"Gas. 1-12b tells you how much power leaves a black object per unit area, per unit time, and involves a constant called ..."
Of course, it's not how much POWER leaves, but how much ENERGY leaves. (The units of sigma should have made it clear, even if my poor choice of words did not)
Worse yet, in problem 5, I wrote that the energy of a harmonic oscillator is
p^2/2m + k^2 r/2. Of course, the second term is supposed to be
kr^2/2!! Sorry about that.
>so, here's the story: > >after some trig, I got 8206 K for the temperature of the sun... which >eventually gave me lambda max of 3534 A-- blue light!! alarm bells, >shouldn't the sun's dominant wavelength be yellow? or is that not what >lambda max is? i guess i'm confused... > >please help clear the air. > >thanks,Good observation! I guess you made an algebra mistake, because you should have gotten a temperature of the sun around 6000K or so. (Also *not* yellow, but a lot closer) There is of course a whole spectrum being emitted, so the perceived color will depend in part on our eyes, and absorption of some colors in the atmosphere, etc. (But as you guessed, I think an object at 8206 K would indeed look bluer than the sun does) Cheers, Steve
>I am confused with problem # 4 on the homework set.
>
>You said that the kinetic energy = (E^2 - (mc^2)^2)^.5 = pc. THis all
>makes since and life seems good. I still am not shure if what I am
>doing is correct.
>
>I am taking p/(c*m{rest}) = v and then taking the change in time to
>be: t = d/v {d is the distance the packet travels and is given}.
>
>If this is fine I am close but if you go about the problem in this
>fashion I
>
>don't see how relitivity comes into play, I know these relations are
>realitvistic but the work seems to be the same in each case.
>
>please enlighten me,
>
Let see... I didn't say that kinetic energy = (E^2-(mc^2)^2)^.5 = pc.
In fact, kinetic energy (for a free particle) is always defined to be
E - m_0c^2 (i.e. total energy minus rest energy, leaving behind...
kinetic energy) This is quite different from pc!
Next, I don't understand why you take p/(c*m_0)=v, (e.g., the units
aren't right - your formula would yield p = m_0 v c which can't be
correct?) Was that just a typo?
The formula for p in terms of v is p = mv = m_0 v / (1-v^2/c^2)^.5)
If you want to find v in this problem, you can use the above (having first
found p from E^2 = p^2c^2 + m_0^2c^4, and E = Ekin + m_0c^2)
Or, you can find v more directly from E = mc^2 = m_0 c^2/ (1-v^2/c^2)^.5,
skipping p entirely, and again getting E from E=Ekin+m_0c^2.
Of course, since in this (relativistic) case, E is so much larger than
m_0c^2, you better end up getting a v which is VERY close to c (so the
time is going to be roughly t = d/c)
Relativity comes into play in this problem in two ways. First, the
velocity will turn out to be just about c (if you calculated v
non-relativistically with such a large energy it would be >>c)
Second, the formula for spreading involves this parameter "beta". Beta
is given by d^2(omega)/dk^2, which is the same as hbar*( d^2 E/dp^2)
(Because E= hbar*omega , and p = hbar* k) So, to get beta you need the
formula for E in terms of p only, i.e. E = (p^2c^2 + m_0^2c^4)^.5, and
then take two derivatives. It's a bit of a pain. (The argument I made
in office hours was that for ultrarelativistic particles, E is quite
close to pc, so d^2E/dp^2 is close to zero, so beta is close to zero.
Your job is to find out HOW close to zero, but you know the answer has
to turn out small) Again, very different from the nonrelativistic
case, where beta = hbar/m_0, a constant.
Does that enlighten at all?
On number 5, is the problem really as ugly as I am making it? I have Schroeding and his conjugate, and I solve for psi(x,t) by dividing by V(x), so now I have V(x) = somthing Next, I need the derivative of psi(x,t) wrt t and wrt x in different combinations, so I take all the derivatives (that aren't already done in Schroedinger) by differentiating my psi(x,t). Then I put all the derivatives in my formula forand it is truly a monster. And now I (try to) solve it. Please show me the true path, honorable master. Grasshopper.
