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
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.
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?
Edit: Went over the Wikipedia article, I think that cleared it up a bit - it’s not actually about a single particle being inverted in an otherwise unchanged system, but the whole system that you’re observing being inverted, is that correct? In that case, it would actually not matter what point is chosen as the origin, as the relative positions of everything would work out to be the same no matter the origin of the inversion. That makes a bit more sense then.
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
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.
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?
Edit: Went over the Wikipedia article, I think that cleared it up a bit - it’s not actually about a single particle being inverted in an otherwise unchanged system, but the whole system that you’re observing being inverted, is that correct? In that case, it would actually not matter what point is chosen as the origin, as the relative positions of everything would work out to be the same no matter the origin of the inversion. That makes a bit more sense then.