Register to reply 
Unclear approximation in demonstration regarding neutrino oscillations 
Share this thread: 
#1
Jul314, 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? 


#2
Jul314, 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? 


#3
Jul314, 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).



#4
Jul314, 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? 


#5
Jul314, 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 


#6
Jul1014, 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]. 


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 