As far as I can tell, there’s actually no mathematical difference between a vector field in 3D and a 3-scalar field that assigns a 3D scalar to each point.
The difference is in how they transform under coordinate changes. To physicists, a vector field is defined by how it transforms. So this:
You can do replace the vector field V(x) with a 3-scalar field and see the same thing
is not correct; by definition, a 3-scalar field should transform trivially under coordinate changes.
Reading the wikipedia page on scalar field, I think I understand the confusion here. Scalar fields are supposed to be invariant under changes in reference frame assuming a canonical coordinate system for space.
Take two reference frames P(x) and G(x). A scalar field S(x) needs to satisfy:
S(x) = P’(x)S(x)P(x) = G’(x)S(x)G(x)
Where P’(x) is the inverse of P(x) and G’(x) is the inverse of G(x).
Meaning the inference of S(x) should not change with reference frame. A scalar field is a vector field that commutes with perspective transformations. Maybe that’s what you meant?
I wouldn’t use the phrase “transforms trivially” here since a “trivial transformation” usually refers to the identity transformation. I wouldn’t use a head tilt example either since a lot of vector fields are going to commute with spatial rotations, so it’s not good for revealing the differences. And I think you got the association backwards in your original explanation: scalar fields appear to represent quantities in the underlying space unaffected by head tilts, and so they would be the ones “transforming in the opposite direction” in the analogy since they would remain fixed in “canonical space”.
I wouldn’t use the phrase “transforms trivially” here since a “trivial transformation” usually refers to the identity transformation
No, I do mean the identity transformation. Scalar fields do not transform at all under coordinate changes. To be precise, if we have a coordinate change matrix P, a scalar field f(x) transforms like f′(x)=f(Px)
Whereas a vector field v(x) transforms like
v′(x)=P−1v(Px)
Ah. Thank you, that is perfectly clear. The Wikipedia page for Scalar Field makes sense with that too. A scalar field is a function that takes values in some canonical units, and so it transforms only on the right of f under a perspective shift. A vector field (effectively) takes values both on and in the same space, and so it transforms both on the left and right of v under a perspective shift.
Interesting. That seems to contradict the explanation for Lie Algebras, and it seems incompatible with commutators in general, since with commutators all operators involved need to be compatible with both composition and precomposition (otherwise AB—BA is undefined). I guess scalar fields are not meant to be operators? That doesn’t quite work since they’re supposed used to describe energy, which is often represented as an operator. In any case, I’ll have to keep that in mind when reading about these things.
The difference is in how they transform under coordinate changes. To physicists, a vector field is defined by how it transforms. So this:
is not correct; by definition, a 3-scalar field should transform trivially under coordinate changes.
Reading the wikipedia page on scalar field, I think I understand the confusion here. Scalar fields are supposed to be invariant under changes in reference frame assuming a canonical coordinate system for space.
Take two reference frames P(x) and G(x). A scalar field S(x) needs to satisfy:
S(x) = P’(x)S(x)P(x) = G’(x)S(x)G(x)
Where P’(x) is the inverse of P(x) and G’(x) is the inverse of G(x).
Meaning the inference of S(x) should not change with reference frame. A scalar field is a vector field that commutes with perspective transformations. Maybe that’s what you meant?
I wouldn’t use the phrase “transforms trivially” here since a “trivial transformation” usually refers to the identity transformation. I wouldn’t use a head tilt example either since a lot of vector fields are going to commute with spatial rotations, so it’s not good for revealing the differences. And I think you got the association backwards in your original explanation: scalar fields appear to represent quantities in the underlying space unaffected by head tilts, and so they would be the ones “transforming in the opposite direction” in the analogy since they would remain fixed in “canonical space”.
No, I do mean the identity transformation. Scalar fields do not transform at all under coordinate changes. To be precise, if we have a coordinate change matrix P, a scalar field f(x) transforms like f′(x)=f(Px)
Whereas a vector field v(x) transforms like v′(x)=P−1v(Px)
For more details check out these wikipedia pages.
Ah. Thank you, that is perfectly clear. The Wikipedia page for Scalar Field makes sense with that too. A scalar field is a function that takes values in some canonical units, and so it transforms only on the right of f under a perspective shift. A vector field (effectively) takes values both on and in the same space, and so it transforms both on the left and right of v under a perspective shift.
I updated my first reply to point to yours.
Interesting. That seems to contradict the explanation for Lie Algebras, and it seems incompatible with commutators in general, since with commutators all operators involved need to be compatible with both composition and precomposition (otherwise AB—BA is undefined). I guess scalar fields are not meant to be operators? That doesn’t quite work since they’re supposed used to describe energy, which is often represented as an operator. In any case, I’ll have to keep that in mind when reading about these things.