# The space-time interval definition in special relativity

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• albertrichardf
In summary, you have tried to derive the space-time interval by considering that c is an invariant speed, and that T, the proper time is also an invariant time interval. You then found that s^2 = t^2(c^2 - v^2), and that if K is zero or the space-time interval depends on the change in time and the change in position, then x'=vt'.
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?

Albertrichardf said:
Anyway, I have tried "deriving" the space-time interval in the following way:
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##

Dale said:
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##
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?

Albertrichardf said:
And what could motivate the ##-dt^2 + dx^2 + dy^2 + dz^2## as a combination?
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.

Alright. Thank you for the answer

You are welcome, it was a very enjoyable question.

## 1. What is the space-time interval definition in special relativity?

The space-time interval in special relativity refers to the distance between two events in space and time. It takes into account the fact that space and time are relative, meaning they can appear differently to different observers depending on their relative motion.

## 2. How is the space-time interval calculated?

The space-time interval is calculated using the Minkowski metric, which is a mathematical formula that takes into account both space and time components. It is represented as Δs^2 = c^2Δt^2 - Δx^2 - Δy^2 - Δz^2, where c is the speed of light and Δt, Δx, Δy, and Δz are the differences in time and space between the two events.

## 3. What is the significance of the space-time interval in special relativity?

The space-time interval is significant because it is the only quantity in special relativity that is invariant, meaning it has the same value for all observers regardless of their relative motion. This allows for a consistent understanding of space and time between different frames of reference.

## 4. How does the space-time interval relate to the concept of causality?

The space-time interval is closely related to the concept of causality, which states that cause must precede effect. In special relativity, if the space-time interval between two events is positive, it means that the first event must have occurred before the second event, and thus maintains the principle of causality.

## 5. Can the space-time interval ever be negative?

No, the space-time interval cannot be negative. In special relativity, a negative space-time interval would imply that the second event occurred before the first event, which would violate the principle of causality. Additionally, a negative space-time interval would also imply that one of the components (time or space) would have to have an imaginary value, which is not physically possible.

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