michel123456 Posted December 18, 2010 Posted December 18, 2010 (edited) In this article of Wikipedia, about spacetime interval, from which I post some extract here below, it is said (in bold the part concerning my question): "After Einstein derived special relativity formally from the (at first sight counter-intuitive) assumption that the speed of light is the same to all observers, Hermann Minkowski built on mathematical approaches (...) (...)In the Minkowski space, one needs four real numbers (three space coordinates and one time coordinate) (...)The distance between two different events is called the spacetime interval. A path through the four-dimensional spacetime, usually called Minkowski space, is called a world line. Since it specifies both position and time, a particle having a known world line has a completely determined trajectory and velocity. This is just like graphing the displacement of a particle moving in a straight line against the time elapsed. The curve contains the complete motional information of the particle. In the same way as the measurement of distance in 3D space needed all three coordinates we must include time as well as the three space coordinates when calculating the distance in Minkowski space (henceforth called M). In a sense, the spacetime interval provides a combined estimate of how far two events occur in space as well as the time that elapses between their occurrence. But there is a problem. Time is related to the space coordinates, but they are not equivalent. Pythagoras's theorem treats all coordinates on an equal footing (see Euclidean space for more details). We can exchange two space coordinates without changing the length, but we can not simply exchange a space coordinate with time: they are fundamentally different. It is an entirely different thing for two events to be separated in space and to be separated in time. Minkowski proposed that the formula for distance needed a change. He found that the correct formula was actually quite simple, differing only by a sign from Pythagoras's theorem: where c is a constant and t is the time coordinate.[Note 2] Multiplication by c, which has the dimensions L T −1, converts the time to units of length and this constant has the same value as the speed of light." My question is: where does the negative sign come from ? Edited December 18, 2010 by michel123456
IM Egdall Posted December 19, 2010 Posted December 19, 2010 (edited) Einstein's light postulate says that the speed of light is the same for all observers. In other words, no matter what speed the source of that light is going at, no matter what speed you are going at; you will always measure the speed of that light as the same value, about 671 million miles an hour (in a vacuum). This is the core principle behind special relativity. In his famous 1905 paper , Einstein proposed this light postulate. And then he came up with a new transformation formula to go from at rest to uniform motion. It is called the Lorentz transform because unbelknownst to Einstein, Lorentz and others had already come up with the formula (but didn't understrand its physical significance). Anyway, then in 1908, Einstein's former math teacher Minkowski took the Lorentz transform and from it, came up with the formula for the spacetime interval. So to make a long story short: the light postulatre leads to the Lorentz transfom, which leads to the spacetime interval. Why the minus sign? Because that is what comes out of the Lorentz transform mathematics. See links: http://galileo.phys.virginia.edu/classes/252/lorentztrans.html If link doesn't work, google Michael Fowler Lorentz transform Edited December 19, 2010 by I ME
michel123456 Posted December 19, 2010 Author Posted December 19, 2010 Thanks for the link. But still, I don't understand the origin of the negative sign. From your link, this extract: The obvious difference between the green equation & the red one (the Lorentzian) is the negative sign. How was this imported?
John Cuthber Posted December 19, 2010 Posted December 19, 2010 By the same token you might ask why the equations are all in terms of x^2. Why not cubes or x^2 and a bit? The simple answer is that, if you don't use a negative sign, you get the wrong answer.
michel123456 Posted December 19, 2010 Author Posted December 19, 2010 By the same token you might ask why the equations are all in terms of x^2. Why not cubes or x^2 and a bit? The simple answer is that, if you don't use a negative sign, you get the wrong answer. Do you mean it is an ad hoc?
D H Posted December 19, 2010 Posted December 19, 2010 Do you mean it is an ad hoc? Yes and no. The distance driven by a taxicab in Manhattan (or any other city with a nice rectangular grid of streets) from point A to point B is not the same as the distance that a crow would fly in getting from point A to point B. The number of moves taken by a king on a chess board illustrates of yet another definition of "distance". There are an infinite number of mathematical definitions all of which describe in some way the concept of "distance". The number of choices drops dramatically when one requires that distance be generated by an inner product. In fact, ignoring isomorphisms, there is only one. Relax the constraint that the inner product has to be positive definite, instead requiring it to be non-degenerate, and the number of possibilities opens up again. However science is constrained to mathematics that matches reality. Given the experimental confirmations that space is locally Euclidean, that the equivalence principle is true, that the speed of light is the same to all observers, and that motion obeys certain physical laws. of the physics of motion, and of the the observed constancy of the speed of light, there just aren't many choices left. In fact, there is only one (to within isomorphisms). This is the Minkowski metric.
michel123456 Posted December 19, 2010 Author Posted December 19, 2010 What you mean is that probably, following Pythagorean logic, Minkowski made a first try with plus signs everywhere and saw that something was not corresponding to reality. Putting a minus sign solved the problem, and that's it?
imatfaal Posted December 20, 2010 Posted December 20, 2010 Michel - Leonard Susskind's series of lectures "The Theoretical Minimum" have a term dedicated to Special Relativity. These lectures are given at Stamford for a mature/amateur audience, but are still pretty taxing and are not 'pop-science'. From your comments on this board I am sure you would be able to get a great deal from them - and Prof Susskind goes through the mathematics of Lorentz transformations in a fairly step-by-step basis. You can get the lectures from Stamford or from Itunes
Sisyphus Posted December 20, 2010 Posted December 20, 2010 What you mean is that probably, following Pythagorean logic, Minkowski made a first try with plus signs everywhere and saw that something was not corresponding to reality. Putting a minus sign solved the problem, and that's it? No. The following link is an excerpt from Einstein's own book explaining relativity to the layperson. It's a derivation of that equation. http://www.bartleby.com/173/a1.html
swansont Posted December 20, 2010 Posted December 20, 2010 No. The following link is an excerpt from Einstein's own book explaining relativity to the layperson. It's a derivation of that equation. http://www.bartleby.com/173/a1.html To amplify this a little, it's a result from saying that a point source of light will give you a sphere of radius r at a time t in any coordinate system, since the speed of light is constant. Thus, r = ct.
