Spacetime interval of zero

1. Aug 15, 2007

morrobay

When an object is travelling at c the spacetime interval is zero.
I can follow the algebra that gets to this conclusion.
Is there a conceptual, mentally conceived image, explanation for a
spacetime interval of zero ?

2. Aug 15, 2007

robphy

Suppose light travels between two events, from P to Q.
On a spacetime diagram,
that the spacetime interval between those two events is zero
can be visualized by the fact that
Q lies on the future-light-cone of P, and
P lies on the past-light-cone of Q.

$$\] \begin{picture}(200,200)(0,0) \unitlength 1mm \put(20,0){\circle*{1} \put(-2,-5){ P }} \qbezier[200](20,0)(20,0)(70,50) \qbezier[200](20,0)(20,0)(0,20) \put(60,40){\circle*{1} \put(-2,5){ Q }} \qbezier[200](60,40)(60,40)(10,-10) \qbezier[200](60,40)(60,40)(110,-10) \end{picture} $$$ Last edited: Aug 15, 2007 3. Aug 15, 2007 ice109 how about an intuitive example of interval that is analogous to distance in 3space, got one of those? 4. Aug 15, 2007 JesseM The problem is that in 3-space two points at different locations can't have zero distance between them. If you use the pythagorean theorem to define distance in the complex plane, though, that is pretty closely analogous to the spacetime interval (for instance, the distance between the point (0,0) and (1,i) would be zero). 5. Aug 15, 2007 robphy Let me add to the original diagram... inserting a step in the construction. Forget about Q for now. Consider an event T that is in the timelike-future of P---say, at proper-time 3 ticks after P [along PT]. The set of all future events 3 ticks from P trace out a hyperbola centered at P... with equation t^2-x^2=3^2. This is analogous to a circle in ordinary Euclidean geometry with "radius" 3.. or square-interval 3^2. The asymptotes of that hyperbola are determined by t^2-x^2=0.. that is, (t-x)(t+x)=0 or (either x=t or x=-t)... the future light cone of P. Now, consider Q in my first reply. $$$ \def\qb#1#2{\qbezier(#1)(#2)(#2)} \begin{picture}(200,200)(0,0) \unitlength 1mm \put(20,0){\circle*{1} \put(-2,-5){ P }} \qbezier[200](20,0)(20,0)(70,50) \qbezier[200](20,0)(20,0)(0,20) \put(60,40){\circle*{1} \put(-2,5){ Q }} \qbezier[200](60,40)(60,40)(10,-10) \qbezier[200](60,40)(60,40)(110,-10) \put(20,0){ \put(7,12.2066){\circle*{1} \put(-2,-5){ T }} \qb{-22.0000,24.1661}{-25.0000,26.9258} \qb{-19.0000,21.4709}{-22.0000,24.1661} \qb{-16.0000,18.8680}{-19.0000,21.4709} \qb{-13.0000,16.4012}{-16.0000,18.8680} \qb{-10.0000,14.1421}{-13.0000,16.4012} \qb{-7.0000,12.2066}{-10.0000,14.1421} \qb{-4.0000,10.7703}{-7.0000,12.2066} \qb{-1.0000,10.0499}{-4.0000,10.7703} \qb{2.0000,10.1980}{-1.0000,10.0499} \qb{5.0000,11.1803}{2.0000,10.1980} \qb{8.0000,12.8062}{5.0000,11.1803} \qb{11.0000,14.8661}{8.0000,12.8062} \qb{14.0000,17.2047}{11.0000,14.8661} \qb{17.0000,19.7231}{14.0000,17.2047} \qb{20.0000,22.3607}{17.0000,19.7231} \qb{23.0000,25.0799}{20.0000,22.3607} \qb{26.0000,27.8568}{23.0000,25.0799} \qb{29.0000,30.6757}{26.0000,27.8568} \qb{32.0000,33.5261}{29.0000,30.6757} \qb{35.0000,36.4005}{32.0000,33.5261} \qb{38.0000,39.2938}{35.0000,36.4005} \qb{41.0000,42.2019}{38.0000,39.2938} \qb{44.0000,45.1221}{41.0000,42.2019} \qb{47.0000,48.0521}{44.0000,45.1221} \qb{50.0000,50.9902}{47.0000,48.0521} \qb{53.0000,53.9351}{50.0000,50.9902} \qb{56.0000,56.8859}{53.0000,53.9351} \qb{59.0000,59.8415}{56.0000,56.8859} \qb{62.0000,62.8013}{59.0000,59.8415} \qb{65.0000,65.7647}{62.0000,62.8013} \qb{68.0000,68.7314}{65.0000,65.7647} } \end{picture} \[$$

6. Aug 15, 2007

ice109

ahh makes sense. ok now answer this, why is c the conversion factor between distance and time? more like why is the norm of a velocity 4-vector c?

