On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> >
> > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary.. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > Pulling numbers out of your ass does not constitute proof of anything.. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.

That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!

On Friday, May 12, 2023 at 5:25:43 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:01:40 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> >
> > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary.. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > Pulling numbers out of your ass does not constitute proof of anything.. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> No, mr. smartass, it is not. The error you introduce by rounding off the exact value is orders of magnitude larger than the relativistic effects. Lightspeed is exactly 299,792,458 m/s. The time it takes light to travel 10 m is 3.335640952e-8s, not 3.333333333e-8s. That's an error of more than 0.08%, orders of magnitude greater than the relativistic correction.
You are very silly because it doesn't matter what the numbers are exactly because they necessarily result in a speed increase exactly sufficient to compensate for the lengthening of the hypotenuse. If you are so smart why don't you show the calculation that do not? The square root of C^2 + 30km/sec^2 is a speed that exactly compensates for the lengthened hypotenuse.

On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > >
> > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.

On Friday, May 12, 2023 at 8:31:03 PM UTC-4, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 5:25:43 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 8:01:40 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > >
> > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > No, mr. smartass, it is not. The error you introduce by rounding off the exact value is orders of magnitude larger than the relativistic effects. Lightspeed is exactly 299,792,458 m/s. The time it takes light to travel 10 m is 3.335640952e-8s, not 3.333333333e-8s. That's an error of more than 0.08%, orders of magnitude greater than the relativistic correction.
> You are very silly because it doesn't matter what the numbers are exactly because they necessarily result in a speed increase exactly sufficient to compensate for the lengthening of the hypotenuse. If you are so smart why don't you show the calculation that do not? The square root of C^2 + 30km/sec^2 is a speed that exactly compensates for the lengthened hypotenuse.

You still ignore the fact that the speed is slower than the c+v you assert in the direction of the Earth's orbit. How is there not a fringe shift? And if it exactly compensates for the longer hypotenuse, then the time is the same as if the path were not angled and the propagation velocity was only c.. Still a fringe shift if the straightline velocity is c+v.

On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > >
> > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.

In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.

On Friday, May 12, 2023 at 5:36:57 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:31:03 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 5:25:43 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:01:40 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > >
> > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > No, mr. smartass, it is not. The error you introduce by rounding off the exact value is orders of magnitude larger than the relativistic effects.. Lightspeed is exactly 299,792,458 m/s. The time it takes light to travel 10 m is 3.335640952e-8s, not 3.333333333e-8s. That's an error of more than 0.08%, orders of magnitude greater than the relativistic correction.
> > You are very silly because it doesn't matter what the numbers are exactly because they necessarily result in a speed increase exactly sufficient to compensate for the lengthening of the hypotenuse. If you are so smart why don't you show the calculation that do not? The square root of C^2 + 30km/sec^2 is a speed that exactly compensates for the lengthened hypotenuse.
> You still ignore the fact that the speed is slower than the c+v you assert in the direction of the Earth's orbit. How is there not a fringe shift? And if it exactly compensates for the longer hypotenuse, then the time is the same as if the path were not angled and the propagation velocity was only c. Still a fringe shift if the straightline velocity is c+v.
You obviously missed my comment correcting your idea that I was saying the speed was C + V. Why would I mean that? Obviously, the square root of C^2 + V^2 is smaller, and it exactly compensates for the lengthening of the hypotenuse. Remember, there are two hypotenuses on the way up and down.

On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > >
> > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other.. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.

On Friday, May 12, 2023 at 5:50:45 PM UTC-7, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > >
> > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
Kookfight

On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > >
> > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.

What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift.. Your explanation fails.

On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > >
> > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift.. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
It is silly when you cannot consider the other side without intruding your own assumptions. Without your (1-V^2/C^2), it is just fine. You don't need any relativity. It's useless.

On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > >
> > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift.. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
You just said I cannot explain it my way because you explained it your way. Silly.

On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > >
> > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift.. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.

On Saturday, May 13, 2023 at 8:14:51 AM UTC+5, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > >
> > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > It is not smart ass to say that relativity is a useless pseudoscience.. It is a fact.
> > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.

Until it reaches you. Then it magically changes it velocity to c relative to you, which means it originated much further away than it actually did. Correct?

