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Flaws in Derivation of the Lorentz Transformation
- tvanflandern
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17 years 5 months ago #19605
by tvanflandern
Reply from Tom Van Flandern was created 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 />In all cases it turns out that the derivation is mathematically inconsistent unless the velocity v=0.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The transformations have been proved to be mathematically consistent. Just get rid of your assumption that remote time is the same for all frames and you will see the consistency immediately. All "paradoxes" are the result of assuming either a "universal instant of 'now'" or that remote time is the same for frames in relative motion. Neither is true in SR.
Then get up to speed on LR so you understand what the modern discussion is all about, now that SR has been falsified on experimental grounds. See the primer on LR at this web site: metaresearch.org/cosmology/gravity/LR.asp -|Tom|-
<br />In all cases it turns out that the derivation is mathematically inconsistent unless the velocity v=0.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The transformations have been proved to be mathematically consistent. Just get rid of your assumption that remote time is the same for all frames and you will see the consistency immediately. All "paradoxes" are the result of assuming either a "universal instant of 'now'" or that remote time is the same for frames in relative motion. Neither is true in SR.
Then get up to speed on LR so you understand what the modern discussion is all about, now that SR has been falsified on experimental grounds. See the primer on LR at this web site: metaresearch.org/cosmology/gravity/LR.asp -|Tom|-
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17 years 5 months ago #19713
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Thomas, I wonder what your take on the comments of Reichenbach is,
"We find ourselves in a vicious circle: in order to determine the simultaneity of distant events, we must know a velocity; and in order to measure the velocity, we must be capable of judging the simultaneity of events separated by distance."
"Einstein's theory of simultaneity has a presupposition without which it could not be maintained: it is nothing other than the assumption that no velocity greater than the velocity of light can occur in nature."
"We find ourselves in a vicious circle: in order to determine the simultaneity of distant events, we must know a velocity; and in order to measure the velocity, we must be capable of judging the simultaneity of events separated by distance."
"Einstein's theory of simultaneity has a presupposition without which it could not be maintained: it is nothing other than the assumption that no velocity greater than the velocity of light can occur in nature."
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17 years 5 months ago #17969
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 />In all cases it turns out that the derivation is mathematically inconsistent unless the velocity v=0.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The transformations have been proved to be mathematically consistent.
Just get rid of your assumption that remote time is the same for all frames and you will see the consistency immediately. All "paradoxes" are the result of assuming either a "universal instant of 'now'" or that remote time is the same for frames in relative motion. Neither is true in SR.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Then you seem to have some rather relaxed ideas about 'mathematical consistency'. The simply fact is that any velocity dependent transformation of coordinates is inconsistent with the invariance of c. The latter requires for instance that two light signals sent into opposite directions from the origin have the same distance at any time in a given reference frame. This means that if the first signal is at position x , the other must be at position -x. Likewise in the other reference frame, if the first signal is at x', the other must be at -x'. So if you consider the coordinates of a light signal, a change of the sign of x must be reflected in the change of the sign of x'. This condition can not be fulfilled by the Lorentz transformation x'=γ*(x-vt) unless v=0. This holds independently of whether you propose different system times or not.
It would be misleading and wrong to try to falsify SR on experimental grounds. Any experiments in this respect are nonsense anyway given the incorrect (and indeed inconsistent) interpretation of the invariance of the speed of light.
Thomas
<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 />In all cases it turns out that the derivation is mathematically inconsistent unless the velocity v=0.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The transformations have been proved to be mathematically consistent.
Just get rid of your assumption that remote time is the same for all frames and you will see the consistency immediately. All "paradoxes" are the result of assuming either a "universal instant of 'now'" or that remote time is the same for frames in relative motion. Neither is true in SR.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Then you seem to have some rather relaxed ideas about 'mathematical consistency'. The simply fact is that any velocity dependent transformation of coordinates is inconsistent with the invariance of c. The latter requires for instance that two light signals sent into opposite directions from the origin have the same distance at any time in a given reference frame. This means that if the first signal is at position x , the other must be at position -x. Likewise in the other reference frame, if the first signal is at x', the other must be at -x'. So if you consider the coordinates of a light signal, a change of the sign of x must be reflected in the change of the sign of x'. This condition can not be fulfilled by the Lorentz transformation x'=γ*(x-vt) unless v=0. This holds independently of whether you propose different system times or not.
It would be misleading and wrong to try to falsify SR on experimental grounds. Any experiments in this respect are nonsense anyway given the incorrect (and indeed inconsistent) interpretation of the invariance of the speed of light.
Thomas
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17 years 5 months ago #19714
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 Stoat</i>
<br />Hi Thomas, I wonder what your take on the comments of Reichenbach is,
"We find ourselves in a vicious circle: in order to determine the simultaneity of distant events, we must know a velocity; and in order to measure the velocity, we must be capable of judging the simultaneity of events separated by distance."
