- #1

etotheipi

##\mathcal{P}## is a point in Minkowski spacetime ##M##, and ##\varphi_1: U \in M \mapsto \mathbb{R}^4## and ##\varphi_2: U \in M \mapsto \mathbb{R}^4## are two coordinate systems on the spacetime. A scalar field is a function ##\Phi(\mathcal{P}): M \mapsto \mathbb{R}##, and we can define coordinate representations of this scalar field as the compositions ##\phi = \Phi \circ \varphi_1^{-1}## and ##\phi' = \Phi \circ \varphi_2^{-1}##. Since the value of a scalar field at some point ##\mathcal{P}## is coordinate independent,$$\Phi(\mathcal{P}) = \phi(x) = \phi'(x')$$where ##x \in \mathbb{R}^4## and ##x' \in \mathbb{R}^4## are the coordinate vectors of ##\mathcal{P}## w.r.t. the two different coordinate systems. The Lorentz transformation between the coordinates of ##\mathcal{P}## in the two different frames is$$x' = \Lambda x \iff \Lambda^{-1} x' = x$$we can just substitute that into obtain$$\phi'(x) = \phi(\Lambda^{-1} x)$$Is that an active or a passive transformation? What would the other type look like? And most importantly, why is there any difference between them: don't we just always need to satisfy ##\Phi(\mathcal{P}) = \phi(x) = \phi'(x')##?

Maybe for an active transformation you need to actually move the manifold, and by extension the point ##\mathcal{P}##, underneath the coordinate systems? I have no idea, it's quite confusing.

Maybe for an active transformation you need to actually move the manifold, and by extension the point ##\mathcal{P}##, underneath the coordinate systems? I have no idea, it's quite confusing.

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