You and a friend just landed on Mars and are responsible for mapping out the area so that engineers back on Earth can design a new city there. After being dropped off, you both decide that it would be easiest for you to split up and record the locations of major landmarks separately and then put together a complete map.
.. tikz::
\centering
\node[anchor=south west, inner sep=0] (image) at (0,0) {\includegraphics[width=10cm]{/home/tlatorre/Documents/phy237/tensor/mars.png}};
\draw[step=1cm,gray,very thin] (0,0) grid (10,10);
\foreach \i in {0,...,10}
{
\draw[dashed] (0,\i) -- (10-\i,10);
\draw[dashed] (\i,0) -- (10,10-\i);
}
\draw[->,very thick] (0,0) -- (1,0);
\draw[->,very thick] (0,0) -- (0,1);
\draw[->,very thick,dashed] (1,1) -- (2,2);
\draw[->,very thick,dashed] (1,1) -- (1,2);
\tikzstyle{every node}=[font=\footnotesize\bfseries];
\draw[very thick] (5,5) node[right] {Arsia Mons Mosaic};
\draw[very thick] (7.5,7.5) node[right] {Ascraeus Mons};
\draw (3,9) node[above] {Mount Eunostos};
\draw (0.5,7) node[right] {Amazonis Planitia};
You decide to use an orthonormal system (shown as solid arrows), while your friend decides the landscape is more naturally suited to his choice (the dashed arrows). You agree on a common origin at the bottom left hand of the image and begin surveying the land.
After a week, your table looks like this:
| Landmark | x | y |
|---|---|---|
| Arsia Mons Mosaic | 5 | 5 |
| Ascraeus Mons | 7 | 7.5 |
and your friend's table looks like this:
| Landmark | x' | y' |
|---|---|---|
| Amazonis Planitia | 0.5 | 6 |
| Mount Eunostos | 3 | 6 |
You decide it would be best to put all the information in a single table, and your friend is convinced that his coordinate system is the best, so you give in and decide to work in his units. You tell him that transforming coordinates is easy: you know that a vector \begin{pmatrix}1 \\ 0\end{pmatrix} in your coordinate system is equal to the vector \begin{pmatrix}1 \\ -1\end{pmatrix} in his coordinate system and that you both have the same y coordinates, so evidently the transformation matrix must be
\Lambda^\nu_{\;\mu} = \begin{pmatrix}1 & 0 \\-1 & 1\end{pmatrix}
so that you can transform coordinates as
x^{\prime\nu} = \Lambda^\nu_{\;\mu} x^\mu
And you compile the complete table
| Landmark | x' | y' |
|---|---|---|
| Arsia Mons Mosaic | 5 | 0 |
| Ascraeus Mons | 7 | 0.5 |
| Amazonis Planitia | 0.5 | 6 |
| Mount Eunostos | 3 | 6 |
The next day you both feel pretty good about a job well done and decide to go hiking to Mount Eunostos. You want to figure out how long it will take, and so your friend says "No problem, I'll calculate the distance"
\left|x^\mu\right| &= \sqrt{x^{\prime\mu} x^{\prime\mu}} \\
&= \sqrt{3^2 + 6^2} \\
&= 6.7
"Woah, hold on there a minute, You aren't using an orthonormal basis, you can't calculate like that. I took the liberty of making the complete table in my coordinate system last night."
| Landmark | x | y | d |
|---|---|---|---|
| Arsia Mons Mosaic | 5 | 5 | 7.1 |
| Ascraeus Mons | 7 | 7.5 | 10.2 |
| Amazonis Planitia | 0.5 | 6.5 | 6.5 |
| Mount Eunostos | 3 | 9 | 9.5 |
"You see, the correct distance is 9.5," you say.
"But why was I wrong?"
"Let me show you. First, we'll start with the distance in my coordinate system, and then switch to your coordinate system."
d^2 &= x^\mu x^\mu \\
d^2 &= \left(\left(\Lambda^{-1}\right)^\nu_{\;\mu} x^{\prime\mu}\right)
\left(\left(\Lambda^{-1}\right)^\nu_{\;\alpha}
x^{\prime\alpha}\right) \\
d^2 &= \left(\Lambda^{-1}\right)^\nu_{\;\mu}
\left(\Lambda^{-1}\right)^\nu_{\;\alpha} x^{\prime\mu} x^{\prime\alpha} \\
d^2 &= g^\prime_{\mu\alpha} x^{\prime\mu} x^{\prime\alpha}
"What is that g_{\mu\alpha}?"
"That's the metric. It tells you how to properly measure distances in your coordinate system."
"What do you mean by proper?"
"I mean it tells you how to compute distances which everyone regardless of their coordinate system can agree on."
"So, to calculate distances in your coordinate system, we need to compute the metric."
g^\prime_{\mu\alpha} &= \left(\Lambda^{-1}\right)^\nu_{\;\mu}
\left(\Lambda^{-1}\right)^\nu_{\;\alpha} \\
&= \sum_\nu \left(\Lambda^{-1}\right)^\nu_{\;\mu}
\left(\Lambda^{-1}\right)^\nu_{\;\alpha} \\
&= \left(\Lambda^{-1}\right)^0_{\;\mu} \left(\Lambda^{-1}\right)^0_{\;\alpha} +
\left(\Lambda^{-1}\right)^1_{\;\mu} \left(\Lambda^{-1}\right)^1_{\;\alpha} \\
&= \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}
Since
\left(\Lambda^{-1}\right)^\nu_{\;\mu} =
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
"So, to compute the distance to Mount Eunostos in your coordinate system:"
d^2 &= g^\prime_{\mu\alpha} x^\mu x^\alpha \\
&= x^\mu \left(g^\prime_{\mu\alpha} x^\alpha\right) \\
&= \begin{pmatrix}3 & 6\end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix}3 \\ 6\end{pmatrix} \\
&= \begin{pmatrix}3 & 6\end{pmatrix} \begin{pmatrix}12 \\ 9\end{pmatrix} \\
&= 90
"Therefore, the distance is \sqrt{90} = 9.5."
