Monday, December 10, 2018

Stellar Absorption and Emission Processes Revisited (2)

In approaching stellar line formation we will be looking at a number of related equations, including:
1)      The Boltzmann equation
 
2)     The Saha equation
 
3)     Combined Boltzmann and Saha equations

 

These will enable us to form a picture of spectral line formation which can then be generalized for different atoms and energy transitions.  We start then with the Boltzmann equation, which we already introduced in the previous chapter:

 

N2 / N1   =     [g2 / g1 ]   exp (- E2 – E1) / kT

 

That is, for the atoms of a given element in a specified state of ionization, the ratio of the number of atoms N2 with energy E2, to the number of atoms N1   with energy E1, in different states of ionization is given by the above formula. The same form of the equation can also be used to find the ratio of probabilities, i.e. that the system will be found in any of the  g2   degenerate states with energy E2 to the probability that the system is in any of the g1   degenerate states E1, viz.

 

P(E2) / P(E1)   =     [g2 / g1 ]   exp (- E2 – E1) / kT

 

Thus, the Boltzmann equation can be posed in two forms.  In statistical mechanics we could have also seen the partition function:

 

Z  =   å j   exp ( - e j )/ t

 

Which is just the summation over the Boltzmann factor (exp ( - e j )/ t ) for all states j for which the number of particles (N) is constant. We will find it useful to rewrite it:

 

Z = g1 +  å¥ j = 2   g j  exp (- E j – E1) / kT

 
Of interest now are the relative numbers of atoms in ionization stage i, which is written:

 

N e N i + 1 / N i  =  2 Z i + 1 / Z i (2 p m e kT/ h 2) 1.5  e - c i/ kT

 

This is the Saha equation, named after the Indian astrophysicist who first derived it.  Here,  N e  is the number of free electrons per unit volume and c i  is the ionization potential of the ith ionization stage. Thus, the equation relates the number of atoms in two successive  ionization stages to the quantities that are relevant. As per our introduction to quantum mechanics, the factor ‘2’ in the equation refers to the two possible spins of the free hydrogen election with spin quantum number:

m s =  +½.


 

 Recall that for thermodynamic equilibrium, the rate of ionization cannot exceed the rate of recombination[1].  In other words, the rate at which atoms in the ith stage are ionized (i.e. to the i +1st stage) must equal the rate at wich ions in that i +1st stage are recombining with free electrons to form ions in the ith stage. The latter depends on N e N i + 1   and the former on N i . Hence, Saha’s equation simply expresses the fact these two processes must occur at the same rate.
 
One can also rewrite the equation in a more manageable logarithmic form if one substitutes the numerical constants:

log(N e N i + 1 /N i)  = 

 

15.38 + log (2 Z i + 1 / Z i ) + 1.5 log T – 5040 c i/ T

 

The units here are important to note, and are consistent with the ionization potential being measured in electron volts (eV). Therefore N e  must be in particles per cubic centimeter.

 

    Yet another way to express the Saha equation is via the electron pressure, P e . This acknowledges that each separate species of particle makes its own contribution to the total gas pressure. The free electrons in a gas therefore produce a pressure given by: P e = N e kT.

 

Then we may write another log form of the Saha equation:

 

log(P e N i + 1 /N i)  = 

 

-0.48  + log (2 Z i + 1 / Z i ) + 2.5 log T – 5040 c i/ T

 

There are also two distinct processes by which lines can be formed:

 
1)      Bound-bound transitions
2)     Bound-free transitions

In the first case, the photon goes from one bound atom to another. Also, in case (1) the photon has a good chance of being scattered, i.e. emitted in the same downward transition.  This may be at the same frequency as the absorption, in which case we say the scattering is coherent, or not, in which case it is non-coherent.
Details of bound-bound transitions differ in significant ways from bound-free transitions. In the first case, the transitions are also affected by a broadening function which is not so important for continuous emission. If we write out the equation for absorption in more detail we get:

a u  =  [1 - e -  h u o / kT] (p e2/ mc) f f u
 
which yields units in  cm2 / atoms at lower level. Two other absorption derivative values are possible from the preceding:

 
i)                   The absorption coefficient per unit length (cm-1)
ii)                 The mass absorption coefficient  k u .

