Re: transforming coordinates
- From: rbwinn <rbwinn3@xxxxxxxx>
- Date: Mon, 24 Nov 2008 19:13:44 -0800 (PST)
On Nov 24, 5:50�pm, Dono <sa...@xxxxxxxxxxx> wrote:
On Nov 24, 3:06�pm, YBM <ybm...@xxxxxxxx> wrote:
rbwinn a �crit :
On Nov 24, 10:54 am, YBM <ybm...@xxxxxxxx> wrote:
rbwinn a crit :
On Nov 23, 7:13 pm, YBM <ybm...@xxxxxxxx> wrote:This have been explained to you for ages by numerous people.
rbwinn a crit :So now you are claiming that scientists did not say there was a slower
On Nov 19, 5:09 pm, YBM <ybm...@xxxxxxxx> wrote:It is meaningless and contradictory as such. As usual, you didn't read
rbwinn a crit :Well, YBM, you have misinterpreted the equations. n' is the time it
On Nov 19, 6:33 am, PD <TheDraperFam...@xxxxxxxxx> wrote:You have a huge problem with basic logic, Robert:
On Nov 18, 7:54 pm, rbwinn <rbwi...@xxxxxxxx> wrote:Well, here is a proof.
On Nov 18, 5:51 pm, PD <TheDraperFam...@xxxxxxxxx> wrote:Well, you apparently make no distinction between proof and assertion.
On Nov 18, 5:54 pm, rbwinn <rbwi...@xxxxxxxx> wrote:I already proved everything that needs to be proven about them. I
can be proven wrong, something will be posted here inI disagree. As a case in point, note that no one -- not a single soul
sci.physics.relativity.
-- will prove wrong the meaningless equations I made up. If you think
you can prove them wrong, I invite you to do so.>
said that they were just meaningless equations that had nothing to do
with momentum, and you admitted that they were meaningless equations
that you had made up that in fact did not have anything to do with
momentum.
However, I completely understand that you consider what you assert to
be sufficient proof, and you usually consider what others assert to be
requiring further proof. I'm sure this suits you, and perhaps other
welders, just fine.
x'=x-vt
y'=y
z'=z
t'=t
w=velocity of light
x = wt
x'=wn'
x' = x-vt
wn' = wt-vt
n'=t(1-v/w)
Why is it a proof? It is a proof because it cannot be disproven..
What you just wrote is that IF a slow clock made light appear to
propagate at speed c for any light ray on the (Ox) line (so limiting
yourself to velocity -c and +c, to the contrary of LTs which works
for any direction in 3D), THEN the slow clock would mark t(1-v/w)
when the "not slow clock" (in S) would mark t.
A little funny fact is that if you take w=-c, the "slow clock" in
indeed a fast clock.
This is, as a matter of fact, the beginning of a proof that the
"slow clock" hypothesis is absurd and contradictory: the "slow
clock" rate would depend on a arbitrary parameter (w=+c or -c)
unrelated to the relative movement of the frames, t(1-w/c) DOES
NOT specify a clock independantly of the events you consider.
The easy way to check this, no easy enough for a your dead brain
unfortunately, is to consider two light rays in the same experiment,
on at velocity (c,0,0), the other one at velocity (-c,0,0). If they
are simultaneously travelling your "slow clock" would have to mark
at the same time t(1-v/c) AND t(1+v/c), which is absurd. If the
two light ray are travelling at different interval of times (like
in M&M experiment), then the clock would have to change its rate
miraculously in the middle of the experiment, which is absurd as
well.
A even easier way to check that the "slow clock" hypothesis is
absurd is to consider a light ray with velocity (0,c,0), then
your "clock" fail to make it appear travelling at speed c (when
LTs succeed).
The most amazing thing is not even that you are unable to get
the point, given how obvious and simple it is, but that you were
unable during all these years of publishing nonsense you was
unable to figure this out by yourself, especially given all the
people who've kindly tried to help you on this.- Hide quoted text -
- Show quoted text -
takes for a photon to travel a distance of x'.
or, if you did, you are unable to understand the simpliest point.
Keep in mind when you talk about a slower clock, scientists were theWrong.
ones who said there was a slower clock.
You have the same problem with so-called "time dilatation" that you
have with "length contraction": you don't know what it means in SR..- Hide quoted text -
clock that they had proven experimentally.
I did it months ago too. You didn't read and are unwilling to
do so. Why should I bother ? I'm not a psychiatrist.- Hide quoted text -
- Show quoted text -
Well, from your posts here, YBM, your greatest concerns with what I
say lie in the fact that a photon directed in the negative direction
does not result in the same times that results from a photon directed
in the positive direction. �For example, if c is positive,
� � � � � �cn'=ct-vt
� � � � � � n'=t(1-v/c)
But if c is negative,
� � � � � �-cn'=-c-vt
� � � � � � �n'=t(1+v/c)
which you say means that n' indicates there is a faster clock in S',
not in S.
This never was your conclusion in using the Lorentz equations
� � � � � � t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
See: t' is the same whatever light ray or no light ray or whatever
events you consider, It is a function of (x,y) parametrized by v.
Compare with n'=t(1-w/c) which depends on "w" the (non constant)
of a specific light ray you choose to consider, this is basically
meaningless.
� If x is positive,
if x is positive : t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
� � � � � � t'=(t-vct/c^2)/sqrt(1-v^2/c^2)
� � � � � � t'=t(1-v/c)/sqrt(1-v^2/c^2)
You assume that x=ct, this has nothing to do with the case when
x is positive, this has to do with the case when the set of
events (x,t) you consider happens to be the trajectory of
a light ray with velocity (c,0,0). So you obtain the trajectory
in S'. so what ?
The point is that you don't understand the relations between
events, trajectories, coordinates, transformation and velocity.
Even when this has been explained to you thousands of times for
more than ten years. Nobody but a psychiatrist could help you
now.
� If x is negative,
if x is negative : t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
� � � � � �t'=(t-v(-c)t/(-c)^2/sqrt(1-v^2/(-c)^2)
� � � � � �t'=t(1+v/c)/sqrt(1-v^2/c^2)
You assume that x=-ct, this has nothing to do with the case when
x is positive, this has to do with the case when the set of
events (x,t) you consider happens to be the trajectory of
a light ray with velocity (-c,0,0). So you obtain the trajectory
in S'. so what ?
The point is that you don't understand the relations between
events, trajectories, coordinates, transformation and velocity.
Even when this has been explained to you thousands of times for
more than ten years. Nobody but a psychiatrist could help you
now.
So what about a clock measured by a photon going in the -x direction?
The expression "a clock measured by a photon" is utterly meaningless!
Your objection applies to the Lorentz equations as much as it does to
mine. �
No: t' depend only on x,t and v. Nothing specific to whatever light ray
you choose or not to consider.
Now, take your phone and call a psychiatrist.- Hide quoted text -
- Show quoted text -
You have better chances teaching special relativity to your chair :-)- Hide quoted text -
- Show quoted text -
Thank you for sharing, Dono.
Robert B. Winn
.
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