $B$ is the binding
energy of the level of interest (which has principal quantum number $n$ and orbital angular momentum quantum number $l$).
Note that, because of the smallness of the binding energies, $E$ is only slightly less than $m_e$.
Thus $\tilde{E}$ is less than (but very close to) 1, which is why
we factored out a minus sign
from the third term in Eq.\ (\ref{eq:Vv_feqnx}), relative to that in Eq.\ (\ref{eq:Vv_feqn}).
We will use the binding energy $\tilde{B} = B/m_e$ instead of $\tilde{E}$ as one of
the two parameters to be determined in the matching of the inwards and outwards integrations.

\subsection{Boundary Conditions for Inward Integration \label{inBCs}}

For asymptotically large $x$ the ODEs reduce to
\begin{eqnarray}
	g'(x) &=& (1 + \tilde{E}) \; f(x)  \ , \nonumber \\
	f'(x) &=& (1 - \tilde{E}) \; g(x) \ . \label{eq:asympteqsH}
\end{eqnarray}
The solutions of these asymptotic equations at $x = x_{\rm max}$ are
\begin{eqnarray}
	g(x_{\rm max}) &=& a_1 \; e^{-\mu x_{\rm max}}  \ , \nonumber \\
	f(x_{\rm max}) &=& -a_1 \left( \frac{1-\tilde{E}}{1+\tilde{E}}\right)^{1/2}
		e^{-\mu x_{\rm max}}  \ ,  \label{eq:asymptsolnsH}
\end{eqnarray}
which will be used as the BCs (starting values) for the inwards integration with $a_1$
as another parameter to be determined later by the normalization condition.
The coefficient in the exponential decay is small:
\begin{equation}
	\mu = \sqrt{1 - \tilde{E}^2} = \sqrt{2\tilde{B} - \tilde{B}^2}
	\approx\sqrt{2\tilde{B}} \; \leq \; 0.00730 \approx \alpha
	\ . \label{eq:mudef}
\end{equation}
The Coulomb wave functions are quite long-ranged, and are all the more so the
smaller the binding energy.
Equation (\ref{eq:asymptsolnsH}) also shows that, asymptotically, $f(x)$ is very much
smaller than $g(x)$.
The smallness of the lower component is an indication of how non-relativistic
the hydrogen atom is.

\subsection{Boundary Conditions for Outward Integration}

As before, to avoid the singularity at $x=0$, we integrate outwards from
$x=\epsilon$, where $\epsilon$ takes an appropriately small value.
In this case, one might even consider taking $\epsilon$ as the radius of the
nucleus providing the Coulomb potential (but see Sec. \ref{muonic} for a
better approach).
For the boundary conditions for integrating outwards from $x = \epsilon$ when
$j = l + \frac{1}{2}$, i.e., $l' = l+1$ and $k = -(l+1)$, we can again assume
\begin{equation}
	g(\epsilon) = a_0 \;\epsilon^{l+1} \ , \quad f(\epsilon) = b_0 \;\epsilon^{l+2} \ .
\end{equation}
Substituting these in the second ODE, we can solve for $b_0$, finding
\begin{equation}
	f(\epsilon) = - a_0 \frac{Z\alpha}{2 l + 3}\;\epsilon^{l+1} + {\cal O}(\epsilon^{l+2})
	\ . \label{eq:Vvlower_f}
\end{equation}
Noticeably different is the lowering of the power behavior of $f(x)$ due to the
$1/x$ behavior of the potential.
Note that, because of the factor of $\alpha \approx 1/137$, the magnitude of $f(x)$
is much smaller than that of $g(x)$, also here near the origin.

For the $j = l - \frac{1}{2}$ case, i.e., $l' = l-1$ and $k = l$, we let $f(x)$ determine the
nature of the boundary conditions [as in Eq.\ (\ref{eq:BC34_l-1/2})], i.e.,
\begin{equation}
	f(\epsilon) = a_0 \;\epsilon^l \ , \quad\quad g(\epsilon) = b_0 \;\epsilon^{l+1} \ .
\end{equation}
Substituting these in the first ODE, we again solve for $b_0$ and find
\begin{equation}
	g(\epsilon) = a_0 \frac{Z\alpha}{2 l + 1} \epsilon^l + {\cal O}(\epsilon^{l+1}) \ .
\end{equation}
Again, for this case, $g(x)$ and $f(x)$ start off with the same slope, as in
Eq.\ (\ref{eq:gBE4}).
And again, there is a lower power behavior of $g(x)$ due to the $1/x$ in the potential,
as in Eq.\ (\ref{eq:Vvlower_f}).

\subsection{The $1s$ Ground State \label{sec:H1sgs}}

The natural unit of length for the hydrogen atom problem is the Bohr radius,
$a_B = \hbar ^2/m_e \alpha = 0.529$ \AA $\ = 0.529 \times 10^5$ fm.
Thus a natural scale for $x$ is in units of the dimensionless Bohr radius,
$x_B = a_B m_e/\hbar c$ which is $1/\alpha$ in $\hbar = c = 1$ units.

