To derive the formula for the position vector $\vec{r}$ in polar coordinates $(r, \theta)$ using only the metric tensor $ds^2 = dr^2 + r^2 d\theta^2$ and without referring to Cartesian coordinates, we can follow an intrinsic approach.

The position vector $\vec{r}$ points from the origin to a point $P(r, \theta)$. We want to express $\vec{r}$ in terms of the local coordinate basis vectors $\vec{e_r} = \partial_r \vec{r}$ and $\vec{e_\theta} = \partial_\theta \vec{r}$. Let this expression be:

Our goal is to find the functions $A(r, \theta)$ and $B(r, \theta)$. The user suggests the target is $\vec{r}=(r,0)$, which implies $A=r$ and $B=0$.

There are two main intrinsic ways to derive this:

Method 1: Solving Partial Differential Equations¶

This method uses the definition of the basis vectors and their derivatives (Christoffel symbols).

1. Basis Vectors and their Derivatives: The basis vectors are defined as $\vec{e_r} = \frac{\partial \vec{r}}{\partial r}$ and $\vec{e_\theta} = \frac{\partial \vec{r}}{\partial \theta}$. The derivatives of the basis vectors are given by $\partial_j \vec{e_i} = \Gamma^k_{ij} \vec{e_k}$, where $\Gamma^k_{ij}$ are the Christoffel symbols. The metric tensor components are $g_{rr}=1$, $g_{\theta\theta}=r^2$, and $g_{r\theta}=0$. The non-zero Christoffel symbols for this metric are:

• $\Gamma^\theta_{r\theta} = \Gamma^\theta_{\theta r} = \frac{1}{r}$

• $\Gamma^r_{\theta\theta} = -r$

All other Christoffel symbols are zero (e.g., $\Gamma^k_{rr}=0$, $\Gamma^k_{r r}=0$, $\Gamma^r_{r\theta}=0$, $\Gamma^\theta_{\theta\theta}=0$). This leads to the following derivatives of the basis vectors:

• $\partial_r \vec{e_r} = \Gamma^k_{rr} \vec{e_k} = 0$

• $\partial_r \vec{e_\theta} = \Gamma^k_{r\theta} \vec{e_k} = \Gamma^\theta_{r\theta} \vec{e_\theta} = \frac{1}{r} \vec{e_\theta}$

• $\partial_\theta \vec{e_r} = \Gamma^k_{\theta r} \vec{e_k} = \Gamma^\theta_{\theta r} \vec{e_\theta} = \frac{1}{r} \vec{e_\theta}$

• $\partial_\theta \vec{e_\theta} = \Gamma^k_{\theta\theta} \vec{e_k} = \Gamma^r_{\theta\theta} \vec{e_r} = -r \vec{e_r}$

2. Setting up the PDEs: Substitute $\vec{r} = A \vec{e_r} + B \vec{e_\theta}$ into the definitions of $\vec{e_r}$ and $\vec{e_\theta}$: For $\vec{e_r} = \partial_r \vec{r}$:

Comparing coefficients of $\vec{e_r}$ and $\vec{e_\theta}$: (1a) $\frac{\partial A}{\partial r} = 1$ (1b) $\frac{\partial B}{\partial r} + \frac{B}{r} = 0$

For $\vec{e_\theta} = \partial_\theta \vec{r}$:

Comparing coefficients: (2a) $\frac{\partial A}{\partial \theta} - rB = 0$ (2b) $\frac{A}{r} + \frac{\partial B}{\partial \theta} = 1$

3. Solving the PDEs: From (1a), $A(r, \theta) = r + f(\theta)$ for some function $f(\theta)$. From (1b), $\frac{dB}{B} = -\frac{dr}{r} \implies \ln|B| = -\ln|r| + \ln|g(\theta)| \implies B(r, \theta) = \frac{g(\theta)}{r}$ for some function $g(\theta)$.

