It might be a naive question but i was wondering what a derivative can do regarding spin. If there is a Riemann scalar it is clear that its an invariant object under tensor transformation and it does not have any spin, i.e. spin-0. However, if i consider its derivative, its (tensor) transformation properties should change rendering it an indexed form, and its parity should also change. Hence, i am guessing that it should be a spin-1 object. So the question is; is it really the case that a (1st) space derivative of a spin-0 form should be spin-1.
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$\begingroup$ It really depends on your definitions. As always, there are aspects of “spin” that are objective / physically measurable, and other aspects that are an arbitrary convention. It’s not always easy to differentiate between the two. The derivative transforms as a covariant vector under GCTs and Lorentz transformations. $\endgroup$– Prof. LegolasovCommented Jun 14, 2023 at 6:37
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$\begingroup$ I think this is an interesting question. For example there are fractional derivatives in mathematics such as Riemann-Liouville, I wonder if there could be an analogous derivative that increases or decreases spin in steps of 1/2. $\endgroup$– FridoCommented Jun 14, 2023 at 7:19
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$\begingroup$ @Frido clearly not. For starters, the number of components doesn't match. $\endgroup$– Prof. LegolasovCommented Jun 15, 2023 at 15:36
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$\begingroup$ @Prof.Legolasov You're very likely right, but if I have a scalar function of objects that transform as spinors I was wonderinf if I could define an appropriate derivative such that the derivative acting on the scalar function gives a spinor. $\endgroup$– FridoCommented Jun 15, 2023 at 16:47
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