Why do electrons have to fall on the nucleus in the Rutherford atomic model?

As I read on Wikipedia, the Rutherford atomic model is not correct according to classical electrodynamics, as it states that electron must radiate electromagnetic waves, lose energy and fall onto the nucleus.

I don't understand this explanation.

It is clear to me that with given acceleration directed to nucleus and proper speed, electron can move around the nucleus.

I don't understand explanation about energy, but I understand that there must be some force directed to nucleus. Also this force must not be constant because if it is, a larger speed could keep electron moving around the nucleus.

So what is that force? Why does this explanation on Wikipedia and on other resources operate with energy, not with force?

Well, I dont see a problem in any of those answers here, but, since you want in force terms... lets go.

The Lorentz force is: $$\mathbf F = q(\mathbf E + \mathbf v\times\mathbf B)$$

Lets assume the nice and simple atom of hydrogen. A single electron is classically orbiting it. Lets say there is no magnetic field. Only electric. The electric field is a central field, meaning it is pointing only radially, meaning it will result in an orbit. And more: Its a kepler orbit (same of the planets).

$$\mathbf F = q\mathbf E$$

But then, the electron when accelerated irradiates electromagnetic energy. Conservation of energy must apply, such that the irradiation takes away the energy of the electron. The electron loses then its energy. Energy is proportional to the momentum (kinetic energy). Thus, electron loses momentum. Changing in momentum is force. If we take Larmor Formula and make this process, we will arrive at Abraham-Lorentz force.

Now the complete force of this is: $$\mathbf F = \frac{d\mathbf p}{dt} = m\frac{d^2\mathbf r}{dt} = q\mathbf E(\mathbf r) + \frac{\mu_0 q^2}{6\pi c}\frac{d^3\mathbf r}{dt^3}$$

Note that, for a circular orbit in xy-plane: $$\mathbf r = r(\cos\omega t, \sin\omega t, 0)$$, and thus: $$\omega^2\mathbf r = -\frac{d^2\mathbf r}{dt^2} \quad\Longrightarrow\quad \omega^2\frac{d\mathbf r}{dt} = -\frac{d^3\mathbf r}{dt^3} \quad\Longrightarrow\quad \mathbf F = q\mathbf E - \frac{\mu_0 q^2}{6\pi c}\omega^2\mathbf v$$

Meaning, the third order derivative has a relationship with the speed. And not only that: Has a minus sign over there, indicating a drag force: A force always opposite to the velocity, and thus will tend to stop the motion. So, an electron orbiting a proton with no magnetic field present, will drag because this force, spiral in, and collapse into the proton.

• Yes, this is what I was looking for - explanation in terms of force, thank you. And now I realized what made and still makes me confused. I thought that Lorentz force is the only one that affects charged particles in electromagnetic field. You see, I thought, that only this force, no more. And now it turns out to be more foreces. – Alex Velickiy Oct 13 '15 at 17:25

The phenomenon responsible for this is bremsstrahlung (literally, "braking radiation"), which is emitted whenever a charged particle is accelerated in an electric field (in this case, caused by the nucleus). The forces at work are the electric and magnetic forces produced by the nucleus.

Specifically, this change in energy is given by the Larmor formula: $$\frac{dE}{dt}=\frac{q^2a^2}{6\pi\varepsilon_0c^3}\to\frac{dE}{dt}\propto a^2$$ The acceleration can be computed via the Lorentz force equation: $$\mathbf{F}=m\mathbf{a}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$$

• As I understand, magnetic part of Lorentz force causes electron to fall, as it directly proportional to velocity so there is no velocity to correspond radial acceleration, right? But if nucleus doesn't move, what is the cause of magnetic field that affect electron? – Alex Velickiy Oct 12 '15 at 8:53
• @AlexVelickiy The nucleus has a nuclear magnetic moment arising from spin. – HDE 226868 Oct 12 '15 at 12:32
• @AlexVelickiy No. The magnetic part does not causes the electron to fall. The magnetic force executes no work upon a charged particle, and thus it has zero power contribution. Larmor formula on the other hand, indicates a non-zero power loss. – Physicist137 Oct 13 '15 at 13:16

As the other answer explains, the electron looses energy constantly. This energy is taken from its kinetic energy.

As kinetic energy is reduced, it's speed is reduced. If it speeds down, it will not be able to sustain it's orbit, because a certain speed $v$ corresponds to a certain orbit radius $r$ at that specific radial acceleration, which the attraction force causes:

$$a_{rad}=\frac{v^2}{r}$$

Smaller $v$ gives smaller $r$, and so the radius will slowly decrease as the speed decreases following the gradual kinetic energy loss. According to the model the electron should therefore spiral towards the centre until it collides.

• I got everything you said but first paragraph - it sound for me similar to Wikipedia's one. Could you explain it not in terms of energy but in terms of force? – Alex Velickiy Oct 12 '15 at 8:39
• @AlexVelickiy Why not in energy terms? If an electron radiates energy away from itself constantly, then it looses energy constantly. Then the total energy of the electron decreases. This is the kinetic energy. Decreasing its kinetic energy means decreasing it's speed, $K=\frac12mv^2$. Decreasing speed in the orbit means decreasing it's orbit radius as explained, and so on... Could you point out where exactly you don't follow the explanation? – Steeven Oct 12 '15 at 9:13
• I'm zero in understanding physics in energy terms. There are other energies, not only kentic one. Why does it decrease? Also, I thought that moving in electomagnetic field can by fully described by Lorentz force, now it appears, that there are extra things with energy. – Alex Velickiy Oct 13 '15 at 17:20
• The Lorentz force explains the cause of the motion due to magnetic and electic forces. But it doesn't say anything about the radiation that happens constantly from the moving electron. The radiation is energy that is sent away from the electron and when it looses energy it has less energy to sustain the speed, and thus I always consider this in energy terms. I'm sorry, but I cannot explain it without talking energy - I hope someone else can do that for you – Steeven Oct 13 '15 at 18:35