Download Question 9: Starting from the ground state, press two and only two

E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 1 of 29 pages
5th class Feb. 25, 2013 Outline for PHS 207 – The Stars – Spectra and Luminosity
1. Quiz on the Sun
Label the 6 layers of the Sun and the 2 features
Picture from
http://t0.gstatic.com/images?q=tbn:ANd9GcRHKEB7P1935hoRjAGBYSjgbh8NYRZOXxTFQycAjiiudvHMglI4_w
Part
1
2
3
4
5
6
Corona
Temperature (oK in Kelvin
degrees)
2,000,000
Description/Function/Result
Outermost; seen only during
solar eclipse
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 2 of 29 pages
2. Introduction to NAAP ClassAction Questions
#4 Number line – EM Spectrum
Spectra or spectral lines indicate what elements are present in a
star and its overall color its temperature.
3
The light seen from a star is typically an absorption spectrum. The light produced by the
surface of the star – the photosphere – is a continuous spectrum meaning that all
wavelengths of light are present. However, certain wavelengths of light are redirected
(absorbed and re-emitted in random directions) in the cooler low-density layers above the
surface (the chromosphere) of a star. This occurs because electrons in Hydrogen (and other
atoms) absorb light as they jump to higher orbitals and then re-emit the light as they drop
back down. Thus, we see absorption spectra from stars – rainbows with dark gaps known as
spectral lines – because a lot of the light is missing at particular wavelengths
4. . Hydrogen Energy Level Transitions
Absorbing Photons
Because an electron bound to an atom can only have certain energies the
electron can only absorb photons of certain energies.
According to the theory quantum mechanics, an electron bound to an atom can not have
any value of energy, rather it can only occupy certain states which correspond to certain
energy levels. The formula defining the energy levels of a Hydrogen atom are given by the
equation: E = -E0/n2, where E0 = 13.6 eV (1 eV = 1.602×10-19 Joules) and n = 1,2,3… and so
on. The energy is expressed as a negative number because it takes that much energy to
unbind (ionize) the electron from the nucleus. It is common convention to say an unbound
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 3 of 29 pages
electron has zero (binding) energy. Because an electron bound to an atom can only have
certain energies the electron can only absorb photons of certain energies exactly matched
to the energy difference, or “quantum leap”, between two energy states.
When an electron absorbs a photon it gains the energy of the photon. Because an electron
bound to an atom can only have certain energies the electron can only absorb photons of
certain energies. For example an electron in the ground state has an energy of -13.6 eV.
The second energy level is -3.4 eV. Thus it would take E2 − E1 = -3.4 eV − -13.6 eV = 10.2 eV
to excite the electron from the ground state to the first excited state.
If a photon has more energy than the binding energy of the electron then the photon will
free the electron from the atom – ionizing it. The ground state is the most bound state and
therefore takes the most energy to ionize.
Emitting Photons
When an electron drops from a higher level to a lower level it sheds the
excess energy, a positive amount, by emitting a photon.
Generally speaking, the excited state is not the most stable state of an atom. An electron has a
certain probability to spontaneously drop from one excited state to a lower (i.e. more negative)
energy level. When an electron drops from a higher level to a lower level it sheds the excess
energy, a positive amount, by emitting a photon.
The energy of the emitted photon is given by the Rydberg Formula. This formula is
essentially the subtraction of two energy levels. It is:
– Rydberg Formula
1
1
Ephoton = E0 n 2 − n 2
1
2
(
)
where n1 < n2 and (as before) E0 = 13.6 eV. With the restriction n1 < n2 the energy of the
photon is always positive. This means that the photon is emitted and that interpretation
was the original application of Rydberg. It also works if the n1, n2 restriction is relaxed. In
that case the negative energy means a photon (of positive energy) is absorbed.
For Hydrogen some of the (emitting photon transitions) are named for
Lyman, Balmer, and Paschen (Lines) those who discovered them.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 4 of 29 pages
Long before the Hydrogen atom was understood in terms of energy levels and transitions,
astronomers had being observing the photons that are emitted by Hydrogen (because
stars are mostly Hydrogen). Atomic physicist Balmer noted, empirically, a numerical
relationship in the energies of photons emitted. This relationship was generalized and
given context by the Rydberg Formula. But the various discrete photon
energies/wavelengths that were observed by Balmer were named the Balmer series.
It was later understood that the Balmer lines are created by energy transitions in the
Hydrogen atom. Specifically, when a photon drops from an excited state to the
second orbital, a Balmer line is observed. The Balmer series is important
because the photons emitted by this transition are in the visible regime. The
Balmer series is indicated by an H with a subscript α, β, γ, etc. with longest wavelength
given by α.
As there are other transitions possible, there are other “series”. All transitions which
drop to the first orbital (i.e. the ground state) emit photons in the Lyman
series. All transitions which drop to the 3rd orbital are known as the Paschen series. The
graphic to the right shows some of the Lyman and Balmer transitions graphically.
Spectra Lines
Lyman
Balmer
Transitions to
Ground state
1st Excited state
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 5 of 29 pages
Paschen
2nd Excited state
Hydrogen Atom Simulator
http://astro.unl.edu/classaction/animations/light/hydrogenatom.html
Hydrogen Atom Simulator – Introduction
The Hydrogen Atom Simulator allows one to view the interaction of an idealized
Hydrogen atom with photons of various wavelengths. This atom is far from the influence
of neighboring atoms and is not moving. The simulator consists of four panels. Below
gives a brief overview of the basics of the simulator.