michel123456 Posted December 20, 2010 Author Posted December 20, 2010 (edited) Thanks. [math] r=\sqrt{x^2+y^2+z^2}=ct [/math] That is understandable. Edited December 20, 2010 by michel123456
michel123456 Posted December 21, 2010 Author Posted December 21, 2010 (edited) from the same link (my post in blue fonts for clear differenciation) (green and red notations mine) There is something I don't understand: when both parts are null, how do you conclude σ=1? In equation (8a) which is mentionned, both parts are also null, isn't it? ([math] x^2-c^2t^2=0 [/math]) and ([math] x'^2-c^2t'^2=0 [/math]) from the link: Edited December 21, 2010 by michel123456
zheng sheng ming Posted December 22, 2010 Posted December 22, 2010 I think this show the limit of people!
michel123456 Posted December 26, 2010 Author Posted December 26, 2010 (edited) Nobody answered, no matter, that was not the main question. The question is the following: We have [math] r=\sqrt{x^2+y^2+z^2}=ct [/math] On which we all agree. By squaring, we obtain [math] r^2=x^2+y^2+z^2=(ct)^2 [/math] And thus [math] x^2+y^2+z^2-(ct)^2=0 [/math] How came the idea that [math] x^2+y^2+z^2-(ct)^2=S^2 [/math] ? Edited December 26, 2010 by michel123456
IM Egdall Posted December 26, 2010 Posted December 26, 2010 (edited) Nobody answered, no matter, that was not the main question. How came the idea that [math] x^2+y^2+z^2-(ct)^2=S^2 [/math] ? Because x^2 +y^2 +z^2 - (ct)^2 is invariant with uniform motion. That is for any two events, the space interval squared (x^2 +y^2 + z^2) MINUS the square of c times the time interval (ct^2) is the same value for all observers in uniform motion. We call this the spacetime interval (S^2) Mathematically, this is demostrated by the Lorentz transform for uniform motion. Does this help? Edited December 26, 2010 by I ME
michel123456 Posted December 27, 2010 Author Posted December 27, 2010 (edited) Because x^2 +y^2 +z^2 - (ct)^2 is invariant with uniform motion. That is for any two events, the space interval squared (x^2 +y^2 + z^2) MINUS the square of c times the time interval (ct^2) is the same value for all observers in uniform motion. We call this the spacetime interval (S^2) Mathematically, this is demostrated by the Lorentz transform for uniform motion. Does this help? No it does not help much. Of course zero is invariant. Since light moves in straight line, we can always choose a XYZ system of axis so that X corresponds to the direction of light, obtaining the elementary equation [math] x=ct [math] or [math] c=x/t [math] which is the definition of speed. For any other random system of axis, you take Pythagore and you get the squared equation, what's the big deal? What we did is α=βγ, rearranging into α-βγ=0, stating then that zero is invariant. My question is how did we call zero a "spacetime interval"? Edited December 27, 2010 by michel123456
swansont Posted December 27, 2010 Posted December 27, 2010 S^2 is only zero when you are referring to light.
michel123456 Posted December 27, 2010 Author Posted December 27, 2010 S^2 is only zero when you are referring to light. I have no doubt. Why is this called a spacetime interval?
swansont Posted December 27, 2010 Posted December 27, 2010 Why that particular name? I don't know. We could call it "George," but that's not as descriptive. Why name it at all? Because it's useful. It's invariant — all inertial frames will agree on it.
michel123456 Posted December 27, 2010 Author Posted December 27, 2010 It means that the displacement of light in spacetime is null.
D H Posted December 27, 2010 Posted December 27, 2010 Why is this called a spacetime interval? Euclidean distance is but one measure of the distance or "interval" between two points in some space. One generalization of the concept of distance is based on a metric, a function of two variables that is symmetric, positive definite, and obeys the triangle inequality. (There are other generalizations such as the taxicab norm that cannot be expressed in terms of a metric). The Lorentz metric is not really a metric because it is not positive definite. It is instead merely non-degenerate. It perhaps would have been better to call it something else. However, it does share many other things in common with a metric, so for lack of a better concise term it is still called a metric and the result of applying the Lorentz metric is still called an "interval" (aka distance).
michel123456 Posted December 28, 2010 Author Posted December 28, 2010 You are trying to flood the fish. (to cloud the issue). Duration can be represented in a Minkowski diagram without the use of elaborated ST.
D H Posted December 28, 2010 Posted December 28, 2010 You asked why [math]ds^2=dx^2+dy^2+dz^2-c^2dt^2[/math] is called the spacetime interval. I answered that question. If you wanted the answer to be "because that's what its called" then why didn't you just say so?
michel123456 Posted December 28, 2010 Author Posted December 28, 2010 (edited) No, I asked why [math] s^2=x^2+y^2+z^2-c^2t^2 [/math] I am patient. You will say I am stubborn. My question is not about the meaning of "interval", it about zero. "spacetime interval" has a meaning. [math] s^2=0 [/math] and of course [math] ds^2=0 [/math] Why do you call zero "spacetime interval"? How did that come out? When the result of an operation is zero, It is usually called "null result", not "something". Is that more clear as a question? Edited December 28, 2010 by michel123456
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