Star Trek
I was researching for a new problem, but decided that the following link was more interesting: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
OK, I've thought of a question: how fast must the rocket be going in order that the microwave background temperature rises to 10000 Kelvin? Ignore GR considerations, treat it as a SR doppler shift problem, use simple black-body radiation concepts only (unless you can do better).
OK, I've thought of a question: how fast must the rocket be going in order that the microwave background temperature rises to 10000 Kelvin? Ignore GR considerations, treat it as a SR doppler shift problem, use simple black-body radiation concepts only (unless you can do better).
Labels: funphysics, thinktank, Your Opinion
posted by Chris at
5:29 PM
11 Comments:
This post has been removed by the author.
This post has been removed by the author.
Hi Chris.
If f is frequency, then Dopler shift is f'=f*rt((1+v)/(1-v)) when you move precisely opposite to the direction of wave and
f'=f*rt((1-v)/(1+v)) if in the same direction. Microwave backround is nearly isotropic, thus when you move in some arbitrary direction, some frequencies will increase, some decrese. Temperature is the statistical characteristic, not just T=hf/k. But in average freq. will increase anyway, so in the order of value
rt((1-v)/(1+v))=10000/3=a.
v=(1-1/a)/(1+1/a)=1-1/(2a), where c=1.
Did I really read that whole page? Now I'm exhausted!
Thanks Chris.. ;)
Hi qantense,you'reback, phew! I was surprised when you introduced hf=kT, as I was thinking of the Wien law, λmax = b/T where b is Wien's displacement constant = 2.9 *10^-3 mK approx. This leads to c*T = b*fmax. So I would have guessed that f/T = k/h = c/b, except the k/h <> c/b by a large amount. However that doesn't matter as only ratios are being considered. I'll admit I wasn't considering the all round view, only the view in the direction of motion. Thanks for coming through. Oh, I agree wih your answer (no surprise there).
Did you read the article?
Hi EB, did you enjoy that page. I should have said, you only need to look at just how far you can go in your own lifetime.
Sadly, the article didn't give the results for a 2g acceleration - which is thought to be well within human capabilities. Then you can travel some seriously large distances ;) Shame you have to melt first.
Is there a similar article on 2G acceleration?
Also, I was wondering how you would be able to maintain such an acceleration for so long without surpassing c(italics wont work). I didn't notice if they stated what their speeds would be traveling relative to Earth either during a certain portion of the journey.
If you could please clear this up for me, it would be greatly appreciated.
Hi Chris, I didn't introduce T=hf/k! Read my previous post. The thing is that if you have gas of photons in equlibrium and want to find it's temperature, you should compare it's density(frequensy) distribution with some Planck-type or simply find its maximum as you correctly noted.
Hi Anonymous, 11:44. Take another look at the article. The speed is always less than c. e.g in his first table for T=12, v =0.99999999996c
I might post the 2g versions of (some) of the journeys later today.
Hi quantense. My bad, I jumped to a conclusion. But I was surprised that k/h <> c/b because the equality would have been nice.
If did the trips at 2g (= 2.06 lyr/yr²), then the journeys would be more like in the following table. This corresponds to the first table on the linked website.
T, t, d, v, γ
1, 1.87343, 1.44986, 0.968, 3.98671
2, 14.9376, 14.4601, 0.99947, 30.7877
5, 7216.66, 7216.17, 0.99999999773763, 14866.3
8, 3485590, 3485590, 0.9999999999999903, 7180310
12, 1.32088*10^10, 1.32088*10^10, 0.9999999999999999999993246781, 2.72101*10^10
NB for T=12 years, the Earth would have aged by almost the present age of the universe. Also, you will have travelled most of the way across the known universe (or something in that neighbourhood). GR will definitely modify that result.
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