The panel in the upper left shows the Bohr Model: the proton, electron, and the
first six orbitals with the correct relative spacing.
o The electron can absorb photons and jump higher energy levels where it
will remain for a short time before emitting a photon(s) and drop to lower
energy level (with known probabilities fixed by quantum mechanics).
o The electron can also be ionized. The simulator will a short time later
absorb an electron.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 6 of 29 pages
o For convenience you can drag the electron between levels. Once it is
released it will behave “physically” once again as if it had gotten to that
present level without being dragged.

The upper right panel labeled “energy level diagram” shows the energy levels
vertically with correct relative spacing.

The “Photon Selection” panel (bottom left) allows one to “shoot” photons at the
Hydrogen atom. The slider allows the user to pick a photon of a particular
energy/wavelength/frequency.
o Note how energy and frequency are directly proportional and energy and
wavelength are inversely proportional.
o On the slider are some of the energies which correspond to levels in the
Lyman, Balmer, and Paschen series.
o Clicking on the label will shoot a photon of that energy.
o If the photon is in visual band, its true color is shown. Photons of longer
wavelengths are shown as red and shorter wavelengths as violet.

The “Event Log” in the lower right lists all the photons that the atom has
encountered as well as all the electron transitions.
o The log can be cleared by either using the button or manually dragging the
electron to a particular energy level.
Hydrogen Atom Simulator – Exercises
For any particular level of the Hydrogen atom one can think of the photons that interact
with it as being in three groups:
Range 1
Increasing Energy →
Range 2
Range 3
All the photons have
enough energy to ionize
the atom.
Some of the photons have
the right energy to make the
electrons to jump to a higher
energy level (i.e. excite
them).
Note that the ranges are different for each energy level. Below is an example of the
ranges for an electron in the ground state of a Hydrogen atom.
None of the photons have
enough energy to affect the
atom.
Range 1
Ground State electron of H
Range 2
Range 3
0eV to 10.2 eV
10.2 to 13.6
>13.6 eV
(10.2 eV needed to excite
electron to 1st orbital)
(some will excite, some
won’t)
(anything greater than this
will ionize the electron)
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 7 of 29 pages
When the simulator first loads, the electron is in the ground state and the slider is at 271
nm.

Fire a 271 nm photon. This photon is in range 1.

Gradually increase the slider to find a photon which is between range 1 and range
2 (for a ground state electron). This should be the Lyman-α line (which is the
energy difference between the ground state and the second orbital).

Increase the energy a bit more from the Lyman-α line and click “fire photon”.
Note that nothing happens. This is a range 2 photon but it doesn’t have the “right
energy”.

