Re: Uncle Dickhead, So fucking what? LOL

In sci.physics, Sam Wormley
<swormley1@xxxxxxxxx>
wrote
on Mon, 02 May 2005 02:31:29 GMT
<5igde.40694$NU4.7186@attbi_s22>:
> The Ghost In The Machine wrote:
>
>>>>>Sam Wormley wrote:
>>> Show us your "correct" space-time diagrams. Don't you wish
>>> everybody did?
>>
>> You asking me or eightwings? :-)
>>
>
> Eightwings isn't worth having a conversation with as he is too busy
> fixating on other people's asses. However I do think that many posters
> who criticize SR without a good understanding of the subject, would
> benefit from spacetime diagrams accurately representing the arguments
> at hand.

Ah...well, that's clear as mud. :-) Then again, you do have a point.
I think Androcles got confused (and Eleaticus is showing similar
confusion) as to exactly how a rod contracts (or is observed to
contract) while moving past an observer.

I can't really draw worth a darn in ASCII. However, I can describe
it.

First, assume O has a rod 1m in length (as determined
in his reference-frame). He places this rod on a line.
It just sits there. Or does it?

If we tilt the observation point mathematically through a
timespace-rotation so that the vertical direction of the
view is time, and the horizontal direction is the line,
we would see (or graph) a *band*. This band represents
the rod, and the edges of the band represent the rod's
endpoints, and are vertical as well on the view/graph --
at least after the placement of the rod, which I assume
occurs at time 0, for simplicity.

Now the observer A, who happens to be uniformly moving
along the line at a velocity v and is such that O and
A are coincident at time t_O = t_A = 0, also has a rod,
of length 1m (again, in his reference-frame). He places
the rod at his local origin on the same line.

We tilt the observation as before, in O's coordinate space.
What does O see now?

O will now see a band which is *not* vertical. We can
describe the band in A's coordinate space thusly:

t_A >= 0
0m <= x_A <= 1m

The Lorentz applies, giving us two lines. The first line is
parametrically described:

x_O = (0 - v * t_A) * g
t_O = t_A * g

We solve for t_O and get

x_O = 0 - v * t_O

The second line is more interesting. I'll use primes here for
clarity, as these are new variables:

x_O' = (1m - v * t_A') * g
t_O' = (t_A' - v/c^2) * g

If we solve for t_A' we get

t_A' = t_O'/g + v/c^2

and we can now substitute:

x_O' = (1m - v * (t_O'/g + v/c^2)) * g

At this point it's far from clear how exactly to reconcile these
two expression, as t_O and t_O' are not the same. However, we
can still perform the measurement, if we define our endpoints
*in time* properly. One attempt, for example, is to simply
use t_O = t_O' = 0; this gives us a measured rod length of

(1m - v * (v/c^2)) * g = (1 - v^2/c^2) * g = 1/g

Obviously, the rod did not grow -- it *shrank*. Then again,
one could make the claim that this is more of a measurement
artifact than anything else.

If O and A both have their rods placed, where are the endpoints?

Well, the origin's covered, so we need merely worry about
where the endpoints are observed. If O is continually
observing the far endpoint of the rod (using a light
source or other such if need be), what and when does he
see A's rod doing?

First, O's rod has a timelag; anything O sees at his rod's
far end is going to lag 1m / c seconds behind. This isn't
a big problem but illustrates an issue with simultaneity;
basically, if O has a mirror arrangement such that he can
simultaneously observe both of his rod's endpoints, it's
not going to work quite right, though one possible way
around this is to place a mirror at the exact midpoint of
his rod, and look at the mirror as well as the other end.

It turns out that with this modification A's rod comes up short,
as it's the same situation as noted above.

O might try something else. For instance, at the exact moment
A's rod's origin passes by O, O can fire a laser pulse at the
far end of A's rod (which has a convenient mirror), measure the
time it takes for the pulse to come back, and simultaneously
measure the velocity of A's rod relative to his own (which
*can* be done accurately), and compensate appropriately.
Will this work?