The bottom line was to use the Schrod eq. to get rid of d(psi)/dt in the (two) places it appears, replacing with stuff that is all purely in terms of x. You should end up with an integral of four terms, two of them involving V's, two of them involving only psi's. Of the latter ones, (involving only psi's) one clever partial integration should allow you to recognize what's left as the integral of a total derivative (see my hint sheet) which thus integrates away to zero. Of the two former ones, (the terms involving V's), one of the terms needs to be expanded (chain rule) giving you THREE terms involving V's. But now two of them will exactly and immediately cancel, leaving you behind with the one piece you want. Hope it works... Steve
A Sin^5(Pi x/a),
i.e. the a is inside the argument. Also, when he asks about Delta x in (my) 3rd problem, he gets the formula wrong:
Delta x = Sqrt[Ave(x^2) - (Ave(x))^2]
(he left off the second term, which is not zero for a box that runs from 0 to a)
(Question on problem 1): >What exactly do you mean by if a function starts out even? Do you want us >not to consider part a, an odd function? Please clarify- a function that >starts even will stay even, but if we act the commutator on the wave >function we get an odd function in part a. Is zero even or odd? Hmmm... >Thanks, This one is a little hard to explain in writing, sorry. Consider part b), where the potential is V(x)=k x^2/2 (a harmonic oscillator). So, you have a Hamiltonian which is H = (stuff involving d^2/dx^2) + V(x). The time independent Schrodinger equation says H psi = E psi. This is a differential equation, and if you solve it you will find a SET of eigenfunctions (the u_n's for the harmonic oscillator) which satisfy it. Each eigenfunction has its own distinct energy. (Never mind what they look like, they're sort of spiritual equivalents of the u_n's and E_n's for the square box. You don't have to solve *anything* for this problem, just think about it in general) Now, if I put a particle in a harmonic oscillator well, just like when I put it in the square box, it does NOT have to be in an eigenfunction! I.e., the particle does NOT have to have a definite energy. In fact, I can create an initial wave function which looks like just about anything I want. I.e., psi(x,t=0) can be pretty much anything, and certainly does not HAVE to be one of the u_n's (eigenfunctions) So, the question here is, suppose I start off with an initial wave function, psi(x,t=0) which is even. Will it stay even? The answer is very simple, and I don't expect a proof or even *too* much discussion. (You can pretty much find the answer in Gas, pp. 68-70) Remember, the time evolution of an *eigenfunction* is simple (it's just that exponential phase), but the time evolution of a general psi is in general not so simple. (You've done another problems this week where you looked at the time dependence, and you saw from my little MMA demo in office hours that time evolution can be quite fantastically messy! But of course, the psi(x,t=0) for that demo was NOT initially even, was it?) (Same question for part a, where the potential is linear, so the eigenfunctions are a totally different set of u_n's. But again, I can give you pretty much any initial wave function, psi(x,t=0) that I want. Suppose it is even at t=0. Will it stay even? Does this help? Cheers, Steve
>In your notes when you solve for the odd functions of the double >delta potentials you have--> coth(ka)= 1/((lambda/y) - 1) and I think >it should be tanh(ka).Oops, looks like I screwed up in my lecture notes!! Thanks. (To make matters worse, I first wrote an answer in this space where I was only looking at Gas, so I was arguing that my *wrong* answer in my notes was right, double sorry!) It *is* coth (with the same right hand side as before), or tanh (with the inverse of the old right hand side). See Gas Eq between 5-87 and 5-88. He has it right.
>1- Now for the big allmighty homework question. In problem #3 (gas >gives you the "range" of the potential well; Is this range 'a' or '2a' >in our standard box from -a to a?
Good question. Ambiguous answer. I'd argue that the range is "a" in our standard box from -a to a, but it's open to a little semantic debate.
>2- How many licks does it take to get to the center of a Tootsie-Roll pop?
Classical or quantum? Relativistic licks or non-relativistic? Non-Cartesian geometry of space time? Do we assume a spherical pop? And you thought it was a simple question :-)
>3- In problem #5 part a)-ii My drawing looks like the drawing for the >energy in the double delta energy drawing, kinda like the roof to DIA. >So it seems that the energy should be higher here than when b=0. But >if I extend b out to infinity in this case it seem that the line >between the boxes would be strait which would give the highest energy >and tunneling would still be occuring which I know can't be right. >This is hard to explain over the computer but if you understand what I >am saying please clarify.
I'm not quite getting it. In no case does it look like the roof to DIA, because there are no Delta functions, so there should not be any "cusps". (When b=0, you might as well ignore the blip in the the potential at x=0, because it does NOT go up to infinity, so has zero integrated strength. Might as well just be a well with width 2a in the case b=0). So I claim when b=0, the energy is LOWEST. If you extend b out to infinity, the line between the boxes IS straight, and DOES give the higher energy. Tunneling will occur *infinitesimally* Make sense?
