- #1
Motocross9
- 12
- 4
- Homework Statement
- Given $$\varepsilon^{'}=\varepsilon+i\epsilon_{0}[G]$$
$$(\varepsilon^{'}$$ is a hermitian, second rank tensor) show that
$$\varepsilon^{'-1}=\varepsilon^{-1}-i\epsilon_{0}\varepsilon^{-1}[G]\varepsilon^{-1}$$, note that $$[G]$$ is small. Also, $$\varepsilon$$ is a diagonal second rank tensor, and $$[G]$$ is a real antisymmetric matrix.
- Relevant Equations
- $$\varepsilon^{'}=\varepsilon+i\epsilon_{0}*[G]$$
Clearly, they used the binomial expansion on this; however, I cannot figure out why [G] is sandwiched by the epsilon inverses:
$$\varepsilon^{'-1}=1/(\varepsilon+i\epsilon_{0}[G])\approx(1-i\epsilon_{0}[G]\varepsilon^{-1})\varepsilon^{-1}$$
$$\varepsilon^{'-1}=1/(\varepsilon+i\epsilon_{0}[G])\approx(1-i\epsilon_{0}[G]\varepsilon^{-1})\varepsilon^{-1}$$