Hydrostatic Equilibrium and the Virial Theorem

Introduction

Here we consider how to show the equation for hydrostatic equilibrium. The methodology used is a force balance followed by some algebra.

Solutions

Question: Find the differential equation describing hydrostatic equilibrium.

a. Write an expression for the differential element dM contained in the cell.

Considering a mass shell with thickness dr and density p(r), we can write the equation for the differential element contained in this cell by finding the volume of the shell and multiplying this by the density, as follows:

b. What is the gravitational force, F_g, acting on the shell?

We know the differential element for the mass (that is, the mass of an infinitely thin shell). We can use the general equation for gravitational force:

When M and m describe the two masses. However! We do not have two masses here. We have a mass, M, and the differential element dM that represents the mass of the shell. So, we shold rewrite the expression for F_g as:

When M(r) describes the mass of the sun up to a radius r.

c. Assuming there is a pressure P(r) acting to balance F_g, write the net force equation in terms of the relevant qualities.

We know:

We want to show that the forces balance, so we can set this equal to the expression we found for F_g in part b. Solving for the F_p, force of pressure, and substitute this expression in:

It is correct that the signs are opposite, because the forces are acting in opposite directions.

d. Show the equation of hydrostatic equilibrium.

To show that this equation fits the form of the equation of hydrostatic equilibrium, we can substitute in the value of dm that we found in part a, and then manipulate the equation algebraically:

This is the equation of hydrostatic equilibrium!!

Conclusion

We have shown the derivation for the equation of hydrostatic equilibrium, using only simple physics and algebra. Pretty cool!!

Acknowledgements

Thanks to Iryna and Monica for working with me on this worksheet – see their blogs for the discussions of other questions!

Also, I would like to thank my classmates whose blogs I have been reading - they gave me the idea to use an online Latek editor instead of typing equations in plain text. It looks way better this way! Thanks guys!

Stellar Properties from Afar

Introduction

Here we consider a series of questions related to the sun and its properties. Topics for consideration include the luminosity of the sun and furthest distance at which a similar star could be seen, and what this distance is in various units of measurement. We are addressing questions 5-8 from the worksheet.

Solutions

Question: Given the that eye must receive 10 photons in order to see something, how far away would a solar twin have to be for it to be barely visible from Earth?

We will need to make a few assumptions for this problem:

· The “read rate” of the eye (in what length of time the photons must arrive to be visible) is 1/30 second (this is the frame rate for most movies)

· The sun emits light at 500 nm.

· The human eye has a surface area of .0625 square centimeters that can receive photons.

· The sun’s luminosity is 4*10³³ ergs.

Using this assumption, we can calculate the minimum flux needed to see something. If the sun emits light at 500 nm, we can calculate the energy of 10 photons, and then find what this energy implies for the flux needed to see something.

E=hv and c=vl

3*10¹⁰= v*500*10^-7

v=6*10¹⁴

E=(6.6*10^-27)( 6*10¹⁴)

E=3.96*10^-12

This is the energy for one photon. So, the energy for ten photons is 3.96*10^-11 ergs. Also, if we want to find the minimum flux that would bring this energy to the eye, we must consider the surface area of the eye.

Minimum flux = (3.96*10^-11 ergs)/(.0625 cm²)

Flux = 6.336*10^-10 ergs/ cm²

Now, we have the minimum flux that the eye needs to see something. If we find the distance such that a star with the sun’s luminosity will have this flux at Earth, we will have our answer.

Flux=(luminosity)/(area)

What we can take this equation to mean for this problem is that given the luminosity of the sun, if we divide it by the surface area of the sphere with radius r, we will get the flux at some point r away from the sun. We can do this because photons are emitted isotropically from the sun. Now, in a previous problem on this worksheet (go see Iryna’s blog for the solution to that problem!), we found the luminosity of the sun to be roughly 4*10³³ ergs. So:

6.336*10^-10 ergs/ cm² = (4*10³³ ergs)/(4p(distance in cm)²)

Distance = 7.087*10²⁰ cm

We can find this in AU (again, this is based on the answer to a previous problem! See Iryna’s post for details…):

7.087*10²⁰ cm * (1 AU)/(1.5*10¹³cm) = x AU

X = 4.7*10⁷ AU

For fun, let’s convert to lightyears.

