Alexander Masterov Posted September 8, 2011 Author Posted September 8, 2011 I came to the forum not in order to my a knowledges demonstrate, but in order to get answers to questions. I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Einstein had entitled (bestowed the absoluteness to time), but no did it. Why?
ajb Posted September 8, 2011 Posted September 8, 2011 I'm looking for a professional who understands the relativism and can answer the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis? Can I ask you to define, or at least write down in some coordinates what you mean by cross-scale?
uncool Posted September 8, 2011 Posted September 8, 2011 (edited) Can I ask you to define, or at least write down in some coordinates what you mean by cross-scale? I'm guessing he means the transverse dimensions - y and z are absolute under x-boosts, while t is not. Alexander: So observer A sees an object which follows the path (t, 0, 0, 0). That is, according to observer A, at time t, the object is at the point (0, 0, 0) for any t. Observer B is moving at velocity v relative to A. If observer A thinks that the object is at (t, 0, 0, 0), what will observer B think the position is? =Uncool- Edited September 9, 2011 by uncool
Alexander Masterov Posted September 9, 2011 Author Posted September 9, 2011 (edited) Can I ask you to define, or at least write down in some coordinates what you mean by cross-scale? Compare: Master Theory: SRT: __________________________________________ H - is cross-scale. what will observer think the position is?If the distance between observers (A and B) is zero (for t=0), then: [math]x=Vt[/math] - real distance [math]x'=V't[/math] - visual distance ____________________________________ If: Real coordinates: [math] x(t) =Vt [/math] Then: Visual coordinates: [math] x'(t) =vt'=\frac{vt}{1+V/c}=\frac{2Vt}{(\sqrt{1+4V^2/c^2}+1)(1+V/c)} [/math] [math]V'=\frac{2V}{(\sqrt{1+4V^2/c^2}+1)(1+V/c)}[/math] _______________________________________________________________ Do I understand your questions: you (just as me) do not see any reason for the absoluteness of the cross-scale (transverse dimensions)? Edited September 9, 2011 by Alexander Masterov
ajb Posted September 9, 2011 Posted September 9, 2011 Do you mean the width of an object? The length contraction only occurs only in the direction parallel to the direction in which the observed body is travelling. A simple coordinate change will always render it possible to just consider length contraction, though that might not be what you mean by "length". It is a consequence of the Lorentz transformations. Unless I have completely missed what you mean by H?
Alexander Masterov Posted September 9, 2011 Author Posted September 9, 2011 Do you mean the width of an object?Width and height.
ajb Posted September 9, 2011 Posted September 9, 2011 Width and height. Good, then I have answered your question.
uncool Posted September 9, 2011 Posted September 9, 2011 If the distance between observers (A and B) is zero (for t=0), then: [math]x=Vt[/math] - real distance [math]x'=V't[/math] - visual distance ____________________________________ If: Real coordinates: [math] x(t) =Vt [/math] Then: Visual coordinates: [math] x'(t) =vt'=\frac{vt}{1+V/c}=\frac{2Vt}{(\sqrt{1+4V^2/c^2}+1)(1+V/c)} [/math] [math]V'=\frac{2V}{(\sqrt{1+4V^2/c^2}+1)(1+V/c)}[/math] _______________________________________________________________ Do I understand your questions: you (just as me) do not see any reason for the absoluteness of the cross-scale (transverse dimensions)? I do see a reason. Have you ever read Einstein's original paper? http://www.fourmilab.ch/etexts/einstein/specrel/www/ Now, your post doesn't answer my question. I am looking for the actual transformation. What is the full transformation? In special relativity, the transformation is: [math] t' = \gamma (t - \frac{vx}{c^2})[/math] [math]x' = \gamma (x - vt)[/math] [math] y' = y[/math] [math] z' = z[/math] What is your version of the transformation? =Uncool-
