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| Tags: correct, interpretation, ktx, mmx, null, results |
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#1
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The Correct Interpretation of the MMX and KTX Null Results.
The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. The correct interpretation is as follows: 1. There exits a stationary aether. 2. The direction of absolute motion of the earth in this stationary aether is in the vertical (UP or DOWN) direction. This sentence appears to be self contradictory. It is not. If one side of the earth is defined as vertical up then the other side of the earth is defined vertical down. However, these two directions of motion (vertical up and verticle down) are in the same direction in the stationary aether. 3. This means that the light path length from the mirrors at the ends of the horizontal arms to the re-combining mirror will remain the same for all the orientations of the horizontal arms. That's why the null result for both the MMX and KTX. There is a proposed experiment to test the validity of this interpretaion and it is outlined in the following link: http://www.journaloftheoretics.com/L...apers/Seto.pdf Ken Seto |
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#2
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"kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. But your theory has LET and SR as subsets, so your theory is wrong: | DRT includes SR/LET as subsets and its equations | are valid in all environments--including gravity. http://groups.google.com/groups?&q=d...+author%3Aseto How sad. Dirk Vdm |
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#3
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"kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. The correct interpretation is as follows: 1. There exits a stationary aether. 2. The direction of absolute motion of the earth in this stationary aether is in the vertical (UP or DOWN) direction. This sentence appears to be self contradictory. It is not. If one side of the earth is defined as vertical up then the other side of the earth is defined vertical down. However, these two directions of motion (vertical up and verticle down) are in the same direction in the stationary aether. 3. This means that the light path length from the mirrors at the ends of the horizontal arms to the re-combining mirror will remain the same for all the orientations of the horizontal arms. That's why the null result for both the MMX and KTX. There is a proposed experiment to test the validity of this interpretaion and it is outlined in the following link: http://www.journaloftheoretics.com/L...apers/Seto.pdf Ken Seto You've forgotten the truly correct one, Ken. The null result of MMX is entirely due to the velocity of light bullets (photons) being constant with respect to the source. No magical length contractions or anything like that. Just like two marbles that are released together, follow different paths along guiding chutes and arrive together, and it just doesn't matter which way the apparatus is turned, or if it takes place on an airplane, or anything like that. They'll go on arriving together time and time again if they are released together. Of course, if you blow some wind over them, they won't. One will be slowed more than the other. So no aether wind for light, either. Hence no aether. Simple, isn't it? Androcles |
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#4
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"AndroclesInEngland" wrote in message ... "kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. The correct interpretation is as follows: 1. There exits a stationary aether. 2. The direction of absolute motion of the earth in this stationary aether is in the vertical (UP or DOWN) direction. This sentence appears to be self contradictory. It is not. If one side of the earth is defined as vertical up then the other side of the earth is defined vertical down. However, these two directions of motion (vertical up and verticle down) are in the same direction in the stationary aether. 3. This means that the light path length from the mirrors at the ends of the horizontal arms to the re-combining mirror will remain the same for all the orientations of the horizontal arms. That's why the null result for both the MMX and KTX. There is a proposed experiment to test the validity of this interpretaion and it is outlined in the following link: http://www.journaloftheoretics.com/L...apers/Seto.pdf Ken Seto You've forgotten the truly correct one, Ken. The null result of MMX is entirely due to the velocity of light bullets (photons) being constant with respect to the source. No magical length contractions or anything like that. Just like two marbles that are released together, follow different paths along guiding chutes and arrive together, and it just doesn't matter which way the apparatus is turned, or if it takes place on an airplane, or anything like that. They'll go on arriving together time and time again if they are released together. Of course, if you blow some wind over them, they won't. One will be slowed more than the other. So no aether wind for light, either. Hence no aether. Simple, isn't it? Androcles Wrong....your magic light bullets should have a speed of c+v or c-v relative to the observer but they are not. So it is not that simple after all.:-) Ken Seto |
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#5
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"Dirk Van de moortel" wrote in message ...
"kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. But your theory has LET and SR as subsets, so your theory is wrong: | DRT includes SR/LET as subsets and its equations | are valid in all environments--including gravity. http://groups.google.com/groups?&q=d...+author%3Aseto How sad. Dirk Vdm xxein: I see nothing wrong with sr-let being subsets of a theory, but Ken can't rationally claim that for his theory. |
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#6
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"kenseto" wrote in message ... "AndroclesInEngland" wrote in message ... "kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. The correct interpretation is as follows: 1. There exits a stationary aether. 2. The direction of absolute motion of the earth in this stationary aether is in the vertical (UP or DOWN) direction. This sentence appears to be self contradictory. It is not. If one side of the earth is defined as vertical up then the other side of the earth is defined vertical down. However, these two directions of motion (vertical up and verticle down) are in the same direction in the stationary aether. 3. This means that the light path length from the mirrors at the ends of the horizontal arms to the re-combining mirror will remain the same for all the orientations of the horizontal arms. That's why the null result for both the MMX and KTX. There is a proposed experiment to test the validity of this interpretaion and it is outlined in the following link: http://www.journaloftheoretics.com/L...apers/Seto.pdf Ken Seto You've forgotten the truly correct one, Ken. The null result of MMX is entirely due to the velocity of light bullets (photons) being constant with respect to the source. No magical length contractions or anything like that. Just like two marbles that are released together, follow different paths along guiding chutes and arrive together, and it just doesn't matter which way the apparatus is turned, or if it takes place on an airplane, or anything like that. They'll go on arriving together time and time again if they are released together. Of course, if you blow some wind over them, they won't. One will be slowed more than the other. So no aether wind for light, either. Hence no aether. Simple, isn't it? Androcles Wrong....your magic light bullets should have a speed of c+v or c-v relative to the observer but they are not. So it is not that simple after all.:-) Why? Is the observer moving relative to the source? No... so v = 0 Is the observer moving relative to the detector? No... so v = 0 Is the observer moving relative to the medium (air)? No.. so v = 0. Since v = 0 in this case, c+0 = c. So they DO have a speed of c+v after all. It's just that in MMX, v happens to be zero. I thinks that's pretty simple, don't you? The rolling marbles have a velocity of u, and if you move along one of the chutes with one it will have a velocity of u-v, or zero, with respect to you, and you'll stay alongside it. Stand still and it's velocity with respect to you will be once again be 'u'. There's nothing magic about it. All matter is composed of molecules or individual atoms with electrons, all visible light is emitted by individual electrons falling to a lower energy state, hence all visible light is comprised of individual photons and they don't need any aether to transport them. They are self supporting electomagnetic phenomena, the magnetic field collapsing and producing an electric field, which in turn collapses to produce a magnetic field, ad infinitum. they ar packets of constant energy that cannot be created or destroyed, only converted to or from some other form, which can be mass. Is the observer moving relative to the medium (aether)? Yes, according to you. But you can't detect it, can you? Magical things happen, like the apparatus shrinking along its length. At least, that's what Lorentz said (Dover, pp 3 - 7). Androcles |
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#7
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"xxein" wrote in message om... "Dirk Van de moortel" wrote in message ... "kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. But your theory has LET and SR as subsets, so your theory is wrong: | DRT includes SR/LET as subsets and its equations | are valid in all environments--including gravity. http://groups.google.com/groups?&q=d...+author%3Aseto How sad. Dirk Vdm xxein: I see nothing wrong with sr-let being subsets of a theory, I can imagine *you* see nothing wrong with wrong theories being subsets of a theory. but Ken can't rationally claim that for his theory. Ken? Rationally? Theory? Dirk Vdm |
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#8
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"AndroclesInEngland" wrote in message ... "kenseto" wrote in message ... "AndroclesInEngland" wrote in message ... "kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. The correct interpretation is as follows: 1. There exits a stationary aether. 2. The direction of absolute motion of the earth in this stationary aether is in the vertical (UP or DOWN) direction. This sentence appears to be self contradictory. It is not. If one side of the earth is defined as vertical up then the other side of the earth is defined vertical down. However, these two directions of motion (vertical up and verticle down) are in the same direction in the stationary aether. 3. This means that the light path length from the mirrors at the ends of the horizontal arms to the re-combining mirror will remain the same for all the orientations of the horizontal arms. That's why the null result for both the MMX and KTX. There is a proposed experiment to test the validity of this interpretaion and it is outlined in the following link: http://www.journaloftheoretics.com/L...apers/Seto.pdf Ken Seto You've forgotten the truly correct one, Ken. The null result of MMX is entirely due to the velocity of light bullets (photons) being constant with respect to the source. No magical length contractions or anything like that. Just like two marbles that are released together, follow different paths along guiding chutes and arrive together, and it just doesn't matter which way the apparatus is turned, or if it takes place on an airplane, or anything like that. They'll go on arriving together time and time again if they are released together. Of course, if you blow some wind over them, they won't. One will be slowed more than the other. So no aether wind for light, either. Hence no aether. Simple, isn't it? Androcles Wrong....your magic light bullets should have a speed of c+v or c-v relative to the observer but they are not. So it is not that simple after all.:-) Why? Is the observer moving relative to the source? No... so v = 0 Is the observer moving relative to the detector? No... so v = 0 Is the observer moving relative to the medium (air)? No.. so v = 0. Since v = 0 in this case, c+0 = c. Are you sure about that? Are you sure c+0 is not -c? After all, since sqrt(1) can be -1 and sqrt(1) can be 1, we can have c + 0 = c = 1 * c = sqrt(1) * c = (-1) * c = -c You see, c+0 has two answers: c and -c Dirk Vdm |
