This question arose in me after watching https://youtu.be/mmtLgYVEuJs?t=394. The link has a time in it so it takes you to the part I am talking about in the video.
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1Possible duplicates: Is anti-matter matter going backwards in time? and links therein. – Qmechanic Jun 02 '22 at 10:47
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1@Qmechanic I don't think this is a duplicate I didn't say anything about time. Couldn't it be possible that there are more dimensions beyond just time? Did you watch the video? – Quinten C Jun 03 '22 at 09:14
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1What does "inverted in a higher dimension" mean? – WillO Jun 03 '22 at 11:59
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@WillO see the video. It's a rotation through the higher dimension which is basically a parity transformation. – John Rennie Jun 03 '22 at 15:51
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@JohnRennie A rotation and a parity transformation are unrelated sorts of transformation, so it's difficult to give any meaning to this. A rotation preserves the orientation of space, while a parity transformation inverses it. – Miyase Jun 03 '22 at 18:07
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The simple answer is "no", though it's an interesting question and there is more to it than this simple answer would suggest.
Matter and antimatter are related by a charge parity transformation, often abbreviated to $CP$. The charge transformation changes the sign of the charge, and the parity transformation is reflection in a mirror. So to change an electron into a positron you need to both change the charge from $-e$ to $+e$ and also swap the parity.
In the video you link the presenter shows how in two dimensions the letter R can be flipped by lifting it up out of the two dimensions and rotating it:
This is indeed a parity transformation, so it does the $P$ part, and it would also work with a 3D object lifted into a fourth spatial dimension and rotated. However it wouldn't turn matter into antimatter because the process does not do the charge transformation.
John Rennie
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As for you last paragraph, this would be possible in any dimension: here you are looking for a rotation (determinant 1) in $\mathbb R^4$ that restricts to $(x,y,z,0)\mapsto (x,y,-z,0)$ (or any other element of $O(3)$ of determinant -1) on $\mathbb R^3 \times {0}$. This can be realized by moving through $(x,y,z,w)\mapsto(x,y,z\cos\alpha + w\sin\alpha, z\sin\alpha - w\cos\alpha)$ – doetoe Jun 03 '22 at 17:34
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