"Einstein's theory of simultaneity has a presupposition without which it could not be maintained: it is nothing other than the assumption that no velocity greater than the velocity of light can occur in nature."
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
The vicious circle only exists if you try to apply our usual concept of 'speed' (which implies a velocity dependent coordinate transformation) to light (which has an invariant velocity). This is a logical contridiction in terms and thus bound to lead to paradoxes. As should be evident from my pages I quoted above, there can not be any transformation equation for the coordinates of a light signal. You have to determine the coordinates in each frame separately by making corresponding measurements.
As an illustration of this, consider the following thought experiment (taken from my page Speed of Light and Theory of Relativity ) (this doesn't even need two systems of detectors as the light source is simply assumed to be either stationary or moving with regard to the detectors):
<font face="Comic Sans MS">
To clarify the meaning of the invariance of the speed of light and the velocity dependence in the 'transformation' equation, consider first the propagation of a massive body e.g. a bullet fired from a gun (in the absence of any external forces). Assume you have a number of detectors distributed along the path of the bullet, each associated with a synchronized clock that is stopped when the bullet passes the detector at the particular location. Define furthermore the location of the first detector as x=0 and the time any other detector shows as the difference time t to this detector.
Now, as case (a), assume first the gun is resting with regard to the detectors when the bullet is fired. If you then go and have a look at what time each detector at a particular distance x is stopped, you will find the relationship x=ut, where u is the velocity of the bullet relative to the gun.
Then, as case (b), assume the gun is receding from the detectors with velocity v. Again, if you go and have a look at what time the detector clocks at a particular distance were stopped, you will find the relationship x'=(u-v)*t (where the primed quantity shall indicate that the corresponding location is associated with the moving gun).
Now do the same experiment not with a gun firing a bullet but a light pulse.
Again, as case (a), assume first that the light gun is resting with regard to the detectors. If you check the clock times at the detectors, you will find the relationship x=ct (with c the speed of light).
Now the question is what relationship between the locations and trigger times of the detectors will you observe if the light gun is receding from the detectors with velocity v (case(b))? This is were the principle of the invariance of c comes in. Let's assume that such a principle exists, which means that the trigger time at a given detector must be the same as in case (a), i.e. trivially we have x'=x=ct. So it is obvious that not only is there no further 'transformation' needed to find the coordinates of a light signal in case of a moving light source, but it would indeed logically contradict the invariance principle (which by itself provides already the coordinates).
Of course, both the 'massive particle' and the light case are completely separate and are incompatible with each other (x'=(u-v)*t is incompatible with x'=ct, as otherwise one would have u-v=c, which would contradict the invariance of c ).
</font id="Comic Sans MS">
Thomas
<br />Hi Thomas, I wonder what your take on the comments of Reichenbach is,
"We find ourselves in a vicious circle: in order to determine the simultaneity of distant events, we must know a velocity; and in order to measure the velocity, we must be capable of judging the simultaneity of events separated by distance."
"Einstein's theory of simultaneity has a presupposition without which it could not be maintained: it is nothing other than the assumption that no velocity greater than the velocity of light can occur in nature."
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
The vicious circle only exists if you try to apply our usual concept of 'speed' (which implies a velocity dependent coordinate transformation) to light (which has an invariant velocity). This is a logical contridiction in terms and thus bound to lead to paradoxes. As should be evident from my pages I quoted above, there can not be any transformation equation for the coordinates of a light signal. You have to determine the coordinates in each frame separately by making corresponding measurements.
As an illustration of this, consider the following thought experiment (taken from my page Speed of Light and Theory of Relativity ) (this doesn't even need two systems of detectors as the light source is simply assumed to be either stationary or moving with regard to the detectors):
<font face="Comic Sans MS">
To clarify the meaning of the invariance of the speed of light and the velocity dependence in the 'transformation' equation, consider first the propagation of a massive body e.g. a bullet fired from a gun (in the absence of any external forces). Assume you have a number of detectors distributed along the path of the bullet, each associated with a synchronized clock that is stopped when the bullet passes the detector at the particular location. Define furthermore the location of the first detector as x=0 and the time any other detector shows as the difference time t to this detector.
Now, as case (a), assume first the gun is resting with regard to the detectors when the bullet is fired. If you then go and have a look at what time each detector at a particular distance x is stopped, you will find the relationship x=ut, where u is the velocity of the bullet relative to the gun.
Then, as case (b), assume the gun is receding from the detectors with velocity v. Again, if you go and have a look at what time the detector clocks at a particular distance were stopped, you will find the relationship x'=(u-v)*t (where the primed quantity shall indicate that the corresponding location is associated with the moving gun).