In view of the slow exponential decay in Eq.\ (\ref{eq:asymptsolnsH}) (i.e.,
the smallness of $\mu$), for our calculations we chose $x_{\rm max} = 7 x_B$, along with
$x_{\rm match} = 0.5 x_B$.
Fixing $a_1 = 10$ (recall, it is arbitrary before normalization)
and the starting values of the parameters to be tuned,
$B = 12\times10^{-6}$ MeV (less than the expected 13.6 eV) and $a_0 = 0.015$,
we found the $2 \times 2$ matrix of partial derivatives to be
\begin{equation}
	M = \left[ \begin{array}{rr}
	  -108877 & 107.348 \\
	  -320810 & 107.494  \\
	\end{array} \right] \ . \label{eq:Hpartials}
\end{equation}
This shows a big sensitivity of the hydrogen atom wave functions to the
choice of binding energy.
Nonetheless, the iterative process of refining $\tilde{B}$ and $a_0$ proceeds nicely to
a solution with $B = 13.6059$ eV and $a_0 = 0.0104306$.
The wave functions $g(x)$ and $f(x)$ are displayed as the solid curves in
Figs.\ \ref{fig:Huppers} and \ref{fig:Hlowers}.
As expected, $f(x)$ is much smaller than $g(x)$.
Their shapes differ from those shown for the scalar linear potential shown
in Fig.\ \ref{fig:finalVs1s}.

Figures \ref{fig:Huppers} and \ref{fig:Hlowers} also display, as dashed curves,
the corresponding $\psi_a(x)$ and $\psi_b(x)$ radial wave functions.
These are, in fact, both purely decaying exponentials proportional to $e^{-\mu x}$.
This is not surprising for the upper component, $\psi_a(x)$, since that
is the just what the Schr\"odinger equation predicts for the hydrogen atom ground state.
It may not be so obvious, however, that the lower component, $\psi_b(x)$, has the
same form.
That it has no node at the origin (in contrast to Fig.\ \ref{fig:finalVs1sPsis}) is a
consequence of the $Z\alpha/x$ potential changing the presumed $x^{l+2}$-dependence
of $f(x)$ near the origin to an $x^{l+1}$-dependence.

\subsection{Some Excited States}

The $n=2$ excited states of hydrogen are done in a very similar manner to the
$1s$ ground state calculation.
The major difference is that, from the Balmer formula \cite{Eisberg}, we expect the binding
energy $B$ for these states to be about $1/n^2 = 1/4$ times the ground state binding energy.
Also, based on Eq.\ (\ref{eq:mudef}),
we expect the wave functions will extend outward about twice as far,
so, for our calculations we chose $x_{\rm max} = 16 \; a_B$ instead of $7 \; a_B$.

For the $2s\;\frac{1}{2}$ state, starting with initial parameters $B = 3.2$ eV (presumably low),
$a_0 = 0.0007$, and (fixed) $a_1 = -10.0$,
the program converges in four iterations to give final $B = 3.40085$ eV and
$a_0 = 0.00071905$.
The $2s$ wave functions are plotted in Figs.\ \ref{fig:H2suppers} and \ref{fig:H2slowers}.
As expected on general grounds, unlike the ground state, here $\psi_a(x)$ and
$\psi_b(x)$ each have one node.

For the $2p\;\frac{1}{2}$ state, starting with initial parameters $B = 3.2$ eV (also presumably low), $a_0 = 0.0004$, and (fixed) $a_1 = 100.0$,
the program converged in four iterations to give final $B = 3.40086$ eV and
$a_0 = 0.000393546$.
In fact, the $2s\;\frac{1}{2}$ and $2p\;\frac{1}{2}$ levels are degenerate \cite{Eisberg}
as a result of an $O(4)$ symmetry hiding in the equations.
The small difference in the converged $B$'s we find here is within the
numerical precision of our Mathematica program.
However, the $2s\;\frac{1}{2}$ and $2p\;\frac{1}{2}$ wave functions are {\it very different}.
The  $2p\;\frac{1}{2}$ wave functions are shown in Figs.\ \ref{fig:H2p12uppers} and \ref{fig:H2p12lowers}.
Note that, here also, $\psi_a$ and $\psi_b$ each have one node.

For the $2p\; \frac{3}{2}$ state, starting this time with initial parameters $B = 3.2$ eV,
$a_0 = 0.000015$, and $a_1 = 100.0$, the program converged in four iterations to
give final $B = 3.40144$ eV and $a_0 = 0.0000141876$.
The difference we find between the $j= \frac{3}{2}$ and $j= \frac{1}{2}$ energy levels,
0.58 meV, is the spin-orbit splitting, an intrinsically relativistic effect.
The analytic value of this splitting is 0.453 meV \cite{BetheSal,Eisberg}
and the difference here is due to the limited machine precision used in our
Mathematica program.
The $2p\;\frac{3}{2}$ wave functions are shown in Figs.\ \ref{fig:H2p32uppers} and \ref{fig:H2p32lowers}.
The $\psi_a(x)$'s for the $2p\;\frac{3}{2}$ and $2p\;\frac{1}{2}$ states are {\it very similar},
but the lower component $\psi_b(x)$'s are quite different, reflecting the fact
the lower components and the difference are both entirely due to relativity.
For the $2p\;\frac{3}{2}$ state also, $\psi_a$ and $\psi_b$ each have one node.

\subsection{Hydrogen-like Atoms with $Z>1$ \label{Zgt1}}

There are no big surprises here, as Table I shows.
From Eq.\ (\ref{eq:mudef}), the exponential fall-off of the wave functions
is faster, as $\mu \sim \sqrt{B} \sim Z$, so we adjust the values of $x_{\rm match}$ and
$x_{\rm max}$ accordingly by dividing the hydrogenic values by $Z$.
Running the code for the $1s$ ground state for $V_v(r) = Z \alpha/r$
works well for $Z \leq 100$.
As $Z$ grows larger, the relativistic corrections to the energy become increasingly
important.

