# Unclear approximation in demonstration regarding neutrino oscillations

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1. Jul 3, 2014

### Daaavde

I'm stucked in a passage of Particle Physics (Martin B., Shaw G.) in page 41 regarding neutrino oscillations.

Having defined $E_i$ and $E_j$ as the energies of the eigenstates $\nu_i$ and $\nu_j$, we have:

$E_i - E_j = \sqrt{m^2_i - p^2} - \sqrt{m^2_j - p^2} \approx \frac{m^2_i - m^2_j}{2p}$

It can be useful to know that here natural units are used ($c=1$) and that the masses of the neutrino are considered much smaller than their momenta ($m << p$)
Still, I can't understand where the $\frac{m^2_i - m^2_j}{2p}$ comes from.

Does anyone have any idea?

2. Jul 3, 2014

### maajdl

$$E_i - E_j = \sqrt{m^2_i + p^2} - \sqrt{m^2_j + p^2} \approx \frac{m^2_i - m^2_j}{2p}$$

since up to the second order:

$$\sqrt{m^2 + p^2} \approx p + \frac{m^2}{2 p}$$

I just don't understand why the formula involves only one momentum.
Why is it not:

$$E_i - E_j = \sqrt{m^2_i + p_i^2} - \sqrt{m^2_j + p_j^2} \approx p_i - p_j + \frac{m^2_i - m^2_j}{2p}$$

Any idea?

Last edited: Jul 3, 2014
3. Jul 3, 2014

### ChrisVer

because you consider that the only difference in energies comes from the mass differences - or in other words you consider $p_{i}=p_{j}$ (momentum conservation).

4. Jul 3, 2014

### maajdl

But I don't see how that is defined by the supposed experimental setup.
Or is it simply because of the translation symmetry along the beam?

Along this line, which symmetry would imply rest mass conservation?

5. Jul 3, 2014

### ChrisVer

I don't think it's a symmetry...
I think it has to do with the fact that the momentum is described by the flavor and not by the mass eigenstates...
in other words, when you expand a flavor eigenstate:
$v_{f}$ it has to have some momentum $p$
then the expanded ones should keep the same momentum...and all the differences are supposed to come from the masses

6. Jul 10, 2014

### Daaavde

I'm sorry, but something is missing for me.

If we expand, we get: $\sqrt{m^2 + p^2} + m \frac{2m}{2\sqrt{m^2 + p^2}}$

and considering $p>>m$: $p + \frac{m^2}{p}$

So, I'm missing the factor 2 next to $p$.

7. Jul 10, 2014

### George Jones

Staff Emeritus
Use the first two terms of a binomial expansion for the last line of

\begin{align} \sqrt{m^2 + p^2} &= p \sqrt{1 + \frac{m^2}{p^2}} \\ &= p \left(1 + \frac{m^2}{p^2}\right)^{\frac{1}{2}} \end{align}