# Interval between two events

1. Sep 5, 2008

### ibc

Hello
in SR, the interval between two events is defined to be:
s^2=x^2 + y^2 + z^2 - t^2

so my question is, why does the time get a minus sign? what differs it from the other 3 dimensions, what's the physical meaning of this difference and what is the physical meaning of this interval s^2

I thought I understood it at first, but then, when derived the time dilation and lorenz contraction using coordinate systems, when I tried to do it with normal Pythagoras' theorem, I obviously got 1/(1+v^2)^0.5 instead of 1/(1-v^2)^0.5 each time, so again, where does this minus sign come from?

2. Sep 5, 2008

### Staff: Mentor

Re: Interval

Hi ibc, welcome to PF.

Consider the normal Euclidean distance formula from plane geometry:

s² = x² + y²

This is the definition of distance, but it is also the definition of a circle. In other words, a circle is the set of all points located a given distance from another point.

But what if you defined distance using hyperbolas instead circles? If you did that then your formula for distance would be:

s² = x² - y²

Just like the circular measure for distance is invariant under rotation, this hyperbolic measure for intervals is invariant under boosts. And if you start with this measure for intervals things like time dilation and length contraction come automatically. So it seems that the geometry of the universe is fundamentally hyperbolic in this sense.

3. Sep 6, 2008

### ibc

Re: Interval

So does it seem like the geometry of our universe is hyperbolic, or is there a solid reason for that?

And why do we use euclidean coordinate systems to draw the space-time diagram in relativity, if it actually means nothing, since you can't take sins and cosins or pythagoras theorem when you have 90° triangle?

4. Sep 6, 2008

### HallsofIvy

Staff Emeritus
Re: Interval

Fundamentally, it is because the wave equation is
$$\frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2}+ \frac{\partial^2 \phi}{\partial z^2}- \frac{\partial^2 \phi}{\partial t^2}= 0$$

5. Sep 6, 2008

### atyy

Re: Interval

In Euclidean space, we label points by (x,y,z) or (x',y',z'). But no matter what coordinates we choose, the actual distance d is the same:
x2+y2+z2=x'2+y'2+z'2=d [Equation 1]
Since the distance is invariant under coordinate changes, it describes a property of Euclidean space that is independent of coordinates.

In special relativistic spacetime, for a sphere of light sent out from the origin:
x2+y2+z2=c2t2

Rearranging:
x2+y2+z2-c2t2=0 [Equation 2]

From the Michelson Morley experiment, we know this is true, no matter what coordinates we choose. Thus Equation 2 is invariant for a light ray in Minkowski space, just as Equation 1 is the invariant distance in Euclidean space. So the LHS of Equation 2 measures a property of Minkowski space that is independent of coordinates for photons. We make a guess and generalise this to be an invariant property of Minkowski space for all particles, not just photons. And miraculously it works - and we call it the "interval". (As usual, I'm giving a quick and dirty method)

6. Sep 6, 2008

### ibc

Re: Interval

I think I see your point though I don't really see a direct connection between the phase changing of the electromagnetic wave and the time-space dimensions, but it would be nice if you explain it, or perhaps I should learn GR and I'll find it there?

anyway, what is still bothering me, is that I have a time-space diagram like this:
http://img135.imageshack.us/img135/4567/25459015kx4.png
(from the book Schutz B.F. - A first course in general relativity)
so when you take the parabola equation and the equation of observer O' time axis, and find their meeting point, you get the right time dilation equation (get the coordinates of point B which gives the right time dilation equation), but if I try simple geometry there, taking sins and cosins, I get 1/(1+v^2)^0.5 , so, what does this diagram means, if I have to add a weird minus sign if I want to take simpls sins and cosins, or pythagoras?

7. Sep 6, 2008

### atyy

Re: Interval

In Euclidean space the distance is the dot product of a vector with itself:
(x,y,x).(x,y,z)=x2+y2+z2.

The dot product is used to define angles u.v=|u||v|cos(a)

In Minkowski space, the dot product is the interval instead of the distance, so we should expect some difference in the formulas for angles.

However, note that any one inertial frame of Minkowski space still behaves normally. It's the relationship between inertial frames that is abnormal.

8. Sep 6, 2008

### ibc

Re: Interval

So in the real spacetime (not the low-speed newtonian one) the geometry is hyperbolic and not euclidian, therefore we cannot draw a real space-time diagram on an euclidian coordinate system, because the coordinates and angles shown on the diagram will mean nothing, yet books do draw the space-time diagram of relation between two observers on an euclidian coordinate system just as an attemp to make it visual?

(the question part is if it's actually as I said, or is this space-time diagram (such in the link before) does have some geometric meaning even on euclidian coordinate system)

9. Sep 6, 2008

### Staff: Mentor

Re: Interval

The geometry using circles to define distances is called Euclidean geometry. The geometry using hyperbolas to define intervals is called Minkowski geometry.