As $Z$ approaches $1/\alpha = 137.036$, however, the numerical solutions become
inaccurate and, eventually, unstable.
This can be seen in the rapid increase in the value of the parameter $a_0$.
The value of $Z = 1/\alpha$ is where the Klein paradox \cite{Klein} comes into play,
since the analytic result for this state is \cite{BjD}
\begin{equation}
	E(1s) = m (1 - Z^2 \alpha^2)^{1/2} \label{eq:analyticEgs}
\end{equation}
and the eigenenergy becomes complex beyond that point.
The resolution of this paradox is, as has been well-known for a long time,
the creation of electron-positron pairs from the Dirac-Fermi sea when
$Z > 1/\alpha$.\cite{Dombey}

\subsection{Muonic Atoms \label{muonic}}

When a negatively-charged muon slows down in matter, sometimes it is captured by the
Coulomb potential of an atomic nucleus before it decays.
If so, it then quickly cascades down through its hydrogen-like levels to the
$1s$ ground state, emitting x-rays along the way.
From there it then either decays or can be be captured by the nucleus,
both of which are weak interaction processes.
The capture process depends sensitively upon the value of the $1s$ wave function
at the origin.

The numerical coding for muonic atom ground states for values of $Z$ of interest
goes pretty much as for the electronic atom case for such a $Z$.
The differences come from replacing the mass of the electron (0.511 MeV) with that of
the muon (105.66 MeV).
This must be done for the ``muonic Bohr radius'', $a_{B\mu} = a_B (m_e/m_\mu)$
= 255.8 fm, and consequently in the choices of $x_{\rm match}$ and $x_{\rm max}$
which are now proportional to $a_{B\mu}/Z$.
Also, the non-relativistic binding energy $B$ is now $Z^2 \;(m_\mu/m_e)\;$Ry.

Calculating, as above, the ground state for muonic $^{40}$Ca ($Z=20$),
starting with $B = 1.1$ MeV and $a_0$ = 0.1,
the iterations converge to final values of $B = 1.13136$ MeV and $a_0 = 0.239462$.
However, this initial calculation used a value of $\epsilon = 10^{-4}$ (dimensionless),
or a cut-off at the origin of 0.000187 fm, {\it rather smaller} than the size of the
$^{40}$Ca nucleus,\cite{Evans}
\begin{equation}
	R_{\rm Ca} \approx r_0 A^{1/3} = 1.2 \times 40^{1/3} = 4.1 {\rm \ fm}.
\end{equation}

The nuclear charge is not a point charge, of course, but has a charge density
distribution, $\rho (r)$, which is spread out over the volume of the nuclear sphere.
In fact, one of the major reasons for having studied muonic atoms in the past,
through the last $2p \rightarrow 1s$ x-ray, is for determining this charge density.\cite{mu_xray}

As a simple example, suppose the nucleus is a sphere of radius $R$ and charge $eZ$
with a charge density $\rho$ which is constant out to its surface and zero for $r>R$:
We need to find the potential seen by the negative muon for $r<R$.
If it were sitting exactly at the center of the nucleus, it would (by symmetry)
feel no force.
That is, there is no term in the potential linear in $r$, as the force at the
origin is $-dV_{\rm nuc}/dr]_{r=0} = 0$.
This suggests that we can take $V_{\rm nuc}(r) = A + B r^2$.
The coefficient $B$ can be fixed from the requirement that the slopes
$dV_{\rm nuc}/dr$ and $dV_{\rm Coul}/dr$ match at $r=R$.
The coefficient $A$ in turn is fixed by the requirement of continuity at $r=R$.
Thus a potential reasonably appropriate for such a muonic atom is
\begin{equation}
	V_v(r) = \left\{\begin{array}{ll}
		\alpha Z \;(r^2 - 3 R^2)/{2 R^3} & \quad\mbox{for $r \leq R$,}
		\\
		-\alpha Z/r                & \quad\mbox{for $r > R.$}
		\end{array} \right.  \label{eq:unisphere}
\end{equation}

This potential no longer has the $1/r$ singularity at the origin, so the BCs
for the outgoing integration for the $1s$ state are more like those in
Eqs.\ (\ref{eq:BC34_l+1/2}) and (\ref{eq:fBC4}) for $l=0$:
\begin{equation}
	g(\epsilon) = a_0 \; \epsilon \ , \quad \mbox{and }
	f(\epsilon) = a_0 (1 - \tilde{E}) \; \epsilon^2/3 \ , \label{eq:muonicout}
\end{equation}
bearing in mind that we here are making the equations dimensionless by setting $m_\mu = 1$
and that the eigenenergy of the bound state is less than $m_\mu$.
Unlike the pure Coulomb case, here $f(r)$ will have curvature near
$r=0$, where it will vanish.

Running the numerical integrations with potential $V_v(r)$ for $^{40}$Ca gives
a binding energy $B = 1.07006$, some 60 keV smaller than the pure Coulomb result
for this nucleus.
Figure \ref{fig:CaPsis} displays the difference between the $\psi_a(x)$'s and $\psi_b(x)$'s
for the case with this nuclear charge distribution and the pure Coulomb case.

\section{Combining the Two Potentials \label{VsAndVv}}

We now consider the solution of the radial Dirac equations, Eqs.\ (\ref{eq:VvVs_geqn})
and (\ref{eq:VvVs_feqn}), when both potentials, the Lorentz scalar $V_s(x)$
and Lorentz vector $V_v(x)$, come into play.
We discuss two different situations.

\subsection{When $V_v(x)$ Is Coulombic \label{VvCoul} }

As our first case, we take the scalar potential $V_s(r)$ to be the same as in
Eq.\ (\ref{eq:GMSSpotnl}) and the vector potential as $V_v(x) = -\alpha_s/x$.
Here $\alpha_s$ is the (strong) gluon-quark coupling constant of quantum
chromodynamics (QCD), having a value of ${\cal{O}}(1)$ at low energies and
becoming small at high energies (asymptotic freedom).

