text: Swihart: Quantitative Astronomy --class1-- Read: ch.1 in Swihart on the web: http://www.physics.csbsju.edu/astro/ follow the following links to the quiz. Take each quiz. # tutorial: sky vocabulary # tutorial & quiz: sky map # tutorial & quiz: star finder # tutorial & quiz: SC001 star map # tutorial & quiz: SC002 star map # compare: SC and Sky Map # tutorial: latitude & your sky using the web site: http://www.fourmilab.to/yoursky/ make a sky map for tonight at 9pm (CDT) for here (requires figuring out the latitude and longitude of "here") consider the four stars that make up the bowl of the big dipper. Find four names for each of the four stars (for example: Bayer greek letter, Flamsteed number, HR [bright star], HD, Hipparchos, etc.) The Bright Star Catalogue, Sky Catalogue 2000.0, and the multi-volume Hipparchos catalogue are around in hardcopy to help but you might also try the CDS on-line database named in the syllabus. In addition to the names, record all the digits for the 2000.0 position (RA and dec)--you will need these for future problems. Use these RAs and decs to find the angular size of the lip of the bowl of the big dipper and the angular distance across the bowl of the big dipper. (A rule of thumb would say the answers are 5 and 10 degrees...give me three sigfigs.) --class2-- Find the offset today between sidereal time and our clock time (which is on daylight saving). 1-5 2-3, 2-8 web read: Tutorials for Astrophysics * Orbital Elements & Calculations http://www.physics.csbsju.edu/orbit/orbit.2d.html http://www.physics.csbsju.edu/orbit/orbit.3d.html * 3-body Problem: an example by movie http://www.physics.csbsju.edu/orbit/meissel.mov --class3-- HW: (three problems) 2-6 (correction: 12 AU/yr -> 1.2 AU/yr) old exam (364t199.pdf): #5 You have probably forgotten the formula for the radius of curvature you learned in calculus (I did!). But if you apply it to an ellipse you'll find that the ends of the major axis have radius of curvature r_C = b^2/a. That means that, when the planet is at the ends, i.e., at aphelion or perihelion, the planet feels like it's going in a circle with radius r_C and so should undergo the usual centripetal acceleration. Thus F=ma reads: GMm/r^2 = m v^2/r_C where r is the actual distance the planet is from the Sun. Pick aphelion or perihelion, separately calculate both sides of the above equation, and show the above equality is true. --class4-- 4-2 old exam (364t199.pdf): #9 --class5-- 4-6 old exam (364t199.pdf): #11, 13 Sun: conditions at core (From Allen p.342) density=151 g/cm^3 X=.355 (in the core, H has been burned into He, hence the low value) Z=.019 FYI: Sun's central temperature=15.7E6 K --class6-- 5-8 Consider a cube (of side 1 meter), filled with oxygen (O2) at T=300 K. If, after one year, half of the oxygen has escaped, what is the area of the leak? Consider a cube (of side 1 meter), filled with light at T=3000 K. If, after one year, half of the light energy has escaped, what is the area of the leak? --class7-- 6-1, 6-5, 6-6 old exam (364t199.pdf): #14 a,c,f --class8-- 6-8 old exam (364t199.pdf): #2, #14 b,d,e,g --class9-- midterm Ch. 1-6 --class10-- stellar atmospheres, spectype --class11-- Consider a problem similar to #1 on the midterm... (600 seconds, 100% efficient 1 m diameter space telescope) Find in this directory the files: 1000_lambdaC.dat --- 1000 measured wavelengths in units of micro m (micro=1E-6) 1000_freqC.dat --- the frequency of the same 1000 photons in units of PHz (P=1e15) Note: the second file is generated from the first via: .3/x Using the first file, estimate the value of F_lambda near lambda=.5 micro m; include units! Using the second file, estimate the value of F_nu near frequency=.6 PHz; include units! Note: a .6 PHz photon has a wavelength of .5 micro m. Note: your two Fs should have different units! Using the mean value of the second file, find the average photon energy, and from that the temperature of this blackbody source. (see Eq. 4.4-9) Notice that .3/(average lambda) does not equal average frequency If these data were displayed as a histogram (which is essentially what F_lambda and F_nu are) you would see (in either case) a bell shaped curve. The peak of this curve might be said to be the most commonly emitted wavelength (or frequency). It turns out to be pretty difficult to estimate these maximum positions with out a lot more data. (Extra Credit: try