4.7*10⁷ AU * (8 lightminutes/1 AU)*(1 hour/60 minutes)*(1 day/24 hours)*(1 year/365 days) =715 lightyears

So, the farthest star we could see is 4.7*10⁷ AU away. This is about 715 lightyears.

Question: The assumption in the last problem that all light from the sun is received at 500 nm is incorrect. In fact, light is processed by the eye at a “passband” of 450-600nm. Given this, how does the farthest distance at which a sun-strength star can be seen change?

Knowledge from earlier in the problem set and assumptions that we will use for this problem include:

· The “read rate” of the eye (in what length of time the photons must arrive to be visible) is 1/30 second (this is the frame rate for most movies)

· The sun emits light from 450 to 600 nm – however, not necessarily equally at all wavelengths.

· The human eye has a surface area of .0625 square centimeters that can receive photons.

· The sun’s luminosity is 4*10³³ ergs.

No longer can we use the equations used in the previous problem, since now we have a band of wavelengths to consider. Instead, we can consider the equation for energy per wavelength, which was derived in a previous problem set:

To make this solvable, we will use a Taylor expansion around our e term. If we integrate this expression, from 450nm to 600nm:

The answer to this integral is 1.01*10^-10. We can use as the total energy for all wavelengths. Then, we divide this by 150 to get the energy for one photon – an average photon. This value is 6.7*10^-13. Then, we can multiply this by 10 to get the energy for ten photons, which is 6.7*10^-12. Then, we can use this to find the distance the star would need to be.

Minimum flux = (6.7*10^-12 ergs)/(.0625 cm²)

Flux = 1.077*10^-10 ergs/ cm²

Using this flux, we can use the same method as in the previous problem.

Flux=(luminosity)/(area)

1.077*10^-10 ergs/ cm² = (4*10³³ ergs)/(4p(distance in cm)²)

Distance = 1.72*10²¹ cm

We can find this in AU:

1.72*10²¹ cm * (1 AU)/(1.5*10¹³cm) = x AU

X = 1.14*10⁸ AU

Again, for fun:

1.14*10⁸ AU * (8 lightminutes/1 AU)*(1 hour/60 minutes)*(1 day/24 hours)*(1 year/365 days) =1735 lightyears

So, a more accurate measurement of the farthest star we could see is 1.14*10⁸ AU away. This is 1735 lightyears.

Question: How far away is this star in centimeters?

We actually found this figure mid-way through our calculation in the previous problem, so lifting that answer gives us:

Distance farthest star we can see is from Earth: 1.72*10²¹ cm

Question: How far is this is parsecs?

Knowledge from earlier in the problem set and assumptions that we will use for this problem include:

· The farthest a star of sun luminosity could be from the Earth for us to see it is 1.14*10⁸ AU

First, we should find the distance of a parsec. A parsec is, as the problem explains, the distance from Earth a star must be in order to have a parallax of 1 arcsecond (recall, 1 arcsecond = 1/3600 degrees) from the sun.

Using the small angle approximation, and converting our 1/3600 into radians (p/648000)

tan(p/648000) = 1 AU / (x)

1/(p/648000)= 1/x

1 parsec = x =1 / (p/648000) = 2.06*10⁵ AU

Now we know how many AU there are in a parsec, we can calculate how far the farthest star we can see is. Using the value for distance we previously calculated,

1.14*10⁸ AU * (1 parsec/ 2.06*10⁵ AU) = 5.52*10² parsecs = 552 parsecs

Conclusion

We have found that the farthest distance a star of sun luminosity could be from the Earth for us to still be able to see in with our naked eye is 1.14*10⁸ AU. This is equivalent to 1.72*10²¹ cm, or 552 parsecs, or 1735 lightyears. Our preliminary estimate was 4.7*10⁷ AU (which is about 715 lightyears), and we have also found the distance of a parsec (2.06*10⁵ AU).

Acknowledgements

Thanks to Iryna Butsky for significant help on these problems! See her blog for problems 1-4.

How far to the earth?

Abstract

By taking visual data from the TRACE satellite, which tracked the position of Mercury as measured against the sun, we will determine the distance between earth and the sun (the AU) and other qualities about the sun, Earth, Mercury.