The modifications to the program are minor.
The two BC's at $x_{\rm max}$ are the same as in the pure $V_s$ case, Eq.\ (\ref{eq:BC12}).
The BC's at $x=\epsilon$ are slightly modified from that of
Eq.\ (\ref{eq:BC34_l+1/2}) because here it is $V_v$ that dominates.
For $l = 0$,
\begin{equation}
	g(\epsilon) = a_0\;\epsilon, \quad
	f(\epsilon) = -a_0\;(V_v(\epsilon)/3)\;\epsilon^2 = a_0\;\alpha_s\;\epsilon/3
	\ , \label{eq:VvVsBC4}
\end{equation}
as in the hydrogen atom case, but with $\alpha_s$ generally much larger than $\alpha$.

Running the code for $\alpha_s = 0.4$ gives the wave functions $\psi_a$ and  $\psi_b$
shown in Fig.\ \ref{fig:VsVv1sab_04}.
For comparison, the wave functions for case of $\alpha_s = 0$ (i.e., only using
the scalar potential) are also shown.
The eigenenergies for these two cases are 0.306 GeV ($\alpha_s = 0$) and
0.251 GeV ($\alpha_s = 0.4$).
That is, adding the Coulomb attraction lowers the ground state energy.

The sharp rise in $\psi_a(x)$ and rapid falloff in $\psi_b(x)$ near the origin is another
effect of the point Coulomb attraction.
This behavior is unphysical, as the motions of the quark and anti-quark in the $q\bar{q}$
meson will smear out the point Coulomb potential somewhat.
One might think that a Coulomb divergence remains in the relative coordinate,
but this is a non-relativistic prejudice.
Relativistic retardation effects and pair creation and annihilation both act to ameliorate
this divergence, significantly so in this case, as $\alpha_s \gg \alpha$.
This will give a potential $V_v(x)$ that is similar to the potential used to describe the
nuclear size effect in muonic atoms, Eq.\ (\ref{eq:unisphere}).
We will come back to such a calculation presently.

Before doing that, however, we present the results of running the massless quarks code
for a variety of choices for $\alpha_s$.
The $1s$ ground state energies and the inital slope parameters $a_0$
are shown in the second and third columns of Table II.
As the Coulomb attraction increases, the eigenenergy falls off fairly rapidly
and eventually goes through zero at a value near $\alpha_s = 1$.
At this point, like the Klein paradox situation discussed in
Sec.\ {\ref{Zgt1}, it is possible to produce $q\bar{q}$ pairs at liberty.
In fact,  what is called a ``quark condensate'' forms, a modification of the
vacuum state.\cite{GOR}

To take into account the smearing of the point Coulomb potential by the motions of the
quarks in the meson, we again use the potential $V_v(x)$ given in Eq.\ (\ref{eq:unisphere}),
but with $\alpha$ replaced by $\alpha_s$ and $R$ taken as one half of the electromagnetic
radius of the pion, i.e., $R = \frac{1}{2}\times 0.672 = 0.336$ fm.\cite{PDG}\
We choose this fraction on the basis that the root-mean-square electromagnetic radius
represents the average separation of the quark and antiquark, but since they are oppositely
correlated, each should be at about half of this distance from the center of mass of
the meson.

The boundary conditions (BC's) at the origin for this smeared Coulomb potential
(for angular momentum $j = l + \frac{1}{2}$) now become
\begin{equation}
	g(\epsilon) = a_0\;\epsilon^{l+1}, \quad
	f(\epsilon) = -a_0\;(\tilde{E}- V_v(\epsilon) - V_s(\epsilon) - \tilde{m}) \;
		\epsilon^{l+2}/(2l+3) \ . \label{eq:VvVsBC4smeared}
\end{equation}
Running the program with $\alpha_s = 0.4$ and the above $R$
gives curves for $\psi_a(x)$ and $-\psi_b(x)$ with similar upturns and falloffs near the
origin as in Fig.\ \ref{fig:VsVv1sab_04}, but less sharply so.
One should be aware that, while interesting, these differences in the $\psi_{a,b}(x)$
near the origin make only minor differences in expectation values of operators because of
the factor of $x^2$ in the integration over $d^3x$; see, e.g., the normalization condition,
Eq.\ (\ref{eq:normzn}).

For this smeared potential, the eigenenergies and final $a_0$ parameters for various
values of $\alpha_s$ are also tabulated in Table II.
As expected from the above remark, the energy expectation value is basically unchanged
from the point Coulomb value until one reaches the point where the quark condensate
takes over.
However, the initial slope parameters $a_0$ do differ from those of the point Coulomb
case, being less pathological as $E$ goes through zero.

\subsection{$V_v$ as the Cornell Potential and the Spin-orbit Force. \label{CornellPot}}

In this subsection we will assume both potentials $V_s(x)$ and $V_v(x)$ are linearly
confining and have the same slopes (i.e., have the same string tension $\kappa^2$).
We assume the two potentials are displaced from each other with $V_s(x)$ lying above
$V_v(x)$, as in Fig.\ \ref{fig:VsVvplot}.
That is, we assume
\begin{equation}
	V_s(x) = x + x_s, \quad V_v(x) = -\alpha_s/x + x - x_v\ , \label{eq:VsVvCornell}
\end{equation}
or, more appropriately, the smeared version of $V_v(x)$ as described above in
Sec.\ \ref{VvCoul}.
Here $x_s$ and $x_v$ are parameters to be adjusted to get some desired eigenenergy.
They give rise to a separation between the two asymptotic potentials of $x_d = x_s + x_v$.

