# Why Does the Minus Sign Matter in Space-Time Interval?

• komodekork
In summary, the minus sign in the space-time interval of s^{2}=t^{2}-(x^{2}+y^{2}+z^{2}) comes from the need to differentiate time from the three spatial dimensions in 4-dimensional space-time. This was necessary to avoid a cause and effect paradox and maintain the invariance of the laws of physics in all reference frames. The specific transformations and metrics of Lorentz and Minkowski were derived from the requirement of a finite and invariant speed of light in all reference frames.
komodekork
What is the resoning or the motivation for the minus sign in the space-time interval?

Which particular minus?

Thinking about $$s^{2}=t^{2}-(x^{2}+y^{2}+z^{2})$$

If you want, you can have it plus, provided you will put minus in front of $$t^2$$.
It all depends on what $$t,x,y,z$$ and $$s$$ are for you.

Ofcourse, but that's not what I am asking about. I'm asking why treat time differently? Is there some reasoning other than "we do it because it works". In eucledian geometry one would just add everything, but in Lorentz geometry there is this minus. I guess my question is, how/why did Lorentz "invent" this geometry. By trial and error or was there som reason other than "to make it all fit"?

There are several ways of answering your question. But since you are asking about the "geometry", one can start with the pre-Lorentz and pre-Minkowski geometry of space-time. That was Galilei-Newton geometry with the Galilei group behind it. You then look at whether it can be considered as a deformation of another geometry, and you end up with the Minkowski geometry.

This approach is of course less known and less popular than the approach from physics, light propagation, experiments and extrapolations from experiments.

Another geometrical approach is to search for the invariance group of Maxwell's equations. Depending on your assumptions you may end up with Minkowski or with conformal geometry. In both cases the minus sign appears automatically - from math alone.

komodekork said:
Thinking about $$s^{2}=t^{2}-(x^{2}+y^{2}+z^{2})$$

It comes from the original attempt by Einstein to discover frame invariants. He was the first to show that , if you define :

$$ds^2=(cdt)^2-dx^2-dy^2-dz^2$$

then

$$ds^2=ds'^2$$

As it happens, komodekork, I have discussed this issue on another thread, in a slightly different context, but it comes down to the same basic issue. The explanation I read made sense to me. But I am a layman and what I read is a popular science book that many contributors to this forum would probably disdain. When I outlined this explanation before, there were those who disagreed with aspects of it. But of course, that doesn’t necessarily mean that the book that I read was wrong, only that I might have interpreted it imperfectly. You should understand that the book was written by two highly respected professors of physics. In any case, having provided that context, let me outline it again for you, and see what responses it generates.

So, if you are going to represent two events on a space / time diagram, with the three dimensions of space on the x-axis in units of distance, and time on the y-axis in units of time multiplied by c, then, with the first event on the origin, if you made all terms positive, the second event would lie on the circumference of a circle for all observers, with the centre of the circle on the origin. The problem with this is that it means that different observers will see the two events in a different order, and if one event is the cause of the other, there is a possibility of a cause and effect paradox.

If the time term is negative, then instead of a circle, what you get is a hyperbola. Different observers may have different viewpoints on the spatial and temporal distances between the two events, but they all see the first event happening first and the second event happening second – no cause and effect paradox. There are greater complexities to this, but that will do for a first point for you to consider.

komodekork said:
Ofcourse, but that's not what I am asking about. I'm asking why treat time differently? Is there some reasoning other than "we do it because it works". In eucledian geometry one would just add everything, but in Lorentz geometry there is this minus.

Let us say we have two spatial dimensions (x and y) defining a plane, then by Pythagoras the length of the diagonal connecting a point at the origin to a point with coordinates (x,y) is the familiar $\sqrt{(x^2+y^2)}$.

Now in 3 dimensions, the separation of the point with coordinates (x,y,z) from the origin is $\sqrt{(x^2+y^2+z^2)}$.

Now if we go to 4 dimensions we have a problem because we have run out of physical spatial dimensions, so we can't add the t coordinate simply as another spatial dimension. This problem is solved by making time an extra "imaginary" spatial dimension and being imaginary it does not have to occupy physical space. This is done mathematically by multiplying the time coordinate by c and by the imaginary number i. We can now say the separation of the point with coordinates (x,y,x,t) is $\sqrt{(x^2+y^2+z^2 +(ict)^2)}$.

Now the square of the imaginary number i is -1 by definition, so the last expression can be written as $\sqrt{(x^2+y^2+z^2 - (ct)^2)}$.

The negative number signifies that time is not simply another dimension like the other 3 spatial dimensions, but is in fact special.

I am sure a lot of people here will object to the above, but it just meant as informal, philosophical way of looking at it that might be intuitive to some.

I am also pretty sure that is not how Lorentz or Minkowski arrived at the correct geometry and that they more likely arrived at the correct transformations and metrics, simply by making it a requirement that the speed of light is finite and the same in all reference frames independent of the speed of the light source. There is a generic "transformation" that applies to any coordinate system that requires the laws of physics are the same in all reference frames and the maximum speed in any reference frame is the same. If you assume an infinite speed as the maximum possible speed in any frame you end up with the Galilean transformation and if you assume a finite invariant maximum speed you end up with the Lorentz transformation or Minkowski metric. This is more likely the sort of approach they took.

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## 1. What is the significance of the minus sign in the space-time interval equation?

The minus sign in the space-time interval equation represents the difference between the space and time components. It accounts for the fact that time and space have opposite signs in the equation, indicating that they are fundamentally different quantities.

## 2. Why is the space-time interval equation used in relativity?

The space-time interval equation is used in relativity because it provides a way to calculate the distance between two events in space and time. This is important in understanding the effects of gravity and the curvature of space-time.

## 3. Does the minus sign in the space-time interval equation change our understanding of time?

Yes, the minus sign in the space-time interval equation changes our understanding of time by showing that it is not an absolute quantity, but rather a dimension that is intertwined with space. This is a fundamental concept in Einstein's theory of relativity.

## 4. How does the minus sign affect the measurement of space and time?

The minus sign affects the measurement of space and time by indicating that they are not independent quantities, but rather are intertwined and dependent on each other. It also shows that the measurement of time can be affected by factors such as gravity and velocity.

## 5. Can the minus sign in the space-time interval equation be replaced with a plus sign?

No, the minus sign in the space-time interval equation cannot be replaced with a plus sign. This would change the equation and its fundamental meaning, as it would no longer accurately represent the relationship between space and time in the context of relativity.

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