>4- How much wood would a wood chuck chuck if a wood chuck could chuck wood?
17 cords.
Cheers, Stever
(Number 5) >I think I got it. If there are two eigenvalues (there's that word!), >and we are making our operator O act like the hamiltonian, then where we >used to have >H u(x) = E u(x) we will now have >O u(x) = E u(x) but that looks just like the eigenvalue equation we had >before, with O u(something) = lambda u(something). So E is just like >lambda and we have two values for lambda so there are two allowed >energies. >True? Precisely! >Dr. Pollock- >One question for you. In problem 5, do we know that the components of the >eigenvectors are real? I get that they could be 1/sqrt2 or >1/(i)sqrt2. If you don't get a chance to respond by noon, don't bother >because I have class from noon till our class. >Thanks for your time. Well, I'd say we DON'T know whether the components of the eigenvectors are real. They don't have to be, only the eigenVALUE has to be real. However, I don't exactly see where your 1/(i)sqrt2 comes from. Is that an *overall* factor on the outside? If so, you're right - in fact, you can take whatever eigenvector you got, and multiply it by ANY complex number with magnitude one (e.g. 1, or i, or -i, or -1, or (1+i)/Sqrt(2), or e^(i phi)...) and it'll still be a valid eigenvector with the same eigenvalue. Help?
Steve: The article on quantum mechanics and lasers I mentioned yesterday was not in Physics Today but in Scientific American, November issue, page 72. The claim seems to be that the laser can be used to discern what is in the box without disturbing the contents. I was not clever enough to determine whether this is a special case with caveats or what. - Thanks! I discovered that you can read about this online here ------------------------------ Another one of those if-you-get-this-before-12-write-back homework questions. On the last part of the last problem, when I do the commutator for x at the two different times, I do get 0. Dont I just put in for t1 and t2 and then all of the numbers are constants so they cancell? It seems like .5t1 squared +. . . is just a constant so it all cancels when I multiply in differnt orders then subtract. Any hints? I know the question is a bit hard to understand. If you do not see what I am asking, no worries! Thanks, Well, I think I understand your question. Presumably, you have obtained something like: x(t) = .5gt^2 + p(0)*stuff + x(0) And you want to commute [x(t1),x(t2)]. You're absolutely correct that .5gt^2 commutes with anything. But, note that there is still a p[0] and an x[0] in your answer. And, when you commute *those*, [p(0),x(0)] is not zero! (What is it?) Does that help you get going?...
>If I have two Identical bozons (or bozos as I like to call them) In >the ground state of an infinite well, I get an answer for the >eigenfunction with the constant 2*Sqrt2 in front. I think the right >answer is just a 2 though. I know both bozos are in n+1 So given the >Dont I add Sqrt2/a Sin (n Pi x1/a) Sin (n Pi x2/a) + Sqrt2/a Sin (n Pi >x1/a) Sin(n Pi x2/a)= 2 Sqrt2 Sin (...) Sin (....)? I think the answer >is that the ground state for bozos is exactly the same as the ground >state of distinguishable particles (Psy= U1(x1)Un(x2), but I can't >justify it. The answer is that wave functions *must always be normalized to 1*. You are arbitrarily sticking in a factor if 1/Sqrt(2) by hand to normalize things, but it turns out that that factor is not a *general* rule, but merely works out to be right MOST of the time. Indeed, you have just pointed to an exception to the rule: when two bosons are in the same state, (n), the normalized wave function is the same as the wave fn for distinguishable particles, Psi = Un(x1) Un(x2). No root two's required, or allowed. (When in doubt, square your Psi, integrate over all x1 and x2, and make sure the answer is exactly one) >Also, in the same well, Is the first excited state for identicle >fermions (or furryons as I like to call them) where one particle is in >n=1and particle two is in n=3 or is it 2 and three. I know that one >and three is a lower energy, but is it allowed. Why *wouldn't* one and three be allowed? The Pauli Exclusion Principle only says that you can't put two identical fermions in *the same state* (the reason being simply that you cannot write down an antisymmetric wave function in that case!) As long as you can explicitly construct an antisymmetric 2 particle wave function, it's allowed. So, the first excited state for identical fermions is the one with lower energy, states 1 and 3. (Note that being in states 2 and 2 would be lower energy still, but then you'd have two fermions in the same state, no go!) >Sorry for the inconvenience. No inconvenience, no apologies required!
(No questions, guess it was an easy set!)