The reason for giving the potentials the same asymptotic slopes is to check that
the spin-orbit interaction disappears as claimed in Ref.\ \cite{PGG}.
In doing so, we have changed the earlier offset of $-x_0$ in $V_s$ to $+x_s$ since,
in GMSS (where only a Lorentz scalar potential was used), the negative offset $-x_0$
represented a reasonable approximation the one-gluon-exchange attraction,
roughly an average of the two potentials as shown in the figure.

Now, however, the BC's at large distances are changed from those in Eq.\ (\ref{eq:BC12}),
since, in the equation for $g(x)$, Eq.\ (\ref{eq:VvVs_geqn}),
the difference of the dimensionless $V_s(x)$ and $V_v(x) \rightarrow x$ has its
$x$-dependence cancel.
We need to solve the large-$x$ equations
\begin{equation}
	g\;'(x) - (\tilde{E} + x_d + \tilde{m})\; f(x) = 0, \quad
	f\;'(x) - 2x \; g(x) = 0 \ , \label{eq:preAiry}
\end{equation}
from which
\begin{equation}
	g\;''(x) - 2(\tilde{E} + x_d + \tilde{m})\; x \;g(x) = 0 \ . \label{eq:thisAiry}
\end{equation}
This is a minor modification of the differential equation for the Airy functions.\cite{AbStAiry}\ \
Since we require decaying solutions at our starting point $x_{\rm max}$ for the
inwards integrations, we discard the runaway $Bi$ solution for $g(x)$.
The BC's at large $x=x_{\rm max}$ are then
\begin{eqnarray}
	g(x_{\rm max}) &=& a_1 \;Ai\;[2^{1/3} (\tilde{E} + x_d + \tilde{m})^{1/3} x_{\rm max}\;]
		\ , \nonumber \\
	f(x_{\rm max}) &=& a_1 \left[\frac{2} {(\tilde{E} + x_d +  \tilde{m})^2}\right]^{1/3}
		\!\!Ai\;'\;[\;2^{1/3} (\tilde{E} + x_d +  \tilde{m})^{1/3} x_{\rm max}\;] \ . \label{eq:AiryBCs}
\end{eqnarray}
Despite appearances, $g(x_{\rm max})$ and $f(x_{\rm max})$ still
start off with the opposite slopes, as in Eq.\ (\ref{eq:BC12}), since, for
$t > 0$, $Ai(t) > 0$ and $Ai\;'(t) < 0$.

For calculations, we simplified the displacement between the two potentials by
setting $x_v = 0$, giving us one less parameter to worry about.
The first wave function we want to compute is the ground state for the massless
quark case, $m_q = 0$, starting out with $\alpha_s = 0$ and then adjusting both
$x_d = x_s$ and  $\alpha_s$ until we get the desired eigenenergy, noted above in
Eq.\ (\ref{eq:m_ud}).

It turns out that to do so is trickier than in the $V_s$-only case, presumably because of
the cancellation between $V_s(x)$ and $V_v(x)$ in the differential equation for $g\;'(x)$.
Our first iterative search with $\alpha_s = 0$ found a highly excited state with
$\tilde{E} = 7.975$, {\it not} the desired 0.7263, for which $\psi_a(x)$ had six nodes!
After some hand-searching on $x_s$, we found we {\it could} get a ground state solution,
i.e., a $\psi_a(x)$ with no nodes, looking like the $b$-quark $\psi_a(x)$ of
Fig.\ \ref{fig:upperWFs}.
This solution started with $x_s = 4.0$ (very large!) and converged, in six iterations,
to $\tilde{E} = 5.735$ (about eight times too large).

From this $\alpha_s = 0$ solution we were able to march down, in small steps in $x_s$
(still with $\alpha_s = 0$), to $x_s = 0.20$, with the energy now smaller,
$\tilde{E} = 2.779$, but still a factor of four too large.
(For smaller values of $x_s$ the iterative procedure would not converge, with the matrix
of Eq.\ (\ref{eq:2x2partials}) becoming singular.)
The wave function $\psi_a(x)$ for this solution was broader than that for the $x_s = 4.0$
case, and the $\psi_b(x)$ was relatively larger in comparison with $\psi_a$.

At this point we began marching the value of $\alpha_s$ in small steps up from zero,
i.e., we began turning on the Coulomb attraction.
As expected this attraction lowers the energy, and we were able to achieve the desired
massless quark eigenenergy of 0.306 GeV with the following values:
\begin{equation}
	x_s = 0.20, \quad \alpha_s = 1.103, \quad \tilde{E} = 0.7263 \ . \label{eq:VsVv1s_ud}
\end{equation}
The Long Marches did achieve the factor of eight reduction in the ground state eigenenergy.
The wave functions for this massless quark solution are shown in Fig.\ \ref{fig:VsVv1s_ud}.
Note that $\psi_a(x)$ here begins to resemble the hydrogen-atom ground state $\psi_a(x)$
depicted by the dashed curve in Fig.\ \ref{fig:Huppers}, and it is much narrower than
that shown in Fig.\ \ref{fig:finalVs1sPsis} .

