The first thing I noticed when I read Einstein's book was how he claimed that the Lorentz equations satisfied the results of the Michelson-Morley experiment. He said he had extracted two little equations from the Lorentz equations that showed light to be traveling at c in a frame of reference at rest and a frame of reference in motion.
x=ct
x'=ct'
So I put these values into the Galilean transformation equation and came up with the following.

x'=x-vt
ct'=ct-vt
t'=t-vt/c
This is essentially the same as the numerator of Lorentz's equation.
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
But it is incorrect, and Lorentz's equation shows us why. If x=ct, then you would need different variables for x and x' in Galileo's equation because the equation would only be true for one value of x, the distance light would travel in time t, the amount of time it takes for frame of reference S' to move a distance of vt relative to frame of reference S. If we recognize the difference between this distance and all other coordinates for x, then we could say

x1' = x1 - vt
ct1' = ct1 - vt/c
or as Lorentz was trying to express it
t1' = t1 - vx/c^2
t1 would be the time it takes light to travel a distance of x1, and t1' would be the time it takes light to travel a distance of x1'. x in the equation would be one particular value of x, the distance light travels in a time of t.
So now we go back to the original problem Einstein was trying to solve, the problem of two clocks that show different times. For instance, if we have a clock in a flying airplane, and a clock on the ground, Einstein says the clock in the airplane will be slower than the clock on the ground. The time on the clock in the airplane will not be t1' in the above equations. It will be taken from the inverse equations which show the speed of the airplane and the time of the clock in the airplane according to the inverse equations.
x1 = x1' - v't'
This equation does not predict a time for the clock in the airplane. It just shows that if the pilot of the airplane has a clock with a different rate than the clock on the ground that shows t, then the pilot of the airplane will get a different speed for the airplane than an observer on the ground will get using the clock on the ground to time the flight of the airplane. The equation is just as true if the moving clock is faster, which scientists tell us a clock in a GPS satellite would be.
In order to predict the time of the moving clock, you would have to introduce the gravitation of the system, which would be determined by the mass of the earth and the mass of the airplane and possibly how the speed of the airplane relates to orbital speed at that altitude.
In any event, if Maxwell's equations agree with light traveling at c in two frames of reference, it seems to me that the Galilean transformation equations show why, since Maxwell died before the Michelson-Morley experiment took place, and no one was using anything but the Galilean transformation equations to represent relativity before the Michelson-Morley experiment.

On 2/23/23 12:49 AM, Robert Winn wrote:
> In any event, if Maxwell's equations agree with light traveling at c
> in two frames of reference, it seems to me that the Galilean
> transformation equations show why,

No, they don't. The Galilean transform is INCONSISTENT with the velocity
of any object being the same in two relatively-moving inertial frames.

(Most of the equations you wrote are NOT Galilean transforms.
You are just engaged in mathematical masturbation.)

> since Maxwell died before the Michelson-Morley experiment took place,
> and no one was using anything but the Galilean transformation
> equations to represent relativity before the Michelson-Morley
> experiment.

That is the issue that became of prime importance in roughly 1880-1905
-- how could Maxwell's equations be reconciled with the Galilean
transform? We now know that they can't, and one must use Lorentz
transforms. In particular, the Lorentz group is a subgroup of the
invariance group of Maxwell's equations, while the Galileo group is
inconsistent with both.

Nothing you write can possibly change any of that.

[If you don't understand what an invariance group is,
then you do not have enough knowledge to understand
any of this, and you must STUDY.
BTW that is pure math: group theory applied to
partial differential equations.]

Tom Roberts

On Thursday, 23 February 2023 at 18:10:15 UTC+1, Tom Roberts wrote:
> On 2/23/23 12:49 AM, Robert Winn wrote:
> > In any event, if Maxwell's equations agree with light traveling at c
> > in two frames of reference, it seems to me that the Galilean
> > transformation equations show why,
> No, they don't. The Galilean transform is INCONSISTENT with the velocity

And we're FORCED!!!
Because The Shit is THE BEST WAY!!!

On Thursday, February 23, 2023 at 9:40:25 AM UTC-8, Maciej Wozniak wrote:
> On Thursday, 23 February 2023 at 18:10:15 UTC+1, Tom Roberts wrote:
> > On 2/23/23 12:49 AM, Robert Winn wrote:
> > > In any event, if Maxwell's equations agree with light traveling at c
> > > in two frames of reference, it seems to me that the Galilean
> > > transformation equations show why,
> > No, they don't. The Galilean transform is INCONSISTENT with the velocity
> And we're FORCED to eat Shit!!

Bon appetit, janitor!

On Thursday, February 23, 2023 at 10:10:15 AM UTC-7, Tom Roberts wrote:
> On 2/23/23 12:49 AM, Robert Winn wrote:
> > In any event, if Maxwell's equations agree with light traveling at c
> > in two frames of reference, it seems to me that the Galilean
> > transformation equations show why,
> No, they don't. The Galilean transform is INCONSISTENT with the velocity
> of any object being the same in two relatively-moving inertial frames.
>
> (Most of the equations you wrote are NOT Galilean transforms.
> You are just engaged in mathematical masturbation.)
> > since Maxwell died before the Michelson-Morley experiment took place,
> > and no one was using anything but the Galilean transformation
> > equations to represent relativity before the Michelson-Morley
> > experiment.
> That is the issue that became of prime importance in roughly 1880-1905
> -- how could Maxwell's equations be reconciled with the Galilean
> transform? We now know that they can't, and one must use Lorentz
> transforms. In particular, the Lorentz group is a subgroup of the
> invariance group of Maxwell's equations, while the Galileo group is
> inconsistent with both.
>
> Nothing you write can possibly change any of that.
>
> [If you don't understand what an invariance group is,
> then you do not have enough knowledge to understand
> any of this, and you must STUDY.
> BTW that is pure math: group theory applied to
> partial differential equations.]
>
> Tom Roberts
Well, first of all, you need to say what you mean when you say that the Galilean transform is INCONSISTENT with the velocity of any object being the same in two relatively-moving frames.
That is exactly what I first pointed out in the beginning. If the pilot of an airplane has a clock that is slower or faster than a clock on the ground, then the pilot of the airplane will get a different speed for the airplane than an observer on the ground would get if he times the flight of the airplane with a clock on the ground. Scientists have not shown why they want the velocity to be the same in two relatively moving frames. First of all, if we are talking about the airplane, the speed between frames of reference can be the same, but the velocity cannot because the velocity of frame of reference S relative to frame of S' is negative compared to the velocity of S' relative to S. Are you talking about velocity or speed?
What is the object you are referring to?
Is it a photon? Is it an airplane? Is it a ball that is thrown by a passenger on an airplane?
You need to be more specific about what you are trying to say.
Now with regard to Maxwell's equations, I do not pretend to know anything about partial differental equations. I know a little about differential equations. Here is what I noticed about Einstein's theory. Einstein used differential equations to show velocities and came up with the following equations:
u = dx/dt
u' = dx'/dt' = (u-v)/(1-vu/c^2)
Now with Lorentz's incorrect interpretation of the numerator in his equation for time, which is nothing more than an incorrect interpretation of the Galilean transformation equations, as I have shown before, t'=t-vt/c, t' = t-vx/c^2, we get
x'=x-vt
dx'=dx-vdt
dx'=dx/dt - v
dx' = u-v

