# Order-of magnitude estimate fails for angular momentum in decay

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1. Dec 21, 2015

### bcrowell

Staff Emeritus
When a nucleus gamma decays, the gamma has its intrinsic spin of $1\hbar$, but it can also carry away a significant amount of angular momentum in addition to that. Quadrupole radiation is very common, and in exceptional cases you can even get gammas with 5 or $10\hbar$.

Now suppose I do the following naive estimate of how much angular momentum a 1 MeV gamma *should* be able to carry off (ignoring intrinstic spin). It's radiated from the nucleus, which has a spatial extent of, say, $r=3\times10^{-15}$ m. For a massless particle, $p=E/c$, so I estimate the maximum angular momentum as $rE/c\sim10^{-2}\ \hbar$, which is obviously much too small.

Of course this is a naive estimate, and in particular it ignores quantum mechanics except at the very end when I express the result in terms of $\hbar$. The wavelength of the gamma is about $10^{-12}$ m, which is much bigger than the size of the nucleus, so we shouldn't expect classical physics to be a good approximation. (For an alpha, the corresponding estimate comes out to be more like $10\hbar$, and we do have $\lambda\ll r$, so we expect classical physics to work.)

Nevertheless it seems odd that the estimate for a gamma is off by such a vast amount, and I'm unsure as to whether this is a classical issue or a quantum-mechanical issue. Classically, can an antenna of size $r$ emit electromagnetic radiation with energy $E$ and $L\gg rE/c$? As a heuristic, can one let $r$ equal $\lambda$ rather than the size of the antenna? (Doing so provides just about the right result for the gamma ray.)

2. Dec 21, 2015

### samalkhaiat

In a spherical shell between $r$ and $r + dr$, the ratio of angular momentum to energy of a single-$m$ multipole radiation is independent of the radius of the shell. Indeed, classical calculations show that
$$\frac{dM}{dr} = \frac{m}{8\pi \omega k^{2}} \sum_{m} | a(l,m) |^{2} = \frac{m}{\omega} \frac{d \mathcal{E}}{dr} .$$
Quantum mechanical calculations, for $n$ photons $(l,m)$- multipole radiation, show that
$$\frac{M}{\mathcal{E}} = \frac{\sqrt{n^{2}m^{2}+ n l(l+1) - m^{2}}}{n \omega}.$$
The book by Blatt & Weisskopf “Theoretical Nuclear Physics” contains very nice treatment of multipoles radiation. In fact the book is famous for going from bad to good and then to very good treatment for its content.

3. Dec 22, 2015

### Spinnor

Say we have a gamma emitting crystal of iron 57 cooled to near absolute zero. How do I estimate the uncertainty in position of one of the iron 57 nuclei?

Edit, And does that uncertainty in position come into play during gamma emission?

Last edited: Dec 22, 2015
4. Dec 22, 2015

### bcrowell

Staff Emeritus
@samalkhaiat - Thanks for your post, but I don't think that resolves the issue. The independence of r that you discuss plays out analogously in the case of alpha decay, which can be treated using Newtonian mechanics for these purposes. The angular momentum L and momentum p of the alpha particle are independent of r as it flies off, but we still have a constraint $L/p<r$ at the moment when the alpha separates from the nucleus.

@Spinnor - I don't think the uncertainty in position of the nucleus is relevant. You can just define the coordinates relative to the position of the nucleus's center of mass.

5. Dec 22, 2015

### Staff: Mentor

You have the same effect with atoms and electron energy levels.
The size of the process is given by the minimal photon position uncertainty.

6. Dec 22, 2015

### bcrowell

Staff Emeritus
So are you saying it's a quantum-mechanical effect, not a classical one? This seems unlikely to me, since the discussion can be phrased in terms of unitless ratios like L/rp that are the same regardless of whether the intensity of the radiation is high enough for the classical description to apply. But if you know of an analysis that supports this claim, I'd be interested to see it.

7. Dec 22, 2015

### Staff: Mentor

How can the emission of a single photon be classical?
At least imagining a photon as point cannot work. You cannot localize it sufficiently.

8. Dec 22, 2015

### Spinnor

Bcrowell, in your estimate why not use the mass of the nucleus times the speed of light times the radius? In the classical sense it is the nucleus as a whole that is "sloshing" around?

9. Dec 22, 2015

### bcrowell

Staff Emeritus
That's not what I said in #6.

Right, but it's also true classically that you can't localize a wave to less than about half a wavelength.

