Well as they say, no battle plan survives contact with the enemy...
So I get there early, and after several minutes manage to get my computer talking to the overhead projector. The ROE (rules of engagement) say I'm supposed to give the same talk as last time.
I am prepared. I have my talk ready, and I have added two slides to clarify a few points. I have a second talk prepared, discussing the research done this Fall as a backup.
The professors show up.
Dr Smith: We've seen this presentation before haven't we.
Me: Well yes sir, the rules say I;m supposed to give the same talk.
Dr. Smith: Well that's rather silly, we've heard it before. I think we all agreed that Mark did fine on the talk portion didn't we.
Dr. Sankey: Well, yes we did.
Me: Well, this was the research we did last Spring, I have another talk prepared discussing the research we did last semester if you are interested.
Dr. Sankey: No , that's fine, we know you can give a competent presentation. Let's just skip ahead to the Question Portion. Unless the Student Objects..
Me: No sir, it really is essentially the same presentation. I've only added two slide to clarify a few points that were not necessarily the most clear last time.
All right then, go ahead and turn off the projector..
Sankey: Now, why don;t you go ahead and write down Maxwell's equations.
Me: Um ok, it's been a while. I write down Del dot E equals rho over epsilon naught.
I write down Del dot B equals zero. I hem and haw a bit and eventually I managed to write down d/dt B = -Del X E and d/dt E = J + Del X B (technically wrong since there is should be a minus sign in front of J)
Sankey: Ok number those from one to four. Now, what is the physics behind the first one.
Me: Um well is says this (I forget what I said)
Sankey: no that's not what I'm after.
Me: Um, are you after Coulomb's law?
Sankey: Ok, yeah, sure. How do you get Coulomb's law out of that. (it's actually Gauss's law, but we can get Coulomb's law out of it)
Me: well, coulombs law says this, I start writing Coulomb's law, Um, do you want the electric field or the potential version... I end up writing the E-field version.
Sankey: ok, so how do you get from here to there.
Sankey: how about Gauss's law.. Do you remember that.
Me: well Gauss's law states that you can convert a volume integral to a surface integral. (according to Wikipedia this is actually Gauss's theorem, not law)
We then spend several unpleasant minutes, working out a derivation, in which I keep insisting I need a 4Pi, and Sankey insists I do not. We eventually find our missing 4Pi
Dr Bauer: can you tell us about the chemistry of the LEDs you used for this other project.
Me: well, sure. Does everyone know what Dr Bauer is talking about? Ok, well then let me give you a little background so we are all on the same page. I then manage to spend several minutes discussing the research from last semester, thus killing around 5 minutes.
Me: So, for the Blue, Green, and white LEDs I'm certain the chemistry was InGaN, because that is the usual composition for those LEDs, I;m not certain what the che3mestry for the amber LED was.
Bauer: yes but, what do you know about the electron bands of an LED.
Me: well a modern high brightness LED builds in a quantum well structure to trap the electrons, thus increasing the output. (I'm more than happy to draw the electron band structure for the quantum well)
Bauer: No, I mean E vs. k.
Me: Um ok. I draw a generic E vs k diagram for a direct bandgap semiconductor. I proced to discuss that for a substance with an even number of valence electrons the valence band will be completely full and the conduction band completely empty, I discuss why we might dope the semiconductor, and then I draw an E vs density of states diagram. (take that)
Dr Bauer says something about electrons and photons, so I then discuss the difference between direct bandgap semiconductors and direct band gap semiconductors, and that optical semiconductors need to be of the direct type.
Me: because this is an E vs. k diagram, so in this case over here, a photon can make a direct transition, in the indirect case, the electron has to interact with a phonon, because there is a momentum shift.
Bauer: but doesn't the photon have momentum?
Me: Well yes.
Bauer: well what is the momentum of the electron?
Me: well the momentum of the electron is inversely proportional to... Wait, let me be clear here, the effective mass of the electron is inversely proportional to the curvature of the band.
Bauer: Yes that is correct.
Me: That is why here, at the top of the band, you can't add any energy to the electron, it's effective mass is infinite. you cant just kick it a little and have it move, you have to provide a large quanta of energy, in the form of a photon, then the electron makes a vertical transition.
Bauer: why vertical?
Me. Because a vertical transition involves zero change in momentum. Over here (points to indirect bandgap sketch) you need to interact with a phonon. That means in order to make the transition, you have to either add or subtract a vibratory more to the entire lattice. That, interaction os far less favorable than in the direct case.
(At this point, I swear Dr. Bauer actually similes at me)
Dr Sankey to Bauer: You happy with that answer?
Bauer: Oh yes.
Bauer: now, you mentioned black body radiation in your paper, can you tell us about that?
Me: well, any material object that is heated, such as a tungsten filament, gives off radiation the spectrum of which is known as black body. The traditional example is you take a metal box, punch a hole in it to let the photons out, and then heat up the box. The spectrum that comes out is known as black body or Plankian radiation.
Bauer: can you write the formula for the distribution:
Me: well, not from memory. I can derive it though.
