- #1
Stalafin
- 21
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I am trying to understand the derivation of the covariant derivative in Peskin/Schroeder (chapter 15.1, page 483).
This is the important stuff:
[tex]n^\mu\partial_\mu\psi=\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}\left[\psi(x+\epsilon n)-\psi(x)\right][/tex]
Scalar quantity: U(y,x):
[tex]U(y,x) \rightarrow e^{i\alpha(y)} U(y,x) e^{-i\alpha(x)}[/tex]
Conditions for zero separation and pure phase:
[tex]U(y,y)=1[/tex]
[tex]U(y,x)=exp[i\phi(y,x)][/tex]
Covariant derivative:
[tex]n^\mu D_\mu \psi = \lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon}\left[\psi(x+\epsilon n)-U(x+\epsilon n,x)\psi(x)\right][/tex]
Expansion of U(y,x) in the separation of the two points:
[tex]U(x+\epsilon n,x) = 1 - i e\epsilon n^\mu A_\mu(x) + \mathcal{O}(\epsilon^2)[/tex]
A_\mu(x) is a new vector field and is the coefficient of the displacement \epsilon n^\mu. Why?! I don't see how the author gets there.
This is the important stuff:
[tex]n^\mu\partial_\mu\psi=\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}\left[\psi(x+\epsilon n)-\psi(x)\right][/tex]
Scalar quantity: U(y,x):
[tex]U(y,x) \rightarrow e^{i\alpha(y)} U(y,x) e^{-i\alpha(x)}[/tex]
Conditions for zero separation and pure phase:
[tex]U(y,y)=1[/tex]
[tex]U(y,x)=exp[i\phi(y,x)][/tex]
Covariant derivative:
[tex]n^\mu D_\mu \psi = \lim_{\epsilon\rightarrow 0} \frac{1}{\epsilon}\left[\psi(x+\epsilon n)-U(x+\epsilon n,x)\psi(x)\right][/tex]
Expansion of U(y,x) in the separation of the two points:
[tex]U(x+\epsilon n,x) = 1 - i e\epsilon n^\mu A_\mu(x) + \mathcal{O}(\epsilon^2)[/tex]
A_\mu(x) is a new vector field and is the coefficient of the displacement \epsilon n^\mu. Why?! I don't see how the author gets there.
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