t'=t-vx/c^2
dt' = dt - vdx/c^2
dt' = 1 - v (dx/dt)/c^2
dt' = 1 - vu/c^2

These equations would only be true for something moving at the speed of light, as I have shown numerous times, or in other words, a photon. Einstein's interpretation cannot be true for anything moving at some other velocity. So we are back to the Galilean transformation equations again, which show that two clocks with different rates of time cannot both result in the same speed between frames of reference. What the Galilean transformation equations show is that regardless of what the clocks show, S travels the same distance relative to S' as S' travels relative to S, which is shown by the equation v't' = -vt, where v' is the velocity as seen from S', and t' is the time on the clock in S'. Since the Galilean transformation equations require that t'=t, the time of one clock has to be converted to the time of the other before using the Galilean transformation equations. I know how difficult this can be for scientists to understand because it is called reality.

On Thursday, February 23, 2023 at 10:23:58 AM UTC-8, Robert Winn wrote:
> On Thursday, February 23, 2023 at 10:10:15 AM UTC-7, Tom Roberts wrote:
> > On 2/23/23 12:49 AM, Robert Winn wrote:
> > > In any event, if Maxwell's equations agree with light traveling at c
> > > in two frames of reference, it seems to me that the Galilean
> > > transformation equations show why,
> > No, they don't. The Galilean transform is INCONSISTENT with the velocity
> > of any object being the same in two relatively-moving inertial frames.
> >
> > (Most of the equations you wrote are NOT Galilean transforms.
> > You are just engaged in mathematical masturbation.)
> > > since Maxwell died before the Michelson-Morley experiment took place,
> > > and no one was using anything but the Galilean transformation
> > > equations to represent relativity before the Michelson-Morley
> > > experiment.
> > That is the issue that became of prime importance in roughly 1880-1905
> > -- how could Maxwell's equations be reconciled with the Galilean
> > transform? We now know that they can't, and one must use Lorentz
> > transforms. In particular, the Lorentz group is a subgroup of the
> > invariance group of Maxwell's equations, while the Galileo group is
> > inconsistent with both.
> >
> > Nothing you write can possibly change any of that.
> >
> > [If you don't understand what an invariance group is,
> > then you do not have enough knowledge to understand
> > any of this, and you must STUDY.
> > BTW that is pure math: group theory applied to
> > partial differential equations.]
> >
> > Tom Roberts
> Well, first of all, you need to say what you mean when you say that the Galilean transform is INCONSISTENT with the velocity of any object being the same in two relatively-moving frames.
> That is exactly what I first pointed out in the beginning. If the pilot of an airplane has a clock that is slower or faster than a clock on the ground, then the pilot of the airplane will get a different speed for the airplane than an observer on the ground would get if he times the flight of the airplane with a clock on the ground. Scientists have not shown why they want the velocity to be the same in two relatively moving frames. First of all, if we are talking about the airplane, the speed between frames of reference can be the same, but the velocity cannot because the velocity of frame of reference S relative to frame of S' is negative compared to the velocity of S' relative to S. Are you talking about velocity or speed?
> What is the object you are referring to?
> Is it a photon? Is it an airplane? Is it a ball that is thrown by a passenger on an airplane?
> You need to be more specific about what you are trying to say.
> Now with regard to Maxwell's equations, I do not pretend to know anything about partial differental equations. I know a little about differential equations. Here is what I noticed about Einstein's theory. Einstein used differential equations to show velocities and came up with the following equations:
> u = dx/dt
> u' = dx'/dt' = (u-v)/(1-vu/c^2)
> Now with Lorentz's incorrect interpretation of the numerator in his equation for time, which is nothing more than an incorrect interpretation of the Galilean transformation equations, as I have shown before, t'=t-vt/c, t' = t-vx/c^2, we get
> x'=x-vt
> dx'=dx-vdt
> dx'=dx/dt - v
> dx' = u-v
>
> t'=t-vx/c^2
> dt' = dt - vdx/c^2
> dt' = 1 - v (dx/dt)/c^2
> dt' = 1 - vu/c^2
>
> These equations would only be true for something moving at the speed of light, as I have shown numerous times, or in other words, a photon. Einstein's interpretation cannot be true for anything moving at some other velocity.. So we are back to the Galilean transformation equations again, which show that two clocks with different rates of time cannot both result in the same speed between frames of reference. What the Galilean transformation equations show is that regardless of what the clocks show, S travels the same distance relative to S' as S' travels relative to S, which is shown by the equation v't' = -vt, where v' is the velocity as seen from S', and t' is the time on the clock in S'. Since the Galilean transformation equations require that t'=t, the time of one clock has to be converted to the time of the other before using the Galilean transformation equations. I know how difficult this can be for scientists to understand because it is called reality.

Galilean relativity is a fall curve that has original motion ahead.
Drop something from a moving course and its fall path
moves ahead with it. The fall path is moving from sharing
an original speed of the horse where it was dropped at.

