I can't get my head around relativity and the equivalence of different frames of reference. I was wondering if anyone could help clear the issue up for me, by answering the following problem. Let's assume that a group of four missiles are travelling in a diamond pattern each one moving forward at 0.5c towards a point X. Another group of missiles is travelling downward at 0.5c towards X, such that both sets of missiles collide (and explode producing a blast that travels at 0.9c in all directions) with each other simultaneously from an observer's POV. My question: from the POV of an observer on one of the misiles: do any of the missiles explode sooner/later than others? Cheers, Please Register or Log in to view the hidden image!
I don't quite follow the geometry you are describing here, but you can calculate this by using the Lorentz Transform. Simply determine the spacetime coordinates for the missles exploding in the observer's frame. Then transform them into the missile's frame. If the time coordinates are not equal in the missile's frame then they did not explode simultaneously in that frame. -Dale
If the four missiles in each group are all have the same, say, x coordinate and are all travelling in the x direction, all the collisions will take place simultaneously in any reference frame also moving in the x direction.
Regarding the link provided by Dale: I would like to better understand the equations under the heading "Lorentz Transform for Frames in Standard Configuration". For now I am letting gamma = 2 (therefore v = .866c = 259.634 km /sec). When I let x = 0, t' = 2t which I understand to mean that there are twice as many units on the t' time axis. To obtain a tick rate (frequency) of a clock, I assume that I should use 1/t' to represent a measure of actual time dilation. Am I on the right track so far? Likewise, when I let t = 0, x' = 2x which I understand to mean that there are twice as many units on the x' axis. So, should I use 1/x' to represent a measure of actual length contraction? Something like "units per kilometer"? It makes some sense so far, but what I still can't figure out is how I can use these equations to measure loss of simultanaety. I would like to calculate the readings of synchronized clocks in the x' frame as viewed from the x frame where they are not synchronized. I am not sure what to use for t when I plug in various x values, etc. Any help in understanding what these formula actually represent would be greatly appreciated!
I don't see the point of the second set of 4 missiles. All you need to say is that the first set of missles self-destructs simultaneously in frame O (the frame in which each missle's speed is 0.5c). And even then I don't see the point of having 4 of them. All you need is 2 missles detonating to get the point across. We have 2 events. I'll choose some spacetime coordinates for them in the lab frame (frame O). In Frame O: Event 1: Missile 1 detonates at point (x<sub>1</sub>,y<sub>1</sub>,t<sub>1</sub>)=(1,0,0) Event 2: Missile 2 detonates at point (x<sub>2</sub>,y<sub>2</sub>,t<sub>2</sub>)=(0,1,0) Furthermore in Frame O, Missile 1 has a velocity of u<sub>1</sub>=0.5ci and Missile 2 has a velocity of u<sub>2</sub>=0.5cj. Now the $64,000 question: What are the spacetime coordinates of these events in the frames of Missiles 1 and 2 (call them O' and O'', respectively)? To obtain the spacetime coordinates in Frame O': Use the following transform. x'=γ(x-vt) y'=y t'=γ(t-vx/c<sup>2</sup>) To obtain the spacetime coordinates in Frame O'': Use the following transform. x'=x y'=γ(y-vt) t'=γ(t-vy/c<sup>2</sup>) The calculation is left as an exercise for the reader. Please Register or Log in to view the hidden image!
Here's my calculation, although it's probably not worth $64,000: v = .5c = 149.9 km/sec γ = 1.155 Frame O': (1.155km, 0, -.002sec) Frame O": (0, 1.155km, -.002sec) I'm not sure I really know how to interpret these results. Each missile is always at x=0 and y=0 in their own frame, so the 1.155 coordinates must be relative to frame O. Same with the time dimension, I suppose, but I just can't visualize it for some reason...
