Four-position: not really a four-vector?

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I'm not talking about the instantaneous-displacement four-vector dX=(cdt, dx, dy, dz), which surely is Lorentz-invariant. I'm talking about the so-called position four-vector, X=(ct, x, y, z). Isn't that just an arrow from an arbitrary origin to a fixed point? Unless all frames decide to use the same origin, which isn't needed for all the other four-vectors, then I don't see how it's Lorentz-invariant.
 

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PeterDonis
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I'm not talking about the instantaneous-displacement four-vector dX=(cdt, dx, dy, dz), which surely is Lorentz-invariant.
No, vectors in general are not Lorentz invariant--they are Lorentz covariant, i.e., their components transform in such a way that the spacetime "length" of the 4-vector is an invariant. In this case, the quantity ##c^2 dt^2 - dx^2 - dy^2 - dz^2## is an invariant--it's just the differential interval between two events separated by the displacement ##dX##.

I'm talking about the so-called position four-vector, X=(ct, x, y, z). Isn't that just an arrow from an arbitrary origin to a fixed point?
Yes. And its spacetime "length" will indeed depend on your choice of origin, so it is not Lorentz covariant in the way that the displacement vector above is.

(Note that some texts use the term "Lorentz transformation" only to refer to boosts and rotations, not translations of the origin; with this interpretation, strictly speaking, the "4-position" vector is Lorentz covariant. The broader term for the group of transformations that includes translations as well would be "Poincare transformation", and the 4-position would not be Poincare invariant in this terminology, while the displacement 4-vector would be.)
 
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  • #3
andrewkirk
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Position 'vectors' can only be thought of as vectors in flat space. Their definition is not as a vector but as a point on a manifold which, given a coordinate system, can be identified by four real numbers that are the coordinates. It is only possible to consider that as a vector if the space is flat, and that does not have any practical use.
Displacement vectors, on the other hand, do have a practical use in flat space, as you have noted.
 
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  • #4
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Yes, thanks for the terminological correction, PeterDonis. Of course I meant that the magnitude of the instantaneous-displacement vector is invariant, not the vector itself.

But it is true that, unlike the other common four-vectors introduced in SR (dX, U, P, etc.), the position four-vector has a magnitude that's invariant only if all frames agree on an origin?
 
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PeterDonis
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it is true that, unlike the other common four-vectors introduced in SR (dX, U, P, etc.), the position four-vector has a magnitude that's invariant only if all frames agree on an origin?
Yes. And also, as andrewkirk pointed out, the concept of "position four-vector" only makes sense in the first place in flat spacetime.
 
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Thanks andrewkirk and PD.
 
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pervect
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As far as I know, in special relativity displacement operators are 4-vectors. On a manifold, displacement operators do not in general form a vector space, due to the fact that they do not commute.
 
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