Introduction

For the purposes of this problem, we will consider the angles from the satellite (which we assume is at a distance equal to Earth’s radius from the center of the planet) to some point on the sun (the lines defining these angles are black on the diagram). We will also consider the shadow that Mercury casts on the sun (shown as the blue lines on the diagram). The angle these two lines make from the earth is ‘a’, while the angle from the centroid of earth to the centroid of the sun to the satellite is ‘b’.

Solution

We know that the angle ‘a’ refers to one-half the amplitude of Mercury’s oscillations as seen transposed against the sun. TRACE’s orbit is what causes this oscillation, so the angle it makes with the sun is based on the radius of its orbit (which we defined as equal to the radius of the earth).

We define theta, q, as seen on the diagram.

Consider this section of the diagram, bringing special attention to the triangle outlined in blue:

The angles in this triangle must sum to p. So, we can write the expression:

p=a+b+p-q/2

Which simplifies to:

a+b=q/2

This expression relates one-half the amplitude of Mercury’s oscillations (a) to angles b and q/2. We can measure the oscillations using the image given in class. On his image, we measure the amplitude of the oscillations as about .4 cm (on the paper). Meanwhile, the scale of the sun on the paper is about 11 cm across. We know the angular diameter of the sun from earth is about 0.5° (the same as the moon), so this means that the amplitude of the orbit, as an angle, is 3*10^-5 (.4/11*.05/360*2*p). So a = 3*10^-5

Now, we can look at yet another sub-triangle of the Mercury-Earth-Sun system.

Here, we consider the right triangle that has as its legs the radius of the earth and the distance from earth to mercury. We can set up this triangle as follows:

tan(q/2) = R_earth / R_{earth-mercury}

By a small angle approximation:

(q/2) = R_earth / R_{earth-mercury}

We earlier solved for q/2! So, we can substitute, as follows:

a+b = R_earth / R_{earth-mercury}

We found a, we know R_earth. Based again on our triangles, we can define b as follows:

tan(b) = R_earth / AU

Again, by the small angle approximation:

(b) = R_earth / AU

Substitute!

a +R_earth / AU = R_earth / R_{earth-mercury}

R_earth/( a +R_earth / AU )= R_{earth-mercury}

AU - R_earth/( a +R_earth / AU ) = R_mercury

(P_mercury)²/(P_earth)² = (A_mercury)³/(A_earth)³

We know P_earth=365 days and P_mercury=87 days. Also, we solved for R _mercury =A_mercury earlier. So:

(P_mercury)²/(P_earth)² = (AU - R_earth/( a +R_earth / AU ) = R_mercury)³/(A_earth)³

(87)²/(365)² = (AU - 6378/( 3*10^-5 +6378 / AU ))³/(AU)³

What an equation! Plugging this into Wolfram Alpha (or solving for an approximate solution) give the value:

AU = 1.33 * 10⁸ km

The real value for an AU is (a quick Google search can give us an approximate value):

AU = 1.496 *10^{8 km}

Not only are these values on the same order of magnitude, they are actually pretty close – so you can see that even rough measurements such as measuring the period of Mercury’s projections function using a ruler and a print-out image are sufficient to estimate the AU fairly accurately.

For a final answer, we can convert this value to cm, as the lab asked:

One AU = 1.33 * 10¹³ cm

We can find more values using these same few equations, as well!

(Pearth)²/( A_earth)³ = 4p²/(MG)

(365*24*60*60)²/( 1.33 * 10¹³)³ = 4p²/(M*6.7*10^-8)

So, M = 1.39*10³³ grams (this figure is a little off…)

Actual value: 5.97*10²⁷ grams

Can this be attributed to error? It seems like way too big of an error… could someone else who did the problem tell me what I did wrong here? Thanks!

Epilogue

We have determined one AU to be 1.33 * 10¹³ cm. Additionally, we have found the mass of the sun to be 1.39*10³³ grams. The orbital period of mercury is roughly 87 days.

Acknowledgements

Thanks to Iryna and Monica for working with me on this problem!

A swift dinner...

