| Topic: |
Science > Physics |
| User: |
"Pentcho Valev" |
| Date: |
30 Aug 2006 01:48:45 AM |
| Object: |
DEFINITION OF EINSTEIN |
Einstein can be defined as Newton + the falsehood postulating that the
speed of light is independent of the speed of the light source + the
introduction of a slightly modified version of the principle of
Ignatius of Loyola:
"That we may in all things attain the truth, that we may not err in
anything, we ought ever to hold it a fixed principle, that what I see
white I believe to be black if the Romish Church define it so to be".
Pentcho Valev
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| User: "" |
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| Title: Re: DEFINITION OF EINSTEIN |
31 Aug 2006 07:08:57 AM |
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Pentcho Valev wrote:
Einstein can be defined as Newton + the falsehood postulating that the
speed of light is independent of the speed of the light source + the
introduction of a slightly modified version of the principle of
Ignatius of Loyola:
"That we may in all things attain the truth, that we may not err in
anything, we ought ever to hold it a fixed principle, that what I see
white I believe to be black if the Romish Church define it so to be".
Pentcho Valev
You are right, see
http://groups.google.fr/group/sci.physics.relativity/browse_frm/thread/9e1ecfaed67714e3/72aff1f7ee0c8273?hl=en#72aff1f7ee0c8273
What become the LT transforms if the speed of light depends on
the observers' velocity ?
_______________________
Hereafter is adapted Einstein's derivation of the LT, assuming that
*all observers will always measure the speed of light to be the same
no matter what their state of uniform linear motion is* (Wikipedia):
After supposing that two frames of reference, S and S', are each
in uniform translatory motion relative to the other, the velocity
of S' relative to S being v,
1) Einstein began his derivation with the relations
(1) x' = ax + bt, and
(2) t' = ex + gt
2) Then he claimed that at the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av
3) Now, he supposed that a light signal, starting from the coincident
origins of frames S and S' at t = t' = 0, travels toward positive x.
He claimed that after a time t, it will be at x = ct, and also at
x' = ct', since the speed of light is the same in all frames.
Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', he found
0 = ac + b - ec^2 - gc.
If the signal travels toward negative x, x = -ct and x' = -ct', thus
0 = -ac + b -ec^2 + gc.
Hence, a = g and b = ec^2 (or e = b/c^2).
4) Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c(ev + g) * t.
Equating y' to y, he found
c(ev + g) = sqrt(c^2 - v^2) = c * sqrt(1 - v^2/c^2), and claimed:
"Since, by prior results, e = b/c^2 = -av/c^2 = -gv/c^2,
it follows that cg(1 - v^2/c^2) = c * sqrt(1 - v^2/c^2)
and g = 1/sqrt(1 - v^2/c^2).
All constants thus being determined, the relations (1) and (2)
can be written
x' = g(x - vt) and t' = g(t - vx/c^2)."
TRANSFORMATIONS WITHOUT THE POSTULATE OF THE CONSTANCY OF LIGHT SPEED
Let's consider two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.
Basis relations:
x' = ax + bt
t' = ex + gt
2) At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av
The basis relations are now
(1) x' = a(x - vt)
(2) t' = ex + gt
3) Now, let's suppose that a light signal, starting from the coincident
origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at
x = ct + vt, and also at
x' = ct'.
Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets
(3) a - ec - ev - g = 0
If the signal travels toward negative x,
x = -ct + vt and x' = -ct'
Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets
(4) a + ec - ev - g = 0
From (3) and (4), one gets e = 0
With e = 0, relations (3) or (4) reduce to a = g
Hence, relations (1) and (2) become
(1) x' = g(x - vt)
(2) t' = gt
4) Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * g * t.
Equating y' to y, one gets g = sqrt(1 - v^2/c^2)
The transforms obtained without the *bold* Einstein's postulate that
the speed of light is the same in all frames are thus
(5) x' = sqrt(1 - v^2/c^2) * (x - vt)
(6) t' = sqrt(1 - v^2/c^2) * t
Transform (5) straightforwardly tells us that any body measures shorter
in terms of a frame relative to which it is moving with speed v than
it does as measured in a frame relative to which it is at rest.