Substitute into (2a): $\frac{\partial}{\partial \theta}(r+f(\theta)) - r\left(\frac{g(\theta)}{r}\right) = 0 \implies f'(\theta) - g(\theta) = 0 \implies g(\theta) = f'(\theta)$. So, $B(r, \theta) = \frac{f'(\theta)}{r}$.

Substitute into (2b): $\frac{r+f(\theta)}{r} + \frac{\partial}{\partial \theta}\left(\frac{f'(\theta)}{r}\right) = 1 \implies 1 + \frac{f(\theta)}{r} + \frac{f''(\theta)}{r} = 1$. This simplifies to $f(\theta) + f''(\theta) = 0$. The general solution is $f(\theta) = C_1 \cos\theta + C_2 \sin\theta$.

Thus, $A(r, \theta) = r + C_1 \cos\theta + C_2 \sin\theta$ and $B(r, \theta) = \frac{-C_1 \sin\theta + C_2 \cos\theta}{r}$.

4. Applying Boundary Conditions: The origin of the polar coordinate system is at $r=0$. The position vector $\vec{r}$ should be the zero vector at the origin, i.e., $|\vec{r}| \to 0$ as $r \to 0$. The squared magnitude of $\vec{r}$ is $|\vec{r}|^2 = g_{rr}A^2 + g_{\theta\theta}B^2 = A^2 + r^2 B^2$. $|\vec{r}|^2 = (r + C_1 \cos\theta + C_2 \sin\theta)^2 + r^2 \left(\frac{-C_1 \sin\theta + C_2 \cos\theta}{r}\right)^2$ $|\vec{r}|^2 = (r + f(\theta))^2 + (f'(\theta))^2$. As $r \to 0$, $|\vec{r}|^2 \to (f(\theta))^2 + (f'(\theta))^2$. For $|\vec{r}|$ to be zero at the origin ($r=0$) for all $\theta$, we need $f(\theta) = 0$ and $f'(\theta) = 0$ for all $\theta$. $f(\theta) = C_1 \cos\theta + C_2 \sin\theta = 0$ for all $\theta \implies C_1=0, C_2=0$. Therefore, $f(\theta)=0$ and $g(\theta)=f'(\theta)=0$.

This gives $A(r, \theta) = r$ and $B(r, \theta) = 0$. So, the position vector is $\vec{r} = r \vec{e_r}$. In terms of components in the basis $(\vec{e_r}, \vec{e_\theta})$, this is $(r, 0)$.

Method 2: Using Geodesics¶

This method is somewhat more intuitive for a flat space described by polar coordinates.

1. Origin and Radial Geodesics: The origin is the point $r=0$. The position vector $\vec{r}$ points from the origin to a point $P(r, \theta)$. A natural path to take from the origin to $P(r,\theta)$ is a radial line, where $\theta$ is constant. In polar coordinates with the given metric (which describes flat Euclidean space), radial lines are geodesics.

2. Tangent Vector and Integration: Let $s$ be the arc length along such a radial geodesic. The metric for $d\theta=0$ is $ds^2 = dr^2$, so $s=r$ (choosing $s=0$ at $r=0$). The tangent vector to this geodesic path is $\vec{T} = \frac{d\vec{r}}{ds}$. Since $s=r$ (the coordinate $r$), $\vec{T} = \frac{d\vec{r}}{dr}$. By definition, $\vec{e_r} = \partial_r \vec{r}$. So, the tangent vector to the radial geodesic is $\vec{e_r}(r', \theta)$, where $r'$ is the radial coordinate along the path.

3. Position Vector by Integration: The position vector $\vec{r}(r, \theta)$ can be found by integrating the tangent vector (which is $\vec{e_r}$) from the origin ($r'=0$) to the point ($r'=r$) along this radial geodesic:

4. Constancy of $\vec{e_r}$ along the Path: We need to know how $\vec{e_r}(r', \theta)$ changes with $r'$ when $\theta$ is held constant. This is given by $\partial_{r'} \vec{e_r}(r', \theta)$. As shown in Method 1 (using Christoffel symbols), $\partial_r \vec{e_r} = 0$. This means that for a fixed $\theta$, the basis vector $\vec{e_r}$ does not change as $r'$ changes. So, $\vec{e_r}(r', \theta)$ is constant along the path of integration with respect to $r'$. We can write it as $\vec{e_r}(r, \theta)$ (its value at the endpoint).

