In the left drawing: A is passing Z at .5c at event 1.For convenience A and Z clocks read zero at event 1. Two photons are emitted at event 1. The 1st reflects from a mirror at x (event 2), and returns to A where it is reemitted to return to Z. The mirror is at rest in the Z frame. The 2nd photon reflects from a mirror at the end of a rod of length d (event 2) and returns to A at event 3 where it is reemitted to return to Z. Event 4 is A passing the mirror in the Z frame. The two mirrors are side by side, and the reflections are considered as simultaneous events by A and Z. The right drawing is A's description of events. Here are some values for v=.5c event: (t, x), (t', x')....(primed values are for A) 1: (0, 0), (0, 0) 2: (10, 10), (5.8, 5.8) 3: (13.3, 6.7), (11.5, 0) 4: (20, 10), (17.3, 0) d: (0, 5.0), (0, 5.8) Why can't the difference in length be the result of A measuring the length too long? https://www.qdrive.net/en/home/images/?userid=2104&ky=B2O3NAOItmM5G5xXXtqxmmPvxF6psP7XLXm&id=107076&fn=len%20contraction.gif
Are you familiar with Minkowski diagrams? A times the signal along the rod to the mirror and back (path 123), which equals t'. d'=ct'/2 The relevant issue is comparing event 2 in the measurements by A and Z.
Is this description of the setup correct: In a 2 dimensional Lorentzian space, two inertial observers, and two inertial mirrors zip through space. Each observer measures one of the mirrors to be in the same state of motion as himself. So we have Observers A and Z, and mirrors AM and ZM. (Equivalently, you may assume that AM and ZM are attached to A and Z by long sticks.) At the instant A and Z meet (event 1) a flash of light is sent towards the mirrors and happens to hit at the time the two mirrors meet (event 2). The (reflected) flash hits observer A (event 3) before A hits ZM (event 4). Both observers have the information they need to measure the distance between the worldlines of A and AM. Working with algebra instead of arithmetic, we solve for the relationship between Z's measurement of the distance between Z and ZM and A's measurement of the distance between A and AM. Since the meeting of the mirrors happens just as a light-speed signal from the meeting of the observers reaches the mirrors, there may be one or more constraints on z and a'. (a' is measured by A, so it gets primed to be consistent in methodology.) Coordinates in Z's frame, (t,x) (calculated from Z's measurement of distance z to ZM and movement of A at speed v) 1 : (0, 0) 2 : (z/c, z) 3 : (2z/(c + v), 2zv/(c + v)) 4 : (z/v, z) Coordinates in A's frame, (t',x') (calculated from A's measurment of distance a' to AM and movement of ZM at speed -v) 1 : (0, 0) 2 : (a'/c, a') 3 : (2 a'/c, 0) 4 : (a'/c + a'/v, 0) Coordinates in A's frame (t', x') (calculated from Z via Lorentz transform) t' = γ(t - vx/c²) x' = γ(x - vt) 1 : (0, 0) 2 : (γ(z - vz/c)/c, γ(z - vz/c)) 3 : (2 γ(z - vz/c)/c), 0) 4 : (γ(z/v - vz/c²), 0) Now to check if the Lorentz transform is consistent, we examine the eight equivalences: t'_1 ) 0 ≟ 0 t'_2 ) a'/c ≟ γ(z - vz/c)/c t'_3 ) 2a'/c ≟ 2 γ(z - vz/c)/c) t'_4 ) a'/c + a'/v ≟ γ(z/v - vz/c²) x'_1 ) 0 ≟ 0 x'_2 ) a' ≟ γ(z - vz/c) x'_3 ) 0 ≟ 0 x'_4 ) 0 ≟ 0 Equation x'_2 is key, and assuming it to be true obviously is the same constraint as t'_2 and t'_3 but what about t'_4? a'/c + a'/v = γ(z - vz/c)/c + γ(z - vz/c)/v = γ(z/c - vz/c² + z/v - z/c) = γ(z/v - vz/c²) and it checks. For these two descriptions to be equivalent, a' = γ(z - vz/c) must be true. Checking, you have c = 1, v = 0.5, z = 10 You list a' = 5.8, but I calculate a' = γ(z - vz/c) = 5/sqrt(0.75) = 5.7735+ which is reasonably close. In fact all of your coordinates are in agreement with all means of calculating them. Now Z measures the distance between A and AM as the time between events 2 and 4 times the speed v, so d = (t_4 - t_2)v = (z/v - z/c)v = z(1 - v/c) and A measures the distance between A and AM as 1/2 of the time between events 1 and 3 times the speed c. so d' = (c/2)(t'_3 - t'_1) = (c/2) 2 a'/c = a' = γ(z - vz/c) = γ d But the endpoints A and AM are moving with respect to Z, so the length Z measures should be smaller than the length A measures. 