In mathematics, a relative scalar (of weight w) is a scalar-valued function whose transform under a coordinate transform,

$${\displaystyle {\bar {x))^{j}={\bar {x))^{j}(x^{i})}$$

on an n-dimensional manifold obeys the following equation

$${\displaystyle {\bar {f))({\bar {x))^{j})=J^{w}f(x^{i})}$$

where

$${\displaystyle J=\left|{\dfrac {\partial (x_{1},\ldots ,x_{n})}{\partial ({\bar {x))^{1},\ldots ,{\bar {x))^{n})))\right|,}$$

that is, the determinant of the Jacobian of the transformation.^[1] A scalar density refers to the ${\displaystyle w=1}$ case.

Relative scalars are an important special case of the more general concept of a relative tensor.

Ordinary scalar

An ordinary scalar or absolute scalar^[2] refers to the ${\displaystyle w=0}$ case.

If ${\displaystyle x^{i))$ and ${\displaystyle {\bar {x))^{j))$ refer to the same point ${\displaystyle P}$ on the manifold, then we desire ${\displaystyle {\bar {f))({\bar {x))^{j})=f(x^{i})}$. This equation can be interpreted two ways when ${\displaystyle {\bar {x))^{j))$ are viewed as the "new coordinates" and ${\displaystyle x^{i))$ are viewed as the "original coordinates". The first is as ${\displaystyle {\bar {f))({\bar {x))^{j})=f(x^{i}({\bar {x))^{j}))}$, which "converts the function to the new coordinates". The second is as ${\displaystyle f(x^{i})={\bar {f))({\bar {x))^{j}(x^{i}))}$, which "converts back to the original coordinates. Of course, "new" or "original" is a relative concept.

There are many physical quantities that are represented by ordinary scalars, such as temperature and pressure.

Weight 0 example

Suppose the temperature in a room is given in terms of the function ${\displaystyle f(x,y,z)=2x+y+5}$ in Cartesian coordinates ${\displaystyle (x,y,z)}$ and the function in cylindrical coordinates ${\displaystyle (r,t,h)}$ is desired. The two coordinate systems are related by the following sets of equations:

$${\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2))}\\t&=\arctan(y/x)\\h&=z\end{aligned))}$$

$${\displaystyle {\begin{aligned}x&=r\cos(t)\\y&=r\sin(t)\\z&=h.\end{aligned))}$$

Using ${\displaystyle {\bar {f))({\bar {x))^{j})=f(x^{i}({\bar {x))^{j}))}$ allows one to derive ${\displaystyle {\bar {f))(r,t,h)=2r\cos(t)+r\sin(t)+5}$ as the transformed function.

Consider the point ${\displaystyle P}$ whose Cartesian coordinates are ${\displaystyle (x,y,z)=(2,3,4)}$ and whose corresponding value in the cylindrical system is ${\displaystyle (r,t,h)=({\sqrt {13)),\arctan {(3/2)},4)}$. A quick calculation shows that ${\displaystyle f(2,3,4)=12}$ and ${\displaystyle {\bar {f))({\sqrt {13)),\arctan {(3/2)},4)=12}$ also. This equality would have held for any chosen point ${\displaystyle P}$. Thus, ${\displaystyle f(x,y,z)}$ is the "temperature function in the Cartesian coordinate system" and ${\displaystyle {\bar {f))(r,t,h)}$ is the "temperature function in the cylindrical coordinate system".

One way to view these functions is as representations of the "parent" function that takes a point of the manifold as an argument and gives the temperature.

The problem could have been reversed. One could have been given ${\displaystyle {\bar {f))}$ and wished to have derived the Cartesian temperature function ${\displaystyle f}$. This just flips the notion of "new" vs the "original" coordinate system.

Suppose that one wishes to integrate these functions over "the room", which will be denoted by ${\displaystyle D}$. (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region ${\displaystyle D}$ is given in cylindrical coordinates as ${\displaystyle r}$ from ${\displaystyle [0,2]}$, ${\displaystyle t}$ from ${\displaystyle [0,\pi /2]}$ and ${\displaystyle h}$ from ${\displaystyle [0,2]}$ (that is, the "room" is a quarter slice of a cylinder of radius and height 2). The integral of ${\displaystyle f}$ over the region ${\displaystyle D}$ is^{[citation needed]}

$${\displaystyle \int _{0}^{2}\!\int _{0}^{\sqrt {2^{2}-x^{2))}\!\int _{0}^{2}\!f(x,y,z)\,dz\,dy\,dx=16+10\pi .}$$

${\displaystyle {\bar {f))}$

$${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f))(r,t,h)\,dh\,dt\,dr=12+10\pi .}$$

${\displaystyle {\bar {f))}$

${\displaystyle r}$

$${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f))(r,t,h)r\,dh\,dt\,dr=16+10\pi ,}$$

Weight 1 example

If we had said ${\displaystyle f(x,y,z)=2x+y+5}$ was representing mass density, however, then its transformed value should include the Jacobian factor that takes into account the geometric distortion of the coordinate system. The transformed function is now ${\displaystyle {\bar {f))(r,t,h)=(2r\cos(t)+r\sin(t)+5)r}$. This time ${\displaystyle f(2,3,4)=12}$ but ${\displaystyle {\bar {f))({\sqrt {13)),\arctan {(3/2)},4)=12{\sqrt {29))}$. As before is integral (the total mass) in Cartesian coordinates is

$${\displaystyle \int _{0}^{2}\!\int _{0}^{\sqrt {2^{2}-x^{2))}\!\int _{0}^{2}\!f(x,y,z)\,dz\,dy\,dx=16+10\pi .}$$

${\displaystyle {\bar {f))}$

$${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f))(r,t,h)\,dh\,dt\,dr=16+10\pi .}$$

${\displaystyle {\bar {f))}$

$${\displaystyle \int _{0}^{2}\!\int _{0}^{\pi /2}\!\int _{0}^{2}\!{\bar {f))(r,t,h)r\,dh\,dt\,dr=24+40\pi /3,}$$

Other cases

Weights other than 0 and 1 do not arise as often. It can be shown the determinant of a type (0,2) tensor is a relative scalar of weight 2.