5. Final Result: Since $\vec{e_r}(r', \theta)$ is constant with respect to $r'$ during the integration (it’s $\vec{e_r}(r, \theta)$ or $\vec{e_r}(\text{value at fixed }\theta)$):

Thus, $\vec{r} = r \vec{e_r}$. This means the components of the position vector in the coordinate basis $(\vec{e_r}, \vec{e_\theta})$ are $(r, 0)$.

Both methods, which are “intrinsic” as they rely only on the metric and coordinate definitions, lead to the same result: the position vector $\vec{r}$ expressed in the polar coordinate basis $(\vec{e_r}, \vec{e_\theta})$ is $r \vec{e_r} + 0 \vec{e_\theta}$. This corresponds to the components $(r,0)$.

Okay, let’s use Method 1 to compute the position vector $\vec{R}$ for the metric tensor:

Let the coordinates be $(x^1, x^2) = (u, v)$. The metric tensor components are: $g_{uu} = u^2+v^2$ $g_{vv} = u^2+v^2$ $g_{uv} = g_{vu} = 0$

Let $S = u^2+v^2$. So, $g_{uu}=S$ and $g_{vv}=S$. The position vector is assumed to be $\vec{R}(u,v) = A(u,v) \vec{e_u} + B(u,v) \vec{e_v}$, where $\vec{e_u} = \partial_u \vec{R}$ and $\vec{e_v} = \partial_v \vec{R}$.

Step 1: Inverse Metric and Christoffel Symbols

The inverse metric components are $g^{uu} = 1/S$, $g^{vv} = 1/S$, and $g^{uv}=0$. The derivatives of the metric components are: $\partial_u g_{uu} = 2u$, $\partial_v g_{uu} = 2v$ $\partial_u g_{vv} = 2u$, $\partial_v g_{vv} = 2v$

The Christoffel symbols $\Gamma^k_{ij} = \frac{1}{2} g^{kl} (\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij})$ are: $\Gamma^u_{uu} = \frac{1}{2} g^{uu} (\partial_u g_{uu} + \partial_u g_{uu} - \partial_u g_{uu}) = \frac{1}{2S} (2u) = \frac{u}{S}$ $\Gamma^v_{uu} = \frac{1}{2} g^{vv} (\partial_u g_{uv} + \partial_u g_{uv} - \partial_v g_{uu}) = \frac{1}{2S} (0 + 0 - 2v) = -\frac{v}{S}$

$\Gamma^u_{vv} = \frac{1}{2} g^{uu} (\partial_v g_{vu} + \partial_v g_{vu} - \partial_u g_{vv}) = \frac{1}{2S} (0 + 0 - 2u) = -\frac{u}{S}$ $\Gamma^v_{vv} = \frac{1}{2} g^{vv} (\partial_v g_{vv} + \partial_v g_{vv} - \partial_v g_{vv}) = \frac{1}{2S} (2v) = \frac{v}{S}$

$\Gamma^u_{uv} = \Gamma^u_{vu} = \frac{1}{2} g^{uu} (\partial_u g_{vu} + \partial_v g_{uu} - \partial_u g_{uv}) = \frac{1}{2S} (0 + 2v - 0) = \frac{v}{S}$ $\Gamma^v_{uv} = \Gamma^v_{vu} = \frac{1}{2} g^{vv} (\partial_u g_{vv} + \partial_v g_{uv} - \partial_v g_{uv}) = \frac{1}{2S} (2u + 0 - 0) = \frac{u}{S}$