On Thursday, February 23, 2023 at 11:49:22 AM UTC-7, mitchr...@gmail.com wrote:
> On Thursday, February 23, 2023 at 10:23:58 AM UTC-8, Robert Winn wrote:
> > On Thursday, February 23, 2023 at 10:10:15 AM UTC-7, Tom Roberts wrote:
> > > On 2/23/23 12:49 AM, Robert Winn wrote:
> > > > In any event, if Maxwell's equations agree with light traveling at c
> > > > in two frames of reference, it seems to me that the Galilean
> > > > transformation equations show why,
> > > No, they don't. The Galilean transform is INCONSISTENT with the velocity
> > > of any object being the same in two relatively-moving inertial frames..
> > >
> > > (Most of the equations you wrote are NOT Galilean transforms.
> > > You are just engaged in mathematical masturbation.)
> > > > since Maxwell died before the Michelson-Morley experiment took place,
> > > > and no one was using anything but the Galilean transformation
> > > > equations to represent relativity before the Michelson-Morley
> > > > experiment.
> > > That is the issue that became of prime importance in roughly 1880-1905
> > > -- how could Maxwell's equations be reconciled with the Galilean
> > > transform? We now know that they can't, and one must use Lorentz
> > > transforms. In particular, the Lorentz group is a subgroup of the
> > > invariance group of Maxwell's equations, while the Galileo group is
> > > inconsistent with both.
> > >
> > > Nothing you write can possibly change any of that.
> > >
> > > [If you don't understand what an invariance group is,
> > > then you do not have enough knowledge to understand
> > > any of this, and you must STUDY.
> > > BTW that is pure math: group theory applied to
> > > partial differential equations.]
> > >
> > > Tom Roberts
> > Well, first of all, you need to say what you mean when you say that the Galilean transform is INCONSISTENT with the velocity of any object being the same in two relatively-moving frames.
> > That is exactly what I first pointed out in the beginning. If the pilot of an airplane has a clock that is slower or faster than a clock on the ground, then the pilot of the airplane will get a different speed for the airplane than an observer on the ground would get if he times the flight of the airplane with a clock on the ground. Scientists have not shown why they want the velocity to be the same in two relatively moving frames. First of all, if we are talking about the airplane, the speed between frames of reference can be the same, but the velocity cannot because the velocity of frame of reference S relative to frame of S' is negative compared to the velocity of S' relative to S. Are you talking about velocity or speed?
> > What is the object you are referring to?
> > Is it a photon? Is it an airplane? Is it a ball that is thrown by a passenger on an airplane?
> > You need to be more specific about what you are trying to say.
> > Now with regard to Maxwell's equations, I do not pretend to know anything about partial differental equations. I know a little about differential equations. Here is what I noticed about Einstein's theory. Einstein used differential equations to show velocities and came up with the following equations:
> > u = dx/dt
> > u' = dx'/dt' = (u-v)/(1-vu/c^2)
> > Now with Lorentz's incorrect interpretation of the numerator in his equation for time, which is nothing more than an incorrect interpretation of the Galilean transformation equations, as I have shown before, t'=t-vt/c, t' = t-vx/c^2, we get
> > x'=x-vt
> > dx'=dx-vdt
> > dx'=dx/dt - v
> > dx' = u-v
> >
> > t'=t-vx/c^2
> > dt' = dt - vdx/c^2
> > dt' = 1 - v (dx/dt)/c^2
> > dt' = 1 - vu/c^2
> >
> > These equations would only be true/w for something moving at the speed of light, as I have shown numerous times, or in other words, a photon. Einstein's interpretation cannot be true for anything moving at some other velocity. So we are back to the Galilean transformation equations again, which show that two clocks with different rates of time cannot both result in the same speed between frames of reference. What the Galilean transformation equations show is that regardless of what the clocks show, S travels the same distance relative to S' as S' travels relative to S, which is shown by the equation v't' = -vt, where v' is the velocity as seen from S', and t' is the time on the clock in S'. Since the Galilean transformation equations require that t'=t, the time of one clock has to be converted to the time of the other before using the Galilean transformation equations. I know how difficult this can be for scientists to understand because it is called reality.
> Galilean relativity is a fall curve that has original motion ahead.
> Drop something from a moving course and its fall path
> moves ahead with it. The fall path is moving from sharing
> an original speed of the horse where it was dropped at.
Well, I am not sure what it has to do with horses, but, yes, the Galilean transformation equations do describe what you are describing

On Thursday, 23 February 2023 at 18:45:40 UTC+1, Dono. wrote:
> On Thursday, February 23, 2023 at 9:40:25 AM UTC-8, Maciej Wozniak wrote:
> > On Thursday, 23 February 2023 at 18:10:15 UTC+1, Tom Roberts wrote:
> > > On 2/23/23 12:49 AM, Robert Winn wrote:
> > > > In any event, if Maxwell's equations agree with light traveling at c
> > > > in two frames of reference, it seems to me that the Galilean
> > > > transformation equations show why,
> > > No, they don't. The Galilean transform is INCONSISTENT with the velocity
> > And we're FORCED to eat Shit!!

No, I didn't. Poor idiot Dono is impudently
lying, as expected from relativistic scum.

On Thursday, February 23, 2023 at 12:31:34 PM UTC-8, Maciej Wozniak wrote:
> On Thursday, 23 February 2023 at 18:45:40 UTC+1, Dono. wrote:
> > On Thursday, February 23, 2023 at 9:40:25 AM UTC-8, Maciej Wozniak wrote:
> > > On Thursday, 23 February 2023 at 18:10:15 UTC+1, Tom Roberts wrote:
> > > > On 2/23/23 12:49 AM, Robert Winn wrote:
> > > > > In any event, if Maxwell's equations agree with light traveling at c
> > > > > in two frames of reference, it seems to me that the Galilean
> > > > > transformation equations show why,
> > > > No, they don't. The Galilean transform is INCONSISTENT with the velocity
> > > And we're FORCED to eat Shit!!
> No, I didn't.
But you do eat shit on a daily basis, janitor.

