# Proving Invariance of Spacetime Interval

1. Mar 24, 2006

### emob2p

Hi,
I was wondering if there was a way to prove the invariance of the space-time interval just from postulating a constant speed of light and an isotropic space-time. Most arguments go "from the Lorentz transformations it can be seen that the interval is invariant." Can we show the invariance without first appealing to the Lorentz transformations? Thanks.

I ask because many texts define L.T. as linear transformations that leave the interval invariant. But if we use L.T. to show the invariance, then the observation of invariance is trivial since that's how we definied our L.T. So I guess I'm looking for an independent proof/argument.

2. Mar 25, 2006

### Staff: Mentor

The Lorentz transformations can be derived from the postulates of a constant speed of light and isotropic space-time. It is not necessary to define them in the way you describe.

Perhaps postulating an invariant space-time interval is equivalent to postulating a constant speed of light and isotropic space-time? I wouldn't be surprised if this were the case.

Last edited: Mar 25, 2006
3. Mar 25, 2006

### Galileo

If you derive the Lorentz transformations from the postulates and the invariance of the interval from the Lorentz transformation you, then you have proved the invariance of the interval from the postulates (somewhat indirectly).

Now you say many textbooks define a LT by a transformation which leaves the interval invariant. Then clearly the book should already show that the interval is invariant from the postulates. So I don't really understand your problem.

Anyway, it's very easy to see it directly. Imagine systems S and S'. Person A stands at the origin in S and person B stands in the origin of S', which moves with velocity v relative to S. The origins coincide at t=t'=0. Imagine A lights a spark at t=0. A spherical light wave spreads out in all directions. At any point in time A stands in the center of the sphere which has radius ct. So the wavefront obeys the equation:
$$x^2+y^2+z^2=(ct)^2$$
Same thing holds in S'. B also stands in the center of the sphere at all times (2nd postulate):
$$x'^2+y'^2+z'^2=(ct')^2$$
so whatever happens to the coordinates, it should always be the case that:
$$x^2+y^2+z^2-(ct)^2=x'^2+y'^2+z'^2-(ct')^2$$

Last edited: Mar 25, 2006
4. Mar 25, 2006

### robphy

Use a radar method (which secretly employs the eigenvectors and eigenvalues of the Lorentz Transformation).

See
the references in post #5 in https://www.physicsforums.com/showthread.php?t=88839
where the square-interval is described as the area of certain parallelograms
and

* the Bondi k-calculus (suggested in posts #4 and #6 above and described in https://www.amazon.com/exec/obidos/tg/detail/-/0486240215?v=glance" )
* the radar approach (described in https://www.amazon.com/exec/obidos/tg/detail/-/0226288641?v=glance")
where the square-interval is expressed as the product of two stopwatch-intervals in a radar experiment.

Last edited by a moderator: May 2, 2017
5. Mar 25, 2006

### emob2p

Galileo -
For your argument, aren't you only showing that the interval measured for a light ray is the same in both frames. How does this show that the interval between two general events is invariant?

6. Mar 25, 2006

### Perturbation

If you perform a linear transformation on the interval where each spatial/temporal interval in the new frame becomes a linear combination of those in the old frame, i.e. $(\Delta x^{\mu}')^2=\sum_{\alpha}M^{\mu '}_{\alpha}\Delta x^{\alpha}$, one can show the invariance. Lorentz transformations are then these symmetric linear transformations on the space of four-vectors that leaves the Lorentz norm invariant ($\eta_{\mu '\nu '}\Lambda^{\mu '}_{\alpha}\Lambda^{\nu '}_{\beta}=\eta_{\alpha\beta}$, $(\Lambda^{\mu '}_{\alpha})\in SO^{\uparrow}_{+}(1, 3)$).

There's a proof of the invariance of the interval in Schutz's "A First Course in GR", which shows it before Lorentz transformations are actually introduced.

Last edited: Mar 26, 2006
7. Mar 25, 2006

### emob2p

Maybe I'm missing the logic of Schutz's argument, but in his derivation he assumes that $$\Delta s^2$$ is zero. Like I asked Galileo, what about cases when the two events described isn't a light ray, i.e. $$\Delta s^2$$ is not zero.

8. Mar 25, 2006

### Perturbation

No, he uses a null interval to deduce the relationship between the coefficients of the transformation, since he'd already shown geometrically and from the invariance of c that a null line is null for all inertial frames.