Bauer: well what does it look like?
Me: (I write down the thermodynamic distribution formula for a Bose gas)
Bauer: what's that?
Well, this is the Bose-Einstein distribution. If I solve it for energy, that will give you your answer.
Bauer: no we don't have time for that. Just draw a picture.
(I draw a graph showing spectral power vs. lambda for several temperatures.
Bauer: Do you like lambda?
Me: Generally, yes I prefer to work with wavelength. (either wavelength or frequency is ok actually)
Me: So, these lines would represent different temperatures, the key here is that as the temp increases, this curve gets higher, and the peak shifts this way, towards the visible. The problem with a tungsten lamp at the VATT, or at your house even, is that the visible spectrum, (I draw a line) ends here, so only 10, 15 percent of the energy given off is energy your eye can use, the rest is near and mid IR.
Bauer: You have written power up there, how does the power relate to the photons that exit? I mean their electric and magnetic fields.
Me: well the power is given by the Poynting Vector, (I write S = 1/mu-naught times E X B) um, I may have this constant wrong, I think it's one over, but it could be the other way..
Bauer: we are not interested in the constants... (I had it right by the way)
DR Smith: Ok, well as I recall last time you had a little problem with some derivatives, sankey asked you to solve the wave equation.
Me: actually Dr Sankey asked me if I could prove the Heisenberg Uncertainty Principle. Which, actually I can. It's a rather interesting derivation.
Smith: could you sole the Wave equation for us.
Me: sure. I write down Schrodinger's equation.
Smith: Well, that's one Wave equation, how about the generic wave equation.
Me: Um, you mean Laplace's equation equation. I write down Laplace's eqation.
Sankey: how about the vibration of a string.
I finally write down Del squared Psi + k squared Psi = zero.
Sankey: ok solve that.
I write down a "proposed solution" take the derivative, and successfully solve the equation.
Sankey: Ok so what kind of wave is that?
Me: it's a standing wave. Wait, technically it's only a standing wave if the two constants A-1 and A-2 are the same.
The professors: Hey yeah, he's right.
Sankey: so what's the solution then.
Me: well it's either a series of sines or cosines depending on where you set your zero.
Sankey: look at the equation, that equals 2 Cosine.
Me: (look at the equation.) Oh, yes it is. but we can shift our coordinates to get either sines or cosines.
Sankey: draw a string.
I draw 2 points label them zero and "L", and draw an arc between them.
Sankey: Make it look more like a cosine.
Me: This is correct for the fundamental mode, but ok. (I draw the 2nd harmonic, which looks like a Cosine)
Sankey: so what are we missing from our solution?
Me: we are missing the boundry conditions. by applying periodic boundry conditions we get a solution that is a sum of cosines.
Sankey says something...
Me (forging ahead) no, the wave has to go to zero at the endpoints so the solution is
k^2 Cos (n Pi/L)
Sankey: you forgot the x.
Me: oops, yeah we kind of need an x. (I add the x)
The point I am trying to make is that the cosines form an orthonormal basis set, Sankey is after something else entirely.
I don't see any time dependence in here, how does the string vibrate.
Well, we can add a time dependent term, I ad a minus omega tee to each exponential.
Eventually we get to where we need to replace the K-squared with an omega squared. Sankey asks what that is.
Me: well it's the dispersion relation in the material.
Me: Well, the dispersion relation tells you the relation between the material, the velocity of the wave in the material, and the wave vector.
Sankey: Just write down f = 2Pi omega.
Me: well f = v/lambda
Sankey: where'd that come from?
Me: Um, I know it.
Sankey: laughs, but that's what I wanted you to prove!
Me: oh, well, lets go backward then
(I work the algebra backward to get where he wanted)
Smith: so write down the proper wave equation then.
Me getting frustrated: (I write down the wave equation for Minkowski space: which looks like Square Psi = zero, The square is the De'Lambertian operator)
Sankey: Oh come on.
Me: no, that's correct. Ok fine, I'll expand it out. I write 1/v^2 d^2/dt^2 Psi - Del-squared Psi = zero)
Sankey: make it one dimension.
Me: (I write an x subscript on the Laplacian operator)
Sankey gives me "a look"
Me: ok fine 1 v^2 d^2/dt^2 Psi = d^2/dx^2 Psi
Sankey: you lost a minus sign.
Me: no I didn't, I moved this over to the other side of the equation.
Sankley: oh, you're right.
Sankey to the rest of the group: I think we're done?
at this point there is discussion of whether I need to leave the room, or if they should decamp and discuss stuff elsewhere.
I propose that they leave, so that I can clean up, since the room is supposed to be used for another exam in a few minutes, and that I should erase the board, and get he room ready for the next student.
So, general impression...
I did better this time. First, I was expecting an ambush, so I wasn't thrown for a loop when they asked off the wall questions. Second, I pounded Bauer's question about bandgap materials out of the park, if it's one think I do know, it's bandgaps and semiconductors.
We should know the results in around 2 weeks. I think I passed, barely. Certainly I feel better about round two than round one.