• 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.

      • hikaru755@feddit.de
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        1 year ago

        Thanks for the reply! I’m not sure I fully got that, though. So it seems to be that it’s not actually about position (the absolute coordinates), but about the velocity of the particle? So, you could just always use a coordinate system that has the particle at its origin so that its position doesn’t need to change, and just invert the vector of its velocity to get the same result?