Graduate Vector field vs. four scalar fields ("QFT and the SM", Schwartz)

[Hill]

TL;DR

Four components of the massive spin 1 field

This is the statement in question:

But if they were scalar fields, they would not transform at all. How could they contribute differently if they didn't change?

 

[Demystifier]

In this interpretation you change them by hand, not as a result of vector transformation.

 

[Hill]

In this interpretation you change them by hand, not as a result of vector transformation.

Then, as I understand, they are not "scalar fields", but rather just numbers.

 

[martinbn]

Then, as I understand, they are not "scalar fields", but rather just numbers.

If you have a global coordinate system/frame then you can think of them as scalar fields. But in a different frame there will be a different set of four fields. It is a bit sloppy. Which book is this?

 

[Hill]

If you have a global coordinate system/frame then you can think of them as scalar fields. But in a different frame there will be a different set of four fields. It is a bit sloppy. Which book is this?

"QFT and the SM" by Schwartz.

 

[Demystifier]

Then, as I understand, they are not "scalar fields", but rather just numbers.

Yes, you can say it so.

 

[PeterDonis]

Then, as I understand, they are not "scalar fields", but rather just numbers.

More precisely, a scalar field is a mapping of numbers to points in spacetime. The number assigned to a given point doesn't change when you change coordinates.

The statement you quote in the OP is saying that once you pick a coordinate chart, you can find four scalar fields on spacetime that give the same numbers at each point as the four components of $A_\mu$. But if you change coordinates, as @martinbn said, you now have to find a different set of four scalar fields on spacetime that give the same numbers at each point as the four components of $A_\mu$. The usual vector transformation law, on this view, is just a constraint on how the different sets of four scalar fields that you get in different coordinate charts are related.

 

[Hill]

More precisely, a scalar field is a mapping of numbers to points in spacetime. The number assigned to a given point doesn't change when you change coordinates.

The statement you quote in the OP is saying that once you pick a coordinate chart, you can find four scalar fields on spacetime that give the same numbers at each point as the four components of $A_\mu$. But if you change coordinates, as @martinbn said, you now have to find a different set of four scalar fields on spacetime that give the same numbers at each point as the four components of $A_\mu$. The usual vector transformation law, on this view, is just a constraint on how the different sets of four scalar fields that you get in different coordinate charts are related.

Thank you. I get this. My point is, that such arrangement contradicts the earlier definition,

 

[PeterDonis]

such arrangement contradicts the earlier definition

No, it doesn't. Read this again, carefully:

if you change coordinates, as @martinbn said, you now have to find a different set of four scalar fields on spacetime

 

[Hill]

No, it doesn't. Read this again, carefully:

Lorentz transformation changes coordinates and, by the definition, should not affect the scalar fields.
I understand that your emphasis is: the fields in this case do not change but are rather forgotten and replaced. OK, but why to call them "scalar fields" then, and not just "functions of space-time"? I understood that the rest of the definition, i.e. "... that are Lorentz invariant etc.", distinguishes such functions as being "scalar fields".

 

[PeterDonis]

Lorentz transformation changes coordinates and, by the definition, should not affect the scalar fields.

And it doesn't.

I understand that your emphasis is: the fields in this case do not change but are rather forgotten and replaced.

Yes. Which means the coordinate transformation has not affected any scalar fields, as it shouldn't.

why to call them "scalar fields" then, and not just "functions of space-time"?

Because "scalar fields" is the term that physicists have used for a long time in this context. Yes, it means basically the same thing as "scalar function on spacetime".

I understood that the rest of the definition, i.e. "... that are Lorentz invariant etc.", distinguishes such functions as being "scalar fields".

No. Any scalar function on spacetime, i.e., any mapping of numbers to points in spacetime, has to be Lorentz invariant, by construction: there is simply nothing for a coordinate transformation to change.