Increase the energy more until photons of range 3 are reached. In the simulator
this will be just above the Lε line.
o Technically there are photons which would excite to the 7th, 8th, 9th, etc.
energy levels, but these are very close together and those lines not shown
on the simulator.
o The Lε line has an energy of -13.22 eV and is in range two. The ionization
energy for an electron in the ground state is 13.6 eV and so that is the
correct range 3 boundary.
Question 4: Which photon energies will excite the Hydrogen atom when its electron is in
the ground state? (Hint: there are 5 named on the simulator, though there are more.)
Question 5: Starting from the ground state, press the Lα button twice in succession (that is,
press it a second time before the electron decays). What happens to the electron?
Question 6: Complete the energy range values for the 1st excited state (i.e. the second
orbital) of Hydrogen. Use the simulator to fill out ranges 2 and range 3. The electron can
be placed in the 1st orbital by manually dragging the electron or firing a Lα photon once
when the electron is in the ground state. Note also that the electron will deexcite with time
and so it may need to be placed in the 1st orbital repeatedly.
Range 1
0 to 1.9 eV (anything less
than this energy will fail to
excite the atom)
1st Excited State Electron in H
Range 2
Range 3
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 8 of 29 pages
Question 7: What is the necessary condition for Balmer Line photons (H α , etc) to be
absorbed by the Hydrogen atom? _____________________________________________
Question 8: Complete the energy range values for the 3 rd orbital (2nd excited state) of
Hydrogen. The electron can be placed in the 3rd orbital by manually dragging the electron
or firing an Lβ photon once when the electron is in the ground state. Note also that the
electron will deexcite with time and so it may need to be placed in the 2nd orbital
repeatedly.
56
56
4
4
3
3
Hδ
2
2
Lα
1
1
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 9 of 29 pages
Question 8: Complete the energy range values for the 3 rd orbital (2nd excited state) of
Hydrogen. The electron can be placed in the 3rd orbital by manually dragging the electron
or firing an Lβ photon once when the electron is in the ground state. Note also that the
electron will deexcite with time and so it may need to be placed in the 2nd orbital
repeatedly.
Range 1
3rd Electron Orbital in H
Range 2
Range 3
>1.5 eV (anything more than
this will ionize the atom)
Question 9: Starting from the ground state, press two and only two buttons to achieve the
6th orbital in two different ways. One of the ways has been given. Illustrate your
transitions with arrows on the energy level diagrams provided and label the arrow with the
button pressed. Metthod 1
Method 2
56
56
4
4
3
3
Hδ
2
2
b) Method 2:
Lα
1
1
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 10 of 29 pages
Question 10: Press three buttons to bring the electron from
the ground state to the 4th orbital. Illustrate the transitions as 56
4
arrows on the energy level diagrams provided and label the
3
arrow with the button pressed.
2
1
Question 11: How does the energy of a photon emitted when the electron moves from the
3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron
moves from the 2nd orbital to the 3rd orbital? ____________________________________
Question 12: Compare the amount of energy needed for the following 3 transitions.
Explain why these values occur.

Lα: Level 1 to Level 2 _______________

Hα: Level 2 to Level 3 _______________

Pα: Level 3 to Level 4
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 11 of 29 pages
Black Body Radiation Curve is produced by temperature (Thermal
Distribution) of the atoms of the star.
5.
Line Strength is the intensity of the absorption or
emissions.
When Hydrogen is excited it emits light as photons de-excite. Or conversely, the Hydrogen
will absorb photons of certain energies. The strength of the line from a source of Hydrogen
will depend on how many electrons are in a particular excited state. If only very few
electrons are the first excited state, the Balmer lines will be very weak. If many Hydrogen
atoms are in the first excited state then the Balmer lines will be strong.
The number of Hydrogen atoms are in what state is a statistical distribution that depends
on the temperature of the Hydrogen source. The Thermal Distribution simulator shown
later demonstrates this.
Use the “Thermal Distribution Simulator” below this graph to plot of the number of atoms
with electrons in the 2nd orbital. There should be at least 8 points on your plot. Note also
that the y-axis is in terms of 1015 particles. Thus the point for 15,000 K, which has 3.82 ×
1017 particles in the 2nd orbital, will read as 382 on the graph.

