Origin of negative term $-ct^2$ in the Lorentz invariance of a 4-vector

Question

So I am taking an introductory course in Special Relativity. In my book the spacetime 4-vector is defined as: $$X= \begin{bmatrix}ct \\ x \\ y \\ z \end{bmatrix}$$. Then the book proceeds to say that the squared length of a 4-vector is given by: $$|X|^2 = x^2+y^2+z^2- c^2t^2$$

Now what I don't understand why there is a negative term $-c^2t^2$, isn't the length of a vector defined as the square root of the sum of each component?

Answer 1

This metric is the result of works of a mathematician named Minkowski. He observed that the following metric$$\Delta S^2=\Delta x^2+\Delta y^2+\Delta z^2-c^2\Delta t^2$$is invariant under Lorentz transformation which connects two inertial frames. Each event namely $P$ carries a light cone with itself which is given by$$\Delta x^2+\Delta y^2+\Delta z^2=c^2\Delta t^2$$The events right on the cone are those who can be connected to event $P$ through a light signal. Inside the cone, are those events that can be connected to event $P$ with less that speed of light. There is a very good explanation in Wolfgang Rindler's Relativity: Special, General and Cosmological (which I found it very useful). Hope it also helps you on 4-vectors and 4-tensors.

Answer 2

The length $|X|^2=x^2+y^2+z^2+c^2t^2$ that you’re familiar with is the vector’s length in a Euclidean geometry. The geometry of spacetime in S.R. is non-Euclidean, though, and is basically defined by the metric given in your book. (BTW, calling this a “length” seems a bit off to me since $x^2+y^2+z^2-c^2t^2$ can be negative.)