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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 In vector calculus, the Jacobian matrix (, ) of a vector-valued function of several variables is the matrix of all its first-order partial derivatives. When this matrix is square, that is, when the function takes the same number of variables ...


References

{{Reflist Tensors Tensors in general relativity Scalars