Plot at least 8 points on the graph. More points near “interesting” features is highly
recommended.

Fit (draw) a curve to the plotted points. It should be a smooth curve.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes …\Stars\Document1 12 of 29 pages
The light seen from a star typically contains an absorption spectrum. The
light produced by the surface of the star – the photosphere – is a
continuous spectrum meaning that all wavelengths of light are present.
However, certain wavelengths of light are redirected (absorbed and reemitted in random directions) in the cooler low-density layers above the
surface (the chromosphere) of a star. This occurs because electrons in
Hydrogen (and other atoms) absorb light as they jump to higher orbitals and
then re-emit the light as they drop back down. Thus, we see absorption
spectra from stars – rainbows with dark gaps known as spectral lines –
3500
number of levl 2 atoms (1E15)
3000
2500
2000
1500
1000
500
0
3000 6000 9000 12000 15000 18000 21000 24000 27000 30000
Temperature (K)
because a lot of the light is missing at particular wavelengths.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 13 of 29 pages
Question 13: Consider the strength of the Hβ absorption line in the spectra of stars of various
surface temperatures. This is the amount of light that is missing from the spectra because
Hydrogen electrons have absorbed the photons and jumped from level 2 to level 4. How do you
think the strength of Hβ absorption varies with stellar surface temperature?
A way to measure the average speed of the atoms with respect to
each other is temperature.
Thermal Distribution of Hydrogen atoms
From http://astro.unl.edu/naap/hydrogen/abundances.html
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 14 of 29 pages
One Atom vs. Many Atoms
When an atom is by itself, in isolation, its orbitals behave differently than when in packed tightly
with other atoms. For example, the energy levels of a single carbon atom are slightly different
than a diamond. Similarly, hydrogen gas (H2) is a tiny bit different than a simple H atom. The
difference in energy levels, however, is not much in that case. The Hydrogen Atom Simulator
showed just one H atom.
Astronomically one H atom is never observed. Rather, only vast numbers of Hydrogen atoms
together are observed. Often they atoms (or H2 molecules) are on average far enough apart so
that the orbitals aren't significantly altered. Another way of saying that is the density is “low”.
But even when the density is low (which we will assume here), there are the occasional
collisions between the atoms. When they collide some of the energy goes into them bouncing off
of each other and some of it can go into exciting electrons. How frequently collisions occur and
how much energy typically goes to exciting the atom depends on how fast the low density cloud
of hydrogen atoms are moving on average. A way to measure the average speed
of the atoms with respect to each other is temperature.
Thermal Distribution
The histogram above is of 1025 Hydrogen atoms. The density of the atoms is low so that their
energy levels are very close to what we see for a single atom. The temperature can be varied
from 3000 K to 30,000 K (a range which includes the surface temperature of almost all stars).
Experiment with the slider the histogram to change the temperature for the 8 data points for the
graph on the previous page. The histogram can be downloaded from
http://astro.unl.edu/naap/hydrogen/abundances.html.



Note that at 3000 K almost all of the atoms are in the ground state.
Note also that at 3000 K there are more ionized atoms than there are atoms with electrons
in the 2nd orbital. As the temperature is increased the disparity between the number of
ionized atoms and level 2 atoms will become even larger. This occurs because of the
energy spacing of the levels – an electron in level 1 is much more tightly bound than one
in level 2. Collisions between atoms that are sufficiently energetic to knock the electron
from the ground state to level 2, but not sufficiently energetic to ionize that atom become
rare as the temperature rises.
Note also that the histogram is logarithmic and the relative heights of the bars behave in a
non-intuitive fashion. For example, at 20,700 K the ionized atom bar is only about 20%
higher than the level 1 bar but there are 10,000 ionized atom to every level 1 atom.
http://astro.unl.edu/naap/blackbody/animations/blackbody.html
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 15 of 29 pages
Filters Simulator Overview
The filters simulator allows one to observe light from various sources passing through multiple
filters and the resulting light that passes through to some detector. An “optical bench” shows
the source, slots for filters, and the detected light. The wavelengths of light involved range from
380 nm to 825 nm which more than encompass the range of wavelengths detected by the
human eye.
The upper half of the simulator graphically displays the source-filter-detector process. A graph
of intensity versus wavelength for the source is shown in the leftmost graph. The middle graph
displays the combined filter transmittance – the percentage of light the filters allow to pass for
each wavelength. The rightmost graph displays a graph of intensity versus wavelength for the
light that actually gets through the filter and could travel on to some detector such as your eye
or a CCD. Color swatches at the far left and right demonstrate the effective color of the source
and detector profile respectively.
The lower portion of the simulator contains tools for controlling both the light source and the
filter transmittance.