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#9
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"xxein" wrote in message om... "Dirk Van de moortel" wrote in message ... "kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. But your theory has LET and SR as subsets, so your theory is wrong: | DRT includes SR/LET as subsets and its equations | are valid in all environments--including gravity. http://groups.google.com/groups?&q=d...+author%3Aseto How sad. Dirk Vdm xxein: I see nothing wrong with sr-let being subsets of a theory, but Ken can't rationally claim that for his theory. Why not? My theory includes both LET and SR. For LET: The LET assertion of length contraction is replaced with the longer light path length of a moving rod while a rod remains the same length physically in all frames. For SR: 1.The speed of light is not a universal constant as asserted by SR. It is a constant math ratio as follows: Light path length of rod (299,792,458 m) /the absolute time content for a clock second co-moving with the rod. 2. The SR assertion of apparent length contraction is replaced with the longer light path length of a moving rod. Ken Seto |
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#10
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"AndroclesInEngland" wrote in message ... "kenseto" wrote in message ... "AndroclesInEngland" wrote in message ... "kenseto" wrote in message ... The Correct Interpretation of the MMX and KTX Null Results. The current interpretations of the MMX and KTX null results are as follows: 1. The LET interpretation: The arm that is parallel to the direction of motion is contracted by an appropriate amount and thus cancels out the effect of motion of the earth through the aether. 2. The SR interpretation: There is no aether and the speed of light is isotropic. Both of these interpretations are wrong. The correct interpretation is as follows: 1. There exits a stationary aether. 2. The direction of absolute motion of the earth in this stationary aether is in the vertical (UP or DOWN) direction. This sentence appears to be self contradictory. It is not. If one side of the earth is defined as vertical up then the other side of the earth is defined vertical down. However, these two directions of motion (vertical up and verticle down) are in the same direction in the stationary aether. 3. This means that the light path length from the mirrors at the ends of the horizontal arms to the re-combining mirror will remain the same for all the orientations of the horizontal arms. That's why the null result for both the MMX and KTX. There is a proposed experiment to test the validity of this interpretaion and it is outlined in the following link: http://www.journaloftheoretics.com/L...apers/Seto.pdf Ken Seto You've forgotten the truly correct one, Ken. The null result of MMX is entirely due to the velocity of light bullets (photons) being constant with respect to the source. No magical length contractions or anything like that. Just like two marbles that are released together, follow different paths along guiding chutes and arrive together, and it just doesn't matter which way the apparatus is turned, or if it takes place on an airplane, or anything like that. They'll go on arriving together time and time again if they are released together. Of course, if you blow some wind over them, they won't. One will be slowed more than the other. So no aether wind for light, either. Hence no aether. Simple, isn't it? Androcles Wrong....your magic light bullets should have a speed of c+v or c-v relative to the observer but they are not. So it is not that simple after all.:-) Why? Is the observer moving relative to the source? No... so v = 0 Is the observer moving relative to the detector? No... so v = 0 Is the observer moving relative to the medium (air)? No.. so v = 0. When the observer is moving wrt the source he still measures the light from the source to be c. How do you explain that? Ken Seto Since v = 0 in this case, c+0 = c. So they DO have a speed of c+v after all. It's just that in MMX, v happens to be zero. I thinks that's pretty simple, don't you? The rolling marbles have a velocity of u, and if you move along one of the chutes with one it will have a velocity of u-v, or zero, with respect to you, and you'll stay alongside it. Stand still and it's velocity with respect to you will be once again be 'u'. There's nothing magic about it. All matter is composed of molecules or individual atoms with electrons, all visible light is emitted by individual electrons falling to a lower energy state, hence all visible light is comprised of individual photons and they don't need any aether to transport them. They are self supporting electomagnetic phenomena, the magnetic field collapsing and producing an electric field, which in turn collapses to produce a magnetic field, ad infinitum. they ar packets of constant energy that cannot be created or destroyed, only converted to or from some other form, which can be mass. Is the observer moving relative to the medium (aether)? Yes, according to you. But you can't detect it, can you? Magical things happen, like the apparatus shrinking along its length. At least, that's what Lorentz said (Dover, pp 3 - 7). Androcles |
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