Ghideon Posted June 1, 2019 Posted June 1, 2019 5 hours ago, Jan Slowak said: SC1 and SC2 are two independent experiments and you must NOT mix their results!Ask a professor of mathematics! Can you show the mathematical evidence? I'm not planning to ask a professor of mathematics.
beecee Posted June 1, 2019 Posted June 1, 2019 6 hours ago, Jan Slowak said: I cannot accept your counter-arguments by using words, by saying "it's not true"! Only mathematical arguments! Instead of words, you should use mathematical calculations.It doesn't matter who we are (I don't know who you are, what education you have) we'll stick to the theme: Exactly! So then please explain to me why I should accept the claims and supposed mathematical ability, of someone on a public science forum, open to any Tom, Dick and Harry, or even cranks of many persausions, against more then 100 years of experimental verification and reputable renowned mathematical calculations from accepted mathematicians?
swansont Posted June 1, 2019 Posted June 1, 2019 8 hours ago, Jan Slowak said: If you with Eqn 2.5 refer to page 15 in [7] then I say again:The result B = -Av which you get by solving the equation systemLEx': x' = Ax + BtLEt': t' = Cx + DtSC1: x' = 0, x = vtapplies only to x' = 0. The result B = -Dv which you get by solving the equation systemLEx': x' = Ax + BtLEt': t' = Cx + DtSC2: x' = -vt', x = 0applies only to x = 0. SC1 and SC2 are two independent experiments and you must NOT mix their results!Ask a professor of mathematics! So when you solve y = mx + b by substituting in for y (or x) intercept, the solution is only valid at that one point? Yeah, try getting the math professor to stop laughing. Quote You saidAfter 115 years of mathematical development and experimentation. You are clearly wrong. This is no proof! SeeCosmos - a short story; Stephen Hawking; page 15:"A physical theory is always provisional in the sense that it is merely a hypothesis: one can never prove it. However many times the experimental results are consistent with a particular theory, one can never be sure that the results next will contradict the theory. On the other hand, one can disprove a theory by finding only one that does not conform to the predictions of theory ” You are challenging the math, not the experimental confirmation, so this is irrelevant. (Though it’s curious that experiment confirms this allegedly flawed math)
Jan Slowak Posted June 2, 2019 Author Posted June 2, 2019 20 hours ago, Strange said: Clearly you are not willing to listen to anything that contradicts what you believe. As I said, you can easily derive the same equations without assuming x'=0, it is just a simplification. See the link provided by swansont, for example. You haven't shown that is wrong. I said at the beginning that in this section, I deal with the derivation of LT from [7]. We can discuss the derivation of LT in the link provided by swansont. -1
Strange Posted June 2, 2019 Posted June 2, 2019 34 minutes ago, Jan Slowak said: I said at the beginning that in this section, I deal with the derivation of LT from [7]. We can discuss the derivation of LT in the link provided by swansont. Right. We have shown you where you have gone wrong in your criticism of [7]. There is clearly nothing more to say (because you refuse to acknowledge your error). So lets move on to your analysis of the derivation here: https://thecuriousastronomer.wordpress.com/2013/03/10/derivation-of-the-lorentz-transformations-from-first-principles/
Jan Slowak Posted June 2, 2019 Author Posted June 2, 2019 20 hours ago, Ghideon said: Can you show the mathematical evidence? I'm not planning to ask a professor of mathematics. I can repeat myself again. The derivation of LT in [7]: Special Case 1: we have the following equation systems:LEx': x' = Ax + BtLEt': t' = Cx + DtSC1: x' = 0, x = vtBy replacing SC1 in LEx' we get a relationship between constants.The solution in this case is B = -Av. This solution, this relationship between constants, applies only to x' = 0! Special Case 2: we have the following equation systems:LEx': x' = Ax + BtLEt': t' = Cx + DtSC2: x' = -vt', x = 0By replacing SC2 in LEx', LEt' we get a relationship between constants.The solution in this case is B = -Dv. This solution, this relationship between constants, applies only to x = 0! In [7] you combine these two solutions in one and get D = A. I want to draw the reader's attention to an important thing:Special Case 1 and Special Case 2 are two different problems, two different physical phenomena, two different thought experiments. They should NOT be combined. If you do so, the following applies:Special Case 1 give us B = -Av that applies only to x' = 0!Special Case 2 give us B = -Dv that applies only to x = 0!→ D = A applies only to {x' = 0, x = 0}This is the case when the two reference systems are in the same point! Then no LT is needed. -1
Ghideon Posted June 2, 2019 Posted June 2, 2019 23 minutes ago, Jan Slowak said: This solution, this relationship between constants, applies only to x' = 0! 23 minutes ago, Jan Slowak said: Special Case 1 and Special Case 2 are two different problems, two different physical phenomena, two different thought experiments. They should NOT be combined. Why? What is the mathematical reasons? Can you show, using math, why it is wrong? If the method int the derivation LT in [7] is wrong it seems to apply to a lot more than SR, you would be able to state the issue with more than "They should NOT be combined".
