Maxwell's theory and electrons in orbits

[erisedk]

Homework Statement

Using Maxwell's theory of EM waves, show that an electron while revolving in Bohr's orbit does not radiate any energy. It radiates energy only when it jumps from a higher energy orbit to a lower energy orbit.

Homework Equations

[PLAIN]https://upload.wikimedia.org/math/3/0/c/30c24016df2b868da4e3a8ec58e45ce7.png[PLAIN]https://upload.wikimedia.org/math/9/1/7/917f254fb7d2a63176d640035f6117df.png [PLAIN]https://upload.wikimedia.org/wikipedia/commons/thumb/8/86/OiintLaTeX.svg/25px-OiintLaTeX.svg.png[PLAIN]https://upload.wikimedia.org/math/3/0/c/30c24016df2b868da4e3a8ec58e45ce7.png[PLAIN]https://upload.wikimedia.org/math/4/1/a/41a8ca3e07052a7ec3f34d41ad24b3a6.png

The Attempt at a Solution

Firstly, I thought Maxwell's theory and electromagnetism wasn't applicable to atoms anyway and were discarded because they couldn't justify why the electron didn't collapse into the nucleus as well as radiate energy. Cos electrons are supposed to be probability functions etc.
But, the answer says the following, which I don't understand how it falls out of the above equations. Also I don't understand why this is even a valid question.
Bohr's orbit is a circular orbit in which current I = ev/2πr which is constant. So, magnetic field in the orbit is constant. During transition, frequency of revolution v/2πr changes so magnetic field changes hence radiation is emitted.

 

[tech99]

Bohr's orbit is a circular orbit in which current I = ev/2πr which is constant. So, magnetic field in the orbit is constant. During transition, frequency of revolution v/2πr changes so magnetic field changes hence radiation is emitted.[/QUOTE]
Sorry not an expert but no one answered yet. I don't think that changing a magnetic field per se will produce radiation. It is the acceleration of the charge during transition which radiates, because it distorts the radial electric field lines and produces a transverse E-field component which constitutes the wave.

 

[collinsmark]

Homework Statement

Using Maxwell's theory of EM waves, show that an electron while revolving in Bohr's orbit does not radiate any energy. It radiates energy only when it jumps from a higher energy orbit to a lower energy orbit.

Homework Equations

[PLAIN]https://upload.wikimedia.org/math/3/0/c/30c24016df2b868da4e3a8ec58e45ce7.png[PLAIN]https://upload.wikimedia.org/math/9/1/7/917f254fb7d2a63176d640035f6117df.png [PLAIN]https://upload.wikimedia.org/wikipedia/commons/thumb/8/86/OiintLaTeX.svg/25px-OiintLaTeX.svg.png[PLAIN]https://upload.wikimedia.org/math/3/0/c/30c24016df2b868da4e3a8ec58e45ce7.png[PLAIN]https://upload.wikimedia.org/math/4/1/a/41a8ca3e07052a7ec3f34d41ad24b3a6.png

The Attempt at a Solution

Firstly, I thought Maxwell's theory and electromagnetism wasn't applicable to atoms anyway and were discarded because they couldn't justify why the electron didn't collapse into the nucleus as well as radiate energy. Cos electrons are supposed to be probability functions etc.
But, the answer says the following, which I don't understand how it falls out of the above equations. Also I don't understand why this is even a valid question.
Bohr's orbit is a circular orbit in which current I = ev/2πr which is constant. So, magnetic field in the orbit is constant. During transition, frequency of revolution v/2πr changes so magnetic field changes hence radiation is emitted.

Let me start by saying up front, for what it's worth, I don't particularly like this problem either. You are correct that if you treat the electron as a classical point particle, the system would radiate EM waves, and as you say, you couldn't rely on classical electrodynamics to explain the phenomenon (what's more is that if you were to solve for the velocity v you might find it faster than the speed of light, which opens a whole separate can of worms).

The given answer skirts the issue by treating the electron charge as being evenly distributed about a particular Bohr orbit. In other words, the given answer isn't treating the electron as a classical point particle, but rather it's treating the electron as being an evenly distributed circular charge around the nucleus.