On 2/23/2023 1:49 AM, Robert Winn wrote:
> The first thing I noticed when I read Einstein's book was how he claimed that the Lorentz equations satisfied the results of the Michelson-Morley experiment. He said he had extracted two little equations from the Lorentz equations that showed light to be traveling at c in a frame of reference at rest and a frame of reference in motion.
> x=ct
> x'=ct'
> So I put these values into the Galilean transformation equation and came up with the following.
>
> x'=x-vt
> ct'=ct-vt
> t'=t-vt/c
> This is essentially the same as the numerator of Lorentz's equation.
> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
> But it is incorrect,

Congratulations. You just figured out why a constant speed of light
(x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.

[rest of nonsense snipped]

Volney wrote:
> On 2/23/2023 1:49 AM, Robert Winn wrote:
>> The first thing I noticed when I read Einstein's book was how he
>> claimed that the Lorentz equations satisfied the results of the
>> Michelson-Morley experiment.  He said he had extracted two little
>> equations from the Lorentz equations that showed light to be traveling
>> at c in a frame of reference at rest and a frame of reference in motion.
>> x=ct
>> x'=ct'
>> So I put these values into the Galilean transformation equation and
>> came up with the following.
>>
>> x'=x-vt
>> ct'=ct-vt
>> t'=t-vt/c
>> This is essentially the same as the numerator of Lorentz's equation.
>> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
>> But it is incorrect,
>
> Congratulations. You just figured out why a constant speed of light
> (x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.
>
> [rest of nonsense snipped]

Robert B. Winn is fighting against basic algebra for decades, and
failed. I don't think he's about to figure out how silly he is soon.

On Thursday, February 23, 2023 at 3:24:02 PM UTC-7, Volney wrote:
> On 2/23/2023 1:49 AM, Robert Winn wrote:
> > The first thing I noticed when I read Einstein's book was how he claimed that the Lorentz equations satisfied the results of the Michelson-Morley experiment. He said he had extracted two little equations from the Lorentz equations that showed light to be traveling at c in a frame of reference at rest and a frame of reference in motion.
> > x=ct
> > x'=ct'
> > So I put these values into the Galilean transformation equation and came up with the following.
> >
> > x'=x-vt
> > ct'=ct-vt
> > t'=t-vt/c
> > This is essentially the same as the numerator of Lorentz's equation.
> > t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
> > But it is incorrect,
> Congratulations. You just figured out why a constant speed of light
> (x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.
>
> [rest of nonsense snipped]

On Friday, 24 February 2023 at 00:12:53 UTC+1, Python wrote:
> Volney wrote:
> > On 2/23/2023 1:49 AM, Robert Winn wrote:
> >> The first thing I noticed when I read Einstein's book was how he
> >> claimed that the Lorentz equations satisfied the results of the
> >> Michelson-Morley experiment. He said he had extracted two little
> >> equations from the Lorentz equations that showed light to be traveling
> >> at c in a frame of reference at rest and a frame of reference in motion.
> >> x=ct
> >> x'=ct'
> >> So I put these values into the Galilean transformation equation and
> >> came up with the following.
> >>
> >> x'=x-vt
> >> ct'=ct-vt
> >> t'=t-vt/c
> >> This is essentially the same as the numerator of Lorentz's equation.
> >> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
> >> But it is incorrect,
> >
> > Congratulations. You just figured out why a constant speed of light
> > (x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.
> >
> > [rest of nonsense snipped]
> Robert B. Winn is fighting against basic algebra for decades, and

and your bunch of idiots is fighting basic geometry.

On Thursday, February 23, 2023 at 3:24:02 PM UTC-7, Volney wrote:
> On 2/23/2023 1:49 AM, Robert Winn wrote:
> > The first thing I noticed when I read Einstein's book was how he claimed that the Lorentz equations satisfied the results of the Michelson-Morley experiment. He said he had extracted two little equations from the Lorentz equations that showed light to be traveling at c in a frame of reference at rest and a frame of reference in motion.
> > x=ct
> > x'=ct'
> > So I put these values into the Galilean transformation equation and came up with the following.
> >
> > x'=x-vt
> > ct'=ct-vt
> > t'=t-vt/c
> > This is essentially the same as the numerator of Lorentz's equation.
> > t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
> > But it is incorrect,
> Congratulations. You just figured out why a constant speed of light
> (x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.
>
> [rest of nonsense snipped]
Well, no, Volney. Here is how speed of light relates to the Galilean transformation equations.
S is a frame of reference at rest and S' is a frame of reference in motion with velocity v relative to S. In the Galilean transformation equations, x can be any coordinate on the x axis. But if you are going to say that x =ct, as Einstein and Lorentz say they are doing, then you would have to use different variables for every value of x because the time it takes for light to travel a distance of x is not the same as t unless x is the distance light travels in a time of t. For example, if t is one second, then x is 186,000 miles. But x can be any distance from the origin of S. For example, if x = 1 mile, light can still travel at a speed of c. But x does not equal ct. It equals c times a shorter time, which would be 1/186,000 sec.. If x = 10 miles, the equation would be x = c (1/18,600 sec.) t is the time it takes frame of reference S' to travel a distance a distance of vt relative to frame of reference S. It is not the time it takes for light to travel a distance of x unless v=c.
So we also have the equation x'=ct'. This is also incorrect for the Galilean transformation equations.
If x and x' are positive, then the time it takes light to travel a distance of x' is (x-vt)/c, not t'.
t'=t in the Galilean transformation equations. t is the time it takes for S' to travel a distance of vt relative to S.
If you have a clock in S' with a different rate than a clock in S, then to use the Galilean transformation equations, you would have to convert the time of one clock to the time of the other before using the times in the Galilean transformation equations.
Now you scientists say you have a clock in S' that is slower or faster than a clock in S, and according to the clock in S', light is traveling at c = 186,000 miles per second. So what do we call the time of this clock?
We cannot call it t' because t'=t in the Galilean transformation equations. But the time of this faster or slower clock actually exists according to scientists. So suppose we call the time of this clock in S' by the variable n'.
So to express this correctly using the Galilean transformation equations we do not say x=ct and x'=ct' the way Lorentz and Einstein did. We have to use variables for these smaller times.