In the source panel perform the following actions to gain familiarity.
o Create an blackbody source distribution – the spectrum produced by a light bulb
which is a continuous spectrum. Practice using the temperature and peak height
controls to control the source spectrum.
o Create a bell-shaped spectrum. This distribution is symmetric about a peak
wavelength. Practice using the peak wavelength, spread, and peak height
controls to vary the source spectrum.
o Practice creating piecewise linear sources. In this mode the user has complete
control over the shape of the spectrum as control points can be dragged to any
value of intensity.
Additional control points are created whenever a piecewise segment is
clicked at that location.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 16 of 29 pages
Control points may be deleted by holding down the Delete key and
clicking them.
Control points can be dragged to any location as long as they don’t pass
the wavelength value of another control point.

In the filters panel perform the following actions to gain familiarity.
o Review the shapes of the preset filters (the B,V, and R filters) in the filters list.
Clicking on them selects them and displays them in the graph in the filters panel.
o Click the add button below the filters list.
Rename the filter from the default (“filter 4”).
Shape the piecewise linear function to something other than a flat line.
o Click the add button below the filters list.
Select bell-shaped from the distribution type pull down menu.
Alter the features of the default and rename the filter.
o If desired, click the remove button below the filters list. This removes the
actively selected filter (can’t remove the preset B,V, and R filters). Filters are not
saved anywhere. Refreshing the flash file deletes the filters.
Filters Simulator Questions

Use the piecewise linear mode of the source panel to create a “flat white light” source
at maximum intensity. This source will have all wavelengths with equal intensity.

Drag the V filter to a slot in the beam path (i.e. place them in the filter rack).

Try the B and the R filter one at a time as well. Dragging a filter anywhere away from the
filter rack will remove it from the beam path.
Question 1: Sketch the graphs for the flat white light and V filter in the boxes below. What is the
effective color of the detected distribution? __________________________________________
source distribution
combined filter transmittance
detected distribution
Question 2: With the flat white light source, what is the relationship between the filter
transmittance and the detected distribution? ______________Add a new piecewise linear filter.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 17 of 29 pages

Adjust the filter so that only large amounts of green light pass. This will require that
addition of points.
Question 3: Use this green filter with the flat white light source and sketch the graphs below.
source distribution
combined filter transmittance
detected distribution
Question 4: Use the blackbody option in the source panel to create a blackbody spectrum that
mimics white light. What is the temperature of this blackbody you created? ________________
• Add a new piecewise linear filter to the filter list.

Modify the new filter to create a 40% “neutral density filter”. That is, create a filter
which allows approximately 40% of the light to pass through at all wavelengths.

Set up the simulator so that light from the “blackbody white light” source passes
through this filter.
Question 5: Sketch the graphs created above in the boxes below. (This situation crudely
approximates what sunglasses do on a bright summer day.)
source distribution
combined filter transmittance
detected distribution
Question 6: Remove all filters in the filters rack. Place a B filter in the beam path with the flat
white light source. Then add a second B filter and then a third. Describe and explain what
happens when you add more than one of a specific filter.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 18 of 29 pages
Question 7: Place a B filter in the beam path with the 40% neutral density filter. Then add a V
filter into the beam path. Describe and explain what happens when you add more than one
filter to the filter rack.
purple filter
Question 8: Create a piecewise linear filter that when used with the
flat
white light source would allow red and blue wavelengths to pass and
thus effectively allowing purple light to pass. Draw the filter in the box
to the right.
•
Remove all filters from the filters rack.