Alexander Masterov Posted September 9, 2011 Author Posted September 9, 2011 (edited) I do see a reason. Have you ever read Einstein's original paper?http://www.fourmilab...in/specrel/www/ Link to article (my reading which would take more than one hour) can not accept as an argument of scientific debate. If you have an argument, then show it. ___________________________________________ What is your version of the transformation? Non inertial reference frame Real coordinates: Real speed: Visual speed: Visual coordinates: Visual time: without Doppler effect Edited September 9, 2011 by Alexander Masterov
uncool Posted September 9, 2011 Posted September 9, 2011 (edited) Link to article (my reading which would take more than one hour) can not serve as an argument in scientific debate. If you have an argument, then show it. This is Einstein's original paper. Papers can serve as arguments in scientific debates - that is the entire point of papers in the first place. Non inertial reference frame Real coordinates: Real speed: Visual speed: Visual coordinates: Visual time: Could you write a specific version of what x', y', z', and t' are? In other words, could you give me your exact version of the equations I showed you? =Uncool- Edited September 9, 2011 by uncool
Alexander Masterov Posted September 9, 2011 Author Posted September 9, 2011 This is Einstein's original paper. Papers can serve as arguments in scientific debates - that is the entire point of papers in the first place.I read this article in Russian. (Long ago.) But I have not found an answer to my question at that time. Can you give me a quote from this article, where Einstein substantiate its choice in favor of the cross-scale? Could you write a specific version of what x', y', z', and t' are? In other words, could you give me your exact version of the equations I showed you? I'm doing it sure, but - later on. As long as I need to find an answer to a fundamentally important issue. I hope that we can find an expert who will answer. (In Russia there is no such expert.)
uncool Posted September 9, 2011 Posted September 9, 2011 I read this article in Russian. (Long ago.) But I have not found an answer to my question at that time. Can you give me a quote from this article, where Einstein substantiate its choice in favor of the cross-scale? Read section 3 - it details everything about the Lorentz transformation. =Uncool-
Alexander Masterov Posted September 9, 2011 Author Posted September 9, 2011 (edited) § 3. Theory of the Transformation of Co-ordinates and Times from a Stationary System to another System in Uniform Motion of Translation Relatively to the Former Let us in "stationary" space take two systems of co-ordinates, i.e. two systems, each of three rigid material lines, perpendicular to one another, and issuing from a point. Let the axes of X of the two systems coincide, and their axes of Y and Z respectively be parallel. Let each system be provided with a rigid measuring-rod and a number of clocks, and let the two measuring-rods, and likewise all the clocks of the two systems, be in all respects alike. Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), and let this velocity be communicated to the axes of the co-ordinates, the relevant measuring-rod, and the clocks. To any time of the stationary system K there then will correspond a definite position of the axes of the moving system, and from reasons of symmetry we are entitled to assume that the motion of k may be such that the axes of the moving system are at the time t (this "t" always denotes a time of the stationary system) parallel to the axes of the stationary system. We now imagine space to be measured from the stationary system K by means of the stationary measuring-rod, and also from the moving system k by means of the measuring-rod moving with it; and that we thus obtain the co-ordinates x, y, z, and.... Please specify a location in the text where Einstein justify their choice. Edited September 9, 2011 by Alexander Masterov