Step 2: Derivatives of Basis Vectors Using $\partial_j \vec{e_i} = \Gamma^k_{ij} \vec{e_k}$: $\partial_u \vec{e_u} = \Gamma^u_{uu} \vec{e_u} + \Gamma^v_{uu} \vec{e_v} = \frac{u}{S} \vec{e_u} - \frac{v}{S} \vec{e_v}$ $\partial_u \vec{e_v} = \Gamma^u_{uv} \vec{e_u} + \Gamma^v_{uv} \vec{e_v} = \frac{v}{S} \vec{e_u} + \frac{u}{S} \vec{e_v}$ $\partial_v \vec{e_u} = \Gamma^u_{vu} \vec{e_u} + \Gamma^v_{vu} \vec{e_v} = \frac{v}{S} \vec{e_u} + \frac{u}{S} \vec{e_v}$ $\partial_v \vec{e_v} = \Gamma^u_{vv} \vec{e_u} + \Gamma^v_{vv} \vec{e_v} = -\frac{u}{S} \vec{e_u} + \frac{v}{S} \vec{e_v}$

Step 3: Setting up the PDEs From $\vec{e_u} = \partial_u \vec{R} = \partial_u (A \vec{e_u} + B \vec{e_v})$: $\vec{e_u} = (\partial_u A) \vec{e_u} + A(\frac{u}{S}\vec{e_u} - \frac{v}{S}\vec{e_v}) + (\partial_u B) \vec{e_v} + B(\frac{v}{S}\vec{e_u} + \frac{u}{S}\vec{e_v})$ $\vec{e_u} = (\partial_u A + \frac{u}{S}A + \frac{v}{S}B) \vec{e_u} + (-\frac{v}{S}A + \partial_u B + \frac{u}{S}B) \vec{e_v}$ Comparing coefficients: (1a) $\partial_u A + \frac{uA+vB}{S} = 1$ (1b) $\partial_u B - \frac{vA-uB}{S} = 0$

From $\vec{e_v} = \partial_v \vec{R} = \partial_v (A \vec{e_u} + B \vec{e_v})$: $\vec{e_v} = (\partial_v A) \vec{e_u} + A(\frac{v}{S}\vec{e_u} + \frac{u}{S}\vec{e_v}) + (\partial_v B) \vec{e_v} + B(-\frac{u}{S}\vec{e_u} + \frac{v}{S}\vec{e_v})$ $\vec{e_v} = (\partial_v A + \frac{v}{S}A - \frac{u}{S}B) \vec{e_u} + (\frac{u}{S}A + \partial_v B + \frac{v}{S}B) \vec{e_v}$ Comparing coefficients: (2a) $\partial_v A + \frac{vA-uB}{S} = 0$ (2b) $\partial_v B + \frac{uA+vB}{S} = 1$

Rearranging the equations: (1a) $\partial_u A = 1 - \frac{uA+vB}{S}$ (1b) $\partial_u B = \frac{vA-uB}{S}$ (2a) $\partial_v A = -\frac{vA-uB}{S}$ (Note: $\partial_v A = -\partial_u B$) (2b) $\partial_v B = 1 - \frac{uA+vB}{S}$ (Note: $\partial_v B = \partial_u A$)