5.0 = d < d' = γ d = 5.7735+ So the difference in d versus d' is not any type of measurement error. It is 100% pure Lorentzian length contraction. Z sees the endpoints moving and so Z measures the length as shorter than A. d = z(1 - v/c) = a'/γ d' = a' = γ d = γ z(1 - v/c)
You get away with this because this OP wrote. Are you familiar with Minkowski diagrams? There are ways around this but the OP does not see this. You are therefore correct in this case only.
Yes, I'm familiar with Minkowski diagrams. So, you're asking whether that method of measuring the length of the rod is an invalid method? No. A would get exactly the same result using an accurate tape measure.
In the Z frame, d/x = 5/10 = .50. In the A frame, d'/x' = 5.8/8.7 = .66. If x is Z's measuring tool, then x' is the same tool for A. It's contracted, but there are still 10 divisions on it. If the A rod contracts, it should remain the same ratio, but is doesn't! Why not?
In the A frame, d'/x' = 5.8/8.7 = 2/3 x is contracted in that frame. d is not - d is contracted in the Z frame. How could they remain the same ratio?
rpenner; Thanks for all that effort. My values were rounded to 1 place, ...at least they were correct. pete; I'll give it more thought and get back to you.
In the previous example, I confused the rest length in Z with the measured length in Z, because the end positions were the same in the drawing, or something to that effect (mental block). Here's a simpler but very interesting one. In the Z frame is a rod x meters long, with one end at the origin, pointing in +x direction, with a mirror at the far end. case 1. A passes Z in +x direction at .5c, and at the origin emits a photon in the +x direction that reflects from the mirror to return to A. From Lorentz transform, x'=g(x-vt) If x=10, x'=5.77 case 2. A passes Z in -x direction at .5c, and at the origin emits a photon in the +x direction that reflects from the mirror to return to A. From Lorentz transform, x'=g(x+vt) If x=10, x'=17.32 Please check for errors. Thanks
Um, you are not thinking four-dimensionally. The Lorentz transform of x is not length contraction, because length is the x separation between two events (or equivalently, world-lines) at the same time and since both x and t transform, you are now giving a x' for two events for which t' is not zero.
In both cases, A is measuring his distance from the reflection event. If A is at the origin when the signal is emitted, the 'x' transformation has the same form as doppler shift. x'=g(x-vt)=gx(1-v/c)=x*sqrt[(1-v/c)/(1+v/c)] x'=g(x+vt)=gx(1+v/c)=x*sqrt[(1+v/c)/(1-v/c)] With doppler shift, the sqrt of the product of approaching freq fa and receding freq fr equals the freq in the emitting frame f. In this example, sqrt(xa'*xr')=x. (x is representing wave length.) You can measure the distance (length) x in the Z frame, by making a pass in opposite directions. You would have to call Z and arrange for him to setup the mirror.
This is not how LT works. It expands the stationary length and then expands the distance between the two light emission points, of the stationary frame and the moving frame and subtracts the two. The result is the x' length. x' = ( x - vt )γ x' = xγ - vtγ. 1) xγ is the length in the stationary frame expanded by γ. 2) vtγ is the distance between the two light emission points expanded. The second term is the confession by SR that it must accept multiple light emission points and therefore switch from one to the other.