The ground state solution says nothing about the PGG claim that,
for $V_s(x)$ and $V_v(x)$ having equal slopes, the spin-orbit interaction vanishes.
To check this we need to find, at a minimum, the eigenenergies of the
$2p\;\frac{1}{2}$ and $2p\;\frac{3}{2}$ excited states.
For the same values of $x_s$ and $\alpha_s$ as in Eq.\ (\ref{eq:VsVv1s_ud}) and
after some searching for good starting values for the iterative procedure,
we obtained the three $n=2$ wave functions displayed in Figs.\ \ref{fig:VsVv2s_ud},
\ref{fig:VsVv2p1by2_ud}, and \ref{fig:VsVv2p3by2_ud}.
The wave functions here are narrower and more hydrogen-like
than in the $V_s$-only case (compare with Figs.\ \ref{fig:finalVs2s},
\ref{fig:finalVs2p3by2}, and \ref{fig:finalVs2p1by2}).
The eigenenergies, in GeV, for these states are
\begin{equation}
	E_{2s} = 1.020, \quad E_{2p\;\frac{1}{2}} = 0.674, \quad E_{2p\;\frac{3}{2}} = 1.008 \ .
	\label{eq:VsVvn=2Es}
\end{equation}
The $2p\;\frac{3}{2}$ state is nearly degenerate with the $2s$ state,
some 200 MeV higher than in Sec.\ \ref{VsExciteds}, but this degeneracy is
probably accidental.
The real surprise here is the {\it huge} spin-orbit splitting between the
$2p\;\frac{1}{2}$ and $2p\;\frac{3}{2}$ states, not what we expected from
the claim made by PGG.\cite{PGG}

This difference, it turns out, is due to the presence of
the Coulomb attraction term in $V_v(x)$.
If we set $\alpha_s = 0$, then (again for $x_s=0.2$) we find the following
$n=2$ eigenenergies (in GeV):
\begin{equation}
	E_{2s} = 1.757, \quad E_{2p\;\frac{1}{2}} = 1.523, 	\quad E_{2p\;\frac{3}{2}} = 1.523 \ .
	\label{eq:VsVvn=2EsNoCoul}
\end{equation}
The wave functions in this case are similar to those in Figs.\ \ref{fig:VsVv2s_ud},
\ref{fig:VsVv2p1by2_ud}, and \ref{fig:VsVv2p3by2_ud} but are somewhat
broader and less peaked near the origin.
These results show the expected PGG symmetry (zero spin-orbit splitting).
Note, incidentally, that the $2s$ state is higher in energy than the $2p$ states,
as in the experimental charmonium spectrum.\cite{PDG}
One does conclude, however, that having a significant Coulomb contribution
in $V_v(x)$ will destroy the PGG symmetry.

\section{Summary}

In this paper we have discussed, pedagogically, three increasingly more
sensitive cases for solving the radial Dirac equations numerically:
\begin{enumerate}
\item
    A Lorentz scalar potential, $V_s$, which is linearly confining for quarks.  The
    case of massless quarks is necessarily relativistic.  As the mass of the quark
    increases, the wave function solutions become more and more n0n-relativistic.
\item
    A Lorentz vector potential, $V_v$, whose time-like component is a
    Coulomb potential.  We first treated, numerically, the hydrogen atom, which is
    basically non-relativistic problem.
    We do, however, obtain the relativistic spin-orbit splitting between the
    $2p \frac{1}{2}$ and $2p \frac{3}{2}$ states (at the edge of our machine precision).
\item
    The combination of scalar and vector potentials, aiming at a relativistic
    quark model of mesons.
    Of particular interest is when we do (and do not) get a relativistic
    spin-orbit splitting.
\end{enumerate}

We thank L.\ J.\ Curtis, J.\ L.\ Friar, C.\ J.\ Fontes, M.\ D.\ Scadron, and R. Sharma for
discussions and comments on this work.

redpajama_set_name: RedPajamaArXiv


text:
Jay Reatard's Band Angry Angles' Complete Recordings Compiled
Goner Records to release collection from mid-2000s band also featuring Alix Brown and Destruction Unit's Ryan Rousseau
In 2005, Jay Reatard and Alix Brown started a band called Angry Angles. The band's rotating lineup also featured Destruction Unit's Ryan Rousseau. They released a handful of singles, but before they could complete their album, Reatard and Brown's relationship fizzled. The band broke up, and soon after, Jay released his debut solo album, Blood Visions.
On May 20, Goner will release the complete recorded works of Angry Angles, including some unreleased material. Listen to the unreleased version of "Things Are Moving" above.
Other unreleased tracks include a cover of the Oblivians' "Memphis Creep" and three tracks from a shelved Montreal session: "Can't Do It Anymore," "All the Same," and "Set in Stone." There are also covers of Wire's "The 15th," Devo's "Blockhead," and the Urinals' "Black Hole."
Read our Jay Reatard "In Memorium" feature from 2011, featuring interviews with Brown and more.
Angry Angles:
01 Things Are Moving (Unreleased Version)
02 Apparent-Transparent
03 You Fell In
04 The 15th
05 Crowds
06 She's Dead
07 Blockhead
08 In My Room
09 Memphis Creep
10 Things Are Moving (Single Version)
11 Black Hole
12 Stab You Dead
13 You Call It Love
14 You Lied
15 Can't Do It Anymore
16 All The Same
17 Set In Stone
Watch Reatard perform "Death Is Forming," originally written as an Angry Angles song, for Pitchfork.tv:
Correction: An earlier version of this story indicated that the shelved recording session took place in Toronto. It was Montreal.
Jay Reatard
Destruction Unit
Angry Angles


text:
Q: How does ret, img = cap.read() & similar syntax work in Python? Can someone explain to me how such syntax works in python.
From what I know, ret stores a boolean and img stores the next frame, but how is that facilitated.
What other places can such functionality be utilized in Python?
ret, img = cap.read()