Yesterday at dinner, I sat with two scientists from Oxford, who were at Palomar working on their instrument, SWIFT. When they mentioned that SWIFT uses Integral imagery, I immediately started thinking about x-rays (Swift and Integral are two x-ray telescopes), but the naming similarities were just a coincidence. SWIFT is an instrument created specifically for the Hale 200-inch telescope. There are many different types of instruments, each for a different purpose. Installing an instrument onto the telescope alters the results of the observation. For example, one instrument that had been used in the past is PHARO, which is an adaptive optics instrument. (However, the two scientists told me that adaptive optics is no longer being used at the Hale 200 inch! This was very surprising…)

What does SWIFT do? Best I can understand it, SWIFT is an integral field spectrograph. This means that it records not only the location (in the sky) of an event, but it also records the spectra at this point. This is useful because for larger sources, you can see the differences in spectra from one spot to another (still part of the same source). This technique is used for extended sources; it is not necessary for point sources, where a single spectrum is sufficient. An integral field spectrograph uses a different configuration of slits than a classic long-slit spectrograph (which took observations for the same purpose, seeing spectra over area). In long-slit observations, an observation is made through a (as the name implies) long slit aperture, and then the light is split by a prism. An integral field spectrograph speeds up this process. The result of using an instrument such as SWIFT is that an astronomer can study extended or grouped sources.

While explaining their project to me, the two scientists also ended up explaining a lot of science – I learned a lot! For example, I never knew that Keck wasn’t a monolithic mirror (I didn’t know what monolithic meant, either – it means that the mirror was one solid piece). Instead, Keck is many smaller mirrors aligned into a plane. The Extremely Large Telescope, which is a British telescope not yet built, will be 40 meters in diameter (!!!!) but will similarly be composed of many smaller mirrors instead of one single monolithic mirror. Also, apparently the problem they are having with building the ELT is not that they do not have enough money, but that they “can’t spend it fast enough” – that is, they cannot find enough qualified people to hire! Manufacturing the mirrors is a long process, and apparently not very efficient… car companies are more efficient than the people who build the telescopes!

I was very surprised at the degree to which the SWIFT scientists were aware and even bitter at the difference between the astronomers and the people who build the instruments. They both (although considering themselves scientists) lamented the fact that scientists never respect the people who build the instruments. I never thought about it before, but without the people who build the instruments, the astronomers who take and analyze data could never do their science. All modern science hinges on the fact that someone in history decided it was worth it to build a telescope – visiting the Hale 200-inch and seeing its battleship-like construction made me aware that telescopes are not built in a day, and no astronomer is self-sufficient… behind every paper are thousands of people whose labors have made it possible!

Telescope time!

I met with Dr. Evan Kirby last week (http://www.astro.caltech.edu/~enk/), and he very nicely spent an hour explaining to me various facets of the telescope time acquisition process! Here are some highlights of what I learned from him:

When Keck telescope was built, the costs were divided between the UC system and Caltech. Caltech paid the bulk of the initial costs, but the UCs were to pay for maintenance. Over the course of many, many years, this will eventually even out, so each institution was initially given a half stake in the telescope. The University of Hawaii also received a stake (10%) in the telescope, because the telescope was built on Hawaiian land.

What a ‘stake’ in a telescope means is that out of all the nights of the year, that institution will receive that many nights for their scientists. So, a higher stake means that the institution will have more observing time – which is a good thing! For Keck, this meant that the UC system and Caltech each had equal shares. However, you might notice that Caltech has FAR fewer people vying for this time than the entire UC system – indeed, this means that it is far easier for Caltech scientists to get telescope time. There are also some Keck nights each year available to anyone in the country. Getting time on these nights is extremely competitive, since there are professors and professional scientists from across the country who perhaps have no other way to get the telescope time they need.

Caltech has stakes in other telescopes as well. It operates Palomar telescope here in California. The access that Caltech scientists have to it is actually quite amazing – did you know that grad students can apply for time on Palomar?? That’s pretty awesome.

Wait – so who exactly can apply for time to telescopes? And how do you apply?