Create a very narrow bell-shaped source distribution that is peaked at green
wavelengths (somewhere close to 550 nm). Notice the color.

Expand the spread of the source distribution to maximum. Notice how the color
changes.

Change the distribution source to an blackbody source peaked at green wavelengths (a
temperature close to 5270 K). Again notice the color.
Question 9: Using observations from the above actions, explain why we don’t observe “green
stars” in nature, though there are indeed stars which emit more green light than other
wavelengths. ___________________________________________________________________
6. Absolute and Apparent Brightness Magnitude
Apparent Magnitude From http://www.phys.ksu.edu/personal/wysin/astro/magnitudes.html
Apparent magnitude m of a star is a number that tells how bright that star appears at its great
distance from Earth. The scale is "backwards" and logarithmic. Larger magnitudes correspond to
fainter stars. Note that brightness is another way to say the flux of light, in Watts per square
meter, coming towards us.
On this magnitude scale, a brightness ratio of 100 is set to correspond exactly to a magnitude
difference of 5. As magnitude is a logarithmic scale, one can always transform a brightness ratio
B2/B1 into the equivalent magnitude difference m2-m1 by the formula:
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 19 of 29 pages
m2-m1 = -2.50 log(B2/B1).
You can check that for brightness ratio B2/B1=100, we have log(B2/B1) =log(100)= log(102) = 2,
and then m2-m1=-5, the basic definition of this scale (brighter is more negative m). One then has
the following magnitudes and their corresponding relative brightnesses:
magnitude m
| 0
1
2
3
4
5
6
7
8
9
10
---------------------------------------------------------------relative
| 1 2.5 6.3 16 40 100 250 630 1600 4000 10,000
brightness
|
ratios
|
(Note that the lower row of numbers is just (2.512)m.)
Absolute Magnitude
Absolute magnitude Mv is the apparent magnitude the star would have if it were placed at a distance of
10 parsecs from the Earth. Doing this to a star (it is a little difficult), will either make it appear brighter or
fainter. From the inverse square law for light, the ratio of its brightness at 10 pc to its brightness at its
known distance d (in parsecs) is
B10/Bd=(d/10)2.
Then, like the formula above, we say that its absolute magnitude is
Mv = m - 2.5 log[ (d/10)2 ].
Stars farther than 10 pc have Mv more negative than m, that is why there is a minus sign in the
formula. If you use this formula, make sure you put the star's distance d in parsecs (1 pc = 3.26
ly = 206265 AU).
Distance Determination
The above relation can also be used to determine the distance to a star if you know both its
apparent magnitude and absolute magnitude. This would be the case, for example, when one uses
Cepheid or other variable stars for distance determination. Turning the formula inside out:
d = (10 pc) x 10(m-Mv)/5
For example, for a Cepheid variable with Mv = -4, and m = 18, the distance is
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 20 of 29 pages
d = (10 pc) x 10[18-(-4)]/5 = 2.51 x 105 pc.
7. Measuring Stellar Distance
The Cosmic Distance Ladder
There are at least seven different astronomical distance determination techniques:
-- radar ranging,
-- parallax,
-- distance modulus,
-- spectroscopic parallax,
-- main sequence fitting,
-- Cepheid variables, and
--supernovae
On Monday February 25, 2013 we will only look at parallax while distance
modulus and spectroscopic parallax are also given below.
Parallax
In addition to astronomical applications, parallax is used for measuring distances in many other
disciplines such as surveying.
http://astro.unl.edu/naap/distance/animations/parallaxExplorer.html
Open the Parallax Explorer where techniques very similar to those used by surveyors
are applied to the problem of finding the distance to a boat out in the middle of a large lake by
finding its position on a small scale drawing of the real world. The simulator consists of a map
providing a scaled overhead view of the lake and a road along the bottom edge where our
surveyor represented by a red X may be located. The surveyor is equipped with a theodolite (a
combination of a small telescope and a large protractor so that the angle of the telescope
orientation can be precisely measured) mounted on a tripod that can be moved along the road to
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 21 of 29 pages
establish a baseline. An Observer’s View panel shows the appearance of the boat relative to
trees on the far shore through the theodolite.
map.
is a
Configure the simulator to preset A which allows us to see the location of the boat on the
(This
