Relative scalar

Ordinary scalar

An ordinary scalar or absolute scalar refers to the w = 0 case.

If x i and x ¯ j refer to the same point P on the manifold, then we desire f ¯ ( x ¯ j ) = f ( x i ) . This equation can be interpreted two ways when x ¯ j are viewed as the "new coordinates" and x i are viewed as the "original coordinates". The first is as f ¯ ( x ¯ j ) = f ( x i ( x ¯ j ) ) , which "converts the function to the new coordinates". The second is as f ( x i ) = f ¯ ( 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 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:

and

Using f ¯ ( x ¯ j ) = f ( x i ( x ¯ j ) ) allows one to derive f ¯ ( 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 ) = ( 13 , arctan ⁡ ( 3 / 2 ) , 4 ) . A quick calculation shows that f ( 2 , 3 , 4 ) = 12 and f ¯ ( 13 , arctan ⁡ ( 3 / 2 ) , 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 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 f ¯ 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 [ 0 , 2 ] , t from [ 0 , π / 2 ] and h from [ 0 , 2 ] (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

The value of the integral of f ¯ over the same region is

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 f ¯ included a factor of the Jacobian (which is just r ), we get

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 f ¯ ( r , t , h ) = ( 2 r cos ⁡ ( t ) + r sin ⁡ ( t ) + 5 ) r . This time f ( 2 , 3 , 4 ) = 12 but f ¯ ( 13 , arctan ⁡ ( 3 / 2 ) , 4 ) = 12 29 . As before is integral (the total mass) in Cartesian coordinates is

The value of the integral of f ¯ over the same region is

They are equal. The integral of mass density gives total mass which is a coordinate-independent concept. Note that if the integral of f ¯ also included a factor of the Jacobian like before, we get

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.