To get telescope time, applicants must write a proposal, which is reviewed by a panel of scientists (this panel changes – Dr. Kirby has been on this panel before). Proposals are due each semester, for the following semester of observing time. These proposals take a lot of work, and they must be submitted by someone who has access to the telescope. For example, an undergraduate cannot write and submit a proposal. However, an undergraduate CAN write a proposal and convince someone (a professor, postdoc) with access to the telescope to submit it for them.

At most institutions, there are strict limits to who can apply for telescope time. For example, grad students are not generally allowed to apply for telescope time, and even postdocs cannot apply in most places. At Caltech, there are certainly such limits, although they are less strict than at other places. Either postdocs or Professors can apply for Keck time here, but grad students, postdocs and Professors all can apply for Palomar time.

The national ratio for proposals submitted to proposals accepted is roughly 10 to 1. The UC system’s average is roughly 5:1, and Caltech’s is much, much lower than that!

I never knew just how good Caltech’s access to telescopes is. It is inspiring to know that if you had science you wanted to do, at Caltech you’d be able to actually do that science! It makes me want to write my own proposal…

More on the purposes of varying apertures…

Abstract

The size of a telescope affects what its ideal observation targets are. Larger telescopes will have higher angular resolution for greater wavelengths of light, while telescopes with smaller apertures will have better angular resolution for shorter wavelengths. However, this dependence is only valid to an order of magnitude: increasing the order of magnitude for the wavelength but not the aperture will result in a worse angular resolution.

Introduction

J-band refers to a specific range of wavelengths, from 1.1 to 1.4 microns (a micrometer is equal to 10-6 meters). This value means that light in the J-band spectrum falls into the infrared. Considering CCAT and Keck, two telescopes with different apertures (25 meters and 10 meters respectively), how does the angular resolution compare? The CCAT will detect wavelengths up to 850 microns, while Keck observes in J-band (1.1-1.4 microns). This question elaborates on my earlier blog post concerning the differences between large and small telescopes: it is a practical example of why you might want a telescope with a smaller aperture in some cases.

Solution

The relationship between aperture, wavelength and angular resolution can be defined as follows:

sin q = (1.220)(l /D)

Taking care to convert units (using meters as the standard unit), we can calculate the values for each telescope.

For CCAT, sin q = (1.220)(850*10-6 /25) = 4.15*10-5 and using the small angle approximation, we can say that q = 4.15*10-5. This value is given in radians, so a quick conversion to degrees gives us .0024° as the angular resolution of CCAT at this wavelength. Since there are 3600 arcseconds in a degree, this is an angular resolution of 8.64 arcseconds.

Keck, on the other hand, is observing at a much smaller frequency, at about 1 micron, with a slightly smaller aperture as well. So, sin q = (1.220)(1*10-6 /10) = 1.22*10-7. This is equivalent to 0.000007°, which is equal to 0.03 arcseconds. This is two orders of magnitude smaller than the resolution for CCAT, which makes sense since the wavelength Keck is studying here is two orders of magnitude smaller as well.

Epilogue

The angular resolution for CCAT was better for 850 micrometer wavelengths than Keck’s resolution was for 1 micrometer waves: we can compare CCAT’s angular resolution of 8.64 arcseconds to Keck’s resolution of 0.03 arcseconds. However, if we were to study different wavelengths of light, then the angular resolution for each telescope would also change – this one measurement of angular resolution cannot be considered the standard by which we judge the telescope.

Questions for Further Study (next blog post?)

Is it true that the formula for angular resolution can only work for ground-based telescopes? That is, how broadly can the term ‘aperture’ be defined? Can the resolution of x-ray telescopes be defined this way – do x-ray telescopes even have an angular resolution?

Acknowledgements

Thanks to Iryna Butsky and Monica He for their help!

The ideal size for a telescope...

At the end of class Friday, Professor Johnson made a comment about there being telescopes on earth with much, much larger apertures than the 10-25 meters discussed on the problem set (I still think these are pretty big…) I didn’t really think that he was serious, partially because it’s so hard to imagine a telescope mirror being so large, but in fact I was thinking about it wrong: the creatively named Very Large Array, which consists of 27 radio antennas in New Mexico, is capable of making very high resolution measurements. On a side note - If you think its name is unoriginal, then just consider for comparison the Very Large Telescope (VLT) and Overwhelmingly Large Telescope (OLT).

