Generic formulas for designing special theories of relativity
Albertito
velocities, as
V(v_1, v_2) = v_3,
such that V can be defined through both an unary
operator U and its inverse U',
V(v_1, v_2) = U' (U(v_1) + U(v_2)) = v_3.
where
the binary operator + stands for standard vector
addition.
So, we get
U(v_3) = U(v_1) + U(v_2),
where U(v), for any velocity v, is a vector
in the same direction of v. This kind of vector
U(v) is called a vector rapidity.
Let's now define a generic Doppler formula as
f' = f Exp( - |U(v)| ),
where
|U(v)| is the norm of the vector U(v).
From, these generic formulas for addition of velocities
and Doppler, we can construct SR by defining a vector
rapidity as
U(v) = u arctanh( |v|/c),
where
u is a unit vector pointing in the direction of v,
so, it is u = v/|v|, and
c is the speed of light.
Then, the relativistic Doppler formula is
f' = f Exp( - |U(v)| ),
f' = f Exp( - |u arctanh( |v|/c)| ),
f' = f sqrt((1 - |v|/c)/(1 + |v|/c)).
The Lorentz factor would be then
gamma = Exp( - | u arctanh( |v|/c) | ) / (1 - |v|/c) =
= 1/sqrt( 1 - |v|^2/c^2),
because SR assumes a non-relativistic Doppler
is f' = f (1 - |v|/c), and a relativistic Doppler should
be f' = f gamma (1 - |v|/c)
We also can attain Galilean relativity by defining the vector
rapidity as an identity
U(v) = v,
U(v_3) = U(v_1) + U(v_2),
v_3 = v_1 + v_2.
So, the Doppler formula would look as
f' = f Exp( - |U(v)| ),
f' = f Exp( - |v| / c).
It is easy to see that a gamma in Galilean relativity
must be the invariant gamma = 1, for any velocity v.
The interesting thing of these generic formulas is
that it is not required the relative motion of bobies
be restricted to inertial systems. We can add velocities
or even gravitational fields. Complex problems can be
easily resolved with these generic formulas. Let 's
consider the following example:
The gravitational redshift for an uncharged mass which
is spherically symmetric (a Schwarzschild's solution in GR) is:
z = (1/sqrt(1 - |v|^2/c^2)) - 1,
where |v| = sqrt(2GM/r), is the norm of vector v.
with
G gravitational constant,
M the mass of the object creating the gravitational field,
r the radial coordinate of the observer , and
c the speed of light.
So, the vector v is pointing to the center of that uncharged
mass.
Then, we get
z + 1 = 1/sqrt(1 - |v|^2/c^2)
ln(z + 1) = -(1/2) ln(1 - |v|^2/c^2)
This means, in the Schwarzschild's solution, we have the
norm of a vector rapidity U(v) defined as
|U(v)| = (1/2) ln(1 - |v|^2/c^2),
because the generic Doppler is
f' = f Exp( - |U(v)| ), so f'/f = z +1.
Therefore, we get a vector rapidity as
U(v) = ( v/|v| )(1/2) ln(1 - |v|^2/c^2),
Now, suppose we have another uncharged mass close to
the former, which is also spherically symmetric, such that its
vector rapidity is
U(w) = ( w/ |w| )(1/2) ln(1 - |w|^2/c^2),
We can add those two fields to yield a superposed field as
U(s) = U(w) + U(v),
s = U'(U(w) + U(v)).
For the case where the angle between w and v is phi=0,
it would yield
|U(s)| = |U(w)| + |U(v)|,
|U(s)| = (1/2) ln(1 - |w|^2/c^2) +
(1/2) ln(1 - |v|^2/c^2),
|U(s)| = (1/2) ln((1 - |w|^2/c^2)(1 - |v|^2/c^2)),
(1/2) ln(1 - |s|^2/c^2) = (1/2) ln((1 - |w|^2/c^2)(1 - |v|^2/
c^2)) ,
1 - |s|^2/c^2 = (1 - |w|^2/c^2)(1 - |v|^2/c^2) ,
|s| = ± c sqrt(1 - (1 - |w|^2/c^2)(1 - |v|^2/c^2)),
and a gravitational redshift z for this superposition of fields
(case phi=0) would yield
z = (1/sqrt( 1 - |s|^2/c^2)) - 1 =
= (1/sqrt( (1 - |w|^2/c^2)(1 - |v|^2/c^2) ) ) - 1
In conclusion, these generic formulas can give us a
beautiful and simple formalism to solve n-body problems
under GR.
Eric Gisse
[snip babble]
Neither SR or GR are discussed here in any actual way, just
misconceptions and simplifications resulting from never having
actually studied the subject.
The fact is you simply do not know how the velocity addition formula
is derived. Rather than look at a textbook and learn, you instead opt
to make shit up and this is just the latest in the series titled
"Albertito does not know what he is talking about."
