1. Show that under a coordinate transformation (x1, …, xn) →...

1. Show that under a coordinate transformation $(x_1, \dots, x_n) \to (y_1(\vec{x}), \dots, y_n(\vec{x}))$ in $\mathbb{R}^n$ the $n$-form $dy^1 \wedge dy^2 \wedge \dots \wedge dy^n = J(\vec{y}, \vec{x})dx^1 \wedge dx^2 \wedge \dots \wedge dx^n$, where $J$ is the Jacobian determinant $J(\vec{y}, \vec{x}) = \det \left(\frac{\partial y^i}{\partial x^j}\right)$. Hint: you may need the following determinant formula $\det \left(\frac{\partial y^i}{\partial x^j}\right) = \epsilon_{j_1 \dots j_n} \frac{\partial y^1}{\partial x^{j_1}} \dots \frac{\partial y^n}{\partial x^{j_n}}$ (1) which uses the completely antisymmetric $\epsilon$-symbol. This is the general determinant formula $\det M = \det(M_{ij}) = \epsilon_{j_1 \dots j_n} M_{1j_1} \dots M_{nj_n}$ (2)

Transcript

00:04 We're asked to find the jacobian of the transformation from cartesian rho phi theta space to cartesian x, y, z space.

00:23 Well, what this really is a question is about spherical coordinates.

00:29 So our parametric equations are x equals rho sine fee cosine theta, y equals row sine phi sine of theta and z equals row cosine phi and therefore jocobian of this transformation d x y z d row phi theta is the determinant of the three by three matrix which is well the partial of x with respect to row is sine phi theta is the determinant of the three by three matrix which is while the partial of x with respect to row is sine phi cosine theta, partial of y with respect to row is sine phi.

01:26 No, sorry, that's mistake.

01:28 Partial of x with respect to phi is row cosine phi, cosine theta, and the partial of x with respect to theta is negative row sine phi, sine theta.

01:50 Partial of y, well, this is also going to be sine phi, sine of theta.

02:01 Row cosine phi, sine of theta, and row sine of phi, cosine theta.

02:16 And finally, the partial of z with respect to row is cosine phi, partial of z with respect to phi is negative row sine phi, and the partial of z with respect to theta is zero.

02:31 So to evaluate this terminant, we'll expand across the bottom row, and so this is equal to cosine phi times the determinant of the upper 2x2 matrix, row cosine phi cosine theta, negative row sine phi, sine theta, and row cosine phi, sine theta, row sine theta, row sine phi, cosine theta.

03:18 And then we have minus negative row sine phi, or plus, row sine phi times the determinant of the matrix sine phi cosine theta, negative row sine phi, sine theta, sine phi, sine theta, and row sine phi, cosine theta...

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