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berlinspeed
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Can someone tell me why the proper time between αβ is not t2+x2 but rather
t2-x2?
Background:
t2-x2?
Background:
Last edited:
Yes but why? Can you derive it?Ibix said: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..Pencilvester said: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.berlinspeed said: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.berlinspeed said: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..Nugatory said: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.berlinspeed said: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?Nugatory said: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.berlinspeed said: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..Nugatory said: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.berlinspeed said:But why is light speed the same for all observer?
The historical flow of discovery was more or less:berlinspeed said: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..
A Lorentz Geometry Diagram is a visual representation of the geometric properties of spacetime in special relativity. It is used to illustrate the effects of time dilation and length contraction on an object moving at high speeds.
A Lorentz Geometry Diagram is constructed by drawing two perpendicular axes representing time and space. The axes are then scaled according to the Lorentz transformation equations, which take into account the relative velocity between two frames of reference.
A Lorentz Geometry Diagram helps to visualize the principles of special relativity and understand how time and space are affected by the speed of an object. It also demonstrates the concept of a spacetime continuum, where time and space are interconnected.
A Lorentz Geometry Diagram differs from a regular graph in that it uses a special scaling factor to account for the effects of relativity. In a regular graph, the axes are usually scaled equally, but in a Lorentz Geometry Diagram, the scaling is asymmetric due to the non-Euclidean nature of spacetime.
A Lorentz Geometry Diagram is commonly used in physics and engineering to calculate the effects of time dilation and length contraction in high-speed systems, such as in particle accelerators and spacecraft. It is also used in GPS technology to account for the time dilation effects of satellites orbiting the Earth.