Lecture 7
“The test of a good teacher is not how many questions he can ask his pupils that they will answer readily, but how many questions he inspires them to ask him which he finds it hard to answer.” Alice Wellington Rollins (American writer 1847-1897)
I want to measure the presence of a planet in orbit around a star by measuring the variation in light intensity from that star. What is the smallest variation in intensity or absorption that can be measured? This is a change in intensity due to absorbing or scattering photons out of the beam. To be concrete, assume a planet like Jupiter moves in front of our sun yielding a 0.1 % variation in the solar output.
What is the source of noise? It is that the EM wave is composed of photons whose arrival at the detector is like the number of cars crossing a line across the freeway in a minute.
How do the photons arrive? Through a telescope to one pixel on a detector array. Have the photon beam incident on one photodiode which converts photons to electrons when an electron hole pair is generated in the junction (like the photoelectric effect with a photomultiplier). The output of the photodiode is a current with fluctuations in electron mirroring those of the photons. These electrons charge a cap (feedback cap in an op amp) over time to get a total charge on a cap. The charge is integrated over a time interval like PLC integration in the voltmeter. The time interval is controlled by switches which allow charge to accumulate and then discharge the cap after a measurement is made of the voltage across the cap. A constant current flowing into the cap increases the charge on the cap and the voltage across it via V =Q(t)/C.
How many photons per second arrive and how are they detected? How would you estimate this (often a question asked by a recruiter)?
Example problem: Intensity is 1000 watts/m^2 at Earth from the sun. What do you know that used 1 kW? From a star a distance Rs away the intensity (watts/m^2) is IntSun/IntStar approx 4Pi rE^2/4 Pi rStar^2. This yields about 10^-10 Watts/m^2. HeNe laser emits 1 mW. How many photons/s? Energy/s divided by h nu energy/photon. 1mW about 10^14 photons/s
How are these photons detected? Photoelectric effect or photodiode. Some electrons in a metal are free to move. Hit a copper hammer on a table. What happens? Why don’t the electrons spew out the end of the hammer in a spark? Inertial moves the electrons to one end leaving positive charges at the other (this has been measured). It is hard to remove the electrons due to the left over positive charge. This is a simple model of the work function in the photoelectric effect. One photon liberates one electron which is made to flow across a resistor in a phototube or photomultiplier.
What are those fluctuations? Make an analogy with fluctuations in the flow or current of cars on a freeway. How many cars pass a line in 10 s at different times of the day and what does a histogram plot of this data look like?
For the photon case, sketch an op-amp circuit of a reversed biased photodiode charging a cap (C=Q/V or V=Q/C) for a set duration with controllable input and output switches, say with a mean value of 100 counts in a msec yielding 0.1 V (for this cap value) mean and with stnd dev of 0.01 V. Draw this histogram of number of times a certain number of photons is measured for the 1 msec window assuming a Gaussian distribution (like a distribution of cars on a freeway crossing a line in 10 s). What is the minimum % fluctuation that could be measured? The percent noise is 0.01V/.1V = 10% (while the signal to noise ratio is Vdc/sigma). Draw a voltage vs time series rather than a histogram. Ask what would the signal from the planet occulting the sun would look like? Show that a 10% signal would be buried in the noise. The noise is 10% of the voltage due to the sun. We need to see a 1% noise to sunlight ratio to see the planet.
Point out that the capacitor adds the number of photoelectrons over time to give the total photo-voltage for a certain time interval reading. Draw a histogram of number of times a certain number of photons is measured for the 100 msec window and compare with that of the 1 msec window. This is like the total number of hamburgers eaten in a class. The fluctuations for a short time are somewhat smoothed out over a longer time due to averaging. Illustrate this by drawing a histogram of number vs voltage (AND number vs hamburgers) across a cap for different time intervals.
What could you do to see this small variation in the sun’s output? Charge the cap for 0.1 sec or 100 times longer. The histogram of voltages now has a 10 V mean and stn dev 100x 0.01/Sqrt[100] = 0.1 with a percent fluctuation of 0.1/10 = .01
(aside: When put across a resistor we get a DC voltage Io R = R e N/sec (where e is the charge on an electron in Coulombs) with fluctuations R e sqrt[N]/sec yielding a percent error in the DC voltage = 1/sqrt[N]. This is the percent absorption that can be measured.)
Current in a circuit is (N electrons per sec)
N q +- sqrt[N] q
Percent fluctuation is sqrt[N] q/N q = 1/sqrt[N] which decreases as N increases! The per second is IMPORTANT. Say you measured in 1/10000 of a second. You would have had 10000 less particles and therefore a higher percent fluctuation. Why would you want to measure in a shorter time interval? The signal may be your voice and it amplitude modulates at say from 200 to 2000 times a second. This requires a bandwidth of a kHz and therefore a shorter time given by delta nu delta t = 1 This is the same issue with the voltmeter set to measure 1 or 100 PLC's. During the time interval for 1 PLC there is more sqrt[N] noise than with 100 PLC's.
For a laser beam generating photoelectrons in a photodiode N can easily be 10^14/s. Changes in intensity can be as small as 10^-7 of the DC intensity.
Critical thinking discussion:
what are the assumptions: The sun does not have intensity fluctuations over the time you expect planets to move across its disc and it period. Only one planet crosses the sun at a time. The DC reference voltage for the amplifier and voltmeter do not vary in time at the period measured. No periodic scattered light between the star and earth enters the detector.
logical fallacy: assume that the change in intensity is proportional to planet size. You really need to overlap circles for the planet and sun which may not lead to a linear change in intensity with planet size. Light scattered by the planet back to your detector as the planet moves to the dark side is not detected.
different kinds of evidence: wobble of the center of mass of the sun due to the orbiting planet yields the same period.
Hmwk on “How to Lie with Statistics”: -List two questions that should be asked when given data (who says so, how does he know, what's missing, did somebody change the subject, does it make sense?)
Review pendula questioning exercise.
-Focus on a model about which to ask questions or on how to determine a model. -What is the model? Simplify with 3 pendula, with periods T1, T2, and T3. They will repeat after m T1 = nT2 = p T3 where m, n, and p are integers. If T1=2, T2=3, T3=4 and the three pendula start at the same angle then after 24 seconds they all will be back in the same place. One analogy question was could this be done with sound waves? Here it is done with EM waves and it could also be done with sound.
https://en.wikipedia.org/wiki/Mode-locking
Questioning exercise:
http://www.youtube.com/watch?v=gj1pkyCL75E
Before thinking of questions, think first about a model which describes what you see!