On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > >
> > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > It is not smart ass to say that relativity is a useless pseudoscience.. It is a fact.
> > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.

Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger. This is what is known as proof by contradiction. You assume that relativistic velocity addition is false. I demonstrate that your assumption leads to a contradiction. Logically, this negates your assumption and shit-cans your argument. But you can't handle defeat, so now the fact that a relativistic looking factor is a result of your lunacy, you want to shift the blame to relativity. In relativity, the velocity of light is the same with or without the velocity of the source.. You cooked your own goose. And since you feel compelled to lie about it, there is no point for this conversation to continue. Adios, MF!

On Saturday, May 13, 2023 at 9:14:22 AM UTC+5, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > >
> > > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> > > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger

Light travels with velocity c relative to source, then relative to observer , what happens if the observer changes velocity in mid flight, before the light reaches him?

Light adjusts its speed? What does it do?

On Saturday, May 13, 2023 at 12:20:49 AM UTC-4, gehan.am...@gmail.com wrote:
> On Saturday, May 13, 2023 at 9:14:22 AM UTC+5, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > > >
> > > > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V.. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > > > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> > > > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only.. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> > > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> > Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger
> Light travels with velocity c relative to source, then relative to observer , what happens if the observer changes velocity in mid flight, before the light reaches him?
>
> Light adjusts its speed? What does it do?

Light always travels at the same speed. Its Proper velocity is infinite. The observer's perception is always the real projection of infinity. The observer's velocity has nothing to do with it. Infinity is a difficult concept. For example, an infinitely tall stack of $1 bills is the same value as an infinitely tall stack of $20 bills. Common sense would suggest that the stack of $20's would be worth more. But that is only true of a finite stack of both. Common sense fails because infinity is not part of our common experience.

On Saturday, May 13, 2023 at 6:23:39 PM UTC+5, Tom Capizzi wrote:
> On Saturday, May 13, 2023 at 12:20:49 AM UTC-4, gehan.am...@gmail..com wrote:
> > On Saturday, May 13, 2023 at 9:14:22 AM UTC+5, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > > > > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > > > >
> > > > > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light.. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > > > > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> > > > > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> > > > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> > > Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger
> > Light travels with velocity c relative to source, then relative to observer , what happens if the observer changes velocity in mid flight, before the light reaches him?
> >
> > Light adjusts its speed? What does it do?
> Light always travels at the same speed. Its Proper velocity is infinite. The observer's perception is always the real projection of infinity. The observer's velocity has nothing to do with it. Infinity is a difficult concept. For example, an infinitely tall stack of $1 bills is the same value as an infinitely tall stack of $20 bills. Common sense would suggest that the stack of $20's would be worth more. But that is only true of a finite stack of both. Common sense fails because infinity is not part of our common experience.

You lost me

On Saturday, May 13, 2023 at 11:41:18 AM UTC-4, gehan.am...@gmail.com wrote:
> On Saturday, May 13, 2023 at 6:23:39 PM UTC+5, Tom Capizzi wrote:
> > On Saturday, May 13, 2023 at 12:20:49 AM UTC-4, gehan.am...@gmail.com wrote:
> > > On Saturday, May 13, 2023 at 9:14:22 AM UTC+5, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > > > > > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > > > > >
> > > > > > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > > > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > > > > > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> > > > > > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> > > > > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> > > > Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger
> > > Light travels with velocity c relative to source, then relative to observer , what happens if the observer changes velocity in mid flight, before the light reaches him?
> > >
> > > Light adjusts its speed? What does it do?
> > Light always travels at the same speed. Its Proper velocity is infinite.. The observer's perception is always the real projection of infinity. The observer's velocity has nothing to do with it. Infinity is a difficult concept. For example, an infinitely tall stack of $1 bills is the same value as an infinitely tall stack of $20 bills. Common sense would suggest that the stack of $20's would be worth more. But that is only true of a finite stack of both. Common sense fails because infinity is not part of our common experience.
> You lost me

