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A photon is sent directly from a hydrogen atom?
A photon is sent directly from a hydrogen atom on earth to a solid
sphere of uniform density at rest at the top of an inclined plane
plane. Consequently, the sphere rolls down the rigid incline without
loss. After collision, a single photon is emitted. Said photon
is absorbed by the same hydrogen atom on earth. How much time
has elapsed relative to an observer next to the atom?
Notes: my thought experiment: 1/1/2011
Assume equipotentials for atom, Sirius, and incline remain constant.
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1/2/2011:
@First Grade: thank you for answering.
1. Yes, the field is perpendicular.
2. The top of the sphere cannot move
faster than c, and yes, I am interested
in the relativistic case.
3. You can consider the COM of the solid
sphere and ignore the somewhat paradoxical
nature of geometry and motion along the ramp.
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1/4/2011:
@koshka: thank you for answering.
1. The ball is rolling toward Sirius on the ramp.
2. The ball is accelerating down the Pythagorean ramp.
3. One idea: suppose the mass density of this ponderable
entity remained constant as measured in a certain frame
of reference.
Happy new year to you and First Grade.
FGR, don't bust your butt on this. If changing the angle
or sliding a block down a frictionless incline helps
feel free. Thank you for your time.
koshka's questions were important.
2 個解答
- 1 十年前最愛解答
You do have a constant g field along the whole plane where the the g field is perpendicular to the earth-sirius axis, right?
Can I use a sliding sled rather than a solid ball?
If you have the above g field, the ball would have relativistic speeds and I cannot (or if I can, I don't want to even try to) figure out how to handle relativistic moments of inertia (if the ball is going .9c, the top of the ball would be traveling 1.8 c).
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I have not forgot about you al.
The integral for the moment of inertia + velocity for the sphere does not solve out in terms of ordinary functions, or at least wolfram can't do it. There might, however, be a way to do it backwards which I'll look into.
In the next day or so, I'll solve it out as best I can. Happy new year.
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Ok, here we go.
First, what is the time for a lightbeam to get to the starting point and then high tail it back from Sirius?
Earth to Sirius Distance = 8.6 ly
Earth to Start point = 8.6 ly / √3 = 5.0 ly
Case 1: Solve for Slide:
Energy equation:
m (g/2) x = m c² ( 1/(√( 1 - (v/c)² ) - 1)
g x /2 = c²/(√( 1 - (v/c)² ) - 1
g x /(2 c²) + 1 = 1/(√( 1 - (v/c)² )
( 1 - (v/c)² ) = 1 / ( g x /(2 c²) + 1)² )
v = c √ (1 - 4 c⁴ / ( g x + 2 c²))² )
dx/dt = c √( (g x + 2c²)² - 4 c⁴)/ ( g x + 2 c²)² )
t = ∫ 1/c ( g x + 2 c²) / √ ( g² x² + 4 g x c²) dx
t = √ ( g x ( g x + 4 c²) ) / g c
ref: http://www.wolframalpha.com/input/?i=1%2Fc++integr...
g = 9.8 m/s
c = 3 x 10^8 m/s²
x = 8.6 ly * 2/√3 = 9.4 x 10^16 m
t = 3.7 x 10^8 s = 11.7 yr
Total Time: = 8.6 + 5 + 11.7 yr
= 25.3 yrs.
...................
Case 2: Solve for Rolling Object.
I'm not finishing, but you need to solve using the instantaneous axis of rotation. It turns out that it doesn't resolve into ordinary function. Needs to be solved with a computer. But the gist is that it will accelerate fairly normally to close to 1/2 c. Almost no allowance necessary for relativity, and then it will stay at near 1/2 c for the remainder gaining energy by very little speed until it gets to Sirius.
I'll post my notes so far on this part of the problem sometime tomorrow.
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I'll have some more time on Saturday. Going to have to solve this one using spreadsheet computing. The integral is rather ugly:
E = g/2 x = ∫[0 to 2θ] ∫ [ 0 to R] ( 2 r √ (R^2 - r^2) ) / ( √ ( 1 - 2 v (1 - cos θ ) r^2 / ( c^2 R^2 ) ) dr d θ
Wolfram regurgitates when you type that in. But parts can be solved by computer (but not in terms of elementary functions) as Wolfram demonstrates by a numerical solution to part of the bottom of the problem: http://www.wolframalpha.com/input/?i=integral+1%2F...
- 匿名4 年前
the mass of photons is 0 and photons are by no skill at relax, so i'm uncertain the place you're going with this. there is various semantical confusion on your question. in case you're asking what photon *potential* is had to dissociate a nucleus, you basically ought to require that the photon potential exceeds the binding potential of the nucleus and that this potential is a hundred% equilibrated as nuclear excitation potential. if so, the nucleus will not be certain. B.E./A ~ 8 MeV in prepare the photon potential could prefer to be a *lot* better than this, simply by fact the photon is quite hardly (examine "by no skill") going to be interacting coherantly with the finished nucleus. often it is going to "Deep Inelastically Scatter" with basically some nucleons not coherantly with the finished nucleus. this demands the gamma potential to be many GeV and up (not some thousands of MeV). the DIS technique will produce pions, jets, and so on, with the aid of QED diagrams and effectively the nucleus would be destroyed. yet lower back the photon potential required for it particularly is a lot larger than basically the binding potential of the nucleus. cheers edit: oh crap you stated H *atoms* and that i examine it as *nuclei* in many situations... ***slaps brow***... the clarification i did not realize it simply by fact the better question that that's is using the fact i recognised your manage and that i be attentive to which you be attentive to your stuff. sigh - *fail* !!