On 2/25/2023 11:27 AM, Robert Winn wrote:
> 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.
I said that "u" is not the same as the "u" in Einstein's speed
combination formula.
OK I am changing my equation. Instead of "u" I will use the Cyrillic
letter "ж" for the speed instead of "u". Because ж looks cool. My
equation is now x=жt. Hopefully you won't find any source where Einstein
used "ж".
> 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.
Now deal with x=жt.
<snip nonsense, unread>

On Tuesday, February 28, 2023 at 3:49:23 PM UTC-7, Volney wrote:
> On 2/25/2023 11:27 AM, Robert Winn wrote:
> > 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.
> I said that "u" is not the same as the "u" in Einstein's speed
> combination formula.
>
> OK I am changing my equation. Instead of "u" I will use the Cyrillic
> letter "ж" for the speed instead of "u". Because ж looks cool. My
> equation is now x=жt. Hopefully you won't find any source where Einstein
> used "ж".
> > 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.
> Now deal with x=жt.
>
> <snip nonsense, unread>
OK.

x = (>l<)t
x' = (>l<)t - vt

I still agree with Einstein about u. (>l<) would equal u =dx/dt
t'=t
dx' = u dt - v dt
dx'/dt = u-v
u' = dx'/dt' = dx'/dt
u' = u-v
But here is what I see. x' is not going to equal (>l<)t' in the Lorentz equations. We already established that the Lorentz equations only work if x is the distance light travels in a time of t, which is the time it takes for S' to move a distance of vt relative to S. So you have another unknown in your interpretation of the Lorentz equations.
x=(>l<)t
x'=(?)t'
It does not look to me like (?) = (>l<) unless (>l<)=c. So what does (?) equal for a point on the x axis that is not the distance light travels in a time of t?
I know this probably seems boring and irrelevant when you can use Einstein's equations to make nuclear bombs, but I have always wondered about this. Can you do anything with Einstein's equations other than kill everyone on earth?