It would help if you were a little more specific. The fundamental principle is that the speed of light is constant and invariant with respect to relative velocity. That's the implication of Einstein's 2nd Postulate. Mainstream physics stops there, because they don't answer "Why?" questions. Science by pronouncement. I try to fill in the gaps in their theory with mathematics. There is a detailed derivation of why there is a limiting velocity in the first place, but if this explanation lost you, then that derivation would surely lose you quicker. Instead, let's approach it from a different angle.. Countless experiments noted that a moving particle had far more momentum than could be accounted for by Newton's formula for momentum. It was empirically determined that the relativistic momentum is the Lorentz factor times Newtonian momentum. Over a century ago, it was incorrectly proposed that this was due to the property of relativistic mass. It was observed that as coordinate velocity got closer to c, it got harder and harder to accelerate the particle. This was the reason they speculated that it was because the particle got heavier. Simple Newtonian logic. However, it was conclusively shown that mass was a relativistic invariant of the Lorentz transformation of 4-momentum. The empirical formula for relativistic momentum is γmv. Since relativistic mass, γm, does not exist, the logical conclusion is that p = m(γv). The quantity in parentheses is Proper velocity. While relativistic momentum is not linearly proportional to coordinate velocity, it IS linearly proportional to Proper velocity. As v approaches lightspeed, γ and Proper velocity approach infinity. In the limit, at v = c, Proper velocity is infinite. We can use this information to formally define lightspeed as the mathematical limit of the real, cosine projections of Proper velocity as the magnitude of Proper velocity approaches infinity. This is totally consistent with the fact that at any sub-light coordinate velocity, the Newtonian momentum component is just the cosine projection of relativistic momentum, the same as coordinate velocity is the cosine projection of Proper velocity.

Newtonian velocity addition looks simple, just v3 = v1+v2. But it is actually too simple. It is only valid for velocities that are small fractions of c. Relativistic velocity looks more complicated, but it is because the universe prefers hyperbolic trigonometry over Cartesian. In hyperbolic coordinates, velocity is expressed as a hyperbolic angle, otherwise known as a Lorentz boost. It is an easily verifiable fact that the boost of the product of two Lorentz transformation matrices is just the linear sum of the boosts of the individual factor matrices. Hyperbolic angle addition is simple, and exact. Velocity addition is simple, but approximate. The conversion from hyperbolic trig functions to ordinary trig functions makes velocity addition appear to be complicated. But the truth is, it is a faithful translation of the hyperbolic formula.

The point is, hyperbolic angle addition is linear. The lightspeed limit of coordinate velocity corresponds to infinite Proper velocity. In hyperbolic coordinates, infinite Proper velocity corresponds to infinite boost. When you try to combine finite coordinate velocity with the speed of light, you have to use the complicated velocity addition formula. Or you can convert the two velocities to the equivalent hyperbolic coordinates, and simply add the corresponding boosts. When one of the velocities is lightspeed, its hyperbolic angle is infinite. No matter what the other velocity is, adding a finite amount of boost to an infinite amount of boost has no effect on the total. It is still just infinite. Since only lightspeed is paired with infinite Proper velocity, the result converts to the same 1c as the original velocity of light. In your scenario, the observer frame contributes two different amounts of boost at its two different velocities. However, when they are simply added to the infinite boost of the light ray, the result is the same unchanged infinite amount in both cases. When the limit of the cosine projections is taken, the result is the same coordinate velocity, 1c, in both cases. Light does not change speed, regardless of the observer's speed. The cosine projection of infinity is always the same. And all observer perceptions are limited to cosine projections. This applies to time dilation and length contraction as well, but that's another post.

On Friday, May 12, 2023 at 12:22:20 PM UTC-4, Tom Roberts wrote:
> On 5/12/23 4:46 AM, Tom Capizzi wrote:
> > I do agree that relativity has problems,[...]
>
> The "problems" you mention are all in YOUR head.
> > Thousands of experiments confirm the predictions of relativity.
> Yes. That is how science works.
> > What they don't confirm is the Einstein Interpretation of the
> > results.
> I have no idea what you mean by this.
> > This is a fundamental problem of a "science" that thinks "Why?" is
> > only a question for philosophers and "If the numbers agree, that's
> > good enough."
> This shows that YOU do not understand how science actually works. We are
> building MODELS of how the world works, and that inherently cannot
> answer "why?" questions. This, of course, goes much deeper than mere
> "numbers that agree".
>
> Tom Roberts

For someone who was pretty confident "The "problems" you mention are all in YOUR head.", you seem to be having a problem answering my questions. I am looking forward to your reply.

