Coordinate Systems
The second postulate of special relativity is that physical systems must remain invariant of the coordinate systems used to describe them. This is what we emphasise in this chapter. We will consider the mathematical tools to required to study relativity.Basis Vectors
Let's first consider the cartesian coordiantes \((x,y,z)\). One might come up with another system of coordinates say \((u,v,w)\). Before trying to talk to our \((u,v,w)\) coordiante friends about any physical system, we must know how to convert between the two coordinates.An example could be \((u,v,w) = (r, \theta, \phi)\), the spherical coordinates.
then, we can convert as: \[ \begin{align*} x &= r\sin{\theta}\cos{\phi}\\ y &= r\sin{\theta}\sin{\phi}\\ z &= r\cos{\theta} \\ \end{align*} \] where \(r \ge 0\) and \(0 \le \theta \le 2\pi\) and \(0 \le \phi \le \pi\)
Basis Vectors: Tangent Basis
If we have any vector \(\vec{r} = x(u,v,w)\hat{i}+y(u,v,w)\hat{j}+z(u,v,w)\hat{k}\) in a coordinate system \((u,v,w)\), then we define the unit vectors in such a coordinate system as: \[ \begin{align*} \vec{e}_u &= \frac{\partial{\vec{r}}}{\partial{u}} =\partial_r{\vec{r}}\\ \vec{e}_v &= \frac{\partial{\vec{r}}}{\partial{v}} =\partial_{\theta}{\vec{r}} \\ \vec{e}_w &= \frac{\partial{\vec{r}}}{\partial{w}} =\partial_{\phi}{\vec{r}} \\ \end{align*} , \] Let's take the same example of the spherical coordinates where \((u,v,w)=(r,\theta,\phi)\).So, a general position vector \(\vec{r}\) becomes \(\vec{r} =(r\sin{\theta}\cos{\phi}) \hat{i}+(r\sin{\theta}\sin{\phi})\hat{j}+(r\cos{\theta})\hat{k}\)
We prefer to write in the vector form: $$\vec{r}=\left[\begin{matrix}r \sin{\left(\theta \right)} \cos{\left(\phi \right)}\\r \sin{\left(\phi \right)} \sin{\left(\theta \right)}\\r \cos{\left(\theta \right)}\end{matrix}\right]$$ Now, we apply the above equations to find the basis vectors in the spherical coordiante system. \[ \begin{align*} \vec{e}_r &= \frac{\partial{\vec{r}}}{\partial{r}} \\ \vec{e}_\theta &= \frac{\partial{\vec{r}}}{\partial{\theta}} \\ \vec{e}_\phi &= \frac{\partial{\vec{r}}}{\partial{\phi}} \\ \end{align*} \] We can do this by hand like a caveman, or use SymPy:
Using SymPy
We may verify this in SymPy:
However, these do not form an orthonormal basis as the basis vectors don't have \(|\vec{e}_i|=\sqrt{|\vec{e}_i\cdot\vec{e}_i|}=1\):
We can divide the basis vectors by their norm to finally have an orthonormal basis: \[ \begin{align*} \vec{e}_r&=\left[\begin{matrix}\sin{\left(\theta \right)} \cos{\left(\phi \right)}\\\sin{\left(\phi \right)} \sin{\left(\theta \right)}\\\cos{\left(\theta \right)}\end{matrix}\right] \\ \vec{e}_\theta&=\left[\begin{matrix}\cos{\left(\phi \right)} \cos{\left(\theta \right)}\\\sin{\left(\phi \right)} \cos{\left(\theta \right)}\\- \sin{\left(\theta \right)}\end{matrix}\right] \\ \vec{e}_\phi&=\left[\begin{matrix}- \sin{\left(\phi \right)}\\\cos{\left(\phi \right)}\\0\end{matrix}\right] \\ \end{align*} \]
Remember that the spherical coordiante system has basis vectors that vary at different point in space. The basis vectors are coordinate-dependent unlike cartesian basis vectors which are constant regardless of the position in the coordinate system.
Basis Vectors: Dual Basis
It is also possible to have a set of basis vectors which are formed using the normals to a general point rather than the tangents.
In such a case, we define the basis vectors using the gradients instead of the partial derivatives.
Here, instead of writing \(x(u,v,w),y(u,v,w),z(u,v,w)\), we write \((u,v,w)\) in terms of \((x,y,z)\):
\[
\begin{align*}
u &= u(x,y,z) \\
v &= v(x,y,z) \\
w &= w(x,y,z) \\
\end{align*}
\]
The basis vectors are defined by:
\[
\begin{align*}
\vec{e}^u &= \nabla{u} = \partial_x{u} \hat{i}+\partial_y{u} \hat{j}+\partial_z{u} \hat{k}\\
\vec{e}^v &= \nabla{v} = \partial_x{v} \hat{i}+\partial_y{v} \hat{j}+\partial_z{v} \hat{k}\\
\vec{e}^u &= \nabla{w} = \partial_x{w} \hat{i}+\partial_y{w} \hat{j}+\partial_z{w} \hat{k}\\
\end{align*}
\]
tangent-basis are covariant.
dual-basis are contravariant.
A basis need not be orthonormal in general.
