lorentz contraction from space-time interval invariance

bernhard.rothenstein

Please tell me if it is possible to derive the formula which accounts for the Lorentz contraction from the invariance of the space-time interval.
Thanks

neutrino

I don't know, maybe I'm blabbering (it's pretty late here), but I don't see how we can relate "space-components" (of two different inertial frames) of the spacetime interval this way. For one, the interval relates the space and time coordinates of two distinct events as observed in two different frames. While measuring the length say, of a rod, [itex]\Delta x[/itex] must be measured at one instant in time, and therefore [itex]\Delta t[/itex] must be 0. Since these events are simultaneous in the unprimed frame, they must not be so in the primed frame (Non-zero [itex]\Delta t'[/itex]), which implies [itex]\Delta x'[/itex] was not measured at a "single point" in time.

Someone please tell if that made any sense, and please correct me if I'm wrong. Something tells me that there might be a way to derive the formula, since the invariance is a fundamental property, but as I said, I'm getting ready for some Z's now.

morrobay

Is this the algebra your looking for ?

Let " equal squared
Let 0" -(cT)" equal spacetime interval of stationary observer
Let (vt)" - (ct)" equal spacetime interval of moving observer
-(cT)" = (vt)" -(ct)" invariance of spacetime interval
c"t" = v"t" + c"T" divide by t"
c"=v"+ c"T"/t" divide by c"
1-v"/c" =T"/t" take sq root both sides
(sq rt) 1-v"/c" = T/t
t= T/ (sq rt) 1-v"/c"

bernhard.rothenstein

Thanks for your answer. That is the derivation of the time dilation formula.
My problem is how to find out
L=L(0)sqrt(1-VV/cc)
from the invariance of the space time interval without to use the previous result.
Regards

yuiop

Try this:

[tex] -(c^2T^2) +X^2 = -c^2t^2 + L_0^2[/tex]

Say a clock is transported from A to B (distance [tex]L_0[/tex]) in time t at velocity v relative to your frame.
T is the proper time of the transported clock as measured by itself (by definition).
X is the proper distance the clock moves in its own frame which is zero .

[tex] -(c^2T^2) = -c^2t^2 + L_0^2[/tex]

[tex] T^2 = t^2 - L_0^2/c^2[/tex]

[tex] T^2 = L_0^2/v^2 - L_0^2/c^2[/tex]

[tex] T^2 v^2 = L_0^2 -L_0^2v^2/c^2[/tex]

[tex] T^2 v^2 = L_0^2 (1-v^2/c^2)[/tex]

[tex] vT = L_0\sqrt{1-v^2/c^2} = L[/tex]

L is the distance A to B as measured by an observer comoving with the clock.

country boy

Begin with the unit invariant interval x^2-(ct)^2=1. This traces out a hyperbola on a Minkowski diagram that gives the unit distance along the x-axis and any other x'-axis. Now place one end of a rod of length L at the origin and the other end on the x-axis. This rod is at rest on the x-axis. The world lines of the ends of L cross the x'-axis at the origin and some other point L'. To get the Lorentz contraction, compare L and L' to where the hyperbola crosses the x- and x'-axes (the unit lengths in the two frames).

bernhard.rothenstein

Thanks. Do you know a derivation of the invariance of the space-time interval

1effect

Look up pages 42-43 in "Spacetime Physics" by Wheeler and Taylor.

1effect

That is a postulate, so there is no derivation for it.

yuiop

I notice in another thread http://www.physicsforums.com/archive/index.php/t-115451 you posted this:

..which seems perfectly valid to me, but you were bothered that the final equation is obtained by subtracting (ct)^2 from both sides of the first two equations so that the first two equations equal zero and then saying if two quantities are equal to zero then they are equal to each other.

As far as know using the technique of setting two equations to zero and saying they equal to each other is a perfectly normal mathematical technique.

bernhard.rothenstein

Thanks. Do you consider that starting with
c*t*-x*=d* (1) *2
c"t'*-x'*=d* (2)
a consequence of the invariance of distances measured perpendicular to the direction of relative motion d then
c*t*-x*=c*t'*-x'*
is more convincing?
Regards

country boy

By setting two zero-interval expressions equal to each other you are not demonstrating the in variance for non-zero values. The argument could be completed by deriving the Lorentz transformations from your two light spheres and then showing that all spacetime intervals are invariant.

yuiop

Hi Bernhard. Your symbols are little confusing and I'm not sure exactly what you are getting at. What is the stray *2 at the end of equation (1) ? I'm assuming the bracketed number are equation numbers and not used in the same context as you you L(0) earlier. In equation (2) I'm assuming you meant to put c* instead of c" to represent c^2. I'm not sure where you get d^2 and I'm not sure that starting with the premis that distances measured perpendicular to the direction of relative motion is invarient demonstrates that the interval is invarient in all directions is ... entirely convincing.

Since ct = x and c^2t^2 = x^2 the quantity c^2t^2-x^2 is zero and equating it to d^2 only hides the zero.

The statement by countryboy

would appear to be correct.

yuiop

This is my stab at deriving the invarient interval but it's not very rigorous ;)


Start with speed of light is the same to all observers, so

(1) x/t = c and
(2) x'/t' =c

(1) and (2) are in terms of velocity so re-arrange them in terms of distance (because we are looking for the invarient spatial interval) to get

(3) x = ct and
(4) x' = ct'

Now x can take positive or negative values depending on direction from the origin so square both sides of each equation to remove any pesky negatives. (If we were doing the full version with x, y and z coordinates there would be a sqrt(x^2+y^2+z^2) term that can have a positive or negative root)

(5) x^2 = (ct)^2 and
(6) x'^2 = (ct')^2

Now subtract the left side of (6) from the left side of (5) and right side of (6) from the right side of (5). We are in effect subtracting the same value from each side of equation (5) because the equality of (6)

(7) x^2-x'^2 = (ct)^2 - (ct')^2

Rearrange to get

(8) x^2 - (ct)^2 = x'^2 -(ct')^2 which is what we are looking for.

The full version can be done by starting with

(9) sqrt(x^2+y^2)+z^2)/t = c and
(10) sqrt(x'^2+y'^2+x'^2)/t' =c

but I'm too lazy :P

bernhard.rothenstein

Thanks kev. It is of big help to me.
Please have a critical look at the following:
Consider the well known light clock experiment performed in its rest frame I(0). Let d be the distance between the two mirrors. Being measured along a direction which is perpendicular to the direction of relative motion between the involved inertial reference frames it has the same magnitude in all of them. Consider the same experiment from I relative which moves relative to I(0) in the standard way. Let x be the horizontal displacement of the upper mirror and ct the distance travelled by the light signal between the two mirrors. Pythagoras' theorem requires
d^2=c^2t^2-x^2 (1)
Invariance of d requires the invariance of the right side of (1) and so in a I' reference frame we should have
d^2=c^2t'2-x'^2. (2)
Equating two different from zero relativistic expressions we have
c^2t^2-x^2=c^2t'^2-x'^2 (3)
Do you consider that (3) is a prove for the invariance of the relativistic space-time interval?

yuiop

OK, I see what you are getting at now.
When the light clock is moving relative to you the light pulse goes a distance ct in time t and this is equivalent to the diagonal distance sqrt(d^2+x^2) where the light clock moves a distance x in the same time t. This works for any relative velocity of the light clock so its seems a reasonable approach. However I'm not qualified to say if this would be regarded as a rigorous formal proof ;)

1effect

The invariance of the space-time interval is a postulate. You don't prove postulates since they are ...postulates.