Now do the same experiment not with a gun firing a bullet but a light pulse.
Again, as case (a), assume first that the light gun is resting with regard to the detectors. If you check the clock times at the detectors, you will find the relationship x=ct (with c the speed of light).
Now the question is what relationship between the locations and trigger times of the detectors will you observe if the light gun is receding from the detectors with velocity v (case(b))? This is were the principle of the invariance of c comes in. Let's assume that such a principle exists, which means that the trigger time at a given detector must be the same as in case (a), i.e. trivially we have x'=x=ct. So it is obvious that not only is there no further 'transformation' needed to find the coordinates of a light signal in case of a moving light source, but it would indeed logically contradict the invariance principle (which by itself provides already the coordinates).
Of course, both the 'massive particle' and the light case are completely separate and are incompatible with each other (x'=(u-v)*t is incompatible with x'=ct, as otherwise one would have u-v=c, which would contradict the invariance of c ).
</font id="Comic Sans MS">
Thomas
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- tvanflandern
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17 years 5 months ago #17970
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<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 />The latter requires for instance that two light signals sent into opposite directions from the origin have the same distance at any time in a given reference frame.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Right there, you have shown that you do not understand what special relativity is all about. As I said above, in SR there is no "universal instant of 'now'". So you definitely cannot make that claim for any other frame because remote time in another frame will be different than it is in your frame, and different by different amounts in different places.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">This means that if the first signal is at position x, the other must be at position -x.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is true only in your own inertial frame, according to the postulates of SR.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Likewise in the other reference frame, if the first signal is at x', the other must be at -x'.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Naturally, there will come a moment when the signal is at x', and a moment when the other signal is at -x'. Those two moments will be simultaneous for an observer in the moving frame. However, they are most definitely <b>not</b> simultaneous for an observer in the original frame describing events in the moving frame. If you don't accept that such a thing is possible, then you deny the postulates of SR that make it mathematically consistent.
None of this is true in reality because SR is falsified. But it the postulates of SR had been correct, it would have been true. Hence, SR most definitely is a mathematically consistent theory, just one with some weird physical implications.
The most important consequence of your denial of the logical consequences of SR's postulates is that you will never succeed in communicating with a relativist. They will all consider you hopelessly naive for not realizing that there is no remote simultaneity in SR. If you and I have a high relative motion, at the moment we pass, we will disagree about what time it is right "now" in Tokyo. For me, the noon chimes may have already sounded in Tokyo, whereas for you that event hasn't happened yet. In the opposite direction, you will judge events to have happened already that are still in my future.
Think about this, and hopefully come to understand it. Try not to be one of the numerous people in and out of the mainstream who cannot back off a position once taken publicly, no matter what evidence or reasoning is brought to bear. -|Tom|-
<br />The latter requires for instance that two light signals sent into opposite directions from the origin have the same distance at any time in a given reference frame.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Right there, you have shown that you do not understand what special relativity is all about. As I said above, in SR there is no "universal instant of 'now'". So you definitely cannot make that claim for any other frame because remote time in another frame will be different than it is in your frame, and different by different amounts in different places.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">This means that if the first signal is at position x, the other must be at position -x.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is true only in your own inertial frame, according to the postulates of SR.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Likewise in the other reference frame, if the first signal is at x', the other must be at -x'.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Naturally, there will come a moment when the signal is at x', and a moment when the other signal is at -x'. Those two moments will be simultaneous for an observer in the moving frame. However, they are most definitely <b>not</b> simultaneous for an observer in the original frame describing events in the moving frame. If you don't accept that such a thing is possible, then you deny the postulates of SR that make it mathematically consistent.
None of this is true in reality because SR is falsified. But it the postulates of SR had been correct, it would have been true. Hence, SR most definitely is a mathematically consistent theory, just one with some weird physical implications.
The most important consequence of your denial of the logical consequences of SR's postulates is that you will never succeed in communicating with a relativist. They will all consider you hopelessly naive for not realizing that there is no remote simultaneity in SR. If you and I have a high relative motion, at the moment we pass, we will disagree about what time it is right "now" in Tokyo. For me, the noon chimes may have already sounded in Tokyo, whereas for you that event hasn't happened yet. In the opposite direction, you will judge events to have happened already that are still in my future.
Think about this, and hopefully come to understand it. Try not to be one of the numerous people in and out of the mainstream who cannot back off a position once taken publicly, no matter what evidence or reasoning is brought to bear. -|Tom|-
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17 years 5 months ago #17972
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 />The latter requires for instance that two light signals sent into opposite directions from the origin have the same distance at any time in a given reference frame.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Right there, you have shown that you do not understand what special relativity is all about. As I said above, in SR there is no "universal instant of 'now'". So you definitely cannot make that claim for any other frame because remote time in another frame will be different than it is in your frame, and different by different amounts in different places.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">This means that if the first signal is at position x, the other must be at position -x.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is true only in your own inertial frame, according to the postulates of SR.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Likewise in the other reference frame, if the first signal is at x', the other must be at -x'.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Naturally, there will come a moment when the signal is at x', and a moment when the other signal is at -x'. Those two moments will be simultaneous for an observer in the moving frame. However, they are most definitely <b>not</b> simultaneous for an observer in the original frame describing events in the moving frame. If you don't accept that such a thing is possible, then you deny the postulates of SR that make it mathematically consistent.
