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Lorentz contraction from spacetime interval invariance 
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#1
Jan408, 01:32 PM

P: 997

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


#2
Jan408, 02:28 PM

P: 2,046

I don't know, maybe I'm blabbering (it's pretty late here), but I don't see how we can relate "spacecomponents" (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 (Nonzero [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. 


#3
Jan608, 12:02 AM

P: 380

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" 1v"/c" =T"/t" take sq root both sides (sq rt) 1v"/c" = T/t t= T/ (sq rt) 1v"/c" 


#4
Jan608, 12:58 AM

P: 997

Lorentz contraction from spacetime interval invariance
My problem is how to find out L=L(0)sqrt(1VV/cc) from the invariance of the space time interval without to use the previous result. Regards 


#5
Jan608, 10:01 AM

P: 3,967

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 (1v^2/c^2)[/tex] [tex] vT = L_0\sqrt{1v^2/c^2} = L[/tex] L is the distance A to B as measured by an observer comoving with the clock. 


#6
Jan608, 10:04 AM

P: 228

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 xaxis and any other x'axis. Now place one end of a rod of length L at the origin and the other end on the xaxis. This rod is at rest on the xaxis. 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).



#7
Jan608, 11:53 AM

P: 997




#8
Jan608, 01:05 PM

P: 321




#9
Jan608, 01:06 PM

P: 321




#10
Jan608, 01:08 PM

P: 3,967

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. 


#11
Jan708, 02:34 AM

P: 997

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 


#12
Jan708, 08:42 AM

P: 228

By setting two zerointerval expressions equal to each other you are not demonstrating the in variance for nonzero 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.



#13
Jan708, 01:37 PM

P: 3,967

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


#14
Jan708, 11:10 PM

P: 3,967

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 rearrange 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^2x'^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 


#15
Jan708, 11:32 PM

P: 997

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^2x^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'2x'^2. (2) Equating two different from zero relativistic expressions we have c^2t^2x^2=c^2t'^2x'^2 (3) Do you consider that (3) is a prove for the invariance of the relativistic spacetime interval? 


#16
Jan708, 11:50 PM

P: 3,967

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 ;) 


#17
Jan708, 11:50 PM

P: 321




#18
Jan708, 11:54 PM

P: 3,967




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