to do this and explain your method.) Instead we'll calculate the expected location... According to Wien's displacement law, these peaks can be related to the temperature of the source: lambda_max= b/T where b=.0028978 K m frequency_max= 2.821 kT/h (see Eq. 4.4-6) Use these relationships to find the max location and calculate F_lambda and F_nu at those locations...is the result clearly more than the previously calculated Fs? Extra Credit: Calculate F (with error) at, say, eight points and then fit the blackbody curve to this data with two adjustable parameters: T and normalization. --class12-- ch 8: 3,5,9 --class13-- no homework, but you might look ahead to the next class period --class14-- The aim here is to see plots of radial velocity vs time for a binary star. A nice applet displaying what's going on is: http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm (In this applet initially m2=red=1, m1=blue=3, a=1) The orbital mechanics is detailed in: http://www.physics.csbsju.edu/orbit/orbit.2d.html A problem is that the orbit is best parameterized by the eccentric anomaly (u), and even that relationship is pretty complex. For any given u we can calculate the time (really w*t) from Kepler's equation: wt[e_,u_]=u-e Sin[u] where e is the eccentricity. The angle from perihelion (true anomaly) can also be found from u: phi[e_,u_]=2 ArcTan[ Cos[u/2], Sqrt[(1+e)/(1-e)] Sin[u/2] ] The velocity vector (in ASM units) is scaled by v0=2 Pi Sqrt[(m1+m2)/(a(1-e^2))] where a is the semimajor axis of the orbit and m1 and m2 are the masses of the stars. v[e_,u_]=v0 {-Sin[phi[u]],Cos[phi[u]]+e} [v is actually the relative velocity: v1=(m2/(m1+m2))*v, v2=(m1/(m1+m2))*v, in the applet compare to the low-mass (red, with default values) curve.] If the direction to Earth is at an angle i from the orbit's normal, and at an an angle w from the minor axis, the radial velocity as seen from Earth is: vr[e_,w_,i_,u_]=-v0 {-Sin[phi[u]],Cos[phi[u]]+e}.{Sin[w] Sin[i],Cos[w] Sin[i]} We can now plot this radial velocity vs time using a ParametricPlot: ParametricPlot[{wt[e,u],vr[e,w,i,u]},{u,0, 4 Pi}] plotting two full cycles usually makes seeing the cycle clearer and of course we must put in values for e,w,i You will probably want to copy and paste stuff from this file into Mathematica thus avoiding typos. initial data: m1=3 m2=1 a=1 applet initial values: ./ {e->.5, w->0, i->75*Pi/180} define functions: wt[e_,u_]=u-e Sin[u] phi[e_,u_]=2 ArcTan[ Cos[u/2], Sqrt[(1+e)/(1-e)] Sin[u/2] ] vr[e_,w_,i_,u_]=-2 Pi Sqrt[(m1+m2)/(a(1-e^2))] {-Sin[phi[e,u]],Cos[phi[e,u]]+e}.{Sin[w] Sin[i],Cos[w] Sin[i]} ParametricPlot[{wt[.5,u],vr[.5,0,1.3,u]},{u,0, 4 Pi}] ParametricPlot[Evaluate[{wt[e,u],vr[e,w,i,u]} /. {e->.5, w->0, i->75*Pi/180}],{u,0, 4 Pi}] The easiest way to see how a parameter affects a relationship is to plot that relationship for lots of different values of the parameter: <Identity],{w,0,2 Pi,2 Pi/20}] ShowAnimation[plotw] Here we see how the eccentricity affects the radial velocity plot: plote=Table[ParametricPlot[{wt[e,u],vr[e,.1,.1,u]},{u,0, 4 Pi},DisplayFunction->Identity],{e,0,.95,.05}] ShowAnimation[plote] I hope it's clear that we could fit observed vr date to these functions and determine e and w. Your assignment: Consider the first six non-zero digits of your social security number...say 314159 then let m1=3 m2=1 a=4 ./ {e->.1, w->.5, i->.9} Note: be careful of things like: e=.1 w=.5 i=.9 as Mathematica will remember them forever Make a hardcopy plot of the radial velocity vs time for your social security number orbit. Repeat the process (without making hardcopy) for different values of w, and approximately note the difference between the maximum and minimum radial velocity. Does that velocity difference depend on w? A large semimajor axis and close distance sometimes allows us to see both stars of the binary, and, over the decades, record the path. Of course the apparent orbit we see is a projection of the true orbit and hence depends on the orbit's orientation in the sky (i and w defined above) in addition to its true shape. We can use Mathematica to display this orbit, and I hope you can imagine, determine (fit) the parameters of the true orbit and its orientation in our sky to observed locations. The mathematics of this is detailed in: http://www.physics.csbsju.edu/orbit/orbit.3d.html http://www.physics.csbsju.edu/orbit/binary.html The angle Omega (which