Relative scalar

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

:$\bar^j = \bar^j(x^i)$

on an ''n''-dimensional manifold obeys the following equation

:$\bar(\bar^j) = J^w f(x^i)$

where

: $J = \begin \displaystyle \frac \end ,$

that is, the determinant of the Jacobian of the transformation. A scalar density refers to the $w=1$ 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 refers to the $w=0$ case.

If $x^i$ and $\bar^j$ refer to the same point $P$ on the manifold, then we desire $\bar(\bar^j) = f(x^i)$. This equation can be interpreted two ways when $\bar^j$ are viewed as the "new coordinates" and $x^i$ are viewed as the "original coordinates". The first is as $\bar(\bar^j) = f(x^i(\bar^j))$, which "converts the function to the new coordinates". The second is as $f(x^i)=\bar(\bar^j(x^i))$, 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 $f(x,y,z)= 2 x + y + 5$ in Cartesian coordinates $(x,y,z)$ and the function in cylindrical coordinates $(r,t,h)$ is desired. The two coordinate systems are related by the following sets of equations:
:$r = \sqrt \,$
:$t = \arctan(y/x) \,$
:$h = z \,$
and
:$x = r \cos(t) \,$
:$y = r \sin(t) \,$
:$z = h. \,$

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

Consider the point $P$ whose Cartesian coordinates are $(x,y,z)=(2,3,4)$ and whose corresponding value in the cylindrical system is $(r,t,h)=(\sqrt,\arctan,4)$. A quick calculation shows that $f(2,3,4)=12$ and $\bar(\sqrt,\arctan,4)=12$ also. This equality would have held for any chosen point $P$. Thus, $f(x,y,z)$ is the "temperature function in the Cartesian coordinate system" and $\bar(r,t,h)$ 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 $\bar$ and wished to have derived the Cartesian temperature function $f$. 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 $D$. (Yes, integrating temperature is strange but that's partly what's to be shown.) Suppose the region $D$ is given in cylindrical coordinates as $r$ from ,2/math>, $t$ from ,\pi/2/math> and $h$ from ,2/math> (that is, the "room" is a quarter slice of a cylinder of radius and height 2).
The integral of $f$ over the region $D$ is
$\int_0^2 \! \int_^\sqrt \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 16 + 10 \pi.$
The value of the integral of $\bar$ over the same region is
$\int_0^2 \! \int_^ \! \int_0^2 \! \bar(r,t,h) \, dh \, dt \, dr = 12 + 10 \pi.$
They are not equal. The integral of temperature is not independent of the coordinate
system used. It is non-physical in that sense, hence "strange". Note that if the integral of $\bar$ included a factor of the Jacobian (which is just $r$),
we get
$\int_0^2 \! \int_^ \! \int_0^2 \! \bar(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi,$
which ''is'' equal to the original integral but it is not however the integral of ''temperature'' because
temperature is a relative scalar of weight 0, not a relative scalar of weight 1.

Weight 1 example

If we had said $f(x,y,z)= 2 x + y + 5$ 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 $\bar(r,t,h)= (2 r \cos(t)+ r \sin(t) + 5) r$. This time
$f(2,3,4)=12$ but $\bar(\sqrt,\arctan,4)=12\sqrt$. As before
is integral (the total mass) in Cartesian coordinates is
$\int_0^2 \! \int_^\sqrt \! \int_0^2 \! f(x,y,z) \, dz \, dy \, dx = 16 + 10 \pi.$
The value of the integral of $\bar$ over the same region is
$\int_0^2 \! \int_^ \! \int_0^2 \! \bar(r,t,h) \, dh \, dt \, dr = 16 + 10 \pi.$
They are equal. The integral of mass ''density'' gives total mass which is a coordinate-independent concept.
Note that if the integral of $\bar$ also included a factor of the Jacobian like before, we get
$\int_0^2 \! \int_^ \! \int_0^2 \! \bar(r,t,h) r \, dh \, dt \, dr = 24 + 40 \pi /3,$
which is not equal to the previous case.

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.

See also

*Jacobian matrix and determinant

References

{{Reflist

Tensors
Tensors in general relativity
Scalars