# I Why are scalar fields Lorentz invariant?

Tags:
1. Aug 18, 2016

### voila

Hi. This question most probably shows my lack of understanding on the topic: why are scalar fields Lorentz invariant?

Imagine a field T(x) [x is a vector; I just don't know how to write it, sorry] that tells us the temperature in each point of a room. We make a rotation in the room and now the field is written as T'(x'), where T'(x')=T(x) keeps telling us the temperature at each tile in the room. So, the field is invariant.

Now a different example: picture the same room filled with current densities. We have an scalar field J(x) that gives the modulus of the current density at each point. We make an active Lorentz transformation of the room, so our field now becomes J'(x'). By definition, J'(x')=J(x), but the current densities also change at each tile with a Lorentz transformation, so the field should be giving us different numbers. In other words seems to me that this scalar field is not Lorentz invariant.

2. Aug 18, 2016

### Staff: Mentor

Invariant under rotations, yes. But not necessarily invariant under Lorentz transformations. See below.

No. What this is telling you is that current density is not a scalar field. It's a component of a 4-vector field which includes charge density as well as current density in each direction of space.

(A proper relativistic definition of "temperature" is somewhat problematic; as noted above, the "naïve" definition of temperature, as average kinetic energy of molecules, is not actually Lorentz invariant, because energy itself isn't. You can define a scalar that would be described in ordinary language as something like "temperature of the room as measured by someone at rest relative to the room".)

3. Aug 18, 2016

### dextercioby

Well, speaking about semantics, the question from the thread's title (repeated in the first line of the original post) is tautological: If a field is invariant with respect to a group of (space-time) transformations, it is called/defined as scalar. If, on the contrary, it is not constant, then we can call it tensorial/spinorial field [spinors are a particular type of tensors].

4. Aug 18, 2016

### voila

Oh, I see. So scalar fields are those defined as invariant, not anything that returns a scalar value based on position. Would any function of coordinates (the current density wasn't) be then a scalar field?

5. Aug 19, 2016

### haushofer

No. A scalar field is defined to behave under a (in this case Lorentz) transformation $x \rightarrow x'(x)$ as
$\phi(x) \rightarrow \phi'(x') = \phi(x)$
So you have two effects: the argument of the field is changed, but also the functional form of the field: watch that prime on the field! This is a non-trivial characteristic of a field. Just because something does not have indices does not mean it is a scalar field! The same goes for tensors in general, e.g. the connection component in general relativity has three indices but does not transform as a tensor (per definition).

6. Aug 19, 2016

### voila

What I meant, @haushofer, is whether a scalar field would be any function on the position (event) coordinates that returns a scalar, so that when you transform the space on which it acts, you also transform the field.

Example: f(x,y)=x+y
Now y'=y; x'=2x; f'(x',y')=x'/2+y'

For clearance, what I want to clarify is if what defines a scalar field is that it acts upon coordinates, in contrast with the current density example I wrote above, which assigns values to coordinates but is not a function of them.

7. Aug 19, 2016

### vanhees71

Yes, a scalar [*] field is a quantity that's defined as a function of time and position and is a scalar. An example is temperature (although this is a quite tricky example, as can be seen in another thread in this forum).

[*] Corrected after the hint of a typo by PeterDonis in #8.

Last edited: Aug 20, 2016
8. Aug 19, 2016

### Staff: Mentor

A vector field is not a scalar. (The norm of a vector is a scalar, but that's not the same thing.) Did you make a typo here?

9. Aug 20, 2016

### vanhees71

Argh, of course, I meant to write a scalar field s a quantity that's defined as a function of time and position and is a scalar. I'll correct it in the original posting.

10. Aug 20, 2016

### haushofer

I would say no. Consider e.g. the norm of a three-vector in spacetime. This norm looks like a scalar, but it is not under all of the Lorentz boosts (length contraction!) In your example, you explicitly define your f to be a scalar field, so then I agree with you. If one talks about tensors in general, one should state with respect to what group of transformations. E.g., in Galilean relativity, the norm of a three-vector is a scalar unde the Galilean group.

Hope this helps and does not confuse you.

11. Aug 20, 2016

### voila

Yes, actually just thought about that after posting it but forgot to clarify it in this thread. Thanks for stating it, though. So scalar fields are just functions of coordinates that are invariant under the set of transformations you consider?

12. Aug 21, 2016

### haushofer

Yes. A scalar field as I understand it not 'just' a function, but a field whose functional change (the prime on the field!) is such that every observer (every set of coordinates) agrees on its value. As such it has extra structure compared to functions.