redpajama_set_name: RedPajamaStackExchange


text:
Home Sports Blades and Raiders all square after four games
Blades and Raiders all square after four games
Lucas Punkari
Liam Richards/Saskatoon StarPhoenix Prince Albert Raiders defenceman Jeremy Masella and Saskatoon Blades forward Kirby Dach go into the boards during the first period of WHL playoff action Wednesday.
The best-of-seven battle between the Prince Albert Raiders and the Saskatoon Blades has become a best-of-three affair.
Despite firing 44 shots at Blades netminder Nolan Maier, the Raiders were only able to find the back of the net once over the course of 60 minutes as the hosts came away with a 4-1 win at the SaskTel Centre Wednesday to even up the Western Hockey League Eastern Conference semifinal matchup at two games apiece.
"I thought we were deserving of a better fate," Raiders head coach Marc Habscheid said. "We're getting chances, which is the main thing, but we have to find a way to score a few more goals.
"I have to give Maier credit again as he stood on his head in the second period to keep it from being a 2-0 game. I thought we would have a better push in the third period, but we ran into some penalty troubles on some interesting calls. I'll just leave it at that."
Dante Hannoun got the Raiders on the board at the halfway point of the first period on a shorthanded goal, as the visitors fired 22 shots at Maier in the opening frame.
"I thought we created a lot of scoring chances tonight but we need to bury them when we have those opportunities," Hannoun said. "We need to be stronger on our second and third shots on net."
A few minutes after a shot from Cole Fonstad rolled across the Blades goal line, the hosts tied things up at the 12:02 mark of the second stanza as Max Gerlach sniped a shot past Ian Scott for his first goal of the series.
Gerlach would then dish out a helper just 71 seconds later as he fed the puck to Tristen Robins to give the Blades the lead.
"I know that I'm counted on to provide offence, but when you have two teams like this playing each other, chances are hard to come by out there," Gerlach said.
"I wasn't able to help the guys out there in the first couple of games, so it felt good to get that goal and get things going."
The overager would complete a three-point night in the third period as he got an assist on Kyle Crnkovic's goal.
Eric Florchuk would score into an empty net in the late stages of regulation for the Blades, who got a three-assist night from blueliner Dawson Davidson and a 43-save showing from Maier.
"Nolan's playing some of his best hockey for us right now," Gerlach said. "I can't say enough about him. He's one of the best goalies in the nation."
Scott stopped 34 shots for the Raiders, who lost back-to-back games for just the second time this season.
The series will now shift to the Art Hauser Centre for Game 5 at 7 p.m. Friday, with the sixth game in the series taking place at the SaskTel Centre at 4 p.m. Sunday.
"We've played a lot of good hockey up there this year, especially in the second game on Sunday until they charged back in the third period," Gerlach said. "If we can play like we did in these last two games in their barn, we should be fine."
"Having that home ice advantage is what you want in a series like this, as our fans bring a lot of energy to the rink," Hannoun added. "We know that Saskatoon is going to be battling us hard on Friday."
A seventh and deciding contest, if required, would be held in Prince Albert at 7 p.m. Tuesday.
Post-Game Notes
The only other time that the Raiders have lost consecutive games this season was on Feb. 15 and 16 as they suffered a pair of losses in a home-and-home series to the Brandon Wheat Kings.
After missing two games due to an injury, defender Reece Harsch returned to the lineup on Wednesday for the Blades as he was drawn into the lineup in favour of Majid Kaddoura after the pre-game skate.
Koen MacInnes, Alex Ozar, Randen Schmidt and Colton Dach were also scratched by the Blades, while Loeden Schaufler did not dress for the Raiders.
Wednesday's Scores
Edmonton 6 Calgary 0 (Edmonton wins 4-0 and will play winner of Prince Albert/Saskatoon in Eastern Conference final – Andrew Fyten – One Goal and One Assist – Dylan Myskiw – 18 Saves – First Career Playoff Shutout)
Spokane 3 Everett 1 (Spokane leads 3-0 – Adam Beckman – One Goal and One Assist)
Guelph 4 London 3 (London leads 3-1 – Nick Suzuki – One Goal and Two Assist)
Halifax 4 Moncton 2 (Halifax leads 4-0 – Semifinal Opponent TBD – Antoine Morand – One Goal and Two Assists)
Drummondville 4 Sherbrooke 2 (Drummondville leads 3-1 – Felix Lauzon – Two Goals and One Assist)
Thursday's Schedule
Rimouski vs. Cape Breton (Game 4) – 4 p.m.
Rouyn-Noranda vs. Victoriaville (Game 4) – 5 p.m.
Ottawa vs. Sudbury (Game 4) – 5 p.m.
Niagara vs. Oshawa (Game 4) – 5 p.m.
Saginaw vs. Sault Ste. Marie (Game 4) – 5 p.m.
Vancouver vs. Victoria (Game 4) – 8 p.m.
Friday's Schedule
Rouyn-Noranda vs. Victoriaville (Game 5 – If Necessary) – 5 p.m.
Sherbrooke vs. Drummondville (Game 5) – 5:30 p.m.
Sault Ste. Marie vs. Saginaw (Game 5 – If Necessary) – 5 p.m.
Guelph vs. London (Game 5) – 5:30 p.m.
Saskatoon vs. Prince Albert (Game 5) – 7 p.m.
Everett vs. Spokane (Game 4) – 8 p.m.
Previous articleBlades shutout Raiders in Game 3
Next articlePolice arrest two following ISET search
Lucas Punkari is the Prince Albert Daily Herald sports reporter. He can be reached at sports@paherald.sk.ca or 306-764-4276 ext. 247


text:
Q: Haskell-like `const` in C++ So in the past few weeks, I've been experimenting with functional-programming type solutions to problems in C++11, and from time to time, I've been in a situation where I need a function that returns a constant value.
In Haskell, there is a function
const :: a -> b -> a
const x = \_ -> x

that returns a function that evaluates to const's original argument, no matter what argument is supplied to it. I would like to create something similar in C++11. Such constructions are useful for signifying special behavior in functions (a constant function of true sent to a filter would leave the data intact). Here's my first attempt:
template<class T>
std::function<T(...)> constF(T x) {
    return ([x](...) { return x; });
}