 The value for (i) is just a u  multiplied by the number of absorbing atoms per unit volume. The value for  k u  is just  multiplied by the number of absorbing atoms.  

The value for a o   is just:  a o =  a u  /  f u.  
 
Sometimes referred to as a “fudge factor”, f is known as the oscillator strength or f-value of the line. It is basically the transition probability for the line and is to be computed by quantum mechanics or measured in the laboratory.
 
The broadening function  f u  is:

f u  =   1/ Öp   [exp (u  -u o)  /D u D ]2  D uD
Which can also be rewritten as:   f u  du  =

1/ Öp   [exp (u  - u o)  /D uD ]2  du / D uD

 
This would be the probability that the absorbed photon lies between u  and  u +  du, assuming equal intensities for all frequencies. Thus the integral:

ò f u  du  =  1

 
Where du is over all frequencies.  The value of f u  is larger near  u o  the frequency of the line center, as may be deduced from the line profile diagram below:
No automatic alt text available.

Absorption line profile showing the core and "wings"
 
 
Note here that  D uD  is the Doppler half-width of the line. As can be seen on inspection, f u is very large at line center and falls off in the “wings”, i.e. at larger and smaller frequencies.
 The three important types of line broadening are doppler effect, natural and pressure broadening. We will confine our attention to the first type which is given by the broadening probability equation, provided the velocity is Maxwellian and that the frequency at line center  u o is also observed for some u . The Maxwellian will display the distribution of velocities as shown below where the central line defines the most probable.

No automatic alt text available.
Maxwellian showing distribution of velocity with proportion of particles.

For the Sun, solar physics, it is also necessary to consider the Zeeman effect, a broadening due to strong magnetic fields such as in sunspots. An example of this applied to a sunspot is depicted below:
The left image shows the line-centered sunspot for which the Zeeman effect in classic "triplet" form (right image) is detected and measured. The greater the spectral line splitting the greater the magnitude of the associated magnetic field.  George Ellery Hale, who discovered the effect, posed the quantitative relationship in terms of the original wavelength   lo  (undisturbed line) and the spread of wavelengths, D  l:


D  l =     (lo)2  e H/ 4 π  me c2  


Where H is the intensity of the sunspot magnetic field in gauss (to be found), e is the electron charge in e.s.u.,  me  is the mass of the electron in grams, and c the velocity of light in cm/ sec.

 

Problems:

 
1) For the temperature and conditions of problem (1) of the previous set, find the ratio of the probability that the system will be found in any of the eight degenerate states of energy level E2, to the probability that the system will be in any of the two degenerate states of energy level E1.

2) An H-alpha line undergoes triplet splitting in the vicinity of a sunspot. The undisturbed line is measured at:   lo  =    6.62 x 10  -5 cm. The line shift on either side is: + 0.0025 Å.   Use this information to find the strength of the sunspot magnetic field: a) in gauss, b) in Tesla.

Yes, Trump Is A Felon, Who'll Be Locked Up As Soon As His One Off Term Is Done


"Hmmmm...mebbe I should'na kept calling Mueller's probe a witch hunt."

With last week's  two court filings made by federal prosecutors  out of the Southern District of New York(SDNY).  it's now clear the current resident in the White House is a felon.   Reading the applicable documents (in the link above) one readily sees that Trump  (denoted "individual 1")  directed his then adviser-lawyer, Michael Cohen, to make illegal payoffs to two women who claimed to have had sexual relationships with Trump, implicating the Dotard in the violation of campaign finance laws. They recommended that Cohen receive a prison sentence of about four years.  In the filing, it is also clear that in the process of spelling this out Trump is a so far unindicted co-conspirator.

Let's now get this in perspective: The Dotard's own Justice Department is claiming that he has committed campaign finance violations, paying hush money to two mistresses: Stormy Daniels and Karen McDougal. By virtue of this hush money ploy it thereby followed he was trying to manipulate the 2016 election result by buying the women's silence. (Given that had the story materialized charges would have been filed at the time and Trump's odious campaign would have crashed and burned.)