Jan Slowak Posted June 2, 2019 Author Posted June 2, 2019 18 hours ago, swansont said: So when you solve y = mx + b by substituting in for y (or x) intercept, the solution is only valid at that one point? Yeah, try getting the math professor to stop laughing. You are challenging the math, not the experimental confirmation, so this is irrelevant. (Though it’s curious that experiment confirms this allegedly flawed math) Please check my answer for Ghideon. Please try to familiarize yourself with my logic.The two SC from [7] can be compared to two problems that use the equation you specify y = mx + b. Problem 1: About cars. When setting conditions you get the following equation system:LEy: y = mx + bSC1: x = 0, y = 5→ b = 5 Problem 2: About aircraft. When setting conditions you get the following equation system:LEy: y = mx + bSC2: y = 0, x = 0; in this case, the line LEy goes through the origin of the coordinate system.→ b = 0 These two problems have nothing in common! -2
Jan Slowak Posted June 2, 2019 Author Posted June 2, 2019 9 minutes ago, Ghideon said: Why? What is the mathematical reasons? Can you show, using math, why it is wrong? If the method int the derivation LT in [7] is wrong it seems to apply to a lot more than SR, you would be able to state the issue with more than "They should NOT be combined". Please check my answer for swansont. 59 minutes ago, Strange said: Right. We have shown you where you have gone wrong in your criticism of [7]. There is clearly nothing more to say (because you refuse to acknowledge your error). So lets move on to your analysis of the derivation here: https://thecuriousastronomer.wordpress.com/2013/03/10/derivation-of-the-lorentz-transformations-from-first-principles/ Sorry, I missed a word. I should have written:We can discuss the derivation of LT in the link provided by swansont then. After we finish the derivation of LT in [7]. You write: We have shown you where you have gone wrong in your criticism of [7].You haven't shown anything yet. Check my answers to swansont and Ghideon. -1
swansont Posted June 2, 2019 Posted June 2, 2019 1 hour ago, Jan Slowak said: Please check my answer for Ghideon. Please try to familiarize yourself with my logic. Why? Your logic is flawed. 1 hour ago, Jan Slowak said: The two SC from [7] can be compared to two problems that use the equation you specify y = mx + b. No. 1 hour ago, Jan Slowak said: Problem 1: About cars. When setting conditions you get the following equation system:LEy: y = mx + bSC1: x = 0, y = 5→ b = 5 Problem 2: About aircraft. When setting conditions you get the following equation system:LEy: y = mx + bSC2: y = 0, x = 0; in this case, the line LEy goes through the origin of the coordinate system.→ b = 0 These two problems have nothing in common! Then it’s a bad example. I mean, really bad. “How could a mathematician make such a mistake” bad. It only works if the formula applies to all of the boundary conditions. Both problems have to be about e.g. the same car. You have two data points, from which you can determine the two unknowns. This is first-semester algebra.