Example: Define a coordinate system \((u,v,w)\) in terms of cartesian coordinates as:
\(x = u+v, \quad \quad y=u-v, \quad\quad z = 2uv+w\)
The position vector is thus given by: \[ \begin{align*} \vec{r} &= x\hat{i}+y\hat{j}+z\hat{k} \\ &= (u+v)\hat{i}+(u-v)\hat{j}+(2uv+w)\hat{k} \end{align*} \] Find the tangent (\(\vec{e}_u,\vec{e}_v,\vec{e}_w\)) and dual (\(\vec{e}^u,\vec{e}^v,\vec{e}^w\)) basis for these coordinates.
Solution:
For tangent basis, we use the tangent basis equations (using partial derivatives of a general vector) given above to get the basis vectors. We may use SymPy:
\[ \begin{align*} \vec{e}_u&=\left[\begin{matrix}1\\1\\2 v\end{matrix}\right] \quad \vec{e}_v=\left[\begin{matrix}1\\-1\\2 u\end{matrix}\right] \quad \vec{e}_w=\left[\begin{matrix}0\\0\\1\end{matrix}\right] \end{align*} \]
For dual basis, we first need \((u,v,w)\) in terms of \((x,y,z)\). We can do this easily and get: \[ \begin{align*} u&=\frac{x}{2} + \frac{y}{2} \\ v&=\frac{x}{2} - \frac{y}{2} \\ w&=- \frac{x^{2}}{2} + \frac{y^{2}}{2} + z \end{align*} \] To get the dual basis, we find the gradients. \[ \begin{align*} \vec{e}^u &= \nabla{u} = \frac{\hat{i}}{2} + \frac{\hat{j}}{2} \\ \vec{e}^v &=\nabla{v} = \frac{\hat{i}}{2} - \frac{\hat{j}}{2} \\ \vec{e}^w &=\nabla{w} =- \hat{i} x + \hat{j} y + \hat{k}?. \\ &= - \hat{i} \left(u + v\right) + \hat{j} \left(u - v\right) + \hat{k} \end{align*} \]
\(x = u+v, \quad \quad y=u-v, \quad\quad z = 2uv+w\)
The position vector is thus given by: \[ \begin{align*} \vec{r} &= x\hat{i}+y\hat{j}+z\hat{k} \\ &= (u+v)\hat{i}+(u-v)\hat{j}+(2uv+w)\hat{k} \end{align*} \] Find the tangent (\(\vec{e}_u,\vec{e}_v,\vec{e}_w\)) and dual (\(\vec{e}^u,\vec{e}^v,\vec{e}^w\)) basis for these coordinates.
Solution:
For tangent basis, we use the tangent basis equations (using partial derivatives of a general vector) given above to get the basis vectors. We may use SymPy:
\[ \begin{align*} \vec{e}_u&=\left[\begin{matrix}1\\1\\2 v\end{matrix}\right] \quad \vec{e}_v=\left[\begin{matrix}1\\-1\\2 u\end{matrix}\right] \quad \vec{e}_w=\left[\begin{matrix}0\\0\\1\end{matrix}\right] \end{align*} \]
For dual basis, we first need \((u,v,w)\) in terms of \((x,y,z)\). We can do this easily and get: \[ \begin{align*} u&=\frac{x}{2} + \frac{y}{2} \\ v&=\frac{x}{2} - \frac{y}{2} \\ w&=- \frac{x^{2}}{2} + \frac{y^{2}}{2} + z \end{align*} \] To get the dual basis, we find the gradients. \[ \begin{align*} \vec{e}^u &= \nabla{u} = \frac{\hat{i}}{2} + \frac{\hat{j}}{2} \\ \vec{e}^v &=\nabla{v} = \frac{\hat{i}}{2} - \frac{\hat{j}}{2} \\ \vec{e}^w &=\nabla{w} =- \hat{i} x + \hat{j} y + \hat{k}?. \\ &= - \hat{i} \left(u + v\right) + \hat{j} \left(u - v\right) + \hat{k} \end{align*} \]
Vectors
There are two types of vectors: Covariant and Contravariant.Contravariant vecotrs are identified by superscripts like \(V^{\alpha}\)
Covariant vectors are isentified by subscripts like \(V_{\alpha}\)
We will see the difference between these now. We choose a set of basis vectors, for simplicity in 2D: \(\{\vec{e}_1,\vec{e}_2\}\). We define a vector \(\vec{V}\) by multiplying these basis vecotrs with components \(\{V^1,V^2\}\). $$\vec{V} = V^1\vec{e}_1+V^2\vec{e}_2 = V^{\alpha}\vec{e}_{\alpha} $$ Here we used Einstein summation convention.
These basis vectors need not be orthogonal or normalized in general. As long as they are not parallel to each other, they can be used to define vectors.
We have another set of vectors for a set of basis vectors: the dual-vectors \(\{\tilde{e}^1,\tilde{e}^2\}\). These are also known as one-forms. We have the components for these dual-basis as \(\{P_1,P_2\}\).
We can thus write any vector \(\tilde{P}\) as: $$\tilde{P} = \tilde{P}_1 \tilde{e}^1+\tilde{P}_2 \tilde{e}^2 = \tilde{P}_{\alpha} \tilde{e}^{\alpha}$$
The dual-vectors satifsy the property that they are orthogonal to the basis vectors with a different index. Mathematically, $$\tilde{e}^{\alpha} \vec{e}_{\beta} = \delta^{\alpha}_{\beta}$$ Include a diagram showing how dual basis vectors are orthogonal to normal vectors.