x=cn
x'=cn'
x'=x-vt
cn' = cn - vt
n' = n - vt/c

On Thursday, February 23, 2023 at 4:12:53 PM UTC-7, Python wrote:
> Volney wrote:
> > On 2/23/2023 1:49 AM, Robert Winn wrote:
> >> The first thing I noticed when I read Einstein's book was how he
> >> claimed that the Lorentz equations satisfied the results of the
> >> Michelson-Morley experiment. He said he had extracted two little
> >> equations from the Lorentz equations that showed light to be traveling
> >> at c in a frame of reference at rest and a frame of reference in motion.
> >> x=ct
> >> x'=ct'
> >> So I put these values into the Galilean transformation equation and
> >> came up with the following.
> >>
> >> x'=x-vt
> >> ct'=ct-vt
> >> t'=t-vt/c
> >> This is essentially the same as the numerator of Lorentz's equation.
> >> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
> >> But it is incorrect,
> >
> > Congratulations. You just figured out why a constant speed of light
> > (x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.
> >
> > [rest of nonsense snipped]
> Robert B. Winn is fighting against basic algebra for decades, and
> failed. I don't think he's about to figure out how silly he is soon.
If you think I have made a mistake, Python, go ahead and show the mistake. You say there is an error in my algebra. I have made algebra mistakes in the past. Sometimes I make typographical errors. You say you can see how silly I am. Go ahead and show the algebra mistake you are saying you see.

On 2/24/2023 2:38 AM, Robert Winn wrote:
> On Thursday, February 23, 2023 at 3:24:02 PM UTC-7, Volney wrote:
>> On 2/23/2023 1:49 AM, Robert Winn wrote:
>>> The first thing I noticed when I read Einstein's book was how he claimed that the Lorentz equations satisfied the results of the Michelson-Morley experiment. He said he had extracted two little equations from the Lorentz equations that showed light to be traveling at c in a frame of reference at rest and a frame of reference in motion.
>>> x=ct
>>> x'=ct'
>>> So I put these values into the Galilean transformation equation and came up with the following.
>>>
>>> x'=x-vt
>>> ct'=ct-vt
>>> t'=t-vt/c
>>> This is essentially the same as the numerator of Lorentz's equation.
>>> t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
>>> But it is incorrect,
>> Congratulations. You just figured out why a constant speed of light
>> (x=ct & x'=ct' and t'=t) is inconsistent with the Galilean transformation.
>>
>> [rest of nonsense snipped]
> Well, no, Volney. Here is how speed of light relates to the Galilean transformation equations.
> S is a frame of reference at rest and S' is a frame of reference in motion with velocity v relative to S. In the Galilean transformation equations, x can be any coordinate on the x axis. But if you are going to say that x =ct, as Einstein and Lorentz say they are doing, then you would have to use different variables for every value of x because the time it takes for light to travel a distance of x is not the same as t unless x is the distance light travels in a time of t. For example, if t is one second, then x is 186,000 miles. But x can be any distance from the origin of S. For example, if x = 1 mile, light can still travel at a speed of c. But x does not equal ct. It equals c times a shorter time, which would be 1/186,000 sec. If x = 10 miles, the equation would be x = c (1/18,600 sec.) t is the time it takes frame of reference S' to travel a distance a distance of vt relative to frame of reference S. It is not the time it takes for light to travel a distance of x unless v=c.
What you write is largely gibberish, but it is only true that x=ct if x
is the progress of a beam of light. For anything else with a velocity u,
you get x=ut.
> So we also have the equation x'=ct'. This is also incorrect for the Galilean transformation equations.
Only correct for a beam of light.
> If x and x' are positive, then the time it takes light to travel a distance of x' is (x-vt)/c, not t'.
No, (x-vt)/c would be for something (light) to travel a transformed
distance x'. For anything else, replace c with its actual speed in S.
> t'=t in the Galilean transformation equations. t is the time it takes for S' to travel a distance of vt relative to S.
> If you have a clock in S' with a different rate than a clock in S,
No, it is assumed clocks tick the correct time. Slow clocks are invalid.
> then to use the Galilean transformation equations, you would have to convert the time of one clock to the time of the other before using the times in the Galilean transformation equations.
> Now you scientists say you have a clock in S' that is slower or faster than a clock in S, and according to the clock in S', light is traveling at c = 186,000 miles per second. So what do we call the time of this clock?
> We cannot call it t' because t'=t in the Galilean transformation equations. But the time of this faster or slower clock actually exists according to scientists. So suppose we call the time of this clock in S' by the variable n'.
> So to express this correctly using the Galilean transformation equations we do not say x=ct and x'=ct' the way Lorentz and Einstein did. We have to use variables for these smaller times.
Gobbledygook. You have a good clock or you have useless garbage. Toss
the garbage and use only good clocks.