The VLA looks small compared to the Very Large Baseline Array (VLBA), however. VLBA is a system of similar radio antennas, also 25 meters in diameter, which are spread across the whole United States. To the left, the locations of the dishes are shown - image from the NRAO website (you can also go there for more information): http://www.vlba.nrao.edu/

Why is this useful? The relationship between aperture (D), wavelength (l), and angular resolution (q, measured in radians) is as follows for single telescopes:

sin(q)=(1.220)(l/D)

From this, we can see that aperture of a telescope affects what wavelengths of light it can observe at an optimal resolution. Resolution is important because if a source is larger than the resolution of the telescope observing it, then the observation will not contain the entire source. This is called an extended source, as opposed to a point source, which is smaller than the resolution of the telescope observing it. Also, sources that are too close together (that is, the distance between them is smaller than the angular resolution) cannot be accurately studied, as astronomers would not be able to see where one point source ended and the other began! It would just look like one larger, blurrier point source…

HOWEVER – this equation does not hold for the VLBA, because the VLBA is an array, not a single telescope. Instead, for arrays, the relationship depends upon the separation between radio dishes (B) and wavelength of the source (l). Again, the angular resolution is measured in radians (q):

q=(1.220)(l/B)

Regardless, a larger telescope can observe SOME frequencies of light with higher resolution. Longer wavelengths require a larger aperture, which is why the largest receivers on earth study radio waves, which have a higher wavelength than visible light.

Type of light studied	(Approximate) Wavelength	Approximate aperture width to study this light from Earth	Example of a telescope operating at this wavelength
Radio waves	1 cm – 100 m	Antennas are spaced 8000 km across north America	VLBA
Infrared	1000 nm – 0.1 cm	*	Spitzer
Visible	400 nm – 800 nm	10 meters	Keck
UV	10 nm – 200 nm	*	Extreme Ultraviolet Imaging Telescope (EIT)
X-rays	0.01 nm – 10 nm	*	Chandra

*not ground-based telescopes

So – larger telescopes do exist, but not in the capacity that I was imagining. According to a quick Google search, the largest single telescope in the world is the Gran Telescipo Canarias (don’t think you need to be fluent to translate that one…) in the Canary Islands (http://www.universetoday.com/17652/largest-telescope/). It has a 10.4 meter aperture. No single telescope would actually be 1 km or more across, but what Professor Johnson was referring to was the necessity for much larger values of B to study higher wavelength waves (such as radio waves!) which is why the VLBA was built across such a large distance!

Most important thing I took away from this was that depending on what wavelength of light you are studying, different sized telescopes will have different levels of effectiveness. There is no ideal size for a telescope because what you need depends on what you are studying. Bigger isn’t always better!

Lab at the beach

In what will unfortunately probably be my only trip to the beach for the rest of the year, our Ay class enjoyed our first lab at the Santa Monica beach, measuring the time it takes the sun to set from different altitudes. Next we have to calculate the radius of the Earth using this data collected at the beach.

This lab reminds me of the fact that telescopes are often in fun, fun places - such as Hawaii! To do an Ay lab, we go to the beach. To take measurements for a real project, an astronomer might go to Hawaii. Correlation does not prove causation, but I think I see a theme here...

On the drive back from Santa Monica, Professor Johnson explained the importance of the telescope in astronomy projects. As I understand it, part of his motivation for accepting a professorship where he did was that he could continue to use the same telescope that he had been using previously. This is interesting because it shows how important tools are to an astronomer's work - indeed, I asked Professor Johnson how important telescope time was to success as an astrophysicist, and he said that a lot of telescope time could let anyone publish a decent paper, but that brilliant scientists can get by with less.

At Caltech, these two categories are combined as all the brilliant scientists have a plenthora of telescope time! Caltech has a lot of shares in many telescopes: take a look at the Caltech Astro department's summary, http://www.astro.caltech.edu/observatories/. Also, Caltech has a lot of very good astronomers/astrophysicists publishing papers using data from those telescopes! It seems to me almost like a monopoly of astronomy: Caltech and other leading institutions have all the telescope time, which makes these institutions even better as their scientists can use all this time to publish papers!

I would love to hear more about how the telescope time system works from people familiar with it!