Transform (6) implies that when two physical systems are in uniform
relative translation at speed v, the effects produced by system A
on system B are modified just as if all natural processes on A were
slowed down in the ratio sqrt(1 - v^2/c^2). (i.e., the so-called time
"dilation").
Those transforms, contrary to Einstein's LT, don't allow to claim that
simultaneity is relative (i.e., that events that are considered to
be simultaneous in one reference frame are not simultaneous in another
reference frame moving with respect to the first, cf. Wikipedia).
Marcel Luttgens
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| User: "Paul B. Andersen" |
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| Title: Re: DEFINITION OF EINSTEIN |
05 Sep 2006 09:31:50 AM |
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wrote:
TRANSFORMATIONS WITHOUT THE POSTULATE OF THE CONSTANCY OF LIGHT SPEED
Let's consider two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.
Basis relations:
x' = ax + bt
t' = ex + gt
2) At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av
The basis relations are now
(1) x' = a(x - vt)
(2) t' = ex + gt
3) Now, let's suppose that a light signal, starting from the coincident
origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at
x = ct + vt, and also at
x' = ct'.
Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets
(3) a - ec - ev - g = 0
If the signal travels toward negative x,
x = -ct + vt and x' = -ct'
In other words, the speed of light in S' is isotropic c.
In S, the speed of light is x+v in positive x direction
and x-v in the negative x-direction.
Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets
(4) a + ec - ev - g = 0
From (3) and (4), one gets e = 0
With e = 0, relations (3) or (4) reduce to a = g
Hence, relations (1) and (2) become
(1) x' = g(x - vt)
(2) t' = gt
Quite correct.
4) Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * g * t.
Oooops! :-)
You have now assumed that the speed of light in S is c.
It isn't. The vertical component is c, the horizontal
component is v, The speed of the oblique light in S is
sqrt(c^2 + v^2), not c. Thus y = ct
But, also, y' = ct' = c * g * t.
Equating y' to y, one gets g = 1
The Galilean transform.
Which shouldn't surprise you.
The rest is nonsense.
Paul
Equating y' to y, one gets g = sqrt(1 - v^2/c^2)
The transforms obtained without the *bold* Einstein's postulate that
the speed of light is the same in all frames are thus
(5) x' = sqrt(1 - v^2/c^2) * (x - vt)
(6) t' = sqrt(1 - v^2/c^2) * t
Transform (5) straightforwardly tells us that any body measures shorter
in terms of a frame relative to which it is moving with speed v than
it does as measured in a frame relative to which it is at rest.
Transform (6) implies that when two physical systems are in uniform
relative translation at speed v, the effects produced by system A
on system B are modified just as if all natural processes on A were
slowed down in the ratio sqrt(1 - v^2/c^2). (i.e., the so-called time
"dilation").
Those transforms, contrary to Einstein's LT, don't allow to claim that
simultaneity is relative (i.e., that events that are considered to
be simultaneous in one reference frame are not simultaneous in another
reference frame moving with respect to the first, cf. Wikipedia).
Marcel Luttgens
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| User: "Sorcerer" |
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| Title: Re: DEFINITION OF EINSTEIN |
05 Sep 2006 12:41:16 PM |
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"Paul B. Andersen" <paul.b.andersen@hiadeletethis.no> wrote in message
news:edk1on$rru$1@dolly.uninett.no...
| wrote:
| > TRANSFORMATIONS WITHOUT THE POSTULATE OF THE CONSTANCY OF LIGHT SPEED
| >
| > Let's consider two frames of reference, S and S', each in uniform
| > translatory motion relative to the other, the velocity of S' relative
| > to S being v.
| >
| > Basis relations:
| >
| > x' = ax + bt
| > t' = ex + gt
| >
| > 2) At the origin of S', x' = 0 and x = vt.
| > Hence, 0 = (av+b)t, whence b = -av
| >
| > The basis relations are now
| >
| > (1) x' = a(x - vt)
| > (2) t' = ex + gt
| >
| > 3) Now, let's suppose that a light signal, starting from the coincident
| >
| > origins of frames S and S' at t = t' = 0, travels toward positive x.
| > After a time t, it will be at
| > x = ct + vt, and also at
| > x' = ct'.
| >
| > Substituting these values of x and x' in relations
| > (1) and (2), and eliminating t and t', one gets
| >
| > (3) a - ec - ev - g = 0
| >
| > If the signal travels toward negative x,
| > x = -ct + vt and x' = -ct'
|
| In other words, the speed of light in S' is isotropic c.