On Friday, February 24, 2023 at 4:00:39 PM UTC-7, Volney wrote:
> > Well, no, Volney. Here is how speed of light relates to the Galilean transformation equations.
> > S is a frame of reference at rest and S' is a frame of reference in motion with velocity v relative to S. In the Galilean transformation equations, x can be any coordinate on the x axis. But if you are going to say that x =ct, as Einstein and Lorentz say they are doing, then you would have to use different variables for every value of x because the time it takes for light to travel a distance of x is not the same as t unless x is the distance light travels in a time of t. For example, if t is one second, then x is 186,000 miles. But x can be any distance from the origin of S. For example, if x = 1 mile, light can still travel at a speed of c. But x does not equal ct. It equals c times a shorter time, which would be 1/186,000 sec. If x = 10 miles, the equation would be x = c (1/18,600 sec.) t is the time it takes frame of reference S' to travel a distance a distance of vt relative to frame of reference S. It is not the time it takes for light to travel a distance of x unless v=c.
> What you write is largely gibberish, but it is only true that x=ct if x
> is the progress of a beam of light. For anything else with a velocity u,
> you get x=ut.
Well, so u = dx/dt. But we were talking about the equations Einstein and Lorentz used, x=ct and x'=ct'. As I pointed out, and you seemed to agree, these equations only work for one value of x, the distance light travels in a time of t. So if t = 1 sec. then x has to equal 186,000 miles. But to indicate that light is traveling a distance of x at a velocity of c, we cannot say x=ct. We have to use a different variable for the time it takes light to travel a distance of x. So since x can be any coordinate on the x axis, we say that x=cn, where n is the amount of time it takes for light to travel a distance of x, and x'=cn', where n' is the amount of time it takes for light to travel a distance of x'. So if t = 1 sec., then if x = 1 mile, then n = 1/186,000 sec., if x - 2 mi., then n = 2/186,000 sec., if x = 3 mi., the n = 3/186,000 sec., and so on. Then, in like manner, for each of these values for x, there is an x' that is a shorter distance if the velocity of S' is positive relative to S, and there will be a shorter time n' according to the equation x'=cn'. So once again using the forbidden Galilean transformation equations, we get
x'=x-vt
cn' = cn - vt
n' = n - vt/c
According to your equation x = ut, n' = (u - v)t/c
You might want to recall that Einstein came up with
u=dx/dt
u' = dx'/dt' = (u-v)/(1-vu/c^2)
These equations come from the incorrect interpretation Lorentz and Einstein had in which they claimed that x=ct was true for all values of x. But reality is different.
If x = ut, as you are saying, then
x'= ut - vt
dx' = (u-v)dt
dx'/dt = u-v
Since t=t' in the Galilean transformation equations,
dx'/dt' = u-v

> > So we also have the equation x'=ct'. This is also incorrect for the Galilean transformation equations.
> Only correct for a beam of light.
> > If x and x' are positive, then the time it takes light to travel a distance of x' is (x-vt)/c, not t'.
> No, (x-vt)/c would be for something (light) to travel a transformed
> distance x'. For anything else, replace c with its actual speed in S.
Well, the way to determine if what I say is true would be to set cn' equal to ut-vt. I already proved this with the correct equations for velocity in the Galilean transformation equations.
> > t'=t in the Galilean transformation equations. t is the time it takes for S' to travel a distance of vt relative to S.
> > If you have a clock in S' with a different rate than a clock in S,
> No, it is assumed clocks tick the correct time. Slow clocks are invalid.
Well, not according to Einstein. He plainly says that a moving clock is slower than a clock that is not moving. So, that being the case, the clock in S' does not show t' because t'=t in the Galilean transformation equations. If you want to find out how the time of the clock in S' relates to the time of the clock in S, then you have to do more math than I have done. Whether the clock is slower, as Einstein theorized in Special Relativity, or faster, as scientists say a clock in a GPS satellite is, the times of the clocks will be related by the velocities observed using each clock.

v'(S' clock time) = -vt

> > then to use the Galilean transformation equations, you would have to convert the time of one clock to the time of the other before using the times in the Galilean transformation equations.
> > Now you scientists say you have a clock in S' that is slower or faster than a clock in S, and according to the clock in S', light is traveling at c = 186,000 miles per second. So what do we call the time of this clock?
> > We cannot call it t' because t'=t in the Galilean transformation equations. But the time of this faster or slower clock actually exists according to scientists. So suppose we call the time of this clock in S' by the variable n'.
> > So to express this correctly using the Galilean transformation equations we do not say x=ct and x'=ct' the way Lorentz and Einstein did. We have to use variables for these smaller times.
> Gobbledygook. You have a good clock or you have useless garbage. Toss
> the garbage and use only good clocks.
Einstein said all these clocks were good. So I am assuming that they are all good, since Einstein said they were. Einstein says that a moving clock is slower than a clock that is not moving. Assuming he is correct about that, the Galilean transformation equations can describe what takes place.

On 2/24/2023 8:02 PM, Robert Winn wrote:
> On Friday, February 24, 2023 at 4:00:39 PM UTC-7, Volney wrote:
>>> Well, no, Volney. Here is how speed of light relates to the Galilean transformation equations.
>>> S is a frame of reference at rest and S' is a frame of reference in motion with velocity v relative to S. In the Galilean transformation equations, x can be any coordinate on the x axis. But if you are going to say that x =ct, as Einstein and Lorentz say they are doing, then you would have to use different variables for every value of x because the time it takes for light to travel a distance of x is not the same as t unless x is the distance light travels in a time of t. For example, if t is one second, then x is 186,000 miles. But x can be any distance from the origin of S. For example, if x = 1 mile, light can still travel at a speed of c. But x does not equal ct. It equals c times a shorter time, which would be 1/186,000 sec. If x = 10 miles, the equation would be x = c (1/18,600 sec.) t is the time it takes frame of reference S' to travel a distance a distance of vt relative to frame of reference S. It is not the time it takes for light to travel a distance of x unless v=c.
>> What you write is largely gibberish, but it is only true that x=ct if x
>> is the progress of a beam of light. For anything else with a velocity u,
>> you get x=ut.
> Well, so u = dx/dt. But we were talking about the equations Einstein and Lorentz used, x=ct and x'=ct'. As I pointed out, and you seemed to agree, these equations only work for one value of x, the distance light travels in a time of t. So if t = 1 sec. then x has to equal 186,000 miles. But to indicate that light is traveling a distance of x at a velocity of c, we cannot say x=ct. We have to use a different variable for the time it takes light to travel a distance of x.
Gibberish nonsense. You can pick any time t that you want, and with
x=ct, you will get a corresponding x. For t=1 second, x=300,000 km. If
t=1 mS, x=300 km. And so forth.
> According to your equation x = ut, n' = (u - v)t/c
> You might want to recall that Einstein came up with
> u=dx/dt
> u' = dx'/dt' = (u-v)/(1-vu/c^2)
Einstein's "u" has nothing to do with the "u" I used. I used u to avoid
v, the relative velocity of the two frames.
>>> So to express this correctly using the Galilean transformation equations we do not say x=ct and x'=ct' the way Lorentz and Einstein did. We have to use variables for these smaller times.
>> Gobbledygook. You have a good clock or you have useless garbage. Toss
>> the garbage and use only good clocks.
> Einstein said all these clocks were good.
No, he did not.
> So I am assuming that they are all good, since Einstein said they were. Einstein says that a moving clock is slower than a clock that is not moving. Assuming he is correct about that, the Galilean transformation equations can describe what takes place.
You are combining Galilean transformations and Lorentzian
transformations. That just doesn't make any sense. An observer observing
a clock moving relative to the observer running slow is part of the
Lorentzian transform and cannot happen with a Galilean transformation.