 Note that this is all a separate creature from the Russian conspiracy  (not 'collusion' since there is no federal statute for that) that Robert Mueller is pursuing.  While it is true Mueller said in his own court filing that Cohen had provided him “useful information” on matters at the core of the Trump-Russia investigation,   it is also clear from the separate SDNY filing that Cohen was not as forthcoming with the facts of how and  why Trump had him pay off two mistresses in order to keep his campaign violations secret- and thereby sneak into the White House as a felon.

In their  separate filing, federal prosecutors in New York said Cohen “acted in coordination and at the direction of” Trump when setting up payments to buy the silence of Karen McDougal, a former Playboy model, and Stormy Daniels, a pornographic actress, who were considering making public their allegations of affairs with Trump.  Cohen and Trump paid the women to suppress their damaging stories and “to influence the 2016 presidential election”, according to the filing .  As I noted in a previous post this is a major no-no and indeed, a felony.  A candidate simply cannot hide his campaign finances or redirect them to outside ends to secure his nomination, or election. It's that simple.

Of course, the inbred hillbilly and Trump toady Sarah Sanders, told reporters the filings contained “nothing of value that wasn’t already known”, saying Cohen had “repeatedly lied” and was “no hero”.  But she misses the point as anyone with even non-Mensan IQ can see. That is, that Trump is clearly named as the primary Cohen instigator (and hence felon), directing his thug to buy off the women. A 2-bit Queens thug, grifter and conman hiring another guy of similar pedigree to do his dirty work for him. Who'da thought?

The SDNY prosecutors added that Cohen was motivated by greed and “repeatedly used his power and influence for deceptive ends", noting:

After cheating the [Internal Revenue Service] for years, lying to banks and to Congress, and seeking to criminally influence the presidential election, Cohen’s decision to plead guilty – rather than seek a pardon for his manifold crimes – does not make him a hero.”

Furthermore, Cohen's perjury didn't end with his lies and omissions to the SDNY   Cohen previously pleaded guilty to lying to Congress about Trump’s plans to develop a building in Russia. He admitted the project continued well into Trump’s campaign for the presidency – contradicting Trump’s account – and that Cohen spoke with a Kremlin official about securing Russian government support.

In other words, we have direct evidence of a devious and nefarious synergy between Trump and the Russkies,  that permeated the Trump campaign.  Let's clear the air here and state emphatically that Trump is guilty of at least two felonies and hence two impeachable offenses.  Sadly, the putative future Dem Chairman of the House Judiciary Committee (Jerry Nadler) seems not to grasp that impeaching Trump is not just a matter of political calculation but of holding him accountable.  As MSNBC's  Lawrence O'Donnell' made clear to Nadler last week any failure to impeach Trump now - even if it doesn't lead to indictment (by the Senate) -means that Trump and any followers will believe he's above the law. (Jerry kept telling Lawrence the Dems "need at least twenty Republicans" - but that is only for indictment in the Senate, not impeachment.)

This cannot be allowed, and that also means skewering the rabid distortions and diversion efforts apparent in the WSJ's editorial today ('Wrap It Up, Mr. Mueller'. p.  A16) in which these lackeys for Rupert Murdoch actually compared Trump's violations to those of John Edwards  in 2012 ("for payments made by campaign donors to his mistress.")  But that's comparing rotten apples to fresh oranges.  As Kelly Ann Conway's better half, George put it, directed at Dotard's tweet that he did nothing and nothing was found:

"Except for that little part where the US Attorney’s Office says that you directed and coordinated with Cohen to commit two felonies."

In Edwards' case there was no conspiratorial direction to intentionally commit felonies. Mucho difference, my friends!  And further it isn't a simply reductive case of "lying about sex" as the WSJ's nabobs try to portray it. It is paying hush money to hide the felonies to secure a nomination via fraud, and also an election!   The WSJ editors whining about the length of Mueller's probe is also choice,, i.e.  "Mr.  Mueller has been investigating for 19 months....the country deserves an account of what Mueller knows not more dribs and drabs in sentencing memos."  Aw, boo hoo and hoo, cry me a river.  For reference, Mueller is a professional and is aggregating millions of times more real data - including financial - than Ken Starr used to go after Bill Clinton, for a blow job  And that farce took years, way more than 19 months

Hence, whether an impeachment brings more polarization or division  as Nadler fears (and the WSJ's trolls anticipate), is irrelevant to the need to impeach the vile maggot in the White House.  What we don't need is more bullshit and disinformation from the likes of the WSJ to try to run interference for the Repukes and obscure the issue for too many 'Muricans.