Is there a solid reason why the universe should be Euclidean instead of Minkowski?

The reason for that is that we draw spacetime diagrams on paper and explicitly ignore the timelike dimension. If you take the Minkowski metric (in units where c=1).

s² = x² + y² + z² - t²

and drop the timelike dimension t then you are left with the spacelike Euclidean metric

s² = x² + y² + z²

You can get around it by doing animations on a computer screen, but it is easier just to draw hyperbolas on the paper.

10. Sep 6, 2008

### atyy

Re: Interval

Yes. It is like people who draw a map of the globe on a flat piece of paper. It's distorted, but still useful, provided you know how to use it.

11. Sep 6, 2008

### Fredrik

Staff Emeritus
Re: Interval

It is because that particular model of spacetime predicts stuff that agrees with experiments. If that's not the sort of answer you had in mind, then what would you consider a "solid reason"?

This is a pretty strange question to me. I can resolve all the "paradoxes" that people think they see in SR by drawing a few lines on a piece of paper, and when I solve a problem I almost always start with a spacetime diagram because it makes things so much easier. It doesn't bother me at all that Pythagoras's theorem doesn't work. Why does it bother you?

Edit: D'oh! I'm too slow again. atyy already had the same idea, and it seems that you're starting to understand too.

The answer isn't in GR. I think his point is just that the Minkowski metric is the natural choice if the speed of light is to be the same in every inertial frame. (This takes some work to understand).

12. Sep 6, 2008

### Staff: Mentor

Re: Interval

If by "solid reason" you mean "some deeper underlying principle" as opposed to "agreement with experiments to date," then we do not know of any such principle (yet).

People often refer to "Lorentz symmetry" but I think this is just another name for what we already know about the geometry of the universe.

13. Sep 6, 2008

### ibc

Re: Interval

But we don't draw hyperbolas, we draw a regular euclidian geometry, or we draw hyperbolas on a regular euclidian geometry.
So these hyperbolas come of course from the minus sign of the t^2.
I'm not really sure of what I'm saying here, because I can't express it as a clear question, it's just I don't have a hard to to believe that pythagoras isn't working on all sorts of geometries, what I have difficulties with is how when we draw these Minkowski's space hyparbolas on euclidian geometry, and then it works, I mean, if we could somehow see a 4-dimensional minkowski space, the line s=(x^2+y^2+z^2-t^2)^0.5 would appear "straight" on that geometry? and then when we draw it on euclidian geometry, it looks hyperbolic, then if we disregard euclidian geomtry intuition and just use the formulas, it works?

14. Sep 6, 2008

### Fredrik

Staff Emeritus
Re: Interval

That's not a line at all. It's the 3-dimensional set of points that are spacelike separated from the origin and at a spacetime distance s2 from the origin.

By the way, the Euclidean metric and the Minkowski metric agree about which lines are straight.

15. Sep 6, 2008

### Staff: Mentor

Re: Interval

No, a hyperbola is the Minkowski equivalent of a Euclidean circle, a set of points that are all equidistant from the origin. Neither is a straight line. You do have to "apply formulas" and "disregard intuition" a lot because a hyperbola doesn't look like all the points are the same distance from each other.

16. Sep 6, 2008

### robphy

Re: Interval

Note that a
PHY101 position-vs-time graph is a [Galilean-]spacetime diagram...
in fact, one with a Galilean metric (which is also not Euclidean).
In such a diagram, you must take care to use your Euclidean intuition.
The "equidistant" points lie on a hyperplane instead of a hyperboloid as in Special Relativity.

17. Sep 6, 2008

### snoopies622

Re: Interval

You might like the result of plotting x vs. ti (with $$i^2=-1$$) instead of x vs. t. That way the interval looks Euclidean, and the angles represent rapidities so they can be simply added together.

18. Sep 7, 2008

### ibc

Re: Interval

Well I think my question was answered, so I want to thank you all, this forum seems like a very nice place.

And since we are talking about the interval, I wanted to ask about the invariance of the interval. I wonder if someone has a nice proof for that?
The one I have in my book is very mathematical (I know it's basicly a mathematical issue, but I'd be happy to hear a more physical one, or at least a less algebraic one)

19. Sep 7, 2008

### atyy

Re: Interval

Try posts 3,4. The invariance of the interval is the mathematical expression of the invariance of the speed of light. The invariance of the dot product in Euclidean space is why DaleSpam has being saying distance is measured with circles in Euclidean space.

20. Sep 7, 2008

### ibc

Re: Interval

"So the LHS of Equation 2 measures a property of Minkowski space that is independent of coordinates for photons. We make a guess and generalise this to be an invariant property of Minkowski space for all particles,"

ya I've seen that, but a guess to generalize is not really a solid proof, although I guess it's good enough for a none-mathematical proof =p