Spacetime odyssey

All physical phenomena we see around us take place on something that we call spacetime. Therefore it becomes important to get acquainted with spacetime and understand how it affects the physical phenomena or gets affected by them. I will chronologically introduce the notions associated with spacetime because, usually, everyone gets to learn it in the exact same fashion.

Until 20th century, space and time were thought as different entities. It was thought that there exists a three dimensional euclidean space which acts as a stage for the physical phenomena. This stage was supposed to be rigid, static and flat (rigorous definition will be given later). The space was a three dimensional extension of Euclid’s five axioms for plane geometry which were

1. Between two points in the space, only one line can be drawn.
2. A finite line in a space can be extended indefinitely.
3. A circle with arbitrary radius and center can be drawn in the space
4. All right angles are equal to one another.
5. If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

These axioms were based on the human experience (at the time of Euclid) and considered to be true in every part of the universe.  Even now, based on high-school experience, one doesn’t need enough convincing about the validity of these postulates.

Let us now define our three-dimensional Euclidean space in formal way. It is a set of points with well defined distance between any two points of the set. If you give me two points say $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ belonging to Euclidean space. I will give you the distance between them by computing $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$.

A couple of interesting points can be noted. First, the definition of distance between two points is static i.e the way to find distance doesn’t change with time. We can represent the definition of distance between two points using matrices. Let us represent the points by column matrices $P_1 = \left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right)$ and $P_2 = \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right)$ and define a $3\times 3$ matrix $\eta = \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)$. The distance between two points can now be given by $d^2 = (P_1-P_2)^T.\eta .(P_1-P_2)$ (here ‘.’ represents matrix multiplication). Note that the definition of distance depends on the matrix $\eta$. Physicists and mathematicians call $\eta$ metric of the space. Now we can say that it is the metric which defines the distance between two points of a space. In fact the metric defines the notion of a ‘dot product’ in a space. Take any vector, write its components in column matrix and used the matrix multiplication defined above to find the length of the vector.

The metric of Euclidean space is time independent (build from constant numbers). Although I have used matrix to represent a metric, generally it should be thought as a set of numbers given by $\eta^{\i\j}$ where $i,j\in 1,2,3$.

Next we note that the metric (of Euclidean space) is independent of space itself. In fact this independence is a direct consequence of Euclid’s axiom 5. This independence implies that the space is flat. I will demonstrate this and provide a formal definition of flatness in a later post.

3 thoughts on “Spacetime odyssey”

1. D_noob says:

Nice….! Use of layman language is much appreciated, though I must say as a non physicist, I dint get that matrix multiplication part…I mean how did you just transformed the distance between two points equation into the matrix form. Really sorry if I asked a dumb question.
Cheers 😀

• ravimohan says:

Ok let us see. $P_1 = \left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right)$ and $P_2 = \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right)$. Therefore $P_1-P_2 = \left(\begin{array}{c}x_1-x_2\\y_1-y_2\\z_1-z_2\end{array}\right)$. The transpose of this column matrix is row matrix $\left(\begin{array}{c}x_1-x_2, y_1-y_2, z_1-z_2\end{array}\right)$. Using the definition of distance (in terms of metric), we get the following multiplication of matrices $d^2= \left(\begin{array}{c}x_1-x_2, y_1-y_2, z_1-z_2\end{array}\right)\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{c}x_1-x_2\\y_1-y_2\\z_1-z_2\end{array}\right),$ which leads to $d^2 = (x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2$.

• D_noob says:

Got it ! !….I should have concentrated more on matrices n determinants in 12thies 😀

Anyways, your point is clear , atleast I get the feeling…:P hehehe