Jan Slowak Posted June 2, 2019 Author Posted June 2, 2019 27 minutes ago, swansont said: Why? Your logic is flawed. No. Then it’s a bad example. I mean, really bad. “How could a mathematician make such a mistake” bad. It only works if the formula applies to all of the boundary conditions. Both problems have to be about e.g. the same car. You have two data points, from which you can determine the two unknowns. This is first-semester algebra. You write: Why? Your logic is flawed. SC1, SC2, SC3 in the derivation of LT in [7] are three different problems, tre different physical phenomena, tre different thought experiments. They should NOT be combined in one and the same equation system! That's what is wrong. I suggest you think about this, talk to some other physicists and mathematicians. Think about what SC1 and SC2 are for something. And think properly what gives your right to put them in one and the same equation system. You do not have to be ironic in your answers. We should only talk about mathematics, logic and physics. No personal insults, thank you! -1
Strange Posted June 2, 2019 Posted June 2, 2019 4 minutes ago, Jan Slowak said: SC1, SC2, SC3 in the derivation of LT in [7] are three different problems, tre different physical phenomena, tre different thought experiments. No, it is one problem (the relationship between two frames of reference in relative motion). The three cases are just ways of eliminating one of the variables at a time in order to solve the equations with multiple unknowns. It is a completely standard technique. As any mathematician could tell you. 6 minutes ago, Jan Slowak said: I suggest you think about this, talk to some other physicists and mathematicians. As mathematicians and physicists have been examining this for over 100 years, and none have spotted this "error" I don't think that suggestion is particularly useful. (Apart from the fact that swansont is a physicist, of course.)
Ghideon Posted June 2, 2019 Posted June 2, 2019 (edited) 2 hours ago, Jan Slowak said: Please check my answer for swansont. Can you provide the mathematical details? The car&plane example does not show the problem with the derivation at all. Obviously there can be separate car problems and plane problems that cannot be combined. A trivial example could be the altitude of the plane and the age of passengers in the car, they are probably not related in a scientifically interesting way. But lets say the car has a velocity v1 and the plane has a velocity v2. This property "velocity" that both the car and the plane possess, could be compared in various coordinate systems and models. For instance the relative speed of car and plane could be calculated. Edited June 2, 2019 by Ghideon clarified a sentence
swansont Posted June 2, 2019 Posted June 2, 2019 1 hour ago, Jan Slowak said: You write: Why? Your logic is flawed. SC1, SC2, SC3 in the derivation of LT in [7] are three different problems, tre different physical phenomena, tre different thought experiments. They should NOT be combined in one and the same equation system! That's what is wrong. One problem: that of two frames in relative motion. 1 hour ago, Jan Slowak said: I suggest you think about this, talk to some other physicists and mathematicians. Think about what SC1 and SC2 are for something. And think properly what gives your right to put them in one and the same equation system. I have. So have many others. 1 hour ago, Jan Slowak said: You do not have to be ironic in your answers. We should only talk about mathematics, logic and physics. No personal insults, thank you! Pointing out that you’re wrong, and making a first-semester algebra error, is not a personal insult. I can see that it’s embarrassing, however. 2
beecee Posted June 2, 2019 Posted June 2, 2019 1 hour ago, Jan Slowak said: I suggest you think about this, talk to some other physicists and mathematicians. Which is why I asked you the following.... 23 hours ago, beecee said: Exactly! So then please explain to me why I should accept the claims and supposed mathematical ability, of someone on a public science forum, open to any Tom, Dick and Harry, or even cranks of many persausions, against more then 100 years of experimental verification and reputable renowned mathematical calculations from accepted mathematicians? and as yet you have failed to answer..... 1 hour ago, Jan Slowak said: You do not have to be ironic in your answers. We should only talk about mathematics, logic and physics. No personal insults, thank you! All I see so far, are attempts by others that do have some of the expertise that you mention, showing where you are in error.
Phi for All Posted June 3, 2019 Posted June 3, 2019 18 hours ago, Jan Slowak said: We should only talk about mathematics, logic and physics. No personal insults, thank you! ! Moderator Note The members participating are trying to do just that, but you aren't listening to their mathematical arguments. You're ignoring them, yet insisting they stick to maths and logic, which they're telling you are wrong in this case. Also, we attack ideas here, not people. There have been NO personal insults whatsoever. Disagreeing with you, or calling your ideas wrong, or telling you your logic is flawed is consistent with our rules on civility. None of those is a personal attack. Please engage with the questions posed to you, especially the requests for the maths you believe are wrong. If you can't do that, then you're just soapboxing, which is against the rules, and I'll have to shut the thread down. This is a fantastic opportunity to take advantage of professional knowledge that can help put you back on track with your mathematics education, if you'll acknowledge that.