HAHAHAHA!
Drunk again, huh?
In the real world the length of choo-choos is isometric.
| In S, the speed of light is x+v in positive x direction
| and x-v in the negative x-direction.
Well done! x is velocity now, is it?
u, v and w are more usual, but that's ok, you can use x.
|
| > Substituting these values of x and x' in relations
| > (1) and (2), and eliminating t and t', one gets
| >
| > (4) a + ec - ev - g = 0
| >
| > From (3) and (4), one gets e = 0
| >
| > With e = 0, relations (3) or (4) reduce to a = g
| >
| > Hence, relations (1) and (2) become
| >
| > (1) x' = g(x - vt)
| > (2) t' = gt
|
| Quite correct.
If g = 1, sure.
|
| > 4) Now, a light signal follow the y' axis. Relatively to S,
| > it travels obliquely, for, while the signal goes
| > a distance ct, the y'-axis advances a distance x = vt.
| > Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
| > But, also, y' = ct' = c * g * t.
|
| Oooops! :-)
| You have now assumed that the speed of light in S is c.
| It isn't.
No, of course not. It is x, right?
| The vertical component is c, the horizontal
| component is v, The speed of the oblique light in S is
| sqrt(c^2 + v^2), not c.
So light has a different velocity x to it's vertical velocity c, then.
I see. So that's sqrt(c^2+x^2) obliquely, then?
Thus y = ct
|
| > But, also, y' = ct' = c * g * t.
|
| Equating y' to y, one gets g = 1
| The Galilean transform.
Oh... I thought the Galilean transform was
c' = c-vt
z' = z-vt
| Which shouldn't surprise you.
|
| The rest is nonsense.
Ok... I agree with you.
Androcles
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| User: "Paul B. Andersen" |
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| Title: Re: DEFINITION OF EINSTEIN |
06 Sep 2006 04:52:29 AM |
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Paul B. Andersen wrote:
mluttgens@wanadoo.fr wrote:
3) Now, let's suppose that a light signal, starting from the coincident
origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at
x = ct + vt, and also at
x' = ct'.
Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets
(3) a - ec - ev - g = 0
If the signal travels toward negative x,
x = -ct + vt and x' = -ct'
In other words, the speed of light in S' is isotropic c.
In S, the speed of light is x+v in positive x direction
and x-v in the negative x-direction.
It should obviously be:
In other words, the speed of light in S' is isotropic c.
In S, the speed of light is c+v in positive x direction
and c-v in the negative x-direction.
Paul
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| User: "Sorcerer" |
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| Title: Re: DEFINITION OF EINSTEIN |
06 Sep 2006 05:31:27 AM |
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"Paul B. Andersen" <paul.b.andersen@hiadeletethis.no> wrote in message
news:edm5ov$4eq$1@dolly.uninett.no...
| Paul B. Andersen wrote:
| > wrote:
| >> 3) Now, let's suppose that a light signal, starting from the coincident
| >>
| >> origins of frames S and S' at t = t' = 0, travels toward positive x.
| >> After a time t, it will be at
| >> x = ct + vt, and also at
| >> x' = ct'.
| >>
| >> Substituting these values of x and x' in relations
| >> (1) and (2), and eliminating t and t', one gets
| >>
| >> (3) a - ec - ev - g = 0
| >>
| >> If the signal travels toward negative x,
| >> x = -ct + vt and x' = -ct'
| >
| > In other words, the speed of light in S' is isotropic c.
| > In S, the speed of light is x+v in positive x direction
| > and x-v in the negative x-direction.
|
| It should obviously be:
| In other words, the speed of light in S' is isotropic c.
| In S, the speed of light is c+v in positive x direction
| and c-v in the negative x-direction.
HAHAHAHA!
It should BLATANTLY OBVIOUSLY be:
In other words, the speed of light in S is NOT isotropic c.
In S, the speed of light is v+c in [the] positive x direction
and v-c in the negative x-direction.
You never did understand negative distances or velocities,
too drunk...
Androcles
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