Still on Sunday's CNN interview  - almost as if precognitively channeling the WSJ codswallop published today  - Jerry Nadler continued his verbal parsing by stating, in regard to the SDNY prosecutors' charges:

"They would be impeachable offenses. Whether they are important enough to justify impeachment is a different question."

Actually Jerry is overthinking the issue. IF they are indeed impeachable offenses, then it is manifestly clear that they warrant impeachment, no 'ifs, ands or buts.  What Nadler is really worried about is enraging the Trumpie base, when he spoke on Lawrence O'Donnell's 'Last Word' last week.  But enraging the Trumpie base is what I would call acceptable collateral "damage" if it means refusing to allow the slime ball cretin in the White House to get away with his crimes. And that was also Lawrence O'Donnell's point in his retort to Jerry.

Jerry needs to grow a pair, and grasp that if this foul fungal biped attempted to manipulate the 2016 election result by coercing his mistresses into secrecy then he is guilty.  As a great statesman once put it: "Let justice be done or the heavens fall!"

And hey, at least it's not as violent (or permanent)  as hanging him!

On another sorry note, it appears a Reeptard Senator named Mike Lee is threatening "civil war" if Dems don't toe the line and limit the federal gov't instead of expanding it.   See e.g.

Evidently these Reepo imps like Lee aren't satisfied with repealing the voters' will in WI, NC and MI, they now want a war too. Well, bring it on, Mikey!

See also:

https://brane-space.blogspot.com/2018/08/why-didnt-manafort-jury-find-him-guilty.html

Saturday, December 8, 2018

Selected Questions- Answers From All Experts Astronomy Forum (Estimating No. of Stars In An 8000 Light Year 'Bubble' Around Earth)



Question: How many stars would be in an 8000 light year ''bubble'' around Earth? And how many habitable planets would said area have?


Answer:  First, let's change the units to parsecs instead of light years. Currently, the stellar density near the galactic center is estimated at 100 per cubic parsec - so that in the Earth's vicinity two-thirds of the way out to the galactic edge it is  much more sparse,  like 0.5 per cubic parsec.

Converting light years (radius = 4000 ly) to parsecs we get a radius R = 1.23 x 103 parsecs.


We use this to get the  total volume of interest:

V = 4/3 pi R 3  =  7.8 x 109 cubic parsecs


If there are 0.5 stars on average per cubic parsec then the stellar density is:

(0.5) (7.8 x 109 cubic parsecs)  =  3.9 x 10 9 (approx.)

(This would be the same number as in your "8000 light year ''bubble'' around Earth")

The next question (how many habitable worlds?)  is much more difficult because the data is extremely sketchy.  Assuming a habitable planet requires an early spectral type G star (the Sun is G2) then perhaps only 1 out of 100 stars fit that bill. "Habitable" means within the so-called "Goldilocks" zone - i.e. within a suitable temperature sphere about the star.

So far only about 1 in 100 fit that distribution according to a sampling of the Kepler space telescope exoplanet findings.

So, given these limiting parameters (fractions), the estimated (again 'ball park') number of habitable planets in that volume (not "area")  would be: roughly    40,000

One must take care, however, to distinguish "habitable" from actually inhabited. The latter, may be one thousand times less - again, or only 40.   In the words of Guillermo Torres, an astronomers at the Harvard-Smithsonian Center for Astrophysics:

"These planets do exist. We didn't know that before. What we're really looking for eventually is signs of life. We're not there yet. It will take many years."

Indeed! Which begs the question: 'Can any kind of life on these exo-planets really be determined from a distance?'  I don't believe so.  I suspect it would take actually going there and doing a=careful bio-chemical investigations such as we've done on Mars.  So the most we can do for now is speculate that this or that planet may be life supporting, or potentially "habitable".

