Expansion in a Small Parameter

1. Nov 27, 2009

Dahaka14

When an expression is derived using an "expansion in small parameters," what do they generally mean? I am not too familiar with this term, and the only thing I can think of are Taylor expansions. I have seen expressions that contain functions of these parameters, and after the expansion in small parameters, the answer still contains functions of the parameters, but no nth power terms of the parameters (the $$(x-x_{0})^{n}$$ terms from Taylor expansions). Can someone help me understand how this is done? If needbe, I can give the examples I am referring to. I am posting this in this particular forum since the example has to do with neutrino oscillations. Let me know if I should post somewhere else.

2. Nov 27, 2009

ansgar

expansion in small parameters are general, this is not anything particular for neutrino oscillations, but yes one performs a Taylor expansion anyhow...

3. Nov 27, 2009

Dahaka14

I know it's a general method, but the example that uses it and confuses me is in neutrino oscillation probabilities. In a paper, specifically [A. Cervera et al., Nucl. Phys. B579, 17 (2000)] and other papers that discuss probabilities which include CP-violation and matter effects, the probability for the electron neutrino to antineutrino oscillation, expanded to second order in the small parameters $$\theta_{13},\Delta_{12}/\Delta_{13}~\text{and}~\Delta_{12}L$$, is
$$P_{\nu_{e}\rightarrow\nu_{\mu}}=\sin^{2}\theta_{23}\sin^{2}2\theta_{13}\sin^{2}\left(\frac{\Delta_{13}L}{2}\right)+\cos^{2}\theta_{23}\sin^{2}2\theta_{12}\sin^{2}\left(\Delta_{12}L}{2}\right)+\tilde{J}\cos\left(\pm\delta-\frac{\Delta_{13}L}{2}\right)\frac{\Delta_{12}L}{2}\sin\left(\frac{\Delta_{13}L}{2}\right)$$
where $$\tilde{J}\equiv\cos\theta_{13}\sin2\theta_{12}\sin2\theta_{23}\sin2\theta_{13}$$.
How is it that only functions of these parameters are left after Taylor expansion, and no linear or quadratic terms are left of say $$\theta_{13}$$?

4. Nov 27, 2009

Norman

Correct me if I am wrong, but is the first term in your equation for P, not quadratic in $$\theta_{13}$$?

Have you worked through the derivation yourself? The devil is always in the details. Unless someone familiar with the problem can argue about the probability without doing the derivation.

5. Nov 27, 2009

Dahaka14

I double checked the expression, and it should be quadratic. Take a look at equation 7 on page 7 in the arxiv version http://arxiv.org/PS_cache/hep-ph/pdf/0002/0002108v3.pdf" [Broken]. I took the inner product of the states and took absolute square, which has given me a large mess of terms. When I try to expand parts in $$\theta_{13}$$, I always end up with factors of $$\theta_{13}$$ and $$\sin^{2}2\theta_{13}$$, but there only appear to be factors of $$\sin^{2}2\theta_{13}$$ in their answer. There are other papers that perform similar approximations, and yet they never end up with things being multiplied by $$\theta_{13}$$.

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