On 5/12/23 11:20 PM, gehan.am...@gmail.com wrote:
> Light travels with velocity c relative to source, then relative to
> observer , what happens if the observer changes velocity in mid
> flight, before the light reaches him?

[Everything in vacuum.]

Nothing special. Relative to the inertial rest frame of the source the
light is always traveling with speed c. Relative to the inertial rest
frame of the observer, the light is always traveling with speed c. If
the observer changes inertial rest frame, these statements still hold;
but during an observer's acceleration it is much more complicated....

Note your wording is unacceptably vague: "relative to source" is
insufficiently precise. You MUST discuss measurements relative to the
inertial rest frame of the source. That alone would remove much of your
confusion, as it makes it clear that this is a property of INERTIAL
FRAMES, not sources or observers.

> Light adjusts its speed?

No, it does NOT.

The underlying reason that every inertial frame measures light to travel
with speed c in vacuum is due to the fact that the measuring instruments
of different frames are oriented differently in spacetime. Local Lorentz
invariance ensures that these orientations always make such measurements
yield the value c. The light itself DOES NOTHING -- it just propagates
along its null geodesic through spacetime.

This is really basic and easy to see. Draw orthogonal x and t axes on a
piece of paper; use units with c=1. Now draw a straight line at an angle
such that dx/dt < 1, and consider it to be the worldline of an observer
moving relative to the inertial frame of the original axes. The moving
observer's clock is oriented along that line, not the original t axis;
that is, the clock measures time intervals projected onto its own
worldline [#]. For rulers in the moving frame it's a bit more
complicated, but the same holds: they are oriented differently from the
x axis of the original frame.

[#] EVERY instrument measures quantities projected onto
its own worldline. For a clock this OUGHT to be obvious,
as it clearly cannot measure the time of an event away
from itself.

Tom Roberts

On Friday, May 12, 2023 at 9:10:52 PM UTC-7, gehan.am...@gmail.com wrote:
> On Saturday, May 13, 2023 at 8:14:51 AM UTC+5, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 6:09:37 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 8:50:45 PM UTC-4, Laurence Clark Crossen wrote:
> > > > On Friday, May 12, 2023 at 5:44:02 PM UTC-7, Tom Capizzi wrote:
> > > > > On Friday, May 12, 2023 at 8:33:56 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > On Friday, May 12, 2023 at 5:29:59 PM UTC-7, Tom Capizzi wrote:
> > > > > > > On Friday, May 12, 2023 at 8:15:36 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > > > > On Friday, May 12, 2023 at 5:01:40 PM UTC-7, Laurence Clark Crossen wrote:
> > > > > > > > > On Friday, May 12, 2023 at 4:51:19 PM UTC-7, Tom Capizzi wrote:
> > > > > > > > >
> > > > > > > > > > > None of the ideas you mentioned are necessary to explain the MMX. All that is needed is to accept that light shares the velocity of the source and simple additive velocity calculations. I have just explained to you the MMX using Newtonian physics as you can see as follows: The length of the rod perpendicular to the rod facing the direction of Earth's orbital motion is 10 meters long. It takes light 2/30,000,000th of a second to travel that far and back from the mirror to the detector if the Earth were stationary. Since it does the same when Earth is moving, light necessarily shares the velocity of the source (Earth). Because that perpendicular beam covers a hypotenuse in that period of time it must be moving at C + V. That it is moving faster than C is proven by the fact that the beam moving in the direction of Earths orbit returns in 2/30,000,000ths of a second after covering a total of 20 meters to the mirror and back as shown by the lack of a fringe shift (null result).
> > > > > > > > > > Pulling numbers out of your ass does not constitute proof of anything. You don't even bother to use the correct speed of light. The description is inaccurate, in any case. The distance traveled in the direction of the orbit is shorter than you assume, so the velocity along the hypotenuse is slower than you claim. C+v is meaningless garbage. I don't care that you "explained" the MMX using Newtonian physics. Newtonian physics is only an approximation. It omits relevant information. Even if you were correct about the physics, you are still wrong about the velocity. If the velocity of the Earth is added to the speed of light in the direction of motion, it still wouldn't be c+v along the hypotenuse, because that has to be the vector sum of tangential and normal components. Since the normal component is unaffected by the velocity of the Earth, the velocity along the hypotenuse must be different from c+v. Such a discrepancy would produce a fringe shift. Relativistic calculations eliminate the apparent discrepancies, and confirm the relativistic interpretation.
> > > > > > > > > You could try to understand because 30,000,000ths of a second is the time light takes to travel 10 meters. Try again.
> > > > > > > > Sorry you could not understand something so simple. The added length of the hypotenuse is exactly compensated by the additive velocity which is calculated by the square root of a^2 + b^2 or in the MMX C^2 + 30 km/sec^2.
> > > > > > > That's funny. You claim the velocity along the path of the orbit is c+v, but the velocity along the hypotenuse is only √(c²+v²), which is less than c+v, and the hypotenuse is longer, yet there is no fringe shift!
> > > > > > Obviously, when I said C + V, I was abbreviating the square root of C^2 + V^2.
> > > > > In any case, that is not the correct calculation for perpendicular velocities when one is relativistic. That is the incorrect Newtonian approximation again. Also, the c+v value still applies to the arm that is parallel to the path of the orbit. It still takes less time in one arm than in the other. That means fringe shift. No fringe shift means your elaborate Rube Goldberg scheme fails.
> > > > It is not smart ass to say that relativity is a useless pseudoscience. It is a fact.
> > > What is a fact is that you have failed to explain the MMX. Your blathering about the √(c²+v²) applies to the normal arm only. According to you, the additional length is exactly compensated for by the extra speed. In other words, it is the same as if there were no angle and the light was just moving at c. Or the flight time was 20/c. However, the parallel arm has c+v in one direction and c-v in the other. Flight time is 10/(c+v) + 10/(c-v) = (10(c-v)+10(c+v)/(c²-v²) = 20/c*(1-v²/c²), not 20/c. That's a discrepancy in flight time, and means fringe shift. Your explanation fails.
> > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> Until it reaches you. Then it magically changes it velocity to c relative to you, which means it originated much further away than it actually did. Correct?
It is never C relative to anyone because it has relative motion like everything in the universe, as Pentcho Valev has tried to convey. That it has C + V and - V is proven by the Doppler shift observed in starlight and the limbs of the Sun. This relative motion was necessarily involved in ascertaining the speed of light from the occultations of Jupiter's moons.