Jan Slowak Posted June 3, 2019 Author Posted June 3, 2019 It wasn't me who started with such discussions.I mentioned several times that I only want to talk about mathematics, logic and physics. I felt insulted. I want to ask if this thread is closed.I would like to continue with other things. I have a few more things to discuss regarding the derivation of LT in [7]. Should I continue here or should I start a new thread?
Phi for All Posted June 3, 2019 Posted June 3, 2019 23 hours ago, Jan Slowak said: It wasn't me who started with such discussions.I mentioned several times that I only want to talk about mathematics, logic and physics. I felt insulted. ! Moderator Note Questioning you isn't an insult. In science, an idea needs to be tested rigorously before it's accepted as an explanation for anything. You've claimed well-tested maths are wrong, and it's up to you to show why. You haven't done that here in this thread, so why should anyone discuss your ideas with you? We need to know that you have a point before considering it. So far, your point hasn't been made. EVERYONE here wants you to talk about mathematics and physics, but in a meaningful way, and that means you have to accept that you may be wrong and now have an opportunity to correct that and move on to better intellectual challenges. If you can't resolve the questions in this thread, please don't compound your mistakes in another thread.
DanMP Posted June 4, 2019 Posted June 4, 2019 On 6/2/2019 at 7:03 PM, Jan Slowak said: I can repeat myself again. The derivation of LT in [7]: Special Case 1: we have the following equation systems:LEx': x' = Ax + BtLEt': t' = Cx + DtSC1: x' = 0, x = vtBy replacing SC1 in LEx' we get a relationship between constants.The solution in this case is B = -Av. This solution, this relationship between constants, applies only to x' = 0! Special Case 2: we have the following equation systems:LEx': x' = Ax + BtLEt': t' = Cx + DtSC2: x' = -vt', x = 0By replacing SC2 in LEx', LEt' we get a relationship between constants.The solution in this case is B = -Dv. This solution, this relationship between constants, applies only to x = 0! In [7] you combine these two solutions in one and get D = A. I want to draw the reader's attention to an important thing:Special Case 1 and Special Case 2 are two different problems, two different physical phenomena, two different thought experiments. They should NOT be combined. On 6/2/2019 at 7:32 PM, Jan Slowak said: Please try to familiarize yourself with my logic.The two SC from [7] can be compared to two problems that use the equation you specify y = mx + b. Problem 1: About cars. When setting conditions you get the following equation system:LEy: y = mx + bSC1: x = 0, y = 5→ b = 5 Problem 2: About aircraft. When setting conditions you get the following equation system:LEy: y = mx + bSC2: y = 0, x = 0; in this case, the line LEy goes through the origin of the coordinate system.→ b = 0 These two problems have nothing in common! In the above quote (with y = mx + b) you are right, those two problems have nothing in common, but in the first quote (with x' = Ax + Bt) the "problems" are not independent, they are about exactly the same thing seen from different perspectives (different frames). When x' = 0, x must be vt, because at t=0, x' = x = 0 and the "x' frame" moved with the speed v. In the "Special Case 2", where x = 0, x' must be -vt', for the same reason. This is not random as in your example (SC1: x = 0, y = 5 and SC2: y = 0, x = 0). [I'm not absolutely sure that all I wrote is correct, but I think it may help.]
