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:

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

and

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

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

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

The value of the integral of ${\displaystyle {\bar {f}}}$ over the same region is

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

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 ${\displaystyle {\bar {f}}}$ included a factor of the Jacobian (which is just ${\displaystyle r}$), we get

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

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 ${\displaystyle f(x,y,z)=2x+y+5}$ 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

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

The value of the integral of ${\displaystyle {\bar {f}}}$ over the same region is

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

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

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

which is not equal to the previous case.