You know what? I underspecifed the problem. I didn't fix it so that the missiles were actually at the location of the detonation events at the time of detonation! I'll need to define another event. Also, I'll specify that the spatial units are in light years and the time units are in years. Event 0: Both missiles are launched from the origin of O. The spacetime coordinates of this event in O are: (x<sub>0</sub>,y<sub>0</sub>,t<sub>0</sub>)=(0,0,-2) Now the missiles will be located at the detonation points at t=0. In Frame O the spacetime intervals between launch and detonation for each missile are as follows. Δs<sub>01</sub>=s<sub>1</sub>-s<sub>0</sub>=(1,0,2) Δs<sub>02</sub>=s<sub>2</sub>-s<sub>0</sub>=(0,1,2) Now since the question was about simultaneity, let's transform the time intervals only. Frame O' Δt<sub>01</sub>'=γ(Δt<sub>01</sub>-vΔx<sub>01</sub>/c<sup>2</sup>) Δt<sub>01</sub>'=(2/sqrt(3))(2-(0.5)(1))=sqrt(3)=1.73 y Δt<sub>02</sub>'=γ(Δt<sub>02</sub>-vΔx<sub>02</sub>/c<sup>2</sup>) Δt<sub>02</sub>'=(2/sqrt(3))(2-(0.5)(0))=4/sqrt(3)=2.31 y How do we interpret this? Δt<sub>01</sub>' is the time of flight of Missile O' in Frame O', and Δt<sub>02</sub>' is the time of flight of Missile O'' in Frame O'. They aren't the same, which indicates that the detonations aren't simultaneous in O'. A similar analysis will reveal that they aren't simultaneous in O'' either. In fact the detonations occur in the reverse order in O''.
Drawing this Spacetime Diagram for another thread really helped me understand the Lorentz transform. Here I used v=3/5c so γ=5/4. The units of time are seconds and the units of distance are light-seconds. The black lines represent the "rest" (unprimed) frame. The white lines represent the "moving" (primed) frame. The red and blue lines are light pulses emitted from an emitter moving on the x'=0 line at 1s intervals. You are on the right track, but usually you have to look at several lines in spacetime to really compare things. For example, you really need to compare the line x=0 (a resting clock) with the line x'=0 (a moving clock) to properly visualize time dilation. Also, the Lorentz γ factor is unitless, so it would be t/t' to determine time dilation, not 1/t' which would have units of Hz. Same thing here, it should be x/x' to get a unitless γ, not 1/x'. To see the relativity of simultaneity look at the line t'=0. Notice that t varies. Any two distinct events on that line are simultaneous in the primed frame, but not in the unprimed frame. I really recommend you construct one of these spacetime diagrams. Use the Lorentz transform to accurately place everything as a learning process. Once you have done that you can study a single image to geometrically understand time dilation, length contraction, relativity of simultaneity, the invariance of c, and reciprocity. -Dale
Yes, good idea, I will try and make a spacetime diagram to help me understand. The slopes of the red and blue lines appears to be -1 and 1. Is that by convention, like a light-cone diagram, or is it inherently part of the graph? After I draw one myself I will probably realize the answer to that question was more obvious than I thought. Of course, I should have realized that. I thought I wanted a term with Hz as units because I thought I wanted a frequency shift for the clock. Makes more sense as a unitless ratio. Thanks. I see that in your diagram, and it makes sense, but I have not quite figured out how to get that from the Lorentz Transforms. But I think you're saying that as I use the equations to plot the spacetime diagram, everything will become evident. Thank you for the assistance Dale. To really show reciprocity, should I draw two spacetime diagrams? No I suppose they would be identical, or mirror-image of each other. Alright I think I know what to do. Thanks again.
FYI, chapter 3 of this free textbook will show you how to draw spacetime diagrams. You can also learn how to do and interpret the calculations I did from the book.
