Change of variable in integral using metric

[friend]

What is the formula to evaluate a multi-integral by a change of coordinates using the squareroot of the metric instead of the determinate of the Jacobian? Thanks.

 

[Muphrid]

The square root of the metric absorbs any factors the Jacobian would introduce in a change of coordinates.

 

[friend]

So check my work, let's see if I understand this correctly,

Suppose we want to evaluate a triple integral by going from coordinates (x,y,z) to coordinates (u,v,w), then we start from

$$\int_X {f(\vec x)d\vec x\, = \,\int_{{z_1}}^{{z_2}} {\int_{{y_1}}^{{y_2}} {\int_{{x_1}}^{{x_2}} {f(x,y,z)dxdydz} } } }$$

and this equals

$$\int_{Q(X)} {f(\vec x(\vec q))\sqrt {g(\vec q)} d\vec q\, = \,} \int_{{w_1}}^{{w_2}} {\int_{{v_1}}^{{v_2}} {\int_{{u_1}}^{{u_2}} {f(x(u,v,w),y(u,v,w),z(u,v,w))\sqrt {g(u,v,w)} dudvdw} } }$$

with

$$\int_{Q(X)} {f(\vec x(\vec q))\sqrt {g(\vec q)} d\vec q = \,} \int_{{w_1}}^{{w_2}} {\int_{{v_1}}^{{v_2}} {\int_{{u_1}}^{{u_2}} {f(u,v,w)\left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|dudvdw} } }$$

and with

$$\sqrt {g(u,v,w)} \, = \,\left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|\, = \,\left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial x}}{{\partial w}}}\\ {\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial y}}{{\partial w}}}\\ {\frac{{\partial z}}{{\partial u}}}&{\frac{{\partial z}}{{\partial v}}}&{\frac{{\partial z}}{{\partial w}}} \end{array}} \right|$$

This last term is the determinant of the Jacobian, right?

 

[Muphrid]

Yes, if you're going from flat-space Cartesian coordinates to an arbitrary coordinate system, the square root of the metric determinant is exactly the Jacobian's determinant.

 

[friend]

Yes, if you're going from flat-space Cartesian coordinates to an arbitrary coordinate system, the square root of the metric determinant is exactly the Jacobian's determinant.

And then if we make another change of coordinates from (u,v,w) to coordinates (r,s,t) we would get,

$$\int_{{w_1}}^{{w_2}} {\int_{{v_1}}^{{v_2}} {\int_{{u_1}}^{{u_2}} {f(u,v,w)\left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|dudvdw} } } = \int_{t({u_1},{v_1},{w_2})}^{t({u_2},{v_2},{w_2})} {\int_{s({u_1},{v_1},w)}^{s({u_2},{v_2},w)} {\int_{r({u_1},v,w)}^{r({u_2},v,w)} {f(u(r,s,t),v(r,s,t),w(r,s,t))\left| {\frac{{\partial (u,v,w)}}{{\partial (r,s,t)}}} \right|drdsdt} } }$$
Is this right? Please check the limits of the last integral using the limits in the first integral.

Is there then a relationship between the determinant in the first integral and the determinant of the last integral? Would there be a tensor transformation between the two determinants?

 

[Muphrid]

The original Jacobian determinant from xyz to uvw should still be present--multiplying that and the Jacobian from uvw to rst should give directly the Jacobian determinant from xyz to rst, as if we did it in one single coordinate transformation.

So, on the right hand side, there should be two Jacobian determinants. Or, it should be the Jacobian determinant from xyz to rst, not from uvw.

The usual tensor transformation laws apply. The Jacobian determinant is the unique numerical factor invoked when applying the tensor transformation laws to the Levi-Civita tensor. Integrals over volumes naturally involve this tensor.

 

[friend]

The original Jacobian determinant from xyz to uvw should still be present--

Ahh yes, the determinant from xyz to uvw is part of the integrand that is being converted, right? If that determinant were not included in the left hand side, it would not be needed on the right hand side.

 

[friend]

And can we just as easily use,
$$\int_X {f(\vec x)d\vec x\,} = \int_{Q(X)} {f(\vec x(\vec q))\sqrt {g(\vec q)} d\vec q\,\,}$$
where the metric, g(q), is defined in post 3, if we transfrom from flat space to curved space?

I'm thinking of a transformation from a disk to a hemisphere directly above it. We should be able to map every point on the disk to points of the hemisphere (except for the equator). We can draw lines in flat space on the disk and map them to lines on the hemisphere. However, the metric on the disk will be different from the metric on the hemisphere, and so the length of the line on each will be different. At this point, I'm not sure that the integral on the disk will be equal to the integral on the hemisphere. Any insight into all this would be appreciated. Thanks.

 

[Muphrid]

A hemisphere is still topologically flat. The square root of the metric determinant must appear in both integrals if the space is in fact not topologically flat. No simple coordinate system transformation can convert a topologically flat space to one that isn't (e.g. from a flat space to a complete sphere).

In this way, you can see that the metric has two somewhat distinct pieces of information within it. Part of it tells us about converting between coordinate systems. The other part tells us something unambiguous regardless of coordinate system--whether the space is similar to a flat space, a doughnut, a closed sphere, etc.