Register to reply

Unclear approximation in demonstration regarding neutrino oscillations

Share this thread:
Daaavde
#1
Jul3-14, 12:58 AM
P: 20
I'm stucked in a passage of Particle Physics (Martin B., Shaw G.) in page 41 regarding neutrino oscillations.

Having defined [itex]E_i[/itex] and [itex]E_j[/itex] as the energies of the eigenstates [itex]\nu_i[/itex] and [itex]\nu_j[/itex], we have:

[itex]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}[/itex]

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

Does anyone have any idea?
Phys.Org News Partner Physics news on Phys.org
An interesting glimpse into how future state-of-the-art electronics might work
What is Nothing?
How computing is transforming materials science research
maajdl
#2
Jul3-14, 02:19 AM
PF Gold
P: 370
[tex]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}[/tex]

since up to the second order:

[tex]\sqrt{m^2 + p^2} \approx p + \frac{m^2}{2 p} [/tex]

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

[tex]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}[/tex]

Any idea?
ChrisVer
#3
Jul3-14, 06:28 AM
P: 876
because you consider that the only difference in energies comes from the mass differences - or in other words you consider [itex]p_{i}=p_{j}[/itex] (momentum conservation).

maajdl
#4
Jul3-14, 07:08 AM
PF Gold
P: 370
Unclear approximation in demonstration regarding neutrino oscillations

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?
ChrisVer
#5
Jul3-14, 07:29 AM
P: 876
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:
[itex] v_{f}[/itex] it has to have some momentum [itex]p[/itex]
then the expanded ones should keep the same momentum...and all the differences are supposed to come from the masses
Daaavde
#6
Jul10-14, 11:54 AM
P: 20
I'm sorry, but something is missing for me.

If we expand, we get: [itex]\sqrt{m^2 + p^2} + m \frac{2m}{2\sqrt{m^2 + p^2}}[/itex]

and considering [itex]p>>m[/itex]: [itex]p + \frac{m^2}{p}[/itex]

So, I'm missing the factor 2 next to [itex]p[/itex].
George Jones
#7
Jul10-14, 03:15 PM
Mentor
George Jones's Avatar
P: 6,243
Use the first two terms of a binomial expansion for the last line of

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


Register to reply

Related Discussions
In the rest frame of a solar neutrino would I still measure neutrino oscillations High Energy, Nuclear, Particle Physics 5
Oscillations - A puzzling demonstration General Physics 7
Neutrino oscillations High Energy, Nuclear, Particle Physics 5
Neutrino oscillations Advanced Physics Homework 1