# I The space-time interval definition in special relativity

1. Feb 5, 2017

### albertrichardf

Hello, please note that the following is only about special relativity, not general. Of course, if there are any things to point out that fall in general relativity, feel free to do so, but I don't know GR, so I won't understand arguments based in GR. I also am not great with a geometry-based SR, so I won't understand those either. Thank you.

Anyway, I have tried "deriving" the space-time interval in the following way: consider that c is an invariant speed, which is the speed of an object in space-time. T, the proper time is also an invariant time interval, and distance = speed x time. To define an invariant interval, it would make sense to multiply the two, which gives:

$$s = cT$$

To find out what this would be in space and time coordinates, square the equation, while keeping in mind that
t = T/y, where y is the gamma factor. After expanding y and some algebra, you obtain:

$$s^2 = t^2(c^2 - v^2)$$

Now, suppose you have an object that is at position K in the proper time frame. It should stay at K because its velocity is zero. In another reference frame, moving at v relative to the proper time frame:

$$t' = yT$$

$$x' = y(K - vT) = yK - yvT = yK - vt'$$

If K = 0 the last equation becomes:

$$x' = -vt'$$

If I replace for x' in my space-time interval, I obtain the following equation:

$$s^2 = c^2t^2 - x^2$$

which is how the interval is often shown. But the above only holds if K = 0, or if the space-time interval depends on the change in time and the change in position, which I'm thinking it does because it is an interval, but confirmation would be nice. Furthermore, if this is the case how could I proceed to define the 4-vector for position based on this? Would I have to use the Minkowski Pythagoras theorem or is there another way that I could do so?

Thank you for answering

2. Feb 5, 2017

### Staff: Mentor

Well, I would take the direct approach instead. In units where c=1, define $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$. Then transform to a "primed" coordinate system moving with velocity v relative to the unprimed system giving $dt'=\gamma (dt-v\,dx)$, $dx'=\gamma (dx-v \, dt )$, $dy'=dy$, $dz'=dz$, and calculate $ds'^2 = -dt'^2 + dx'^2 + dy'^2 + dz'^2$. Once you have that you just simplify and show that $ds'^2=ds^2$

3. Feb 5, 2017

### albertrichardf

So then would $ds^2$ be defined as the space-time interval since it is invariant? And what could motivate the $-dt^2 + dx^2 + dy^2 + dz^2$ as a combination? The Minkowski geometry?

4. Feb 5, 2017

### Staff: Mentor

That can be motivated from the second postulate. A flash of light moving spherically outward at c=1 can be written $\Delta t^2=\Delta x^2+\Delta y^2+\Delta z^2$ which is clearly equivalent to $\Delta s^2=0$. Since if it is moving at c in one frame then by the second postulate it must also be moving at c in every other frame, this implies that at a minimum $\Delta s^2$ is invariant for light. That is enough to motivate at least checking to see if $\Delta s^2$ is invariant for other values.

5. Feb 5, 2017

### albertrichardf

Alright. Thank you for the answer

6. Feb 5, 2017

### Staff: Mentor

You are welcome, it was a very enjoyable question.