So we have $\partial_u A = \partial_v B$ and $\partial_v A = -\partial_u B$. These are the Cauchy-Riemann equations for an analytic function $F(z) = A(u,v) + iB(u,v)$, where $z = u+iv$. The derivative $F'(z) = \partial_u A + i \partial_u B$. Substituting the expressions for $\partial_u A$ and $\partial_u B$: $F'(z) = \left(1 - \frac{uA+vB}{S}\right) + i \left(\frac{vA-uB}{S}\right)$ $F'(z) = 1 - \frac{1}{S} [(uA+vB) - i(vA-uB)]$ $F'(z) = 1 - \frac{1}{S} [A(u-iv) + B(v+iu)] = 1 - \frac{1}{S} [A(u-iv) + iB(u-iv)]$ $F'(z) = 1 - \frac{(A+iB)(u-iv)}{S} = 1 - \frac{F(z)\bar{z}}{z\bar{z}}$ (since $S=u^2+v^2=|z|^2$) $F'(z) = 1 - \frac{F(z)}{z}$ This is a linear first-order ordinary differential equation: $z F'(z) + F(z) = z$. The left side is the derivative of $zF(z)$, so $(zF(z))' = z$. Integrating with respect to $z$: $zF(z) = \int z dz = \frac{1}{2}z^2 + C$, where $C$ is a complex constant of integration ($C = C_1 + iC_2$). So, $F(z) = A+iB = \frac{1}{2}z + \frac{C}{z}$. $A+iB = \frac{1}{2}(u+iv) + \frac{C_1+iC_2}{u+iv} = \frac{1}{2}(u+iv) + \frac{(C_1+iC_2)(u-iv)}{u^2+v^2}$ $A+iB = \frac{1}{2}(u+iv) + \frac{(C_1u+C_2v) + i(C_2u-C_1v)}{S}$ Equating real and imaginary parts: $A(u,v) = \frac{1}{2}u + \frac{C_1u+C_2v}{S}$ $B(u,v) = \frac{1}{2}v + \frac{C_2u-C_1v}{S}$

Step 4: Applying Boundary Conditions The origin is $(u,v)=(0,0)$, where $S=0$. The position vector $\vec{R}$ should be well-defined and typically $\vec{R} \to \vec{0}$ as $(u,v) \to (0,0)$. The squared magnitude of the position vector is $|\vec{R}|^2 = g_{uu}A^2 + g_{vv}B^2 = S(A^2+B^2) = S|F(z)|^2$. $|\vec{R}|^2 = S \left|\frac{z}{2} + \frac{C}{z}\right|^2 = S \left|\frac{z^2+2C}{2z}\right|^2 = S \frac{|z^2+2C|^2}{4|z|^2}$. Since $S=|z|^2$: $|\vec{R}|^2 = |z|^2 \frac{|z^2+2C|^2}{4|z|^2} = \frac{1}{4}|z^2+2C|^2$. As $z \to 0$ (i.e., $u,v \to 0$), the magnitude squared $|\vec{R}|^2 \to \frac{1}{4}|2C|^2 = |C|^2$. For the position vector of the origin to be $\vec{0}$, we require its magnitude to be zero, so $|C|^2=0$, which implies $C=0$. Thus $C_1=0$ and $C_2=0$.

This gives: $A(u,v) = \frac{1}{2}u$ $B(u,v) = \frac{1}{2}v$

The position vector is therefore:

The components of the position vector in the basis $(\vec{e_u}, \vec{e_v})$ are $(\frac{u}{2}, \frac{v}{2})$.

When we apply the same method to compute the position vector $\vec{P}$ for the metric tensor of a sphere of radius $R$:

The coordinates are $(x^1, x^2) = (\theta, \phi)$. The metric components are: $g_{\theta\theta} = R^2$ $g_{\phi\phi} = R^2 \sin^2\theta$ $g_{\theta\phi} = 0$

We express the sought position vector as $\vec{P}(\theta,\phi) = A(\theta,\phi) \vec{e_\theta} + B(\theta,\phi) \vec{e_\phi}$, with the conditions $\vec{e_\theta} = \partial_\theta \vec{P}$ and $\vec{e_\phi} = \partial_\phi \vec{P}$.