On Friday, May 12, 2023 at 9:14:22 PM UTC-7, Tom Capizzi wrote:
> On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger. This is what is known as proof by contradiction. You assume that relativistic velocity addition is false. I demonstrate that your assumption leads to a contradiction. Logically, this negates your assumption and shit-cans your argument. But you can't handle defeat, so now the fact that a relativistic looking factor is a result of your lunacy, you want to shift the blame to relativity. In relativity, the velocity of light is the same with or without the velocity of the source. You cooked your own goose. And since you feel compelled to lie about it, there is no point for this conversation to continue. Adios, MF!
No, there is no need for the 1/(1-V^2/C^2) which is relativity and not part of ordinary ballistic calculations of additive velocity. Relativity is completely unnecessary. It wasn't necessary for atomic bombs, for sagnac, for GPS or for anything but to confuse physics for 100 years.

On Saturday, May 13, 2023 at 5:00:18 PM UTC-4, Laurence Clark Crossen wrote:
> On Friday, May 12, 2023 at 9:14:22 PM UTC-7, Tom Capizzi wrote:
> > On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> > Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger. This is what is known as proof by contradiction. You assume that relativistic velocity addition is false. I demonstrate that your assumption leads to a contradiction. Logically, this negates your assumption and shit-cans your argument. But you can't handle defeat, so now the fact that a relativistic looking factor is a result of your lunacy, you want to shift the blame to relativity. In relativity, the velocity of light is the same with or without the velocity of the source. You cooked your own goose. And since you feel compelled to lie about it, there is no point for this conversation to continue. Adios, MF!
> No, there is no need for the 1/(1-V^2/C^2) which is relativity and not part of ordinary ballistic calculations of additive velocity. Relativity is completely unnecessary. It wasn't necessary for atomic bombs, for sagnac, for GPS or for anything but to confuse physics for 100 years.