Strange Posted June 4, 2019 Posted June 4, 2019 On 6/2/2019 at 5:32 PM, Jan Slowak said: Please check my answer for Ghideon. Please try to familiarize yourself with my logic.The two SC from [7] can be compared to two problems that use the equation you specify y = mx + b. Problem 1: About cars. When setting conditions you get the following equation system:LEy: y = mx + bSC1: x = 0, y = 5→ b = 5 Problem 2: About aircraft. When setting conditions you get the following equation system:LEy: y = mx + bSC2: y = 0, x = 0; in this case, the line LEy goes through the origin of the coordinate system.→ b = 0 These two problems have nothing in common! They have nothing in common because you have chosen two different lines (with different values of m and b). As a real mathematician would know, if we have a single straight line, we cannot determine the value of m and b from a single pair of x and y values. However, if we have two pairs of x and y values then we can solve for m and b. One simple way of calculating b is to set x to zero. And then the value of b is just the y-intersect at x=0. Contrary to your ludicrous claim: THIS DOES NOT MEAN THAT THE VALUE OF b IS ONLY CORRECT WHEN x=0; it always has that value in the equation. So this mathematically proves that you claim that we cannot use a "special case" more generally is wrong. We can, of course, calculate the values of m and b using any arbitrary x and pairs (just as we can derive the the Lorentz transform using any combination of frames in relative motion). It just becomes slightly more complicated. 4
Jan Slowak Posted June 4, 2019 Author Posted June 4, 2019 4 hours ago, Strange said: They have nothing in common because you have chosen two different lines (with different values of m and b). As a real mathematician would know, if we have a single straight line, we cannot determine the value of m and b from a single pair of x and y values. However, if we have two pairs of x and y values then we can solve for m and b. One simple way of calculating b is to set x to zero. And then the value of b is just the y-intersect at x=0. Contrary to your ludicrous claim: THIS DOES NOT MEAN THAT THE VALUE OF b IS ONLY CORRECT WHEN x=0; it always has that value in the equation. So this mathematically proves that you claim that we cannot use a "special case" more generally is wrong. We can, of course, calculate the values of m and b using any arbitrary x and pairs (just as we can derive the the Lorentz transform using any combination of frames in relative motion). It just becomes slightly more complicated. This thread is closed by the moderator Phi for All, I refrain from processing it further. -1
Strange Posted June 4, 2019 Posted June 4, 2019 1 hour ago, Jan Slowak said: This thread is closed by the moderator Phi for All, I refrain from processing it further. The thread is not closed. I assume you are just saying that because you have no answer to this simple mathematical explanation of your error. You obviously started out with an assumption that SR was wrong (I don't know why) and then tried to find some pseudo-mathematical reason to reject it. Unfortunately, science doesn't work that way. SR has been subject to over 100 years of analysis and testing. It works. I assume another derivation of the Lorentz transform would be from the Einstein Field Equations, by taking the low energy limit. (But it would probably require someone with more mathematical knowledge than is currently contributing to this thread - apologies to swansont if I am underestimating him! Where is Markus Hanke when you need him !?)
Phi for All Posted June 4, 2019 Posted June 4, 2019 3 hours ago, Jan Slowak said: This thread is closed by the moderator Phi for All, I refrain from processing it further. ! Moderator Note When a thread is closed, we lock the topic. If you can still post, it's an open thread. Now that there's no more confusion about whether or not we want you to answer questions, perhaps you could address some of the concerns already posted in this very open, very active, very curious thread.
DanMP Posted June 5, 2019 Posted June 5, 2019 22 hours ago, DanMP said: in the first quote (with x' = Ax + Bt) the "problems" are not independent, they are about exactly the same thing seen from different perspectives (different frames). When x' = 0, x must be vt, because at t=0, x' = x = 0 and the "x' frame" moved with the speed v. In the "Special Case 2", where x = 0, x' must be -vt', for the same reason. After I wrote the above (something you may have considered redundant/obvious) I wondered why v is the same, I mean when x' = 0, x = vt but when x = 0, x' should be -v't'. Why v' = v ?
Jan Slowak Posted June 5, 2019 Author Posted June 5, 2019 3 hours ago, DanMP said: After I wrote the above (something you may have considered redundant/obvious) I wondered why v is the same, I mean when x' = 0, x = vt but when x = 0, x' should be -v't'. Why v' = v ? I have decided before to not continue to answer this thread because it was moved to Speculations and also other reasons. I only answer you because you are from Cluj-Napoca where I studied mathematics at the University of Babes-Bolyai in 1972-1977. I have analyzed the derivation of LT in [7]. But whatever you read about LT it says that reference systems S and S' move relative to each other at constant speed v. For example, if S' moves to the right with v, S moves to the left with -v.
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