This compiles on its own, but any attempt to use it results in incomplete-type errors. My second attempt was this:
template<class T, class... Args>
std::function<T(Args...)> constF(T x) {
    return ([x](Args...) { return x; });
}

This comes closer, but doesn't allow me to supply any arguments, unless I explicitly state them.
auto trueFunc1 = constF(true);
auto trueFunc2 = constF<bool, int>(true);
cout << trueFunc1() << endl; //works
cout << trueFunc1(12) << endl; //doesn't compile
cout << trueFunc2(12) << endl; //works

Ideally, a correct construction would produce no difference between trueFunc1 and trueFunc2.
Is this even possible in C++?

A: Since C++11 doesn't have generic or variadic lambdas, I'd write a functor template class:
template <typename T>
// requires CopyConstructible<T>
class const_function {
  T value;
public:
  template <typename U, typename = typename std::enable_if<std::is_convertible<U,T>::value>::type>
  const_function(U&& val) :
    value(std::forward<U>(val)) {}

  template <typename... Args>
  T operator () (Args&&...) const {
    return value;
  }
};

and a nice type-deducing wrapper to make them:
template <typename T>
const_function<typename std::remove_reference<T>::type>
constF(T&& t) {
  return {std::forward<T>(t)};
}

In C++1y, I think the simple equivalent is:
template <typename T>
auto constF(T&& t) {
  return [t{std::forward<T>(t)}](auto&&...){return t;};
}


This love story all began in the garden of Eden. There was nothing existing on the face of the earth and so God set out to work, he created all that was needed for survival on earth. Something was missing and that was man God needed a thinking being so he created man not just any kind of man but one just like Him.
After creating man (Adam and Eve),He didn't stop there He created a beautiful garden and put them in there. He loved them so much and He wanted them to be happy and live comfortably. Every evening he would come down to earth to fellowship with man, they were priceless to Him. Man knew no sin they were blameless but eventually they disobeyed God by eating from the tree which He asked them not to eat from THE TREE OF THE KNOWLEDGE OF GOOD AND EVIL. Eating from this tree exposed them to their flaws and they found out they were naked so they were ashamed and they hid from their creator. God killed an animal and used the skin to cloth them. The killing of the animal had a significance which I would explain as we go on.
God's nature isn't the nature of sin and because he is too pure to behold sin they could no longer stay in the garden.
Since the first man and woman now had the sin nature, this nature was automatically passed down to their children and the generation yet to be born.
God had to set rules for man do this don't do this but the more the rules the more man kept on falling short this was because man now had the sin nature and all that the sin nature knew how to do was to sin. The rules kept exposing them to their flaws it just showed them how incapable they were to keep God's commandments based on their strength.
One of those laws was that when someone sinned blood had to be shed for the sin of such a fellow to be forgiven if not such a person would die.
You would ask why blood though? Why not something else? That is because blood plays a very vital role in human existence, medically we have been thought that blood contains all of the oxygen and nutrients needed for ones body to thrive, this nutrients and oxygen will be of no use to the individual if there is no means for it to be transported which is where the role of blood comes into play. So in essence there is life present in blood. Since the human life had to be spared, animals were killed, not any kind of animal the animal had to be without blemish( no defect). meaning the life of an innocent animal was taken in exchange for the life of a guilty human.
The forgiveness gotten through the shedding of animal blood was temporary, the animal blood couldn't give everlasting life. which was why it had to be done over and over.
God didn't want this to keep happening so he decided to send someone, someone who was without blemish someone whose blood would wash away sins once and for all so he decided to send his son Jesus, this is really amazing because he didn't send an angel he chose his most priced possession. WHAT A LOVE. The death of Jesus on the cross signified the death of our old sinful nature and when he arose from the grave we were given a new nature, the nature of righteousness.
All of the shedding of animal blood which started from the very beginning was God preparing man for his redemptive plan. So why didn't God send his son Jesus from the very beginning when man sinned in the garden of Eden? That is because He (God) needed man to know he can never be independent of God, trying to keep a set of laws would never be possible without God's help. God automatically made a way out for man.
How can one access this free gift of redemption?
The only way we can freely access God is through His son Jesus.
Jesus said to him, "I am the [only] Way [to God] and the [real] Truth and the [real] Life; no one comes to the Father but through Me.
Being good doesn't qualify anymore or gives you access into God's kingdom because if it were so jesus wouldn't have died in the first place. What makes us qualified enough to access Gods kingdom is God's grace through the blood of Jesus.
Jesus went through all of the pain and shame on the cross of Calvary just so He could reconcile us back to Himself , he just loves to have a daily and continuous intimacy with each and every one of us. He is ever willing and ever ready to accept us into His arms. Every thing you can ever hope for is in God just open up your heart to Him and receive Him.
Behold, I'm standing at the door, knocking. If your heart is open to hear my voice and you open the door within, I will come in to you and feast with you, and you will feast with me. And to the one who conquers I will give the privilege of sitting with me on my throne, just as I conquered and sat down with my Father on his throne.
Undoubtedly the greatest love story. Blessed week.