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Some dictionaries define a scalar as follows:

A quantity, such as mass, length, or speed, that is completely specified by its magnitude and has no direction. -- The Free Dictionary

However, it is my impression that in many contexts scalars can be signed, in which case their magnitude (their absolute value) does not specify its value. This definition is even used on a test question here. Is it true that this definition is inaccurate?

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Can you give some examples of the negative scalars? –  Kyle Kanos Aug 14 at 14:57
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Pressure, temperature, thermal expansion coefficients - these are all scalars which sometimes take negative values. Perhaps more obvious is the dot product, or the scalar projection. –  Doubt Aug 14 at 15:03
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Negative pressure and negative temperature are already borderline obscure (Temperatures are of course properly expressed in Kelvin, starting at 0K). –  MSalters Aug 14 at 15:17
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Example of negative scalar: charge. –  BMS Aug 14 at 16:57
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@MSalters, Jerry Schirmer: If pressure is defined as −⅓ times the trace of the stress tensor, then negative pressure is tension. You don't need anything as exotic as the Casimir force or dark energy. (Also, re the Casimir force, see hep-th/0503158.) –  benrg Aug 15 at 0:54

4 Answers 4

up vote 10 down vote accepted

The dictionary definition is wrong. For example, time is a scalar in Newtonian mechanics, and time can be negative. That means that time is not completely specified by its magnitude (absolute value). Other examples include charge, energy, and Celsius temperature.

The definition could be improved by cutting "is completely specified by its magnitude" and clarifying "direction" to be "direction in space." We'd then have this definition: a scalar is something that has no direction in space, i.e., if you rotate it, it doesn't change.

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"a scalar is a quantity that has no direction in space, i.e., if you rotate it, it doesn't change" seems even better to me (a sphere is not a scalar). Or, a scalar is an element of $\mathbb{R}$... –  anderstood Aug 15 at 2:50

You must always say with respect to what something is a scalar.

If we are given a group $G$, something is called a scalar if it is a member of the trivial representation of that group, i.e. if the (symmetry) group does nothing to it. Nothing more, nothing less.

In the most common situation, this means that a scalar is a scalar under the rotation group $\mathrm{SO}(3)$, and thus simply a real number instead of a vector or some matrix/tensor.

There are also pseudoscalars, which are scalars w.r.t. $\mathrm{SO}(3)$, but not w.r.t. to the full orthogonal group $\mathrm{O}(3)$.

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I suspect that the asker may not be aware of the usage of the word 'scalar' in more general settings (e.g. field theory), and that just taking scalar = real number is sufficiently accurate for his/her purposes. –  Danu Aug 14 at 15:16
    
Another example: Multiply an eigenvector of some system by a "scalar" (the concept of which is application dependent) and you get an eigenvector. Eigenvectors are typically represented as normalized vectors (magnitude=1). Multiplying by a "scalar" with a magnitude of one (e.g., -1, or i in the case of complex eigenvectors) still yields a normalized eigenvector. –  David Hammen Aug 14 at 15:30
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This is at the wrong level for the OP, who asked a freshman physics question. –  Ben Crowell Aug 14 at 16:59
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@BenCrowell: I do not disagree, yet it is a valid answer to a valid question. The Q&A model of SE means that we should strive to provide answers useful also to others, and not exclusively to the OP. I do not claim this is the "right" way to answer this question, but it is an answer, just as yours is. –  ACuriousMind Aug 14 at 17:16
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@ACuriousMind This isn't quite correct as one can see from an example. A field $\phi:\mathbb R^3\to \mathbb R$ is called an $\mathrm{SO}(3)$ scalar provided it is invariant under the standard action of that group on the set of such fields. The field is not an element of $\mathrm{SO}(3)$ itself. So the definition should really be in terms of an action of a group on a set, and elements of that set that are invariant under the group action. –  joshphysics Aug 14 at 18:11

From a geometric object perspective, a scalar is a rank 0 tensor and, as such, is invariant under rotations of the coordinate system.

The tensor contraction of, e.g., a one-form and a vector is a scalar, i.e., a real number.

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Yes, but this is a qualitatively different definition than the freshman-physics definition that is being asked about. E.g., by this definition, time isn't a scalar. –  Ben Crowell Aug 14 at 16:58
    
@BenCrowell, I started to write a response but I just noticed that ACuriousMind has responded to your similar comment there with essentially what I was about to write. –  Alfred Centauri Aug 14 at 17:21

You're simply reading too much into the word "magnitude." You want to translate it into technical terminology as "absolute magnitude," since the latter is often abbreviated to "magnitude" by physicists anyway.

But in everyday parlance "magnitude" is only trying to convey comparability. You have two things, each with their own magnitude, and the implication is that the magnitude of the one is equal to, greater than, or less than that of the other. This ordering property is exemplified by the real numbers. So "magnitude" → "real number" is better in certain contexts than "magnitude" → "absolute magnitude."

Your dictionary is entirely consistent (and consistent with the way I speak in the English vernacular). Note that for "magnitude" it gives

A number assigned to a quantity so that it may be compared with other quantities.

and

A property that can be described by a real number [...]

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