Flaws in Derivation of the Lorentz Transformation

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17 years 3 months ago #19715 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />If you want to objectively discuss the mathematical validity of the derivation of the Lorentz transformation, then you can't make the presumption that the latter is correct.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The derivation has nothing to do with the mathematical validity. Assume the LT was just guessed out of thin air. Your lack of consistency arguments still do not apply.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">And if you would use a correct logical argumentation in this sense, then you would have to accept that, given the initial postulate of the invariance of c, in any reference frame two light signal sent into opposite directions must have the same distance from the origin at any time<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That just isn't true unless the same time system pervades both frames.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So x' has to change sign whenever x changes sign, but this condition is obviously not fulfilled by the Lorentz transformation x'=ã*(x-vt) unless v=0, which proves that the latter is indeed inconsistent with the invariance of c<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Again, you assume both frames have the same time system, which is simply wrong in SR. And why should the primed frame be forced to adopt a moving origin? We almost never do that in physics.

You'll never "get it" until you drop that "universal time" assumption built into your brain. Nature may have a single time system, but SR does not. In SR, when Frame A looks at Frame B for just a single instant, Frame A sees both a range of space and a range of time (past, present, and future) in Frame B.

Take it or leave it. But that's what the rest of the world is talking about. -|Tom|-

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17 years 3 months ago #17976 by Thomas
Replied by Thomas on topic Reply from Thomas Smid
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />If you want to objectively discuss the mathematical validity of the derivation of the Lorentz transformation, then you can't make the presumption that the latter is correct.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The derivation has nothing to do with the mathematical validity. Assume the LT was just guessed out of thin air. Your lack of consistency arguments still do not apply.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Whether the Lorentz transformation has been guessed out of thin air or not doesn't make any difference. It still would have to be consistent with the initial constraint <i>x=ct implies x'=ct', and x=-ct implies x'=-ct'</i> (as formulated by Einstein himself in his 'algebraic derivation' ( www.bartleby.com/173/a1.html )). There is no way that this condition can be made algebraically consistent with the Lorentz transformation using valid maths, as the former implies that x' changes sign when x changes sign, which is not satisfied by x'=γ*(x-vt) (unless v=0).

It is obvious that there not be any velocity dependent transformation for the coordinates of a light signal at all as it would be a contradiction in terms with the invariance of c. So the trajectory of a light signal in one reference frame can not be determined (like it is possible for material objects) by projection from a different reference frame, but it has to be determined explicitly by measurements in that frame (and it will then turn out that the constraint does indeed already yield the trajectory) .

Thomas

P.S.: I'll be away now for couple of weeks and won't be able to reply to any posts during this time.



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