Recent content by Hazzattack

Hi, thought I'd give your question a bash. Hopefully this helps or at the very least offers you some possible routes to experiment with. So, we have a ring that is presumably rotating on a frictionless pivot. We can therefore say that the total energy of the system (since we are neglecting any...

Thanks for your response and again for all the help you've provided, albeit a bit painful once I saw your solution. I do see exactly what you mean by the steady state solution in this problem now and I certainly made it considerably more difficult for myself by trying to keep a track of every...

Thanks for all the help D H. Do you happen to know of another problem that is similar to this one so I can try all the things you've brought to light?

Also, when you add the 2nd shield, does that mean the middle shield isn't in a steady state with everything else?

I see, so you've used the fact that the shield and probe are in thermal equilibrium (which is why you were mentioning this a few days ago). If that is the case you also know that the shield must be radiating the probe with as much power as the probe is radiating the shield with - 'so a body at...

Here it is, So I start by assuming my probe is a black body. For the first part one can use the Stefan-Boltzmann law to re-arrange for T, since, Pprobe,i = σAprobeT4probe. This is pretty trivial but you find that the temperature of the probe is approximately 131K. Now we add a heat...

Sorry for the delay. I'll put my working on here shortly.

Well, I assumed the power the probe had to 'remit' once a shield was added would be greater and as such the temperature of just the probe would increase. Otherwise, wouldn't it be pretty useless adding insulation? The only time the temperature would decrease is if the initial thermal source...

Thanks for the response. That is only relevant before you add a shield, correct? I originally modeled it so when the power is emitted from the probe (black body) and hits the shield, the shield absorbs the total power then re-emits it in two directions, hence half of the power is 'reflected'...

Thanks Dauto. I realized that shortly after i posted. Interestingly, if you keep adding shields you end up with a geometric sequence. Some of the numbers are bizarre though.

Thanks for any help in advance. I'm currently doing random physics problems in preparation for my exam. I came across this one and thought I'd give it a go. The question at hand is, A space probe has a 100W thermal energy source and is in deep space after some years of travel. The surface of...

Great. I think I've obtained what I was after. Could you verify that I haven't added something (because I want it to be that way...) I shouldn't have. I start by saying; b_n = \Sigma_{m} U_{nm} a_{m} Thus, b=Ua taking the transpose of each side (all entries of U are real)...

Thank you so much for responding. That all makes a lot of sense and was SO helpful. I see so in that case the Kroenicker delta is 'picking' out a single entity. One quick question, the transpose of a unitary operator with only real entries is the same as just flipping the indices right...

I thought I should elaborate as you guys took time to respond and it may be obvious to you where I'm explicitly going wrong. so in the paper they implement a transform such that; b_{n}= \int dkh(k) \pi_{n}(k)a(k) (where \pi_{n}(k) is a normalised nth monic orthogonal polynomial and both this...

Also... When you take the Hermitian conjugate of a matrix do you just flip the indices (since that is what a transpose does), such that; (U^{\dagger})_{nm} = (U)_{mn} Since in this specific case all entries of the unitary operator are real.