uncool Posted September 9, 2011 Posted September 9, 2011 § 3. Theory of the Transformation of Co-ordinates and Times from a Stationary System to another System in Uniform Motion of Translation Relatively to the Former Let us in "stationary" space take two systems of co-ordinates, i.e. two systems, each of three rigid material lines, perpendicular to one another, and issuing from a point. Let the axes of X of the two systems coincide, and their axes of Y and Z respectively be parallel. Let each system be provided with a rigid measuring-rod and a number of clocks, and let the two measuring-rods, and likewise all the clocks of the two systems, be in all respects alike. Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), and let this velocity be communicated to the axes of the co-ordinates, the relevant measuring-rod, and the clocks. To any time of the stationary system K there then will correspond a definite position of the axes of the moving system, and from reasons of symmetry we are entitled to assume that the motion of k may be such that the axes of the moving system are at the time t (this "t" always denotes a time of the stationary system) parallel to the axes of the stationary system. We now imagine space to be measured from the stationary system K by means of the stationary measuring-rod, and also from the moving system k by means of the measuring-rod moving with it; and that we thus obtain the co-ordinates x, y, z, and.... Please specify a location in the text where Einstein justify their choice. Read the entire section. He is specifically showing how to obtain the transform. It starts with two sections related only by a relative velocity v. Using the assumption that light move at speed c (which is required by Maxwell's equations), and the assumption that all inertial frames are equivalent, he then demonstrates the entire transform. Which choice are you referring to? That the transverse dimensions not be scaled? That is not a choice - it is required by the math. =Uncool-
Alexander Masterov Posted September 10, 2011 Author Posted September 10, 2011 (edited) Which choice are you referring to? That the transverse dimensions not be scaled? That is not a choice - it is required by the math.Do you not understand what Einstein wrote? Edited September 10, 2011 by Alexander Masterov
uncool Posted September 10, 2011 Posted September 10, 2011 Do you not understand what Einstein wrote? What I just said is what Einstein wrote - the fact that transverse dimensions are not scaled is required by the theory, not a choice. =Uncool-
Alexander Masterov Posted September 10, 2011 Author Posted September 10, 2011 (edited) What I just said is what Einstein wrote - the fact that transverse dimensions are not scaled is required by the theory, not a choice.This fact is on default, concoct, fabricate and are evident from nothing. Nicely!_________________________ Einstein bestowed the absoluteness to [math]x^2 - (ct)^2 =x'^2 - (ct')^2=0[/math] - on what basis? It is boundary between events that may be a cause and consequence of each other of Einsteinian theory, that is also a consequence of no-proven assumption that nothing can travel faster than light. Edited September 10, 2011 by Alexander Masterov
uncool Posted September 10, 2011 Posted September 10, 2011 This fact is on default, concoct, fabricate and are evident from nothing. Nicely! _________________________ Einstein bestowed the absoluteness to [math]x^2 - (ct)^2 =x'^2 - (ct')^2=0[/math] - on what basis? So you are disputing the second assumption of relativity - that light must move at a velocity of c. That assumption actually does have reasoning; I was assuming that you agreed with it, and did not understand the reasoning that stemmed from that assumption. Do you agree that light is electromagnetic radiation? =Uncool-
Alexander Masterov Posted September 10, 2011 Author Posted September 10, 2011 Do you agree that light is electromagnetic radiation?c=const is basis of Master Theory, but there is no evidence to suggest that matter can not move faster than light. For example: from target of elementary particle accelerator generate unstable particles emitted from the known lifetime (tau leptons - 5 10^-13 seconds) sometimes. During life, they manage to overcome the distance that can not be overcome if the move at the speed of light.
