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| Tags: derive, exists, light, lorentz, never, transformation, universe, which |
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#1
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The Lorentz trasformation expresses time as a function of speed of
moving frame and light speed. After that, people apply Lorentz trasformation to derive several govermenning equations related to special relativity. So many different versions of derivation for Lorentz transformation can be seen presently.These used a lot of different constraint equations to link time, velocity and velocity of light, merely wanted to successfully derive Lorntz Transformation, and were based on comparing speed of a moving frame and light travelling. Could someone tell me how to write the relativity equations if light never exists in our universe? |
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#2
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chenmou:
The Lorentz trasformation expresses time as a function of speed of moving frame and light speed. After that, people apply Lorentz trasformation to derive several govermenning equations related to special relativity. So many different versions of derivation for Lorentz transformation can be seen presently.These used a lot of different constraint equations to link time, velocity and velocity of light, merely wanted to successfully derive Lorntz Transformation, and were based on comparing speed of a moving frame and light travelling. Could someone tell me how to write the relativity equations if light never exists in our universe? Assume that time and space are on equal footing as "spacetime" so that you recoginze that `c' is nothing but a conversion constant that converts seconds to meters. Assume there exists a transformation A, for the vector x = (ct, r) in frame S to a frame S', so that the vector x transforms as: [ct'] [a b] [ct] x' = Ax = [ ] = [ ] [ ] [ x'] [d e] [ x] giving you two equations in 4 unknowns: ct' = act + bx x' dct + ex Require (ct)^2 - x^2 = (ct')^2 - x'^2, so that you have: (ct)^2 - x^2 = (act + bx)^2 - (dct + ex)^2 = (a^2 - d^2)(ct)^2 + (b^2 - e^2)x^2 + 2ctx(ab - de) which requires a^2 - d^2 = 1 b^2 - e^2 = -1 ab - de = 0 Those will be true if a = e = cosh(K) and b = d = -sinh(K), where cosh(K) and sinh(K) are hyperbolic rotations for some angle K, (usually called the rapidity). That gives you the lorentz transforms in hyperbolic form: ct' = ct cosh(K) - x sinh(K) x' = x cosh(K) - ct sinh(K) The relationship between the hyperbolic functions and the velocity may be deduced by noting that a velocity is the slope of a line in the x-t plane so that the slope is: v = c tanh(K) or \beta = (v/c) = tanh(K) From cosh(K)^2 - sinh(K)^2 = 1 you get 1 - tanh(K)^2 = 1/cosh(K)^2 cosh(K) = 1/sqrt(1 - tanh(K)^2) = 1/sqrt(1 - (v/c)^2) so that cosh(K) = \gamma and sinh(K) = \gamma\beta |
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#3
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Bilge wrote: The relationship between the hyperbolic functions and the velocity may be deduced by noting that a velocity is the slope of a line in the x-t plane so that the slope is: v = c tanh(K) or \beta = (v/c) = tanh(K) From cosh(K)^2 - sinh(K)^2 = 1 you get 1 - tanh(K)^2 = 1/cosh(K)^2 cosh(K) = 1/sqrt(1 - tanh(K)^2) = 1/sqrt(1 - (v/c)^2) so that cosh(K) = \gamma and sinh(K) = \gamma\beta Slicker than Harvard Beets on sliced white. I am familiar with this derivation, but I love to see it, each and every time. It is really to bad that the so-called Lorentz Contraction is not called the Lorentz Projection, which would unconfuse many folks. The "contraction" of length under a rotation is no more mysterious than looking at a yardstick from a 45 degree angle. The "contraction" is purely an artifact of projection and nothing more. Turning a yard stick does not stress its material longitudinally. Bob Kolker |
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#5
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Subject: Derive Lorentz Transformation for a universe in which 'Light'
never exists... From: (Perfectly Innocent) Date: 11/25/03 8:26 AM US Mountain Standard Time Message-id: (chenmou) wrote in message .com... Derive Lorentz Transformation for a universe in which 'Light' never exists... Here's a derivation of the Lorentz Transformation without any reference to light: http://www.everythingimportant.org/relativity Eugene Shubert As you've been told before that is not a derivation. |