So far I have determined that, in Dale's diagram, the slope of the light ray path is 1 (or -1) because of the chosen units for x. Apparently, from Pathagorus, the chosen unit for x is sqrt(2) light seconds. (Assuming the units for t are seconds). Later, by edit: Unless the length of the hypotenuse does not actually represent the distance the light travelled. Thanks Tom, that looks like a great resource.
The red and blue lines represent pulses of light traveling through spacetime so their slope is c. Since my units are seconds and light-seconds c=1. Using c=1 just makes constructing everything easier. Yes, at least that is what happened for me. If you get stuck drawing your diagram let me know. Hopefully I can give you pointers without spoiling the value of drawing it. You can certainly do two diagrams and see that they are mirror-images of each other, but you can actually see reciprocity from a single diagram. That is the thing that took me the longest to pull from my diagram; it isn't obvious but it is there. -Dale
This is the data I would use to plot the spacetime diagram for v=.866c (γ=2): Units of length are light-seconds, and units of time are seconds. (x, t, x', t') (0, 0, 0, 0) The origin for both coordinate systems (1.70, 2.00, 0.00, 1.00) The intersection of the x'=0 diagonal line with the t=2 horizontal line. This defines the slope of all x' diagonal lines to be 2/1.7 and the y-intercept of the x'=0 diagonal line is known to be 0. (2.30, 2.00, -1.15, 0.00) The intersection of the t'=0 diagonal line with the t=2 horizontal line. This defines the slope of all t' diagonal lines to be 2/2.3 and the y-intercept of the t'=0 diagonal line is known to be 0. (1.15, 0.00, 2.30, -2.00) The intersection of the t'=-2 diagonal line with the x=0 horizontal line. This defines the horizontal spacing between all t' diagonal lines to be 1.15 so they can be plotted accordingly. (Like Dale's diagram, I would plot t'=0 then t'=-2 without showing t'=-1 in between.) (1.00, 0.00, 2.00, -1.70) The intersection of the x'=2 diagonal line with the x=0 horizontal line. This defines the horizontal spacing between all x' diagonal lines to be 1.00 so they can be plotted accordingly. (Like Dale's diagram, I would plot x'=0 then x'=2 without showing x'=1 in between.) When I have time I'll plot the graph (assuming I have done this correctly?). I'd also like to plot a few different gammas, although I can imagine that it will just be different amounts of "skewing". At least the transforms are making more sense to me now. I can envision synchronized clocks in the primed frame being unsynchronized in the unprimed frame. Cool!
For your third one I got (2.30, 2.00, 1.15, 0.00) instead of (2.30, 2.00, -1.15, 0.00). I could be wrong, but just double-check. All of your other numbers check out according to my quick calculation. I'm glad that I didn't tell you in advance how I did it. I approached it very differently from you, and you probably got the idea better doing it your own way. My approach was to plot e.g. the t'=2 line by taking the Lorentz transform equation for t', setting t'=2, and solving for t to get the slope-intercept form of the line t = m x + b. I would then plot that line and label it t'=2. -Dale
Yes, I agree that it should be (2.30, 2.00, 1.15, 0.00) instead of (2.30, 2.00, -1.15, 0.00). Luckily, this error does not seem to affect the calculation for the slope, y-intercept or horizontal spacing of the diagonal lines to be plotted. I tried to make the graph, but the lines are much closer together at this gamma, and it does not seem to be as illustrative as yours was. It probably would have been more straight-forward to use the y = mx + b approach that you mention, but I wanted to see as many data points as possible, so that I might better understand the implications. While we're on the subject of (sometimes) negative results, I am not sure that I fully understand the implications of negative t' results. Does this mean that the theory allows for certain clocks to run backward during the frame-shifting process of acceleration? I can understand if some clocks move forward at a faster rate than others which are also moving forward, but I am uncomfortable with the idea of a clock running backward. For those who are well-versed in SR theory, what do you envision actually happening (regarding t' clock readings and their positions along the x axis) during an acceleration period which causes the SR effects to actually take form?