None of this is true in reality because SR is falsified. But it the postulates of SR had been correct, it would have been true. Hence, SR most definitely is a mathematically consistent theory, just one with some weird physical implications.
The most important consequence of your denial of the logical consequences of SR's postulates is that you will never succeed in communicating with a relativist. They will all consider you hopelessly naive for not realizing that there is no remote simultaneity in SR. If you and I have a high relative motion, at the moment we pass, we will disagree about what time it is right "now" in Tokyo. For me, the noon chimes may have already sounded in Tokyo, whereas for you that event hasn't happened yet. In the opposite direction, you will judge events to have happened already that are still in my future.
Think about this, and hopefully come to understand it. Try not to be one of the numerous people in and out of the mainstream who cannot back off a position once taken publicly, no matter what evidence or reasoning is brought to bear. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I am afraid it is you who seems to be unable to back off an erroneous position. The links I gave in my opening post show that the derivation of the Lorentz transformation is mathematically flawed, so you shouldn't use the correspondingly flawed conclusions implied by the Lorentz transformation (like the existence of different system times) as an argument here. 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. 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 (whether different systems use the same time units or not is irrelevant for this). So x' has to change sign whenever x changes sign, but this condition is obviously not fulfilled by the Lorentz transformation x'=γ*(x-vt) unless v=0, which proves that the latter is indeed inconsistent with the invariance of c and has been obtained by invalid mathematical reasoning (which I have discussed in detail in the links mentioned).
It may be difficult to realize this if one has been kind of brain-washed into accepting something as true which according to logic can't be true, and I can only suggest that you try to forget that SR exists for a while, and objectively try to discuss the derivations that have led to it.
Thomas
<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 />The latter requires for instance that two light signals sent into opposite directions from the origin have the same distance at any time in a given reference frame.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Right there, you have shown that you do not understand what special relativity is all about. As I said above, in SR there is no "universal instant of 'now'". So you definitely cannot make that claim for any other frame because remote time in another frame will be different than it is in your frame, and different by different amounts in different places.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">This means that if the first signal is at position x, the other must be at position -x.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is true only in your own inertial frame, according to the postulates of SR.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Likewise in the other reference frame, if the first signal is at x', the other must be at -x'.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Naturally, there will come a moment when the signal is at x', and a moment when the other signal is at -x'. Those two moments will be simultaneous for an observer in the moving frame. However, they are most definitely <b>not</b> simultaneous for an observer in the original frame describing events in the moving frame. If you don't accept that such a thing is possible, then you deny the postulates of SR that make it mathematically consistent.
None of this is true in reality because SR is falsified. But it the postulates of SR had been correct, it would have been true. Hence, SR most definitely is a mathematically consistent theory, just one with some weird physical implications.
The most important consequence of your denial of the logical consequences of SR's postulates is that you will never succeed in communicating with a relativist. They will all consider you hopelessly naive for not realizing that there is no remote simultaneity in SR. If you and I have a high relative motion, at the moment we pass, we will disagree about what time it is right "now" in Tokyo. For me, the noon chimes may have already sounded in Tokyo, whereas for you that event hasn't happened yet. In the opposite direction, you will judge events to have happened already that are still in my future.
Think about this, and hopefully come to understand it. Try not to be one of the numerous people in and out of the mainstream who cannot back off a position once taken publicly, no matter what evidence or reasoning is brought to bear. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I am afraid it is you who seems to be unable to back off an erroneous position. The links I gave in my opening post show that the derivation of the Lorentz transformation is mathematically flawed, so you shouldn't use the correspondingly flawed conclusions implied by the Lorentz transformation (like the existence of different system times) as an argument here. 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. 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 (whether different systems use the same time units or not is irrelevant for this). So x' has to change sign whenever x changes sign, but this condition is obviously not fulfilled by the Lorentz transformation x'=γ*(x-vt) unless v=0, which proves that the latter is indeed inconsistent with the invariance of c and has been obtained by invalid mathematical reasoning (which I have discussed in detail in the links mentioned).
It may be difficult to realize this if one has been kind of brain-washed into accepting something as true which according to logic can't be true, and I can only suggest that you try to forget that SR exists for a while, and objectively try to discuss the derivations that have led to it.
Thomas
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