EDIT: This post is incorrect. See the reply chain below. After correcting my misunderstanding, I agree with your explanation.
The difference you’re describing between vector fields and scalar fields, mathematically, is the difference between composition and precomposition. Here it is more precisely:
Pick a change-of-perspective function P(x). The output of P(x) is a matrix that changes vectors from the old perspective to the new perspective.
You can apply the change-of-perspective function either before a vector field V(x) or after a vector field. The result is either V(x)P(x) or P(x)V(x).
If you apply P(x) before, the vector field applies a flow in the new perspective, and so its arrows “tilt with your head.”
If you apply P(x) after, the vector field applies a flow in the old perspective, and so the arrows don’t tilt with your head.
You can do replace the vector field V(x) with a 3-scalar field and see the same thing.
Since both composition and precomposition apply to both vector fields and scalar fields in the same way, that can’t be something that makes vector fields different from scalar fields.
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. It’s just a choice of language. Any difference comes from context. Typically, vector fields are treated like flows (though not always), whereas scalar fields have no specific treatment.
Spinors are represented as vectors in very specific spaces, specifically spaces where there’s an equivalence between matrices and spatial operations. Since a vector is something like the square root of a matrix, a spinor is something like the square root of a spatial operation. You get Dirac Spinors (one specific kind of spinor) from “taking the square root of Lorentz symmetry operations,” along with scaling and addition between them.
As far as spinors go, I think I prefer your Lorentz Group explanation for the “what” though I prefer my Clifford Algebra one for the “how”. The Lorentz Group explanation makes it clear how to find important spinors. For me, the Clifford Algebra makes it clear how the rest of the spinors arise from those important spinors, and it makes it clear that they’re the “correct” representation when you want to sum spatial operations, as you would with wavefunctions. It’s interesting that the intuition doesn’t transfer as I expected. I guess the intuition transfer problem here is more difficult than I expected.
Note: Your generalization only accounts for unit vectors, and spinors are NOT restricted to unit vectors. They can be scaled arbitrarily. If they couldn’t, ψ†ψ would be uniform at every point. You probably know this, but I wanted to make it explicit.
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.
EDIT: This post is incorrect. See the reply chain below. After correcting my misunderstanding, I agree with your explanation.
The difference you’re describing between vector fields and scalar fields, mathematically, is the difference between composition and precomposition. Here it is more precisely:
Pick a change-of-perspective function P(x). The output of P(x) is a matrix that changes vectors from the old perspective to the new perspective.
You can apply the change-of-perspective function either before a vector field V(x) or after a vector field. The result is either V(x)P(x) or P(x)V(x).
If you apply P(x) before, the vector field applies a flow in the new perspective, and so its arrows “tilt with your head.”
If you apply P(x) after, the vector field applies a flow in the old perspective, and so the arrows don’t tilt with your head.
You can do replace the vector field V(x) with a 3-scalar field and see the same thing.
Since both composition and precomposition apply to both vector fields and scalar fields in the same way, that can’t be something that makes vector fields different from scalar fields.
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. It’s just a choice of language. Any difference comes from context. Typically, vector fields are treated like flows (though not always), whereas scalar fields have no specific treatment.
Spinors are represented as vectors in very specific spaces, specifically spaces where there’s an equivalence between matrices and spatial operations. Since a vector is something like the square root of a matrix, a spinor is something like the square root of a spatial operation. You get Dirac Spinors (one specific kind of spinor) from “taking the square root of Lorentz symmetry operations,” along with scaling and addition between them.
As far as spinors go, I think I prefer your Lorentz Group explanation for the “what” though I prefer my Clifford Algebra one for the “how”. The Lorentz Group explanation makes it clear how to find important spinors. For me, the Clifford Algebra makes it clear how the rest of the spinors arise from those important spinors, and it makes it clear that they’re the “correct” representation when you want to sum spatial operations, as you would with wavefunctions. It’s interesting that the intuition doesn’t transfer as I expected. I guess the intuition transfer problem here is more difficult than I expected.
Note: Your generalization only accounts for unit vectors, and spinors are NOT restricted to unit vectors. They can be scaled arbitrarily. If they couldn’t, ψ†ψ would be uniform at every point. You probably know this, but I wanted to make it explicit.
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.