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Can someone tell me why the proper time between αβ is not t^{2}+x^{2} but rather
t^{2}-x^{2}?
Background:
t^{2}-x^{2}?
Background:
Last edited:
Yes but why? Can you derive it?Because this is Minkowski geometry, not Euclidean.
Lorentz, by the way, not Lorenz. Lorenz was a different mathematician and physicist.
I failed to show it graphically..Try constructing an inertial coordinate system ##(x’ , t’)## with the following criteria: events A and B both have ##x’_A =x’_B =0##, and the slope of the line connecting P, Q, and the origin in the primed coordinates is equal to ##- x_B / t_B## (in the unprimed coordinates), and the light rays connecting P to B and B to Q keep their slopes of 1 and -1 respectively. When you do this, which of these quantities remains invariant despite using different coordinates: ##t^2 + x^2## and ##t’^2 + x’^2##, or ##t^2 - x^2## and ##t’^2 - x’^2##?
Yes. If you play with the Lorentz transformations some, you’ll be able to show that they preserve the quantity ##\Delta{t}^2-\Delta{x}^2## but not ##\Delta{t}^2+\Delta{x}^2##. Transforming coordinates doesn’t change any of the real directly observable physics, so this is a pretty strong hint that the former and not the latter quantity is the physically significant one.Yes but why? Can you derive it?
You can't. You're drawing your graph on a sheet of paper that obeys Euclidean geometry so the Pythagorean theorem works and the distance between points is given by ##\Delta{s}^2=\Delta{t}^2+\Delta{x}^2## and that's not the right relationship between distance and displacement in Minkowski geometry.I failed to show it graphically..
Hi do you mind showing me the special cases where the distances come out right? I'm still wrapping my head around it..You can't. You're drawing your graph on a sheet of paper that obeys Euclidean geometry so the Pythagorean theorem works and the distance between points is given by ##\Delta{s}^2=\Delta{t}^2+\Delta{x}^2## and that's not the right relationship between distance and displacement in Minkowski geometry.
You can use the graph to show the relationship between the various intervals (that's what MTW are doing in the image you posted), but you can't make the distances come out right except in a few special cases (and it would be a good exercise to figure out which these are).
I don't mind, but I'm going to give you another day to think about it first. There is a very big hint in post #7.Hi do you mind showing me the special cases where the distances come out right? I'm still wrapping my head around it..
Okay I see now with the help of Minkowski diagram, what I'm seeing is that when a particle is at rest in the primed frame, its ##\Delta t## is exactly the line segment in the unprimed frame given by the Pythagorean theorem ##\Delta x^2 + \Delta t^2## however scaled differently in the primed frame (its value equals to ##\Delta t^2 - \Delta x^2##). So my question now is why and how is ##\Delta t^2 - \Delta x^2## invariant mathematically in the Minkowski space?I don't mind, but I'm going to give you another day to think about it first. There is a very big hint in post #7.
Yes, in that case and it also works when ##\Delta{t}## is zero.Okay I see now with the help of Minkowski diagram, what I'm seeing is that when a particle is at rest in the primed frame, its ##\Delta t## is exactly the line segment
Because the x and t coordinates in one frame are related to the x and t coordinates in another frame by the Lorentz transformations. It's just an exercise in algebra (one that you'll find in any relativity textbook) to show that these transformations change x to x' and t to t' in such a way that ##\Delta{t}^2-\Delta{x}^2=\Delta{t'}^2-\Delta{x'}^2##. Thus your question comes down to asking why the Lorentz transformations describe how the universe we live in works (if they didn't we wouldn't care about them or Minkowski space).So my question now is why and how is ##\Delta t^2 - \Delta x^2## invariant mathematically in the Minkowski space?
But why is light speed the same for all observer? I would think about Maxwell's equations give speed of light as the speed that the EM fields are propagating, but nowhere it gives satisfactory answer there..Yes, in that case and it also works when ##\Delta{t}## is zero.Because if the x and t coordinates in one frame are related to the x and t coordinates in another frame by the Lorentz transformations. It's just an exercise in algebra (one that you'll find in any relativity textbook) to show that these transformations change x to x' and t to t' in such a way that ##\Delta{t}^2-\Delta{x}^2=\Delta{t'}^2-\Delta{x'}^2##. Thus your question comes down to asking why the Lorentz transformations describe how the universe we live in works (if they didn't we wouldn't care about them or Minkowski space).
There are two mathematically consistent possibilities for how our universe works:
1) There is some speed that is the same in all frames and the x and t coordinates of different frames are related by the Lorentz transformations (where the constant ##c## is that invariant speed). There is no speed that is the same in all frames and and the x and t coordinates of different frames are related by the Galilean transformations.
Experiments have proven far beyond any reasonable doubt that #1, with light moving at that invariant speed (which is why we call ##c## the speed of light) is how the universe works. But WHY it is #1 instead of #2? Physics, as an experimental science, cannot answer that question.
Depends on what you mean. The invariance of the speed of light was a postulate of relativity. Modern "one postulate" approaches show that an invariant speed - either finite or infinite - is a consequence of the principle of relativity, and experiment shows we live in a universe that has a finite invariant speed (the infinite invariant speed one is a Newtonian universe). In either case there isn't a "why", really. We just explore the consequences of it being true and find that they are an accurate description of our universe.But why is light speed the same for all observer?
The historical flow of discovery was more or less:But why is light speed the same for all observer? I would think about Maxwell's equations give speed of light as the speed that the EM fields are propagating, but nowhere it gives satisfactory answer there..