Friday, December 7, 2018

SolutionsTo Stellar Emission - Absorption (Revisited) Problems

1) Consider a gas of neutral hydrogen. Using the Boltzmann equation and the information in the table (see April 2 blog post), compute the temperature at which one will expect equal numbers of atoms in the ground state and the first excited state.

Solution:

The Boltzmann equation is:

N2 / N1  =     [g2 / g1 ]   exp (- E2 – E1) / kT

And from the table, g2 = 8 and  g1 = 2

We require the condition that:  N2 =  N1     so:

1 =     [8/ 2 ] exp (- E2 – E1) / kT

But:  E2 =  - 13.6 eV and E1 = -3.4 eV, therefore:

1 = 4 exp [- 13.6eV – (-3.4 eV) ]/ kT

1 =  4 exp (-10.2 eV)/ kT

Taking natural logs:

ln (4)  =    (10.2 eV)/ kT

where: k =   8.6174 x 10 -5 eV/K

Solving for T: 

T =    10.2 eV/  (ln 4) (8.6174 x 10 -5 eV/K)

T =  10.2 eV/ (1.3862) (8.6174 x 10 -5 eV/K)

T= 85 388 K or T = 8.54 x 10 4 K


 
2 ) For the Balmer a line (called H- alpha), we know:

E3 – E2 =  - 13.6 eV ( 1/ 3 2   -  1/ 2 2 )   = 1.88 eV

a)     From this information calculate the ratio N2 / N1    

b)     Obtain the specific intensity from: 

I u  =  2h u 3 / c 2  [1/ exp (hc/lkT]


Solution:

N2 / N1   =     [g2 / g1 ]   exp (- E2 – E1) / kT

From the table in the Dec. 5th  post;  g2 = 8 and  g1 = 2   

k =   8.6174 x 10 -5 eV/K

N2 / N1    =     [8/ 2 ]   exp (- E2 – E1) / kT

N2 / N1      =  4 exp (-1.88 eV)/ (8.61 x 10 -5 eV/K) (10 4 K)

N2 / N1      =  4(0.113) =  0.452

b)I u  =  2h u 3 / c 2  [1/ exp (hc/lkT]

We need to use consistent cgs units.  Planck constant h = 6.62 x 10  -27 erg-s

c=  3 x 10  10 cm/s

l  =  hc/ E =  (6.62 x 10  -27 erg-s) (3 x 10  10 cm/s)/ 3.0 x 10  -12 erg

l  =  6.62 x 10  -5 cm 

k=1.38 x 10  -16 erg/K

[1/ exp (hc/lkT]  =

1/ [exp (6.62 x 10  -27 erg-s) (3 x 10  10 cm/s)/ (6.62 x 10  -5 cm)  (1.38 x 10  -16 erg/K)( 10 4 K)

= 0.113

I u  =   2h u 3 / c 2  [0.113] erg cm -2/s

But  u =   E/h   =  

3.0 x 10  -12 erg/ 6.62 x 10  -27 erg-s  =  4.53 x 10  14 /s

So:
I u  =    0.226(6.62 x 10  -27 erg-s) (4.53 x 10  14 /s) 3 / (3 x 10  10 cm/s) 2

I u  =     1.51  x 10  -4 erg cm -2/s

3)Calculate  the transition probability you get using the Einstein equation:
A 21=   6.67 x 10 16 [g f/g2  l2   Å]
What possible errors might cause the values to diverge? (Take g = f » 1)
  l =  6.62 x 10  -7 m  =  6.62 x 10  -5 cm =  6620 Å

With g2 = 8  and g = f = 1

A 21=   1.93 x 10 8 [Å-2]

Compare to standard form (see e.g. Wikipedia, “Einstein coefficients”)  given in multiple physics papers as:

A 21=  [f g1/g2]{ 2 π  u 3  e 2 }/ e o  me c3  

A 21 =    1.87 x 10 7    or:    0.187  (in defined units of    10 8  s)      

 
Error sources: Imprecise oscillator frequency f

Error in one of the statistical weights.

Uncertainty in line measurements for absorption and extrapolating this for deduction of  A 21.