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#6
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On 11/25/2003 7:07 AM, Robert J. Kolker wrote:
[...] It is really to bad that the so-called Lorentz Contraction is not called the Lorentz Projection, which would unconfuse many folks. Thanks! That's a great idea! The "contraction" of length under a rotation is no more mysterious than looking at a yardstick from a 45 degree angle. The "contraction" is purely an artifact of projection and nothing more. Turning a yard stick does not stress its material longitudinally. Yes. Tom Roberts |
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#8
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"Perfectly Innocent" wrote in message om... (WaiteDavid137) wrote in message ... Perfectly Innocent: Here's a derivation of the Lorentz Transformation without any reference to light: http://www.everythingimportant.org/relativity Eugene Shubert As you've been told before that is not a derivation. David, You've fallen into a closed temporal loop. Perhaps you can find a way out if you can answer my last rebuttal: http://www.everythingimportant.org/v...hp?p=2229#2229 Eugene Shubert Forget it Eugene. Once he starts with the "you've been told" then he's in parrot mode and is unable to stop and think. |
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#9
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"Robert J. Kolker" wrote in message ...
The "contraction" of length under a rotation is no more mysterious than looking at a yardstick from a 45 degree angle. The "contraction" is purely an artifact of projection and nothing more. Turning a yard stick does not stress its material longitudinally. Bob Kolker On the contrary, the Lorentz circumferential contraction of a rotating elastic isotropic disc give rise to a negative spatial curvature : C = 2piR/(1-v^2/c^2)^(1/2) where R is the radius of the rotating disc and C its relative circumference, (see Landau and Lifschitz, Tome 2, §89 the rotation, Mir edition, 4th edition). This Lorentz circumferential contraction give rise to a slight stress-strain field (actually, a very slight perturbation of the classical stress-strain field of a real rotating elastic isotropic disc). When performing the calculations of the stress-stain field stemming from the Lorentz Circumferential contraction of the rotating disc, one finds traction stresses and traction strains in the circumferential direction and compression in the radial one. Radial stress (at radius r) s_r = 3E (e_0(r)-e_0(R))/8 Circumferential stress (at radius r) s_thêta = 3E (3e_0(r)-e_0(R))/8 Radial strain (at radius r) e_r = (s_r – nu s_thêta )/E = 3[(1–3nu) e_0(r) –(1–nu) e_0(R)]/8 Circumferential strain (at radius r) e_thêta = (s_thêta – nu s_r)/E = 3[(3–nu) e_0(r) –(1–nu) e_0(R)]/8 Where * R is the radius of the rotating disc * r and thêta are the polar coordinates of a rotating observer * E is the Young elastic modulus of the disc * nu is the Poisson coefficient of the disc * e_0(r)= 1/(1-v^2/c^2)^(1/2)-1 is the "initial" strain stemming from the Lorentz circumferential contraction effect, with v = omega r and omega = angular speed of the rotating disc Bernard Chaverondier http://perso.wanadoo.fr/lebigbang PS : in a "possible" flat but finite and static spacetime like a T^3 static spacetime for instance (the 3D hypertorus), neither the Lorentz time dilatation of the moving Langevin twin, nor the Lorentz contraction of evenly moving bodies are artifacts (Though I consider this example to be only a mathematical one. I don't believe the spatial part of our universe to have not simply connected geodesics). |
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#10
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"Robert J. Kolker" wrote in message ... The "contraction" of length under a rotation is no more mysterious than looking at a yardstick from a 45 degree angle. The "contraction" is purely an artifact of projection and nothing more. Turning a yard stick does not stress its material longitudinally. Bob Kolker Are you saying that a change in spatial velocity is merely a change in direction within space-time and any measured length contraction is just a result of the altered perspective? |
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