Saying that the wave is delocalized doesn't solve the problem. In classical terms, the question would be how the flux of angular momentum through a spherical surface of radius r can be greater than the flux of energy times r/c.

I suspect that what's going on is that in the near-field region, the wavelength is effectively of the same order of magnitude as the size of the antenna, which is much less than the wavelength of the far field. I'm guessing that a lot of the momentum in the near field region is transverse, rather than longitudinal, so it contributes to the angular momentum but is not related to the (much smaller) radial component of the Poynting vector.

Last edited: Dec 22, 2015
10. Dec 22, 2015

### Staff: Mentor

Let energy flow out at one side and in at the other side. Exchange of angular momentum with the surrounding field doesn't have to include net energy transfer. But I think that is too classical now.

11. Dec 22, 2015

### bcrowell

Staff Emeritus
Sorry, I don't follow you. Radiation patterns don't have energy that flows as you describe.

I'd be interested to see an actual quantitative analysis, such as a reference to a description of this kind of thing in Jackson. Short of that, I don't see any obvious problem with the picture I suggested in #9.

12. Dec 23, 2015

### samalkhaiat

1) Gamma rays are emitted by the tails of the unclear wave-functions, i.e., far from the centre of the nucleus. In fact, for typical energy $E_{\gamma} \sim 2 \mbox{Mev}$ we estimate
$$\ell \hbar \sim \frac{r E_{\gamma}}{c} \Rightarrow r \sim 100 \ell \ \mbox{fm} .$$
2) It is not easy for gamma rays to carry orbital angular momentum. Indeed, transitions with $\ell \neq 0$ are suppressed by a factor $(Rk)^{2\ell}$, where $R$ is the radius of the nucleus. Notice that, $kR \ll 1$ while $kr \sim \ell$.
3) The total angular momentum $L = \ell + s$ carried by gamma photons is restricted by angular momentum conservation and the rule of addition of angular momentum. In terms of the initial and final nuclear spins, you have
$$|J_{i} - J_{f} | \leq L \leq J_{i} + J_{f} .$$

13. Dec 23, 2015

### bcrowell

Staff Emeritus
@samalkhaiat - The numerical estimate you're doing in (1) is the same one I did in post #1. The nuclear wavefunction tails off exponentially, with a characteristic scale of 1 fm, so there are no tails, with any significant probability, reaching out to hundreds of fm.

Re (2), as I pointed out in #1, we do observe gamma rays with angular momenta as high as about 10 hbar. Yes, they are weak, but (1) is not going to work as an explanation of why they exist, because (1) is an exponential fall-off to insanely small values.

I don't understand what point you're making in (3). It's true that if the gamma has an angular momentum of 10 hbar, then only 9 hbar of that needs to be orbital.

I appreciate the suggestions, but at this point I'm pretty sure that my #9 is the correct resolution of the problem.

14. Dec 23, 2015

### samalkhaiat

The slow multipole transition (i.e., higher 1 \hbar) is allowed by angular momentum conservation. And yes, the photon in these transitions can have orbital angular momentum relative to the recoiling nucleus.
The classic text on gamma decay and multipoles transitions is Weisskopf. Jackson text also treat the subject (mostly classical) in chapter 16 in my edition (Multipole Fields).

15. Dec 23, 2015

### bcrowell

Staff Emeritus
Yes. Why is this relevant?

Yes. Why is this relevant?

I'm familiar with this material. My research background is in gamma-ray spectroscopy. I would be interested in a reference that deals specifically with the issues discussed in this thread.

16. Dec 23, 2015

### samalkhaiat

I don’t know what is relevant for you. What I have been trying to say is the following: 1) In dipole transition, $\ell = 0$. So, the gamma photon carries $1\hbar$ which is the photon’s intrinsic spin. In this case, you should not expect $L \sim Er/c$ to give you the spin of the photon.
2) In multipole transition, the photon angular momentum $J_{\gamma} = \ell + s$ is again determined by the spins and parities of the initial and final nuclear quantum states. So, why should you expect $L \sim Er/c$ to give you correct estimate this time?

17. Dec 24, 2015

### bcrowell

Staff Emeritus
It can be the case that A is determined by B, and A is also determined by C. This is one of those cases. "C" here would be the classical operator for the angular momentum density of the electromagnetic field. Thanks for your suggestions, but I'm satisfied at this point that this is a classical issue, and that I understand it to my own satisfaction.