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fixed a typo.
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Frederic Thomas
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A four vector $V^\mu$ has componentsthatcomponents that we can regard as $(\rho, {\bf j})$. Under parity the charge $\rho$ does not change, but the current changes sign: $P:{\bf j}\mapsto -{\bf j}$. If the metric signature is $(+,-,-,-)$ one way to say this is $V^\mu \to V_\mu$. If the metric were $(-,+,+,+)$ we would have to say $V^\mu\to -V_\mu$.

A four vector $V^\mu$ has componentsthat we can regard as $(\rho, {\bf j})$. Under parity the charge $\rho$ does not change, but the current changes sign: $P:{\bf j}\mapsto -{\bf j}$. If the metric signature is $(+,-,-,-)$ one way to say this is $V^\mu \to V_\mu$. If the metric were $(-,+,+,+)$ we would have to say $V^\mu\to -V_\mu$.

A four vector $V^\mu$ has components that we can regard as $(\rho, {\bf j})$. Under parity the charge $\rho$ does not change, but the current changes sign: $P:{\bf j}\mapsto -{\bf j}$. If the metric signature is $(+,-,-,-)$ one way to say this is $V^\mu \to V_\mu$. If the metric were $(-,+,+,+)$ we would have to say $V^\mu\to -V_\mu$.

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mike stone
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A four vector $V^\mu$ has componentsthat we can regard as $(\rho, {\bf j})$. Under parity the charge $\rho$ does not change, but the current changes sign: $P:{\bf j}\mapsto -{\bf j}$. If the metric signature is $(+,-,-,-)$ one way to say this is $V^\mu \to V_\mu$. If the metric were $(-,+,+,+)$ we would have to say $V^\mu\to -V_\mu$.