Question #53a2b

The Minkowski space is constructed to be a 4-dimensional space with parameters coordinates #(x_0,x_1,x_2,x_3,x_4)#, where we usually say #x_0=ct#. At the core of special relativity, we have the Lorentz transformations, which are transformations from one inertial frame to another that leave the speed of light invariant. I will not go into the full derivation of the Lorentz transformations, if you want me to explain that, just ask and I'll go into more detail.

What's important is the following. When we look at Euclidian space (the space in which we have the ordinary definition of length that we are used to #ds^2=dx_1^2+dx_2^2+dx_3^2#), we have certain transformations; spatial rotations, translations and mirrorings. If we calculate the distance between two points in various frames of reference connected by these transformations, we find the distance to be the same. This means that the Euclidian distance is invariant under these transformations.

Now we extend this notion to 4-dimensional spacetime. Before Einsteins theory of special relativity, we connected inertial frames by Galilei transformations, which just replaced a spatial coordinate #x_i# by #x_i-v_it# for #iin{1,2,3}# where #v_i# is the velocity of the observer in the #i# direction relative to the original frame. This transformation didn't leave the speed of light invariant, but it did leave the distance induced by the line element #ds^2=dx_0^2+dx_1^2+dx_2^2+dx_3^2#, simply because there is no change to the time coordinate, so time is absolute.

However, the Galilei transformation does not accurately describe the transformation of one inertial frame to another, because we know the speed of light is invariant under a proper coordinate transformations. Therefore we have introduced the Lorentz transformation. The Euclidian distance extended to 4-dim spacetime as done above is not invariant under this Lorentz transformation, however, the distance induced by #ds^2=-dx_0^2+dx_1^2+dx_2^2+dx_3^2# is, which we call the proper distance. So even though this Euclidian distance where Pythagoras theorem holds is a perfectly decent mathematical structure on the 4 dim space, it doesn't have any physical meaning, since it is dependant on the observer.

The proper distance is not dependant on the observer, therefore we can give it physical meaning, this is done by connecting the arclenght of a worldline through Minkowski space using this distance to the passed time observed by an object travelling along this worldline. Note that if we leave time fixed, the Pythagoras theorem still holds in the spatial coordinates.

EDIT/ADDITIONAL EXPLANATION:
The original asker of this question asked me to elaborate a bit more, he wrote: "Thanks. But, can you please explain the last two paras a little bit more. In a book I saw they had #s^2=x^2−(ct)^2#. Please explain" In essence what we have here is a two dimensional version of what I described above. We have a description of spacetime with one time and one space dimension. On this we define a distance, or more precisely a norm (a distance from the origin to a point) #s# using the formula #s^2=x^2−(ct)^2# where #x# is the spatial coordinate and #t# the temporal coordinate.

What I did above was a three dimensional version of this, but more importantly I used #(ds)^2# instead of #s^2# (I have added parentheses for clarification of what is squared). Without going into details of differential geometry too much, if we have a line connecting two points in space, #ds# is the length of a tiny piece of the line, a so called line element. Via a 2D version of what I wrote above, we have #ds^2=-dx_0^2+dx_1^2#, which relates the length of this tiny piece to the tiny change in the coordinates. To calculate the distance from the origin to a point #x_0=a,x_1=b# in spacetime, we calculate the length of a straight line going from the origin to that point, this line is given #x_0=a/bx_1# where #x_1in[0,b]#, we note that #dx_0=a/bdx_1#, so #ds^2=(1-a^2/b^2)dx_1^2#, so #ds=sqrt(1-a^2/b^2)dx_1#, which we can integrate, giving #s=int_0^bsqrt(1-a^2/b^2)dx_1=bsqrt(1-a^2/b^2)=sqrt(b^2-a^2)#.
Therefore #s^2=b^2-a^2=x_1^2−x_0^2=x^2-(ct)^2# in #(t,x)# coordinates.

So indeed what I wrote above gives what you read in the book. However the line element version allows you to calculate the length of any line, not just straight lines. The story about the Lorentz transformation still holds, this norm #s# is invariant under change of reference frame, while #x^2+(ct)^2# is not.

The fact that Pythagoras theorem does not hold is not that surprising. The Pythagoras theorem holds in Euclidean geometry. This means that the space in which you work is flat. An example of spaces that aren't flat is the surface of a sphere. When you want to find the distance between two points on this surface, you take the length of the shortest path over this surface connecting these two points. If you were to construct a right triangle on this surface, which would look very different from a triangle in Euclidean space, since the lines would not be straight, Pythagoras theorem does not hold in general.

Another important feature of Euclidean geometry is that when you put a coordinate system on this space, every coordinate performs the same role. You could rotate the axes and end up with the same geometry. In the Minkowski geometry above not all coordinates have the same role, since the time axes has a minus sign in the equations and the others have not. If this minus sign wasn't there, time and space would have a similar role in spacetime, or at least in the geometry. But we know that space and time are not the same.