LIAR! The factor you disown is a result of your own assertion that the speed of light can add (or subtract) the velocity of its source. In relativity, this is not possible, so in relativity, there is NO FACTOR at all. The velocity of light is c, regardless of the velocity of its source. It's one thing to misrepresent relativity, which you do constantly. It is another to contradict yourself and then lie about it.

On Sunday, May 14, 2023 at 10:21:32 AM UTC+5, Tom Capizzi wrote:
> On Saturday, May 13, 2023 at 5:00:18 PM UTC-4, Laurence Clark Crossen wrote:
> > On Friday, May 12, 2023 at 9:14:22 PM UTC-7, Tom Capizzi wrote:
> > > On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > > > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> > > Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger. This is what is known as proof by contradiction. You assume that relativistic velocity addition is false. I demonstrate that your assumption leads to a contradiction. Logically, this negates your assumption and shit-cans your argument. But you can't handle defeat, so now the fact that a relativistic looking factor is a result of your lunacy, you want to shift the blame to relativity. In relativity, the velocity of light is the same with or without the velocity of the source. You cooked your own goose. And since you feel compelled to lie about it, there is no point for this conversation to continue. Adios, MF!
> > No, there is no need for the 1/(1-V^2/C^2) which is relativity and not part of ordinary ballistic calculations of additive velocity. Relativity is completely unnecessary. It wasn't necessary for atomic bombs, for sagnac, for GPS or for anything but to confuse physics for 100 years.
> LIAR! The factor you disown is a result of your own assertion that the speed of light can add (or subtract) the velocity of its source. In relativity, this is not possible, so in relativity, there is NO FACTOR at all. The velocity of light is c, regardless of the velocity of its source. It's one thing to misrepresent relativity, which you do constantly. It is another to contradict yourself and then lie about it.

The velocity of light is c,

Relative to what

On Sunday, May 14, 2023 at 8:55:58 AM UTC-4, gehan.am...@gmail.com wrote:
> On Sunday, May 14, 2023 at 10:21:32 AM UTC+5, Tom Capizzi wrote:
> > On Saturday, May 13, 2023 at 5:00:18 PM UTC-4, Laurence Clark Crossen wrote:
> > > On Friday, May 12, 2023 at 9:14:22 PM UTC-7, Tom Capizzi wrote:
> > > > On Friday, May 12, 2023 at 11:14:51 PM UTC-4, Laurence Clark Crossen wrote:
> > > > > The only reason why relativity wants the 1- V^2/C^2 is because it wants to claim that light does not share the velocity of the source. Everything in the universe shares the velocity of the source.
> > > > Now you have gone full troll. The only reason 1/(1- V^2/C^2) is there is YOUR STUPID assertion that the velocity of light should include the velocity of its source, you dim-witted knuckle-dragger. This is what is known as proof by contradiction. You assume that relativistic velocity addition is false. I demonstrate that your assumption leads to a contradiction. Logically, this negates your assumption and shit-cans your argument. But you can't handle defeat, so now the fact that a relativistic looking factor is a result of your lunacy, you want to shift the blame to relativity. In relativity, the velocity of light is the same with or without the velocity of the source. You cooked your own goose. And since you feel compelled to lie about it, there is no point for this conversation to continue. Adios, MF!
> > > No, there is no need for the 1/(1-V^2/C^2) which is relativity and not part of ordinary ballistic calculations of additive velocity. Relativity is completely unnecessary. It wasn't necessary for atomic bombs, for sagnac, for GPS or for anything but to confuse physics for 100 years.
> > LIAR! The factor you disown is a result of your own assertion that the speed of light can add (or subtract) the velocity of its source. In relativity, this is not possible, so in relativity, there is NO FACTOR at all. The velocity of light is c, regardless of the velocity of its source. It's one thing to misrepresent relativity, which you do constantly. It is another to contradict yourself and then lie about it.
>
> The velocity of light is c,
> Relative to what

Relative to any coordinate system. That's the thing about lightspeed. It is c for all observers, regardless of their state of motion. Mathematically, the Proper velocity of lightspeed is infinite. Do you need to ask "infinite relative to what"? It is infinite relative to you, and the limit of its cosine projection is c relative to you. When you add 0 to a number, the result is the same number. When you add a number to infinity, the result is the same infinity. This is counter-intuitive, but a logical property of infinity, nonetheless.