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In 3+1 dimensional spacetime the parity transformation is $$P^\mu_{\;\,\nu}=\begin{pmatrix}+1&&&\\&-1&&\\&&-1&\\&&&-1\end{pmatrix}.$$ This is orthochronous but not proper and thus is not the result of compounding infinitesimals.

However, in $(2n)+1$ dimensional spacetime, the parity transformation will have determinant one and thus be proper orthochronous.

My (naive) question: what physical consequences does this have in evenly spatial dimensional spacetimes? I could not find any references to this seemingly profound phenomenon after preliminary Googling.

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    $\begingroup$ It just means that this is not the proper way to define the parity transformation when there are an even number of space dimensions. You can do -1 for one spatial coordinate and +1 for the others. That works in all dimensions. $\endgroup$ – Brian Bi Mar 13 '15 at 19:01
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Parity and Time reversal are by definition elements of the full lorentz group with which you need to supplement the proper orthochronous subgroup in order to be able to span the entire group.

As noted, the proper way to define parity in any dimension is to flip one of the spatial axes. In even space-time dimensions it so happens that flipping all spatial coordinates and flipping one are related through some sequence of rotations, as should be the case for any two improper transformations, since they both have det$=-1$ and should be connected.

For odd space-time dimensions, flipping all spatial dimension is actually just a rotation (or a sequence thereof). For example take $SO(1,2)$, then flipping all spatial coordinates $$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right)$$ is equivalent to a rotation about the only axis by and angle of $\phi =\pi$ radians, since a rotation is $$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \text{cos }\phi & -\text{sin }\phi \\ 0 & \text{sin }\phi & \text{cos }\phi \end{array}\right)$$ However either $$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \pm 1 & 0 \\ 0 & 0 & \mp 1\end{array}\right)$$ cannot be achieved by a rotation, and has the correct determinant.

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  • $\begingroup$ So in other words, the transformation $$P=\begin{pmatrix}1&&&\\&-1&&\\&&-1&\\&&&-1\end{pmatrix}$$ isn't a parity transformation at all, but a spatial rotation. Interesting - according to Srednicki it is the parity transformation but perhaps the notation is loose. $\endgroup$ – theage Mar 13 '15 at 20:32
  • $\begingroup$ No the one you just wrote is in an even space time dimension, and has negative determinant. So it is a parity transformation. $\endgroup$ – Ali Moh Mar 13 '15 at 20:33

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