"How come you don't need a metric in your coordinate system," your friend says.
"Well, technically I am using a metric, it's just that the metric in my orthonormal basis is just equal to the identity \delta_{\mu\alpha}. In fact, in special relativity you have to use a metric to define useful invariant scalars. It's called the Minkowski metric and is equal to"
g_{\mu\nu} &=
\begin{pmatrix}
1 & & & \\
& -1 & & \\
& & -1 & \\
& & & -1 \\
\end{pmatrix}
After arriving at Mount Eunostos, you decide to hike to the top. You'd like to get there as quick as possible. Luckily you created an elevation map the day before, h(x,y).
"The elevation of this mountain, h(x,y) is pretty complicated, but locally it just looks like this"
h(x,y) &\approx x + y
or, in tensor notation
h(x^\mu) &\approx x^0 + x^1
"Great! Let's calculate the path of steepest ascent from our current spot," your friend says.
\nabla h =
\begin{pmatrix}
\frac{\partial h}{\partial x} \\
\frac{\partial h}{\partial y}
\end{pmatrix}
= \begin{pmatrix} 1 \\ 1 \end{pmatrix}
"Ok, cool, but I'd actually like to know the gradient in my coordinate system. Don't worry, I know how to do it," your friend says.
\nabla h^\prime = \Lambda^\nu_{\;\mu} \nabla h^\mu =
\begin{pmatrix}
1 & 0 \\
-1 & 1 \\
\end{pmatrix}
\begin{pmatrix} 1 \\ 1 \end{pmatrix} =
\begin{pmatrix} 1 \\ 0 \end{pmatrix}
"Wrong," you say.
"Why am I wrong? The gradient of the elevation is a vector, and you showed me how to transform vectors."
"The gradient of the elevation is a vector, but it is a covariant vector. And covariant vectors don't transform in the same way."
"Wait, what?"
"There are two kinds of vectors, contravariant and covariant. You are probably most familiar with contravariant vectors. These are the vectors used in Newton's classic laws for example, and the types of vectors which represent the location of objects. But, there are also covariant vectors which look like a vector but don't transform in the same way. I think the right way to think about it is this: when you describe something as a vector you are describing some real physical object with a set of numbers. When you change coordinate systems, and then look at the same real object in the different coordinate system you will describe it using some other numbers. It turns out that how you transform those numbers from one frame to the next depends on the physical object you are describing. And, in this case, the gradient of a scalar function doesn't transform like location vectors.
"So, how does it transform?"
"Let's see if we can figure it out. We know that the elevation is a scalar function and so we can describe it using your coordinates as follows:"
h(x^\prime,y^\prime) &= h(x^\mu) \\
&= h\left(\left(\Lambda^{-1}\right)^\nu_{\;\mu} x^{\prime\mu}\right) \\
&= h\left(\begin{pmatrix}1 & 0 \\ 1 & 1\end{pmatrix} \begin{pmatrix}x^\prime \\ y^\prime\end{pmatrix}\right) \\
&= h\left(\begin{pmatrix}x^\prime \\ x^\prime + y^\prime\end{pmatrix}\right) \\
&= 2 x^\prime + y^\prime
and therefore, you should have gotten
\nabla h^\prime = \begin{pmatrix}2 \\ 1\end{pmatrix}
We can work out the correct transformation law by looking at the gradient in your coordinate system and expanding with the chain rule.
\nabla h^\prime &=
\begin{pmatrix}
\frac{\partial h^\prime}{\partial x^\prime} \\
\frac{\partial h^\prime}{\partial y^\prime}
\end{pmatrix} \\
&=
\begin{pmatrix}
\frac{\partial h}{\partial x} \frac{\partial x}{\partial x^\prime} +
\frac{\partial h}{\partial y} \frac{\partial y}{\partial x^\prime} \\
\frac{\partial h}{\partial y} \frac{\partial y}{\partial y^\prime} +
\frac{\partial h}{\partial x} \frac{\partial x}{\partial y^\prime} \\
\end{pmatrix} \\
&=
\begin{pmatrix}
\frac{\partial x}{\partial x^\prime} & \frac{\partial y}{\partial x^\prime} \\
\frac{\partial x}{\partial y^\prime} & \frac{\partial y}{\partial y^\prime}
\end{pmatrix}
\begin{pmatrix}
\frac{\partial h}{\partial x} \\
\frac{\partial h}{\partial y}
\end{pmatrix} \\
&=
\left(\Lambda^{-1}\right)^T \nabla h
"So, you see, instead of transforming with a factor of \Lambda, this type of vector transforms with a factor of \left(\Lambda^{-1}\right)^T. These types of vectors, covariant vectors, are written with a subscript instead of a superscript like covariant vectors. So, we should write the gradient of the elevation in tensor notation as \nabla h_\mu. Let's double check that we get the right answer."
\nabla h^\prime_\nu &= \left(\Lambda^{-1}\right)^T \nabla h \\
&= \left(\Lambda^{-1}\right)_\nu^{\;\mu} \nabla h_\mu \\
&=
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
1 \\ 1
\end{pmatrix} \\
&= \begin{pmatrix} 2 \\ 1\end{pmatrix}
"Awesome! It all makes so much sense," your friend says.
And you hike up to the top of Mount Eunostos and enjoy the view!