On Saturday, 25 February 2023 at 08:15:15 UTC+1, Volney wrote:

> transformations. That just doesn't make any sense. An observer observing
> a clock moving relative to the observer running slow is part of the
> Lorentzian transform and cannot happen with a Galilean transformation.

Of course it can, stupid Mike. It can happen even with
completely immobile clocks, and it happens often.

On Saturday, February 25, 2023 at 12:15:15 AM UTC-7, Volney wrote:
> On 2/24/2023 8:02 PM, Robert Winn wrote:
> > On Friday, February 24, 2023 at 4:00:39 PM UTC-7, Volney wrote:
> >>> Well, no, Volney. Here is how speed of light relates to the Galilean transformation equations.
> >>> S is a frame of reference at rest and S' is a frame of reference in motion with velocity v relative to S. In the Galilean transformation equations, x can be any coordinate on the x axis. But if you are going to say that x =ct, as Einstein and Lorentz say they are doing, then you would have to use different variables for every value of x because the time it takes for light to travel a distance of x is not the same as t unless x is the distance light travels in a time of t. For example, if t is one second, then x is 186,000 miles. But x can be any distance from the origin of S. For example, if x = 1 mile, light can still travel at a speed of c. But x does not equal ct. It equals c times a shorter time, which would be 1/186,000 sec. If x = 10 miles, the equation would be x = c (1/18,600 sec.) t is the time it takes frame of reference S' to travel a distance a distance of vt relative to frame of reference S. It is not the time it takes for light to travel a distance of x unless v=c.
> >> What you write is largely gibberish, but it is only true that x=ct if x
> >> is the progress of a beam of light. For anything else with a velocity u,
> >> you get x=ut.
> > Well, so u = dx/dt. But we were talking about the equations Einstein and Lorentz used, x=ct and x'=ct'. As I pointed out, and you seemed to agree, these equations only work for one value of x, the distance light travels in a time of t. So if t = 1 sec. then x has to equal 186,000 miles. But to indicate that light is traveling a distance of x at a velocity of c, we cannot say x=ct. We have to use a different variable for the time it takes light to travel a distance of x.
> Gibberish nonsense. You can pick any time t that you want, and with
> x=ct, you will get a corresponding x. For t=1 second, x=300,000 km. If
> t=1 mS, x=300 km. And so forth.
> > According to your equation x = ut, n' = (u - v)t/c
> > You might want to recall that Einstein came up with
> > u=dx/dt
> > u' = dx'/dt' = (u-v)/(1-vu/c^2)
> Einstein's "u" has nothing to do with the "u" I used. I used u to avoid
> v, the relative velocity of the two frames.
Well, but u would have a meaning if you say x=ut. x can be any coordinate on the x axis. Einstein said u was dx/dt. Now you are saying x is not dx/dt. Here is the breakdown in communication I see. You are defending Einstein's equations. Einstein said that u was dx/dt. Now you want to say that u is something else.

> >>> So to express this correctly using the Galilean transformation equations we do not say x=ct and x'=ct' the way Lorentz and Einstein did. We have to use variables for these smaller times.
> >> Gobbledygook. You have a good clock or you have useless garbage. Toss
> >> the garbage and use only good clocks.
Well, OK. Hafele and Keating had two good clocks. They put one good clock in an airplane and kept the other good clock on the ground. They said that if the airplane went around the earth one way, the clock on the airplane showed less time than the clock on the ground. If they went around the world the other way, the clock on the airplane showed more time than the clock on the ground. Hafele and Keating said that the clocks were good the entire time, the same as Einstein said he was using good clocks in his hypothesis. So, obviously, if the clocks did not read the same when they were reunited on the ground, one clock was faster than the other during the experiment. So common sense tells us that the rate of transitions of a cesium atom in one clock was different than the rate of transitions of a cesium atoms in the other clock. The reason why scientists like you are saying that good clocks always transition the same is because your definition of time is based on transitions of cesium atoms. But Galileo and Newton were not defining time based on transitions of cesium atoms. Their concept of time was based on rotations of the earth on its axis. So going back to the Hafele-Keating experiment, if the airplane flew for a day, and one of the clocks showed that it flew for however many nanoseconds less than a day, then that clock was slower than the other clock that showed exactly a day. The fact that one clock or the other was slower does not change the Galilean transformation equations. t'=t in the Galilean transformation equations. You have to choose one clock or the other and make it your preferred measurement of time to use the Galilean transformation equations. But you can still use the clock on the ground in the Galilean transformation equations as the preferred rate of time and then use the clock in the airplane as the preferred measurement of time and compare the results. I understand that scientists do not want to do that, but it is what I do. If scientists are offended by that, so be it. If scientists want to say the Galilean transformation equations are gobbledegook, that is fine with me. We have freedom of religion here in the United States. Scientists want to worship a miracle they call the length contraction. I just use the Galilean transformation equations because they do not have a length contraction and conform to reality. There is nothing to argue about. Scientists are free to do whatever they choose to do.

>
> > Einstein said all these clocks were good.
> No, he did not.
Sure he did. He spent pages and pages in his book explaining how he was synchronizing all of his clocks.
> > So I am assuming that they are all good, since Einstein said they were. Einstein says that a moving clock is slower than a clock that is not moving. Assuming he is correct about that, the Galilean transformation equations can describe what takes place.
> You are combining Galilean transformations and Lorentzian
> transformations. That just doesn't make any sense. An observer observing
> a clock moving relative to the observer running slow is part of the
> Lorentzian transform and cannot happen with a Galilean transformation.
What happens is that the transitions of cesium atoms do not remain the same, even though scientists say they do. But if scientists want to believe that transitions of cesium atoms always remain the same, we have freedom of religion here in the United States. I choose to believe that if one good clock does not read the same after an experiment as another good clock, then one of the good clocks was running slower than the other.