9. Oct 19, 2007

### bernhard.rothenstein

space tikme interval

Please give me an explanation for the fact that from the first two equations results the third one.
Thanks

10. Oct 19, 2007

### yogi

The invariance of the interval is a direct consequence of Minkowski's unification of space and time - all experimental evidence reduces to, and can be explained by, this simple unification. The one way velocity of light which Einstein relied upon as a postulate was actually unnecessary - more than what is needed to explain MMx - you can get to over and back constant c from Einstein's premise - or you can get there with a more modest proposition of "Space-time" unification. The latter leads directly to a constant two way velocity of light" in other words, Einstein over postulated. I am not by this post intending to say Einsten was wrong - that remains to be determined. All I am saying is that, after 1.5 bottles of wine, if you want to be consistent with two way (over and back) experimental evidence, there is no necessity to impose a c velocity hypothesis on one way experiments.

Last edited: Oct 19, 2007
11. Oct 19, 2007

### Hurkyl

Staff Emeritus
Elementary arithmetic

$$x^2+y^2+z^2-(ct)^2 = (ct)^2 - (ct)^2 = 0 = (ct')^2 - (ct')^2 = x'^2+y'^2+z'^2-(ct')^2$$

12. Oct 19, 2007

### bernhard.rothenstein

space-time interval

Thanks.
My problem is: Are equal to each other two quantities which are eqial to zero?
It would not be better to start with
xx-cctt=A(x'x'-cct't') (1)
and to impose the condition that (1) accounts for the time dilation which can be derived without usung the LT. The result is that A=1.
Soft words and hard arduments

13. Oct 26, 2007

### bernhard.rothenstein

space-time interval invariance

You did not answer my question
Thanks.
My problem is: Are equal to each other two quantities which are eqial to zero?
It would not be better to start with
xx-cctt=A(x'x'-cct't') (1)
and to impose the condition that (1) accounts for the time dilation which can be derived without usung the LT. The result is that A=1.
Soft words and hard arguments

14. Oct 26, 2007

### Hurkyl

Staff Emeritus
Are you seriously asking me if equality is transitive?

15. Oct 26, 2007

### bernhard.rothenstein

space-time interval

I ask why did Einstein consider
x-ct=0
x'-ct'=0
x-ct=a(x'-ct')
and not
x-ct=x'-ct' ?

16. Oct 26, 2007

### Hurkyl

Staff Emeritus
I'm not particularly inclined to review the thread to see what they were talking about, and even if what you've asked is even directly related to it.

17. Feb 3, 2009

### Chephysic

There is a way to show the invarience of space-time in a very straight forward manner:

Given the Lorentz transformations:

dt = Ƴ( dt' + (v*dx')/(c^2) )
dx = Ƴ( dx' + v*dt' )

Ƴ = 1/(1-(v/c)^2)^.5

-> dt and dx denote finite changes in the S frame, dt' and dx' denote changes in the S' frame... ds denotes the space-time interval.

Then given the space-time relation:

ds^2 = (c*dt)^2 - (dx)^2

Substitution for dt and dx, along with some algebra produces something similar to:

ds^2 = Ƴ^2*( 1 - (v/c)^2 )*( (c*dt')^2 - (dx')^2 )

Noting that Ƴ^2 = 1/( 1 - (v/c)^2 )
Readily our equation becomes:

ds^2 = (c*dt')^2 - (dx')^2

The left hand side of our equation did not change, and the choice of S' was arbitrary so this holds for any reference frame. The algebra is takes some getting used to, especially trying to get things similar to gamma (Ƴ) to come out, but the best advice is to factor c, dx' and dt' in terms that would lead you to the result shown. salud

18. Feb 4, 2009

### bernhard.rothenstein

Consider the inertial reference frames I and I’ in the standard configuration,. In both frames the clocks are standard synchronized. I’ moves with velocity V in the positive direction of the overlapped x,x’ axes.
We are in I’ and consider the clocks K’0(x’=0,y’=0) and K’(x’=0,y’=d). Being synchronized by a light signal emitted from O’ at a time t’=0 clock K’ reads d/c. Detect the same synchronization from I. There clock K’ reading d/c is located in front of a clock K(r,θ) which reads r/c. Because clock K’ has advanced with x=r/c in the positive direction of the overlapped x,x’ axes Pythagoras’ theorem leads t
d^2=r^2-x^2=c^2t^2-x^2
The invariance of distances measured perpendicular to the direction of relative motion enables us to conclude
c^2t^2-x^2=invariant
That derivation avoids 0=0!!!
Please be my next room physicist and tell me if you agree.

19. Feb 4, 2009

### Ich

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Last edited: Feb 4, 2009
20. Jun 6, 2010

### qiuhd

I think Invariance of Spacetime Interval come form the invariance of the speed of light.

p1---------------mid---------------p2
------------------A
------------------B-->
Suppose at time t1 there were two light flashes occurred on p1 and p2.
A stayed still at the middle point

at time t2 A saw the flashes

B passed the middle point at t2 at speed of v.

Because of the invariance of the speed of light, B should see the two flashes too.

So the Spacetime Interval of the two flash is the same for A and B.

This is the special case, adjust the point and the time those two flash occurred, any Spacetime Interval can be created.