1
$\begingroup$

When a mathematician says something is a scalar, say on the plane, they mean that it associates to points on the plane real numbers. When a physicist says something is a scalar, they mean that if we put a cartesian coordinate system on the plane and look at the value of this thing, and then rotate the coordinate system and look at the value again, the two values agree. But in the math definition there would be no reason to mention rotations, or the metric, or isometries of the metric, these are irrelevant to the definition of a scalar.

Now I know there are some things that are scalars under rotations but not under rotations + reflections, which would seem to mean that such a thing couldn't be considered a scalar in the math sense, because scalars in math are manifestly coordinate independent. However if we were only interested in rotations of the coordinate system, then a physicist would call it a scalar.

So it seems to me that physicists only talk about scalars with respect to certain group actions, which don't enter the picture when a mathematician talks about scalars, and this leaves me a bit confused as to in what sense they are the same.

$\endgroup$
2
  • $\begingroup$ i would love to see some nice answer to this. I used to think that scalars are numbers, like potential or moment of inertia. Opposite is a vector which, well, a set of scalars. OP: is there real misunderstanding for you, or it is just curiosity? $\endgroup$ Commented Jun 17, 2015 at 6:46
  • $\begingroup$ @aaaaaa I think it's just a terminology issue. $\endgroup$
    – JLA
    Commented Jun 17, 2015 at 6:49

2 Answers 2

1
$\begingroup$

When talking about scalars, mathematicians usually use your definition, that is, something which doesn't vary with coordinate changes. (Basically, that there's some mapping to the actual points in space, in which the scalar is well defined)

When physicists talk about scalars, we usually refer to Lorentz scalars, which requires two things:

  1. Invariance under Lorentz transformations (Rotations and boosts)
  2. Invariance under parity transformation

If an object satisfies requirement #1, but not #2, it's not a scalar, it's a pseudoscalar.

These definitions of scalars in mathematics and physics are the same, but I've never seen a mathematician talk about pseudoscalars.

For some info on this, see this and this and this

$\endgroup$
6
  • $\begingroup$ So suppose an object satisfies 1. but not 2., then it can't be interpreted as a thing which doesn't vary with coordinate changes because otherwise it would have to satisfy 2...but what about every other possible coordinate transformation I can talk about that you haven't mentioned in 1. and 2.? What if object x is a scalar under 1. and 2., hence a scalar by the mathematicians definition, but then I introduce condition 3. which x doesn't satisfy? Then it wouldn't be a scalar under 1. 2. and 3., but then how could it be a scalar with the math definition? $\endgroup$
    – JLA
    Commented Jun 17, 2015 at 6:53
  • $\begingroup$ In addition to Lorentz scalars there are also SU(2) scalars, U(1) scalars, SO(3) scalars (the regular kind, very common), and various others. So I think saying that we usually refer to Lorentz scalars doesn't represent the state of affairs all that well. $\endgroup$
    – David Z
    Commented Jun 17, 2015 at 7:00
  • $\begingroup$ @David But is this not weird? Group actions have nothing to do with scalars in the mathematicians' sense. $\endgroup$
    – JLA
    Commented Jun 17, 2015 at 7:03
  • $\begingroup$ @JLA why would it be weird? We use a certain definition of scalar in physics; mathematicians use a different definition. $\endgroup$
    – David Z
    Commented Jun 17, 2015 at 7:05
  • $\begingroup$ @DavidZ Ok that's my question, are the definitions equivalent? $\endgroup$
    – JLA
    Commented Jun 17, 2015 at 7:08
1
$\begingroup$

Mathematicians don't talk about rotations or isometries or whatnot because they already know if they're talking about a scalar or something else; a physicist has to determine whether a physical quantity has the properties of a scalar or a vector or something else.

The easiest way to do that is to look at transformation laws--to devise some experiment (real or conceptual) that would distinguish between a scalar field and a vector field, or between a scalar field and a pseudoscalar field, and so on. Rotations and reflections are such conceptual "experiments."

It's true that physicists sometimes conflate true scalars and pseudoscalars (or gloss over the differences), but the underlying pure mathematics definition still works fine here. Physicists just tend to emphasize or focus on certain logical consequences of that concept.

$\endgroup$
8
  • $\begingroup$ But wouldn't a physicist have to check all possible coordinate transformations in order to know if the quantity was a scalar? Why just rotations and reflections? I know they are isometries, but what if I do some weird transformation that gives me something I wouldn't expect? $\endgroup$
    – JLA
    Commented Jun 17, 2015 at 7:11
  • $\begingroup$ I think considering more than those transformations is simply overkill. By considering general changes of basis, your experiment only gains information about whether the quantity lives in the framework of pseudo-Riemannian geometry at all. $\endgroup$
    – Muphrid
    Commented Jun 17, 2015 at 7:19
  • $\begingroup$ But doesn't this fact (about general coordinate transformations, or change of basis, or whatever) follow from the mathematician's definition? So how could they be equivalent?... $\endgroup$
    – JLA
    Commented Jun 17, 2015 at 7:24
  • $\begingroup$ Physicists just emphasize aspects of the definition that are most useful to them. I don't see two conflicting definitions here. I see one definition that some people tend to talk about rather formally and others more imprecisely. $\endgroup$
    – Muphrid
    Commented Jun 17, 2015 at 7:33
  • 1
    $\begingroup$ My position is that you should not take the physics "definition" literally, but it is instead a sloppy rephrasing of the mathematics definition. $\endgroup$
    – Muphrid
    Commented Jun 17, 2015 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.