Scalar

In physics, a scalar is often just a real number (element of ℝ) or a complex number (element of ℂ).

Most physical quantities belong to—or are associated with—some linear space with inner product, and their algebraic form or value is—explicitly or implicitly—defined with respect to an orthonormal basis of this space. When another orthonormal basis of the space is chosen, the appearance of the physical quantity expressed with respect to the new reference frame (i.e., basis) is in general different from the algebraic form or value of the quantity with respect to the previous basis. One expresses this by stating that the physical quantity transforms under the change of reference frame.

The second kind of scalars occurring in physics are quantities that are invariant under basis transformation. That is, unlike most physical quantities, scalars do not change their form or value under a basis transformation. It is common to call such invariants proper scalars to distinguish them from pseudo-scalars. The latter are scalars that transform into minus themselves under inversion of the basis; proper scalars are invariant under inversion.

With respect to a given reference frame and expressed in a suitable set of units, such as SI units, a scalar is represented by a single, real or complex, number.

Example

Examples of scalars of the first kind (no reference to a frame) are the number 3.14, 2exp[-i 300] and the elementary charge in SI units.

For an example of scalars of the second kind, we consider two elements, ${\vec {r}},\;{\vec {s}}$ in ℝ³. These vectors may have some physical meaning, direction and strength of an electric and magnetic field, for instance. Let the following be a basis of ℝ³,

$\left({\vec {e}}_{1},\;{\vec {e}}_{2},\;{\vec {e}}_{3}\right)\quad {\hbox{then}}\quad {\vec {r}}=\left({\vec {e}}_{1},\;{\vec {e}}_{2},\;{\vec {e}}_{3}\right)\,\mathbf {r} ,\quad {\hbox{and}}\quad {\vec {s}}=\left({\vec {e}}_{1},\;{\vec {e}}_{2},\;{\vec {e}}_{3}\right)\,\mathbf {s} ,$

where the column vectors (bold) are

$\mathbf {r} \equiv {\begin{pmatrix}r_{1}\\r_{2}\\r_{3}\end{pmatrix}}\quad {\hbox{and}}\quad \mathbf {s} \equiv {\begin{pmatrix}s_{1}\\r_{s}\\s_{3}\end{pmatrix}},\;$

and the matrix-matrix product is implied,

$\left({\vec {e}}_{1},\;{\vec {e}}_{2},\;{\vec {e}}_{3}\right)\,\mathbf {r} \equiv r_{1}{\vec {e}}_{1}+r_{2}{\vec {e}}_{2}+r_{3}{\vec {e}}_{3}.$

The column vectors r and s represent algebraically the physical quantities ${\vec {r}},\;{\vec {s}}$ with respect to the given basis.

Choose now another basis of the same space

${\big (}{\vec {f}}_{1},\;{\vec {f}}_{2},\;{\vec {f}}_{3}{\big )}\quad {\hbox{with}}\quad {\big (}{\vec {e}}_{1},\;{\vec {e}}_{2},\;{\vec {e}}_{3}{\big )}={\big (}{\vec {f}}_{1},\;{\vec {f}}_{2},\;{\vec {f}}_{3}{\big )}\mathbf {B}$

The 3×3 matrix B is assumed to be an orthogonal matrix (transforms orthonormal bases into each other). Clearly

${\vec {r}}={\big (}{\vec {f}}_{1},\;{\vec {f}}_{2},\;{\vec {f}}_{3}{\big )}\,\mathbf {r'} \quad {\hbox{and}}\quad {\vec {s}}={\big (}{\vec {f}}_{1},\;{\vec {f}}_{2},\;{\vec {f}}_{3}{\big )}\,\mathbf {s'} \quad {\hbox{with}}\quad \mathbf {r'} \equiv \mathbf {Br} \quad {\hbox{and}}\quad \mathbf {s'} \equiv \mathbf {Bs} .$

The columns r', s' represent the same vectors ${\vec {r}},\;{\vec {s}}$ with respect to the new basis. They have another form, or more accurately, their three components have new values.

To construct a proper scalar (transformation invariant) kind we consider the inner product. The inner product is not a member of ℝ³, but is clearly associated with it. (Formally it is a bilinear map of the Cartesian product ℝ³×ℝ³ into ℝ). The inner product satisfies

${\vec {r}}\cdot {\vec {s}}=\mathbf {r} ^{\mathrm {T} }\,\mathbf {s} =\mathbf {r'} ^{\mathrm {T} }\,\mathbf {s'} .$

The second equality follows thus,

${\begin{aligned}\mathbf {r} ^{\mathrm {T} }\,\mathbf {s} &={\big (}r_{1}\,r_{2},\,r_{3}{\big )}{\begin{pmatrix}s_{1}\\s_{2}\\s_{3}\\\end{pmatrix}}=\sum _{i=1}^{3}r_{i}s_{i}=\sum _{i=1}^{3}\left(\mathbf {B} ^{-1}\mathbf {r'} \right)_{i}\left(\mathbf {B} ^{-1}\mathbf {s'} \right)_{i}\\&=\sum _{i}\left[\sum _{j}{\big (}B^{-1}{\big )}_{ij}r'_{j}\right]\;\left[\sum _{k}{\big (}B^{-1}{\big )}_{ik}s'_{k}\right]=\sum _{ijk}B_{ki}{\big (}B^{-1}{\big )}_{ij}r'_{j}s'_{k}=\sum _{jk}\delta _{kj}r'_{j}s'_{k}=\sum _{j=1}^{3}r'_{j}s'_{j}=\mathbf {r'} ^{\mathrm {T} }\,\mathbf {s'} \end{aligned}}$

Here it is used that B is orthonormal,

$\mathbf {B} ^{-1}=\mathbf {B} ^{\mathrm {T} }\quad {\hbox{and}}\quad \mathbf {B} ^{\mathrm {T} }\mathbf {B} =\mathbf {E} \quad {\hbox{and}}\quad E_{ij}=\delta _{ij},$

where δ_ij is the Kronecker delta. The inner product of ${\vec {r}},\;{\vec {s}}$ does not depend on the basis used to express the vectors, it is a proper scalar.

As an example of a pseudo-scalar the following triple product may be given

${\vec {r}}\cdot \left({\vec {s}}\times {\vec {t}}\,\right)$

The product ${\vec {s}}\times {\vec {t}}$ stands for a cross product, it is again an element of ℝ³ (sometimes referred to a pseudo vector) and hence the triple product, being an inner product, is a scalar. However, under inversion the three vectors go into minus themselves and the triple product obtains the factor (−1)³ = −1 and in total the triple product is a pseudo-scalar.