• hikaru755@feddit.de
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    1 year ago

    Can you expand on what “invert its spatial coordinates” means? From your example it seems you’re just flipping around the origin of your coordinate system, but since there is no fixed, “natural” reference frame that would provide a “true” origin, isn’t that origin completely arbitrary and the math should then work out with any origin you use? I feel like I’m missing something here

    • krzyz@szmer.info
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      1 year ago

      Not OP, but: it works similarly to looking at the system in a mirror. The clock’s hands turn, well, clockwise, but if you look at the mirror their movement is anticlockwise. Importantly, if you look at that mirror in another mirror, it will be clockwise again. Add yet another mirror and it’s anticlockwise.

      With a single mirror at position x=0 (and YZ plane), you invert “x” position, so (1, 1, 1) becomes (-1, 1, 1). “Inverting” the spatial coordinates ((x,y,z) -> (-x, -y, -z)) is effectively the same as looking at system through 3 mirrors, located at x = 0 (YZ plane), y = 0 (XZ plane) and z = 0 (XY plane), but that is a bit hard to visualize/arrange in practice so usually you would think of it as an equivalent operation of a point reflection around (0, 0, 0). You are right that the point is arbitrary: the important thing is, among others, that clockwise movement becomes anticlockwise.

    • Lvxferre
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      1 year ago

      As the other poster said, it’s like a mirror. If you mirror the system, and swap the particle with its antiparticle counterpart, you’ll get the same effect on the point that you decided “this is my origin”. Like this:

      Here the origin is the tip of the pencil. It’s completely arbitrary, but it should still hold true in most cases. The exceptions are the CP-violations.