only determines rotational orientation of the orbit) is taken here to be zero. <False,AspectRatio->Automatic] ParametricPlot[Evaluate[{aorbit[e,w,i,u],amajor[e,w,i,u],alatusrectum[e,w,i,u]}/. {e->.5, w->0, i->75*Pi/180}],{u,0,2 Pi},Axes->False,AspectRatio->Automatic] here we see how changing w changes the apparent orbit shape: plotw=Table[ParametricPlot[{aorbit[.7,w,1.3,u],amajor[.7,w,1.3,u],alatusrectum[.7,w,1.3,u]},{u,0,2 Pi},Axes->False,AspectRatio->Automatic,DisplayFunction->Identity],{w,0,2 Pi,2 Pi/20}] ShowAnimation[plotw] (This goes so fast, I suggest adjusting the menu on the animation window: View->AnimationControls->Pause and then use the slider to go through the frames a a reasonable pace.) here we see how changing i changes the apparent orbit shape: ploti=Table[ParametricPlot[{aorbit[.7,.5,i,u],amajor[.7,.5,i,u],alatusrectum[.7,.5,i,u]},{u,0,2 Pi},AspectRatio->Automatic,Axes->False,DisplayFunction->Identity],{i,0,2 Pi,2 Pi/20}] ShowAnimation[ploti] Make a hardcopy of your "social security number" orbit. in the directory: http://www.physics.csbsju.edu/364/orbit/ find the files: orbit_X.gif and vr_X.gif where X is the last digit of your social security number. "Chi-by-eye" is the disparaging name for the process of fitting functions just by using trial and error comparison between the data and a plot using a trial set of parameters. Use the applet (i.e., the web site: http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm with the default masses and a, and using the red curves) to display vr and orbit data. Adjust tha applet's (e,i,w) parameters until you think you've got an approximate match. Note that it's the shape you want to match not the values (as the applet is using km/s and the Mathematica plots are in unstated units). Note that the vr plot shape is not affected by i, just the range of vr (which is unknown). Report your best "fit" parameters (e,i,w) --class15--(note "help" on final classday [Monday: 18 Oct = class 18], final, self scheduled around Wed 20-Oct) HW: ch. 10: 1,2 --class16-- cosmology: Ch. 11 no homework --class17-- Recall that the value of the mass density and cosmological constant are usually expressed in dimensionless form: Omega_Lambda = Lambda c^2/(3 H^2) = .7 Omega_m=8 Pi G rho/(3 H^2) =.3 where G=Newton's gravitational constant H = Hubble constant = 70 km/s/Mpc (currently) rho = mass density Lambda = cosmological constant 1) Find the (current) value of rho, and then express the result as the equivalent of so-many H-atoms per m^3 (Note: since 90% of this mass is dark matter -- known not to be H-atoms -- the result is about 10x the actual average number of H-atoms per m^3) 2) Find the value of Lambda. 1/Sqrt[Lambda] has the units of length, express it in units of billions of light years. 1/Sqrt[Lambda*c^2] has the units of time, express it in units of billions of years. (If this seems like two problems...think again.) 3) To the extent that rho*R^3 is a constant (as an aside not related to this problem I report: it is not constant during the first 100,000 years of the Unvierse: at that time most of the "mass" is light and since its energy density is proportional to T^4 and T*R=constant for doppler, rho goes like 1/R^4 rather than 1/R^3), we can write the differential equation for the scale of the Universe R as: The differential equation for R is: (R')^2 = 8 Pi G rho R^2/3 + c^2 Lambda R^2/3 -c^2 k where k=1,0,-1 Let t=0 be now (so the BB is t=-12.7E9 years), and use the measured values for H and Lambda (which produce the dimensionless quantites Omege_m and Omega_L defined above) and k=0... I find it helpful to go to a dimensionless version of time: T=t Sqrt[Lambda c^2] What is the numerical value of the time unit: 1/Sqrt[Lambda c^2] (recall...) we can then write this differential equation: (dR/dT)^2 = (1/3)(Omega_m/Omega_L) (R_0/R)^3 R^2 + (1/3) R^2 (Where I've used rho*R^3=constant) or dR/dT = Sqrt[(Omega_m/Omega_L) (R_0/R)^3 + 1] R/Sqrt[3] Let's start solving the diffeq for current R=R_0=1 solution=NDSolve[{R'[t]==Sqrt[(.3/.7) (1/R[t])^3 + 1] R[t]/Sqrt[3],R[0]==1},R,{t,-2,5}] Plot[Evaluate[R[t]/.First[solution]],{t,-2,5},PlotRange->All] Plot[Evaluate[R[t]/.First[solution]],{t,-1.39711,0},PlotRange->All] Turn in hardcopy of these two plots. Notice that you cannot go back in time before -1.39 or so...a singularity is encountered... How long ago [in years] was this BigBang? --class18-- Final Exam