In case anyone is interested, this is what I get for the Hamiltonian of the ice sliding on the wishing well.hamiltonian

# Category Archives: Problem Sets

# Update on Mathematics and Physics HW set

The Pauli matrix for 3.b) should read like this

That is, there is a minus sign in the lower right corner of the matrix.

Otherwise, the eigenvalues become trivial for an identity.

# Aperture Photometry Test Cases

Here is the image we will be using to check your aperture photometry code this evening:

The following test cases use an aperture radius of 3 pixels, an annulus inner radius of 6 pixels, and an outer radius of 10 pixels. Uncertainties are given not because you need to report them, but to indicate an acceptable range for your answers.

- Object near: (x,y)=(313,371)
- Centroid: (x,y)=(312.94,371.29)
- Total counts in aperture: 19202
- Avg. sky counts in annulus: 333.94
- Signal: 9638.2 +/- 104.0
- Inst. mag.: -9.96 +/- .01

- Object near: (x,y)=(396,521)
- Centroid: (x,y)=(396.15,520.92)
- Total counts in aperture: 18848
- Avg. sky counts in annulus: 332.68
- Signal: 9318.8 +/- 104.5
- Inst. mag.: -9.92 +/- .01

- Object near: (x,y)=(230,432)
- Centroid: (x,y)=(229.75,431.72)
- Total counts in aperture: 10643
- Avg. sky counts in annulus: 332.02
- Signal: 1134.1 +/- 52.3
- Inst. mag.: -7.64 +/- .05

# math/physics problem set 7

Here is the problem set with the full figure for problem 3.mathPhysicsProblemSet7

# Baby OD Test Cases

Self-Check 1:Position [AU, equatorial][2.54076127, 0.84513781, 0.3806295]Velocity [AU/day, equatorial][-0.003577, 0.00943291, 0.00429106] ****************************************Semi-major axis (a) [AU], [km * 1000]3.0 448793.6121Eccentricity (e)0.1Long. of Asc. Node (Omega) [deg], [rad]4.0 0.0698131700798Inclination (i) [deg], [rad]1.0 0.0174532925199Argument of Perihelion (omega; w) [deg], [rad]5.0 0.0872664625997Mean Anomaly (M) [deg], [rad]9.0 0.157079632679 *************************************************************************Self-Check 2:Position [AU, equatorial][1.57, 0.0677, 0.253]Velocity [AU/day, equatorial][-0.00169, 0.00985, 0.00145] ****************************************Semi-major axis (a) [AU], [km * 1000]1.09662428346 164052.657764Eccentricity (e)0.460315249446Long. of Asc. Node (Omega) [deg], [rad]212.790065382 3.71388725645Inclination (i) [deg], [rad]16.0459534852 0.280054719938Arg. of Peri. (omega / w) [deg], [rad]325.338401578 5.6782262907Mean Anomaly (M) [deg], [rad]196.414683647 3.4280829289 *************************************************************************

The TAs will have a third test case to test you against.

# Problem Set 2, problem 2 answer sheet

In case you lose you linear transformation answer sheet, here it is:answerSheet

# baby OD

If the baby is at ( 0, 0, 1.5 ) with the velocity ( 0.7, 0.7, 0 ), the orbital elements include I=90 deg, e=0.47, Omega=-135 deg (+360 deg, I guess), and omega=90 deg. I don’t know about a or M.

For what it is worth, for this year’s unfortunate baby, I get

a=2.7798, e=0.23435, I=12.028 deg, Omega=161.52 deg, omega=322.46 deg, M=327.60 deg.