Well, O fires off a lightpulse x_O = c * t_O, which in A's
coordinate space is *invariant*; for this light pulse x_A = c * t_A
as well. (The wavelength is shifted of course.) The pulse
bounces off the mirror with the equation x_A = 1 - c * (t_A - 1/c),
or x_A = 2 - c * t_A -- in A's space we know exactly where the
mirror is; it's at the rod's far endpoint, 1m away. If we
substitute this into the reverse Lorentz, we get:

x_O = (x_A + v * t_A) * g
= (2 - c * t_A + v * t_A) * g
= (2 - (c - v) * t_A) * g

t_O = (t_A + v * x_A/c^2) * g
= (t_A + v * (2 - c * t_A)/c^2) * g
= (t_A + 2 * v/c^2 - v * t_A / c) * g
= ((2 * v/c^2) + (1 - v/c) * t_A) * g

Solving for t_A:

t_O / g = ((2 * v/c^2) + (1 - v/c) * t_A)
t_O / g - (2 * v/c^2) = (1 - v/c) * t_A
t_A = (t_O / g - (2 * v/c^2) ) / (1 - v/c)

At this point it's probably simpler to set x_O = 0
and then ask what t_O is for this equation.
This yields t_A = 2/(c-v) and

t_O = ((2 * v/c^2) + (1 - v/c) * (2 / (c-v))) * g
= ((2 * v/c) + (c - v) * (2 / (c-v))) * g/c
= 2 * (g/c) * (1 + v/c)

Because g = 1/sqrt(1-v^2/c^2) = 1/(sqrt(1-v/c)*sqrt(1+v/c)),
we can rewrite this as

t_O = 2 * sqrt(1+v/c) / (c * sqrt(1-v/c))

or as

t_O = 2 / (c * g * (1 - v/c))
= 2 / (g * (c - v))

In Galilean math the initial light pulse still has equation

xg_O = c * tg_O

but the mirror is now moving with the equation

xg_O = 1 + v * tg_O

so the light pulse hits the mirror when

c * tg_O = 1 + v * tg_O
(c - v) * tg_O = 1
tg_O = 1/(c-v)

and xg_O = c/(c-v) at that point.

Because the mirror is moving O will see the lightbeam
come back at velocity c-2v. (From A's point of reference
he sees a lightray impinge at velocity c-v; the mirror
changes the direction but not the speed.) Therefore,
the total time taken in BaT theory is simply

tg_O = 1/(c-v) + c/((c-v)*(c-2v))
= (c-2v)/((c-v)*(c-2v)) + c/((c-v)*(c-2v))
= 2(c-v)/((c-v)*(c-2v))
= 2/(c-2v)

This is a very different result from the one predicted by SR.

So, did the rod grow or did it shrink? Depends on how one
measures it, but it's clear that spacetime is getting distorted
in some interesting ways, in SR.

GR is even worse, but I'm not competent in tensors. I'll admit
it looks interesting.

One last thing. In QM it's fairly simple to measure the energy
of a photon -- Compton scattering you already know about.
If we assume the photon has a mass 2E/c^2 (which is admittedly
a BaTism), the kinetic energy of course comes back out, and we
can predict that a photon from a source moving away from us at
velocity v will have an adjusted energy

Eg = Eg_0 * (c-v)^2/c^2 = Eg_0 * (1 - v/c)^2

In a Galilean system, if a "photon gun" fires off point photons
once every second as it's moving at velocity v, the length
between photons will be c+v. Each photon of course is moving
at velocity c-v.

Therefore, lambda[g] = lambda_0[g] * (1+v/c).

AFAIK, SR makes no pronouncements about the mass of a
photon except that when v = c, gamma becomes infinity;
therefore nothing can travel at the speed of light unless
the mass is 0, in which case 0/0 comes in and makes a bit
of a mess of things. (So far, this is well corroborated
by experiment, AFAIK.)

In relativistic QM

E = h * nu = h * c/lambda.

We can calculate the lambda stretch based on the Lorentz,
even if we can't get at the energy directly;
if t_A = lambda_0, x_A = 0, we grind it through and get:

x_O = (v * lambda_0) * g
t_O = (lambda_0) * g

This may be a little strange-looking but remember that O is
at 0, not at v * lambda * g; therefore, he has to wait
until the second pulse catches up to him, for a total time:

lambda = t_O + x_O / c
= lambda_0 * g * (1 + v/c)

so E = E_0 / (g * (1 + v/c))
= E_0 * sqrt(1-v/c) / sqrt(1+v/c)
= E_0 * g * (1-v/c)

Again, a different result, and with a ratio closer to 1 than
the BaT result; if one takes v = 10^-8 c, then

Eg/Eg_0 = 1 - 2 * 10^-8 + 10^-16

but

E/E_0 = 1 - 10^-8 + 5 * 10^-17 - 5 * 10^-25 + 3.75 * 10^-33 - ...

This might actually be measurable as 10^-8 c is just shy of 3 m/s.

--
#191, ewill3@xxxxxxxxxxxxx
It's still legal to go .sigless.
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