helpful simplification to help us get started with this technique – normally the main goal of the
process is to learn the position of the boat on the scaled map.) Drag the position of the surveyor
around and note how the apparent position of the boat relative to background objects changes.
Position the surveyor to the far left of the road and click take measurement which causes the
surveyor to sight the boat through the theodolite and measure the angle between the line of sight
to the boat and the road. Now position the surveyor to the far right of the road and click take
measurement again. The distance between these two positions defines the baseline of our
observations and the intersection of the two red lines of sight indicates the position of the boat.
We now need to make a measurement on our scaled map and convert it back to a distance in the
real world. Check show ruler and use this ruler to measure the distance from the baseline to the
boat in an arbitrary unit. Then use the map scale factor to calculate the perpendicular distance
from the baseline to the boat.
Question 2: Enter your perpendicular distance to the boat in map units. ______________
Show your calculation of the distance to the boat in meters in the box below.
Configure the simulator to preset B. The parallax explorer now assumes that our
surveyor can make angular observations with a typical error of 3°. Due to this error we will
now describe an area where the boat must be located as the overlap of two cones as opposed
to a definite location that was the intersection of two lines. This preset is more realistic in that
it does not illustrate the position of the boat on the map.
Question 3: Repeat the process of applying triangulation to determine the distance to the boat
and then answer the following:
What
is your best estimate
for the
perpendicular distance to the boat?
What is the greatest distance to the boat
that is still
observations?
consistent
with
your
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 22 of 29 pages
What is the smallest distance to the boat
that is still
observations?
consistent
with
your
Configure the simulator to preset C which limits the size of the baseline and has an error of
5° in each angular measurement.
Question 4: Repeat the process of applying triangulation to determine the distance to the boat
and then explain how accurately you can determine this distance and the factors
contributing to that accuracy. ________________________________________________
Distance Modulus
Question 5: Complete the following table concerning the distance modulus for several objects.
Object
Star A
Apparent
Magnitude
Absolute
Magnitude
m
2.4
M
Star B
10
Star D
8.5
Distance
(pc)
10
5
Star C
Distance
Modulus
m-M
16
25
0.5
Question 6: Could one of the stars listed in the table above be an RR Lyrae star? Explain
why or why not. __________________________________________________________
Spectroscopic Parallax
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 23 of 29 pages
Open up the Spectroscopic Parallax Simulator. There is a panel in the upper left
entitled Absorption Line Intensities – this is where we can use information on the types of
lines in a star’s spectrum to determine its spectral type. There is a panel in the lower right
entitled Star Attributes where one can enter the luminosity class based upon information
on the thickness of line in a star’s spectrum. This is enough information to position the star
on the HR Diagram in the upper right and read off its absolute magnitude.
http://astro.unl.edu/naap/distance/animations/spectroParallax.html
Let’s work through an example. Imagine that an astronomer observes a star to have an
apparent magnitude of 4.2 and collects a spectrum that has very strong helium and moderately
strong ionized helium lines – all very thick. Find the distance to the star using spectroscopic
parallax.
Let’s work through an example. Imagine that an astronomer observes a star to have an
apparent magnitude of 4.2 and collects a spectrum that has very strong helium and moderately
strong ionized helium lines – all very thick. Find the distance to the star using spectroscopic
parallax.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 24 of 29 pages
Let’s first find the
spectral type. We can see in
the
Absorption
Line
Intensities panel that for the
star to have any helium lines it
must be a very hot
blue star.
By dragging the
vertical cursor we can see that
for the star to have very
strong helium and moderate
ionized helium lines it must
either be O6 or O7. Since the
spectral lines are all very thick, we can assume that it is a main sequence star. Setting the
star to luminosity
class V in the Star Attributes panel then determines its position on the HR Diagram and