Step 1: Inverse Metric and Christoffel Symbols The inverse metric components are $g^{\theta\theta} = 1/R^2$ and $g^{\phi\phi} = 1/(R^2 \sin^2\theta)$. The non-zero derivatives of the metric components are $\partial_\theta g_{\phi\phi} = 2R^2 \sin\theta \cos\theta$. The non-zero Christoffel symbols are:

• $\Gamma^\theta_{\phi\phi} = \frac{1}{2} g^{\theta\theta} (-\partial_\theta g_{\phi\phi}) = \frac{1}{2R^2} (-2R^2 \sin\theta \cos\theta) = -\sin\theta \cos\theta$

• $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \frac{1}{2} g^{\phi\phi} (\partial_\theta g_{\phi\phi}) = \frac{1}{2R^2\sin^2\theta} (2R^2 \sin\theta \cos\theta) = \frac{\cos\theta}{\sin\theta} = \cot\theta$ All other Christoffel symbols ($\Gamma^\theta_{\theta\theta}, \Gamma^\phi_{\theta\theta}, \Gamma^\theta_{\theta\phi}, \Gamma^\phi_{\phi\phi}$) are zero.

Step 2: Derivatives of Basis Vectors Using $\partial_j \vec{e_i} = \Gamma^k_{ij} \vec{e_k}$:

• $\partial_\theta \vec{e_\theta} = \Gamma^\theta_{\theta\theta} \vec{e_\theta} + \Gamma^\phi_{\theta\theta} \vec{e_\phi} = 0$

• $\partial_\theta \vec{e_\phi} = \Gamma^\theta_{\theta\phi} \vec{e_\theta} + \Gamma^\phi_{\theta\phi} \vec{e_\phi} = \cot\theta \vec{e_\phi}$

• $\partial_\phi \vec{e_\theta} = \Gamma^\theta_{\phi\theta} \vec{e_\theta} + \Gamma^\phi_{\phi\theta} \vec{e_\phi} = \cot\theta \vec{e_\phi}$ (since $\Gamma^\theta_{\phi\theta}=0, \Gamma^\phi_{\phi\theta}=\cot\theta$)

• $\partial_\phi \vec{e_\phi} = \Gamma^\theta_{\phi\phi} \vec{e_\theta} + \Gamma^\phi_{\phi\phi} \vec{e_\phi} = -\sin\theta \cos\theta \vec{e_\theta}$

Step 3: Setting up the PDEs We have $\vec{P} = A \vec{e_\theta} + B \vec{e_\phi}$.

From $\vec{e_\theta} = \partial_\theta \vec{P}$: $\vec{e_\theta} = (\partial_\theta A) \vec{e_\theta} + A(\partial_\theta \vec{e_\theta}) + (\partial_\theta B) \vec{e_\phi} + B(\partial_\theta \vec{e_\phi})$ $\vec{e_\theta} = (\partial_\theta A) \vec{e_\theta} + A(0) + (\partial_\theta B) \vec{e_\phi} + B(\cot\theta \vec{e_\phi})$ $\vec{e_\theta} = (\partial_\theta A) \vec{e_\theta} + (\partial_\theta B + B\cot\theta) \vec{e_\phi}$ Comparing coefficients of $\vec{e_\theta}$ and $\vec{e_\phi}$: (I) $\partial_\theta A = 1$ (II) $\partial_\theta B + B\cot\theta = 0$

From $\vec{e_\phi} = \partial_\phi \vec{P}$: $\vec{e_\phi} = (\partial_\phi A) \vec{e_\theta} + A(\partial_\phi \vec{e_\theta}) + (\partial_\phi B) \vec{e_\phi} + B(\partial_\phi \vec{e_\phi})$ $\vec{e_\phi} = (\partial_\phi A) \vec{e_\theta} + A(\cot\theta \vec{e_\phi}) + (\partial_\phi B) \vec{e_\phi} + B(-\sin\theta \cos\theta \vec{e_\theta})$ $\vec{e_\phi} = (\partial_\phi A - B\sin\theta \cos\theta) \vec{e_\theta} + (A\cot\theta + \partial_\phi B) \vec{e_\phi}$ Comparing coefficients: (III) $\partial_\phi A - B\sin\theta \cos\theta = 0$ (IV) $A\cot\theta + \partial_\phi B = 1$