On Saturday, February 25, 2023 at 8:32:22 AM UTC-8, Robert Winn wrote:
>Scientists are free to do whatever they choose to do.
But you are not a scientist, Robert. You are just a sad sack crank.

On Saturday, February 25, 2023 at 9:42:11 AM UTC-7, Dono. wrote:
> On Saturday, February 25, 2023 at 8:32:22 AM UTC-8, Robert Winn wrote:
> >Scientists are free to do whatever they choose to do.
> But you are not a scientist, Robert. You are just a sad sack crank.

Well, yeah, Dono, but there is a difference between you and me. I can read physics books and see equations and use them in posts in sci.physics.relativity. All you ever do is call people derogatory names. I am actually interested in learning things. You think that if you can say something negative about another person, you are the most intelligent person on earth, so you do not need to learn anything. You are already superior to all other people because you can call them dirty names. You have posted to me for years and years, and not once have you ever posted an equation or said anything about relativity. Everything you have posted to me has been profanity or obscenity.
I would have to say that you are probably the most boring and predictable person I have ever encountered.
But with regard to your objection about me not being a scientist, you should probably read the preamble to this newsgroup. It specifically says that you do not need to be a scientist to post in sci.physics.relativity. Until that gets changed, I will probably continue to post here from time to time..

On Saturday, February 25, 2023 at 12:17:16 PM UTC-8, Robert Winn wrote:
>. I am actually interested in learning things.

No you are not, all you are interested is repeating the same imbecilities

On Saturday, February 25, 2023 at 2:36:23 PM UTC-7, Dono. wrote:
> On Saturday, February 25, 2023 at 12:17:16 PM UTC-8, Robert Winn wrote:
> >. I am actually interested in learning things.
> No you are not, all you are interested is repeating the same imbecilities
Well, the Galilean transformation equations were not considered imbecilities until Einstein came along. You say that the Galilean transformation equations cannot describe the times of two clocks with different rates. I say that they can. If you don't like the equations I post, why don't you find one of these nice Einstein believers to comfort you and console you?
I am sure there are some of them who would be glad to discuss your feelings with you.

On Saturday, February 25, 2023 at 12:15:15 AM UTC-7, Volney wrote:
> On 2/24/2023 8:02 PM, Robert Winn wrote:
> > On Friday, February 24, 2023 at 4:00:39 PM UTC-7, Volney wrote:
> >>> Well, no, Volney. Here is how speed of light relates to the Galilean transformation equations.
> >>> S is a frame of reference at rest and S' is a frame of reference in motion with velocity v relative to S. In the Galilean transformation equations, x can be any coordinate on the x axis. But if you are going to say that x =ct, as Einstein and Lorentz say they are doing, then you would have to use different variables for every value of x because the time it takes for light to travel a distance of x is not the same as t unless x is the distance light travels in a time of t. For example, if t is one second, then x is 186,000 miles. But x can be any distance from the origin of S. For example, if x = 1 mile, light can still travel at a speed of c. But x does not equal ct. It equals c times a shorter time, which would be 1/186,000 sec. If x = 10 miles, the equation would be x = c (1/18,600 sec.) t is the time it takes frame of reference S' to travel a distance a distance of vt relative to frame of reference S. It is not the time it takes for light to travel a distance of x unless v=c.
> >> What you write is largely gibberish, but it is only true that x=ct if x
> >> is the progress of a beam of light. For anything else with a velocity u,
> >> you get x=ut.
> > Well, so u = dx/dt. But we were talking about the equations Einstein and Lorentz used, x=ct and x'=ct'. As I pointed out, and you seemed to agree, these equations only work for one value of x, the distance light travels in a time of t. So if t = 1 sec. then x has to equal 186,000 miles. But to indicate that light is traveling a distance of x at a velocity of c, we cannot say x=ct. We have to use a different variable for the time it takes light to travel a distance of x.
> Gibberish nonsense. You can pick any time t that you want, and with
> x=ct, you will get a corresponding x. For t=1 second, x=300,000 km. If
> t=1 mS, x=300 km. And so forth.
Well, here is the problem I see. If you pick any time t that you want, that is fine, but you have to realize what t means. t is the time it takes for frame of reference S' to travel a distance of vt relative to frame of reference S. The problem is not whether t can be any time. It is whether it can be two different times in the same equation. t can be any time, but x can also be any coordinate on the x axis. If you are saying
x' = (x-vt)/sqrt(1-v^2/c^2)
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
then you are saying that the x in the equation for t' is the same as the x in the equation for x'. x in the equation for x' can be any coordinate on the x axis. But you already conceded that the equation for t' is only true for the distance light travels in time t. What about all of these other values for x that are shown in the equation for x'? I already established that t is the time it takes for S' to move a distance of vt relative to S. That means that if you are saying t = 1 mS then you have also changed the distance S' has moved relative to S. So you are not talking about that value for x when t = 1 sec. because frame of reference S' is the same whether a time of 1 sec. or 1 mS has elapsed. If x is c*1 mS, then that coordinate on the x axis is the same whether 1 mS has elapsed or 1 sec has elapsed.. What Lorentz and Einstein did with the numerator of their equation was to say that there is a value for x' in S' which can result in x = c(1mS) in the equation for x' and x = c (1 mS) in the equation for t'. In other words, S' travels a distance of vt relative to S. If x is c(1 mS), then what is x'?
Well, x'=ct', according to scientists
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
t was defined at the very beginning as the time it takes S' to travel a distance of vt relative to S, which we said was one second. Where your length contraction comes from is by saying that x in the equation for t' is c(1 mS) instead of c(1 sec). So you have two different values for x and two different values for t in the same equations if you are talking about any value for x other than the distance light travels in time t, which in this case would be 186,000 miles and one second, not c(1mS) and 1 mS.
As I have said many times, we have freedom of religion here in the United States. If you want to have a miracle that enables you to create nuclear explosions, then you have what you want. I think that there are probably all manner of other things that do not need to be done that would excite scientists. I was just trying to figure out what was wrong with their mathematics.