identifies its absolute magnitude as -4.1. We can complete the distance modulus
calculation by setting the apparent magnitude slider to 4.2 in the Star Attributes panel.
The distance modulus is 8.3 corresponding to a distance of 449 pc. Students should keep
in mind that spectroscopic parallax is not a particularly precise technique even for
professional astronomers. In reality, the luminosity classes are much wider than they are
shown in this simulation and distances determined by this technique are probably have
uncertainties of about 20%.
Question 7: Complete the table below by applying the technique of spectroscopic
parallax.
Observational Data
Analysis
m
Description of spectral lines
Description of line
thickness
6.2
strong hydrogen lines
moderate helium lines
very thin
13.1
strong molecular lines
very thick
7.2
strong ionized metal lines
moderate hydrogen lines
very thick
Cepheid Variables http://astro.unl.edu/naap/distance/cepheids.html
M
m-M
d
(pc)
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 25 of 29 pages
Cepheid variables are pulsating variable stars similar to the RR Lyraes mentioned earlier on the
distance modulus page. However, Cepheids have longer pulsation periods and they are larger
stars. Cepheids have been extremely important as distance indicators for many years. Although
they don’t all have the same average absolute magnitude as RR Lyraes do, they are more useful
since they are brighter stars and can be observed at greater distances.
Henrietta Leavitt in 1912 was the first to recognize that there was a relationship between the
pulsation periods and the luminosities of Cepheids. She recognized that larger, brighter Cepheids
have longer pulsation periods, although she was unaware of the exact relationship. Harlow
Shapley later calibrated the Cepheids – relating the periods of pulsation to the absolute
magnitudes which led to the first estimate of the size of the Milky Way. The calibration of the
period-luminosity relationship has improved over time and a modern version is depicted in the
figure to the right.
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 26 of 29 pages
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 27 of 29 pages
To the left is a graph of the periodicity of the Type I (high-metallicity) Cepheid variable S Nor.
Cepheid S Nor has a period of pulsation of approximately 10 days and an average apparent
magnitude mV = 6.5, what is its distance? We can use the pulsation period to estimate the
absolute magnitude of the Cepheid. From the chart above a period of 10 days corresponds to an
absolute magnitude of -4. Thus, the distance modulus is m - M = 6.5 - (-4) = 10.5, which
corresponds to a distance of 1260 pc. Note that our estimate is not particularly accurate since we
didn’t take into account many subtleties that research astronomers would consider.
Because of the brightnesses of Cepheids, astronomers can identify them in nearby galaxies.
Observations of Cepheids by the Hubble Space Telescope have recently been used to estimate
the distance to the Virgo Cluster which is about 18 Mpc
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 28 of 29 pages
http://astro.unl.edu/naap/distance/summation.html
The Cosmic Distance Ladder
E:\2012-2013\SSU\PHS 207 spring 2013\Notes…\Stars\Document1 --- 29 of 29 pages
The progression of distance determination techniques surveyed is collectively known as the
Cosmic Distance Ladder.
No one technique is effective at all distances and typically techniques useful at small distances
are used to calibrate those used on objects at greater distances. Thus, one rung on the ladder
allows astronomers to step to the next rung.
As a first example of this, think about RR Lyrae stars. A number of RR Lyraes have been
observed by the Hipparcos satellite which has determined accurate distances to them. It is easy to
observe the apparent magnitudes of these stars and when they are combined with the known
distances, the absolute magnitudes can be determined. This procedure allowed astronomers to
see that all RR Lyrae have absolute magnitudes around 0.5. Astronomers can now calculate the
distance to RR Lyraes beyond Hipparcos' range using the distance modulus. When astronomers
find RR Lyraes in the Large Magallenic Cloud, they can be used to calibrate the PeriodLuminosity relation for Cepheids observed there. Similarly, when Type I supernovae in nearby
galaxies with Cepheids in them (and thus known distances) were observed, the absolute
magnitude of the supernova peak brightness was determined. Thus, a variety of distance
determination techniques allow us to – one step at a time – learn the distances to ever more
distant objects.
THE END