Step 4: Attempting to Solve the PDEs and Identifying Inconsistency From (I): $A(\theta,\phi) = \int 1 d\theta = \theta + f(\phi)$ for some function $f(\phi)$. From (II): $\frac{\partial B}{\partial \theta} = -B\cot\theta$. This is a separable differential equation for $B$ with respect to $\theta$: $\frac{dB}{B} = -\cot\theta d\theta = -\frac{\cos\theta}{\sin\theta} d\theta$ $\int \frac{dB}{B} = -\int \frac{\cos\theta}{\sin\theta} d\theta$ $\ln|B| = -\ln|\sin\theta| + \ln|g(\phi)| = \ln\left|\frac{g(\phi)}{\sin\theta}\right|$ So, $B(\theta,\phi) = \frac{g(\phi)}{\sin\theta}$ for some function $g(\phi)$.

Now, substitute these expressions for $A$ and $B$ into equation (III): $\partial_\phi (\theta + f(\phi)) - \left(\frac{g(\phi)}{\sin\theta}\right)\sin\theta \cos\theta = 0$ $f'(\phi) - g(\phi)\cos\theta = 0$ $f'(\phi) = g(\phi)\cos\theta$ The left side, $f'(\phi)$, depends only on $\phi$. The right side, $g(\phi)\cos\theta$, depends on both $\phi$ and $\theta$ (unless $\cos\theta$ is constant or $g(\phi)=0$). For this equality to hold for all independent values of $\theta$ and $\phi$: \begin{itemize} \item If there is any $\phi_0$ for which $g(\phi_0) \neq 0$, then $f'(\phi_0)/g(\phi_0) = \cos\theta$. This would imply that $\cos\theta$ is a constant (equal to $f'(\phi_0)/g(\phi_0)$), which is false as $\theta$ is a variable coordinate. \item The only way for the equality to hold for all $\theta$ is if $g(\phi) = 0$ for all $\phi$. \end{itemize} If $g(\phi) = 0$ for all $\phi$, then $f'(\phi) = 0 \cdot \cos\theta = 0$. So, $g(\phi)=0 \implies B(\theta,\phi) = 0$. And $f'(\phi)=0 \implies f(\phi) = C_1$ (an arbitrary constant). This means $A(\theta,\phi) = \theta + C_1$.

Now we must check if this potential solution ($A=\theta+C_1$, $B=0$) satisfies the last equation, (IV): $A\cot\theta + \partial_\phi B = 1$ $(\theta+C_1)\cot\theta + \partial_\phi(0) = 1$ $(\theta+C_1)\cot\theta = 1$ This equation must hold for all values of $\theta$ (where $\cot\theta$ is defined). However, this is not possible: \begin{itemize} \item If $\theta = \pi/2$, then $\cot(\pi/2) = 0$. The equation becomes $(\pi/2 + C_1) \cdot 0 = 1$, which simplifies to $0=1$. This is a contradiction. \end{itemize} Since we have reached a contradiction, the system of partial differential equations (I)-(IV) has no solution for $A(\theta,\phi)$ and $B(\theta,\phi)$.

Conclusion: The method used to find a “position vector” $\vec{P} = A \vec{e_\theta} + B \vec{e_\phi}$ such that its partial derivatives $\partial_\theta \vec{P}$ and $\partial_\phi \vec{P}$ are equal to the coordinate basis vectors $\vec{e_\theta}$ and $\vec{e_\phi}$ respectively, does not yield a solution for the metric of a sphere.

This is because the existence of such a vector field is characteristic of flat spaces (or flat space coordinatizations). The sphere is an intrinsically curved manifold (its Riemann curvature tensor is non-zero). On a curved manifold, one cannot generally find such a global vector field whose components satisfy $\nabla_i A^k = \delta_i^k$ (where $\nabla_i$ is the covariant derivative and $\delta_i^k$ is the Kronecker delta), which is the underlying condition that the system of PDEs (I)-(IV) represents. The curvature of the sphere prevents the existence of such a “globally flat” position vector field intrinsic to the manifold’s coordinate system.