uncool Posted September 10, 2011 Posted September 10, 2011 (edited) c=const is basis of Master Theory That requires that if x^2 - (ct)^2 = 0, then x'^2 - (ct')^2 = 0. Do you agree with that? Edited to add: Einstein's theory actually doesn't make the claim that matter cannot move faster than light. The assumption only is that light can only move at c - everything is based off of that. Now, you have claimed that you agree that c is constant. That is enough to derive Einstein's equations, assuming you make the assumption about inertial frames. =Uncool- Edited September 10, 2011 by uncool
Alexander Masterov Posted September 11, 2011 Author Posted September 11, 2011 (edited) c=const is basis of Master TheoryThat requires that if x^2 - (ct)^2 = 0, then x'^2 - (ct')^2 = 0. Do you agree with that? Post 2 no need of it. If [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math] then time must slow down. But two observers can not see the slowdown of each other simultaneously. If one sees a slowdown, then the second sees the acceleration. (Otherwise causality principle come to incorrect.) But such asymmetry violates the equality of inertial reference frames. Time can not slow down. Edited September 11, 2011 by Alexander Masterov
uncool Posted September 11, 2011 Posted September 11, 2011 Post 2 no need of it. If [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math] then time must slow down. But two observers can not see the slowdown of each other simultaneously. If one sees a slowdown, then the second sees the acceleration. (Otherwise causality principle come to incorrect.) But such asymmetry violates the equality of inertial reference frames. Time can not slow down. You are wrong; one seeing a slowdown does not imply that the other sees an acceleration, due to the mixing of space and time. However, that x^2 - (ct)^2 = 0 implies that x'^2 - (ct')^2 = 0 can be directly demonstrated from the assumption that c is constant: If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? Then if we take a transform, the light ray must reach the point x' at time t' if (t', x') is the transformed version of (t, x), correct? In that case, since the speed of light is c in all frames, we then know that x'^2 - (ct')^2 = 0, correct? =Uncool-
Alexander Masterov Posted September 11, 2011 Author Posted September 11, 2011 (edited) You are wrong; one seeing a slowdown does not imply that the other sees an acceleration, due to the mixing of space and time.Experiment: both the observer projected on a screen a films (each from his IRF). What sequence of images we and they see? Edited September 11, 2011 by Alexander Masterov
uncool Posted September 11, 2011 Posted September 11, 2011 (edited) Experiment: both the observer projected on a screen a films (each from his IRF). What sequence of images we and they see? So let's say that each one shows an image once a second from their own points of view. In other words: Observer A sees Observer B's ship moving at velocity v. A projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math], etc. relative to frame A. Correspondingly, Observer B sees Observer A's ship moving at velocity -v. B projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math], etc. relative to frame B. Then relative to frame A, A projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math]..., while B projects a screen at [math](0, 0, 0, 0), (\gamma, \gamma v, 0, 0), (2 \gamma, 2 \gamma v, 0, 0)[/math], ... Which says that the time-ordering (assuming low \gamma) should be the following: [math](0, 0, 0, 0)[/math]: Both project screen 0 [math](1, 0, 0, 0)[/math]: A projects screen 1 [math](\gamma, \gamma v, 0, 0)[/math]: B projects screen 1 [math](2, 0, 0, 0)[/math]: A projects screen 2 [math](2\gamma, 2\gamma v, 0, 0)[/math]: B projects screen 2 etc. However, relative to frame B, A projects a screen at [math](0, 0, 0, 0), (\gamma, -\gamma v, 0, 0), (2\gamma, -2\gamma v, 0, 0)[/math], ... while B projects a screen at [math](0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0)[/math]... which gives the time-ordering of Which says that the time-ordering (assuming low \gamma) should be the following: [math](0, 0, 0, 0)[/math]: Both project screen 0 [math](1, 0, 0, 0)[/math]: B projects screen 1 [math](\gamma, -\gamma v, 0, 0)[/math]: A projects screen 1 [math](2, 0, 0, 0)[/math]: B projects screen 2 [math](2\gamma, -2\gamma v, 0, 0)[/math]: A projects screen 2 etc. The fact that space is mixing with time demonstrates why it is that the order can change in different frames. Now will you answer my question? If x^2 - (ct)^2 = 0, then a light ray sent from point 0 at time 0 will get to point x at time t, correct? Then if we take a transform, the light ray must reach the point x' at time t' if (t', x') is the transformed version of (t, x), correct? In that case, since the speed of light is c in all frames, we then know that x'^2 - (ct')^2 = 0, correct? =Uncool- Edited September 11, 2011 by uncool
Alexander Masterov Posted September 11, 2011 Author Posted September 11, 2011 So let's say that each one shows an image...So. Simpler, easier. Let: time of your interlocutor slow down. Are your time slowed-down or accelerated? (Interlocutor's opinion.)
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