Re: THE PHYSICAL TRANSFORMATION EQUATIONS

On Jun 10, 8:04 am, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On Jun 9, 5:19 am, Alen <a...@xxxxxxxxxxxxxxx> wrote:

On Jun 9, 4:50 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:

On Jun 8, 9:39 pm, Alen <a...@xxxxxxxxxxxxxxx> wrote:
Yes, I agree with that, and what I said doesn't deny it. What
I meant was that the dynamical reality doesn't make it clear
whether or not the static rotation model really exists. That is,
the dynamical reality also allows the possibility of some other
model. I have argued that another model is necessary because
I believe that it can be shown that the rotation model accepted
now for a century is not correct.

You have done no such thing.

Alen

Yes, I have: in my post, and in numerous other posts.

No.

An internal contradiction is equivalent to the destruction of real
analysis and everything that uses it. You have not demonstrated this.
You seem to be unaware that SR is equivalent to Hyperbolic geometry
and is open to mathematical attack as Euclidean geometry. Crank
opinions of SR are actually quite similar to opinions about true
Hyperbolic geometry circa 1860 or so.

An external contradiction [experimental falsification] would put you
on the short list for a Nobel prize. You have not demonstrated this
either.

What you have demonstrated is that you are simply clueless. You whine
about how the rotation explanation is incorrect, but you don't even
get that right. Lorentz transformations are rotations of an imaginary
angle between space and time. This follows almost trivially if you
start out with the group properties of the Lorentz group.

Since you have probably never seen it before, I'll derive the x,t
Lorentz transformations for you.

Start off with the most general symmetric matrix. Since all Lorentz
transformations are symmetric.

[ a, b, c, d]
[ b, e, f, g]
[ c, f, h, j] = A
[ d, g, j, k]

Represent your coordinates as the vector (t,x,y,z). Obviously, I'm
working in c = 1 units.

We want A*(t,x,y,z) = (t',x',y,z). In other words, t and x transform,
y and z do not. A little linear algebra, which I'm sure you are
entirely capable of doing since you think you understand SR enough to
attack it, shows that this is the most general symmetric
transformation matrix that preserves y and z.

[ a, b, 0, 0]
[ b, e, 0, 0]
[ 0, 0, 1, 0] = A
[ 0, 0, 0, 1]

All Lorentz transformations obey the condition that det(A) = +/- 1. If
you have had linear algebra, you would know why. I'll explain if you
don't understand, just ask. Now, all Lorentz transformations have to
transform continuously, so no discrete operations like parity
inversion are allowed - this means det(A) = 1.

That condition gives us a*e - b^2 = 1.

We need one more condition.

All linear transformations have to satisfy transpose(A)*N*A = N, where
N is the metric. In this case, N is the Minkowski metric. Since you
are familiar enough with SR to attack it, you should know what the
Minkowski metric is. Failing that, google it.

This gives us one more condition, a matrix equation:

That...

[ -a^2+b^2, -a*b+b*e, 0, 0]
[ -a*b+b*e, -b^2+e^2, 0, 0]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]

is equal to...

[-1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]

The (1,2) equation is equivalent to the (2,1) equation. It implies
that a = e
The (1,1) equation says that a^2 - b^2 = 1. Equivalent to the det(A) =
1 condition.
The (2,2) equation says what (1,1) says.

Now...thats almost like a trig identity. What does the internets say?

Internets says...http://www.alcyone.com/max/reference/maths/hyperbolic.html

Without any loss of generality, since sinh and cosh are entire and all
that complex analysis crap I try to forget, we can relabel with b =
sinh(theta) and a = cosh(theta).

So the x,t transformations are:

t' = cosh(theta)*t+sinh(theta)*x
x' = sinh(theta)*t+cosh(theta)*x

Gosh, that looks abstract and worthless.

But what do we know about SR? The speed of light has to be 1 in the
primed frame.

In the unprimed frame, dx/dt = 1. We must have dx'/dt' = 1 in the
primed frame.

dt' = cosh(theta)*dt+sinh(theta)*dx
dx' = sinh(theta)*dt+cosh(theta)*dx

dx'/dt' = [tanh(theta)dt + dx] / [dt + tanh(theta)dx]
dx'/dt' = [tanh(theta) + dx/dt] / [dt/dx + tanh(theta)dx/dt]

Use dx/dt = 1

dx'/dt' = [1 + tanh(theta)] / [1 + tanh(theta)] = 1

Excellent.

If you make the identification of theta = arctanh(v), and use a full
8.5x11" piece of paper full of mathematics, you can recover the
standard Lorentz transformations.


Alen

Thank you for taking the trouble to present this
group maths approach. I actually have no argument
with its mathematical validity, and no argument with
the fact that a hyperbolic geometry does describe SR
mathematically. This mathematics, I would say, motivates
people to accept it as the correct interpretation on the
ground, perhaps, that it would be peculiar if it was purely
accidental that it did fit SR so well, but wasn't really true
as a physical reality.

But I have the following problem with it, PHYSICALLY:
If there is a spacetime rotation, a moving length is
projected onto a stationary length, and produces length
contraction. Does it not follow that such a length contraction
exists whatever the method used to measure it?

There are two possible methods of measuring a moving
length, which are different: Method 1 uses some means
that is the equivalent of a moving length sliding past a
stationary ruler.The ruler detects the ends of the moving
length simultaneously and independently. Method 2 sends
light along the moving length, reflecting it off the end,
and calculates the measured length using the
transformation equations. I have the following web page
which, I say, proves that these two methods, in practice,
have to produce different results. Method 2 shows
length contraction, whereas method 1 cannot. On
this basis, a spacetime rotation cannot really exist,
because a spacetime rotation would produce the same
length contraction with either method 1 or method 2.

http://home.westserv.net.au/~alen1/Physics/aSpacelikeExperiments.htm

Alen


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