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Orbiting loses energy 
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#1
Nov1812, 04:42 AM

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The original planetary model of the electron orbiting the nuclear did not make any sense because the electron would lose its energy so quickly and the atom would collapse within a tenth of a second. But I can not see an obvious reason why electron loses its energy while its orbiting the nulear.
And it seems that the earth which is orbiting around the sun is losing its energy every second(a extremely small amount of energy though) as well. Can someone come up with an equation to explain this? 


#2
Nov1812, 06:53 AM

P: 1,020

Do you know that a charge particle under acceleration will emit radiation,this is not the case with things accelerating with no charge.In case,it is having centripetal acceleration.see here
http://en.wikipedia.org/wiki/Larmor_formula 


#3
Nov1812, 07:39 AM

Sci Advisor
P: 2,470

There is a similar loss for planets. Accelerated mass emits gravitational waves. But the rate at which a planet loses energy this way is very, very low.
A larger contribution, for a planet, is due to tidal forces. Deformation of massive bodies due to tidal forces results in slightly uneven gravitational field, which gives rise to an interaction similar to viscosity. This is why Earth's rotation is transferred to the Moon, causing the Moon to drift away from Earth very, very slowly. Similarly, planets' rotation can be transferred to parent star, resulting in loss of energy. 


#4
Nov2012, 05:50 AM

P: 661

Orbiting loses energy
Equivalently to the electron accelerating, you can view the classical orbit as an antenna, where the current due to the electron movement would radiate light.
In QM, orbitals are not movement, because they are stationary solutions, not evolving over time, so they don't radiate. More detailed: the wave's envelope does not spin around the nucleus, hence no radiation, but the phase does  which is possible because the amplitude is complex  and this permits a rotation momentum and a magnetic momentum for the orbitals P, D, F... that have no S symmetry. If you add several stationary solutions (orbitals) you still get an acceptable amplitude for the electron but which isn't stationary. The envelope vibrates over time, at a frequency equal to the energy difference between the levels, and equal to the photon that is presently being absorbed or emitted. Photon interaction supposes certain relations between the orbitals, especially the spin difference must match a photon, to make an "allowed" transition. This is equivalent to the electron vibration in the nonstationary amplitude producing an efficient field, essentially a dipolar field rather than nonpolar (S to S orbital), quadripolar, or other forbidden transitions, which radiate inefficiently. 


#5
Nov2012, 07:29 AM

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#6
Nov2012, 08:34 AM

P: 162




#7
Nov2012, 08:54 AM

P: 162




#8
Nov2012, 09:18 AM

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P: 2,470

And geodesic is only curved if there is gravity involved. On the scale of an atom, all geodesics are straight lines. 


#9
Nov2012, 09:43 AM

P: 162




#10
Nov2012, 10:55 AM

Sci Advisor
P: 2,470

No. Both of these are related to the metric. Geodesics are a generalization of the concept of shortest curve connecting two points. If distance between two points is given by d² = x² + y² + z², the shortest path is always a straight line. When you talk about gravity, you also want to consider time in all of this, which is why Geodesics are a bit more general, but the concept is the same.
So for example, lets say you consider object motion in polar coordinates r = R, θ = ωt. That does look likes "straight line" motion, because one coordinate doesn't change, and the other changes at uniform rate. But lets look at distance element. ds² = dr² + r²dθ². This is what tells you that things are going to be a little more complicated. Specifically, when you have a metric that depends on coordinates, you can't just define a=dv/dt, because you are comparing velocity vectors at two different times at two different locations. This metric doesn't depend on time, so that's not a problem, but it does on location. So you have to take a covariant derivative which takes into account the different locations. The derivation is not that complicated, but it's a bit messy because it involves Christoffel symbols. Once you account for the above metric, you can derive the expression for acceleration vector. a_{r} = dv_{r}/dt  v_{θ}²r a_{θ} = dv_{θ}/dt + 2 v_{θ}v_{r}/r In our case, velocity vector is given by v=(0, ω). From the above, the acceleration vector is given by a=(ω²R, 0). The acceleration is nonzero, and is, in fact, the correct centripetal acceleration. Note that in no part of this did I ever have to consider conversion to (x, y) coordinate system. The moral is that while coordinate systems and velocities are all relative, the accelerations are absolute because we can define a metric. General relativity deals with this rather heavily, but as you can see above, you can use all of these concepts in classical mechanics and get the same results. 


#11
Nov2012, 12:46 PM

P: 162

Interesting.
 "velocities are relative, accelerations are absolute..." Do you have some link that talks specifically about absoluteness of acceleration? 


#12
Nov2012, 01:36 PM

Sci Advisor
P: 2,470

I don't really know any good references. I think, pretty much any GR text will discuss it, but only as it relates to other things.
Basically, it comes down to the fact that you can measure acceleration directly. If you are in a completely closed room, you can still measure acceleration of that room using a mass and a spring. You can't tell acceleration from gravity, of course, but in GR, gravity is just a frame acceleration. It arises the same way centrifugal effect arises in the example above. The motion that causes it, however, is motion through time. So the math is more complex, but idea is exactly the same. You have something that looks like "straight line" in x, y, z, t, but because of the curvature of spacetime, it's not, and you end up experiencing acceleration even if you stand still. What's interesting is that the whole idea behind GR came from an attempt to treat accelerations as relative. Specifically, to adhere to Mach's Principle. That article might actually give you more insight into nature of acceleration, even though it talks specifically about it being relative. Hope that doesn't leave you more confused. 


#13
Nov2012, 01:48 PM

P: 162

Confused as before, but certainly more curious. Thanks for the info, I think I know now what to google for.



#14
Nov2012, 02:15 PM

Mentor
P: 17,526

So, in regular Euclidean geometry y=mx+b gives a straight line. You can express the same line in polar coordinates as r=b/(sin(θ)m cos(θ)). The polar equation no longer looks like a straight line equation, but in both coordinate systems the path is a geodesic. 


#15
Nov2112, 02:12 AM

P: 1,020




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