Sunday, 26 July 2015

general relativity - D'Alembertian for a scalar field


I have read that the D'Alembertian for a scalar field is =gνμνμ=1gμ(gμ).




  1. Exactly when is this correct? Only for ϕ where ϕ is a scalar-field?





  2. How exactly is it shown?




  3. Is it true only on-shell, from the Euler-Lagrange equations of a scalar-field?





Answer



This is based on the observation that, given some vector Vμ,



μVμ=1gμ(gVμ)


We can show explicitly that this is true:


μVμ=μVμ+ΓμμλVλ


Let's examine the last term:


Γμμλ=12gμρ(μgλρ+λgμρρgλμ)

we can cancel the first and third term on the right hand side, yielding Γμμλ=12gμρλgμρ


The idea is to show that this equals 1gλg:


λg=12gλg=12g|gμν+δgμν||gμν|δxλ=12g|gμν||I+(gμν)1λgμνδxλ||gμν|δxλ=12g|gμν|(1+Tr((gμν)1λgμνδxλ))|gμν|δxλ=g12(gμνλgμν)

Here, I used |gμν| to denote the determinant of gμν. Multiplying by 1g and comparing shows that Γμμλ=1gλg=λlng
It follows that μVμ=1gμ(gVμ)
Thus, you see that your formula of the d'Alembertian holds only for scalars: the first covariant derivative reduces to a partial derivative, and for μϕ we appeal to the formula I derived. Note, however, that we never used the equations of motion. This is all just a much longer and more explicit version of what Qmechanic already posted while I was typing this up.


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