7. Aug 15, 2007

robphy

...because distance and time are expressed in different units

in geometry, unit vectors are convenient to work with... 4-velocities are unit timelike vectors in relativity. The c shows up because of my first sentence. If distances and times are expressed in seconds [distances in terms of seconds light has to travel to cover the spatial separation], then the norm of the 4-velocity would be 1 in those units.

8. Aug 16, 2007

ice109

yes v=x/t so vt=x, but why is c used?

9. Aug 16, 2007

robphy

If you choose a speed other than the speed of light for a conversion factor [say the speed of sound], you have a dimensionless constant factor that appears in lots of places in the equations of special relativity. That would be a nuisance... so, it is convenient to choose the speed of light.

In my opinion, there are two roles of the speed of light that one sees in special relativity. One is as a convenient conversion factor which makes calculations simpler. The other is as the maximum signal speed of special relativity, which is the physically important role.

Last edited: Aug 16, 2007
10. Aug 16, 2007

ice109

can you please elaborate on this. i'm assuming you mean that you can set c=1 and everything becomes pretty. essentially in a paper i have about minkowski space it says the c is the conversion factor between time and distance. i don't understand why but i guess that's just restating what i've already said.

11. Aug 16, 2007

robphy

Measure x in light-seconds (=3e8 m) and t in seconds.

MATH
Since we will come to realize that we will do geometry with these, it is convenient if they carried the same units. So, I choose a rescaled variable for distances... call it X=(x/clight), which is measured in seconds.
Explicitly, when x=1 light-second, we have X=(x/clight)=1 second.
If I choose the speed of sound (about 300m/s) for conversions, then for
x=1 light-second, we have X=(x/csound)=10^6 second.

PHYSICS
Now consider the square-interval in SR, where [by experiment] we observe the maximum signal speed to be: cmax=3e8 m/s...
\begin{align*} I^2 &=t^2-\frac{x^2}{c^2_{\mbox{max}}}\\ &=t^2-\left(\frac{x}{c_{\mbox{light}}}\right)^2 \left(\frac{c_\mbox{light}}{c_{\mbox{max}}}\right)^2\\ &=t^2-X^2 \left(\frac{c_\mbox{light}}{c_{\mbox{max}}}\right)^2\\ &=t^2-X^2 \end{align*}
Since, in SR, $$\frac{c_\mbox{light}}{c_{\mbox{max}}} = 1$$, $c_\mbox{light}$ is the best conversion factor for x.
The calculation of the physics looks simple.

If instead I used the speed of sound,
\begin{align*} I^2 &=t^2-\frac{x^2}{c^2_{\mbox{max}}}\\ &=t^2-\left(\frac{x}{c_{\mbox{sound}}}\right)^2 \left(\frac{c_\mbox{sound}}{c_{\mbox{max}}}\right)^2\\ &=t^2-X^2 \left(\frac{c_\mbox{sound}}{c_{\mbox{max}}}\right)^2\\ &=t^2-X^2 \left(10^{-6}\right)^2 \end{align*}
The calculation of the physics is complicated by this nuisance factor of 10^(-6)... because of the way my X is defined.

12. Aug 16, 2007

ice109

in the square interval equation why is x divided by the maximum signal propagation velocity? that seems to be the essence of my question

Last edited: Aug 16, 2007
13. Aug 16, 2007

robphy

For a light ray [an example of a fastest signal], $$\frac{\Delta x}{\Delta t}=c_{max}=\frac{\Delta x'}{\Delta t'}$$, independent of inertial observer.

So, the quantity $$(\Delta t)^2-\frac{ (\Delta x)^2}{(c_{\mbox{max}})^2}$$ is the same value [namely, zero] for this light ray, independent of inertial observer.

14. Aug 16, 2007

bernhard.rothenstein

clock synchronization involvement?

Is there some relationship with the clock synchronization in the involved inertial reference frames?

15. Aug 17, 2007

robphy

To be honest, I haven't thought much about the issues regarding "clock synchronization" [at least at the level that you probably wish to discuss]. At some point, I plan to understand the issue better [since it came up in something else I was working on]. For some reason, I have been able to be not too concerned about it... Maybe it's implicit in the methods that I use to discuss relativity... I'm not sure right now.

16. Aug 22, 2007

ZapperZ

Staff Emeritus
This thread is heading in a rather dubious direction. I will remind everyone involved to re-read the PF Guidelines that you have agreed to. Pay particular attention to speculative theories/ideas. We do not allow them on here, so if that is your intention, you've come to the wrong forum.

Zz.