Albertito
That's your 'respectable' opinion. But, you should ask yourself,
"How can I solve a two-body problem under GR"?. Such a
two-body problem is very easy to solve under Newtonian
gravity. Why is it so difficult under GR? Actually, it is not so
difficult, you only have to avoid the non-linearity of GR, trying
to start from simple solutions as the Schwarzschild's one. I've
given you a beautiful two-body problem solution for a gravitational
redshift in a field composed by two nearby Schwarzschild bodies,
it is
z = (1/sqrt((1 - g_1^2)(1 - g_2^2)) ) - 1,
with
g_1^2 = 2GM_1/r_1, and
g_2^2 = 2GM_2/r_2.
Prove that gravitational redshift z is wrong.
Albertito
Sorry, I forgot to include c. I meant,
z = (1/sqrt( (1 - g_1^2)(1 - g_2^2) ) ) - 1,
with
g_1^2 = 2GM_1/r_1c^2, and
g_2^2 = 2GM_2/r_2c^2.
Eric Gisse
Why should I waste effort explaining to you why your latest babble is
wrong when you clearly do not understand ANYTHING about general
relativity?
Albertito
If I do not know ANYTHING about general relativity,
why have I been able to give an exact two-body solution
for a gravitational redshift in a field composed by two
nearby Schwarzschild bodies, in the specific case
where the masses M_1 and M_2 are aligned along the
sight of line of the observer?
Prove that redshift z is NOT an exact solution, and then
I will have to admit you know more GR than me. I'm not
expecting pedantic dissertations on GR from you, just to
focus on that specific solution of the gravitational
redshift z.
Eric Gisse
Simple - you haven't given an exact two-body solution of anything.
You are doing the physics equivalent of mashing a keyboard and
claiming the result is meaningful. The simple fact is that you are
clueless.
Albertito
Then, give me the exact solution, and I'll see
whether they match or not. The solution I've
provided is an exact solution. You don't believe
that's the solution for that specific two-body problem,
but you can't prove it. A reasonable answer should
be, "Well, I can't prove it, maybe you'are right."
Looking foward some erudite contribution from you.
Eric Gisse
This is also known as "trolling".
We both know you don't know what your talking about. You won't sucker
me into a technical discussion.
Albertito
Then, that only proves the only troll here in this thread
is you. You are unable to reply something coherent,
something like. "GR is hard, we can't solve n-body
problems in it, because GR is non-linear. If your solution
is correct and exact, then, that might be a wonderful
breakthrough, guy!"
Dono
Albertito
gravity, the principle of superposition tells us we can
simply add two gravitational potentials, as V = V_1 + V_2.
So, from the generic formulas provided above, we can
deduce a gravitational redshift z, under Newtonian gravity,
as
z = Exp(-V/c^2) - 1
This means that compared with the redshift z' predicted
under GR for a Schwarzschild body,
z' = 1/sqrt(1 - 2V/c^2) - 1,
we attain a discrepancy of
d = z' / z = 1 + V/c^2, to second-order approximation.
Eric Gisse
Since you are so sure it is an exact solution, why don't you show how
you obtained the solution from the field equations?
Eric Gisse
Bravo. You have transformed being stupid into an art.
What are the units of V / c^2?
Albertito
And you're claiming you have studied physics?
V/c^2 is dimensionless, fuckhead. V is
a Newtonian gravitational potential, which
has dimensions of a velocity squared.
Eric Gisse
Neat, I finally said something braindead.
Nothing you write has any actual meaning associated with it anyway.
Juan R.
That has an easy explanation Albertito.
The 'master' is 'studying' in the only University of the world (and rest
of the Cosmos) where *dimensional analysis* is believed to be some
Chinese food.
Dirk Van de moortel
pan.2008.05...@canonicalscience.com
> Albertito wrote on Mon, 19 May 2008 10:56:33 -0700:
[snip]
>> And you're claiming you have studied physics? V/c^2 is dimensionless,
>> fuckhead. V is a Newtonian gravitational potential, which has dimensions
>> of a velocity squared.
>
> That has an easy explanation Albertito.
>
> The 'master' is 'studying' in the only University of the world (and rest
> of the Cosmos) where *dimensional analysis* is believed to be some
> Chinese food.
Juanita R. González-Álvarez follows guidelines:
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Dirk Vdm
Juan R.
SperM.hotmail.com> wrote:
> Juan R. González-Álvarez <juanREM...@canonicalscience.com> wrote in message
> pan.2008.05.20.06.33...@canonicalscience.com
>
> > Albertito wrote on Mon, 19 May 2008 10:56:33 -0700:
>
> [snip]
>
> >> And you're claiming you have studied physics? V/c^2 is dimensionless,
> >> fuckhead. V is a Newtonian gravitational potential, which has dimensions
> >> of a velocity squared.
>
> > That has an easy explanation Albertito.
>
> > The 'master' is 'studying' in the only University of the world (and rest
> > of the Cosmos) where *dimensional analysis* is believed to be some
> > Chinese food.
>
> Juanita R. González-Álvarez follows guidelines:
> http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
>
> Dirk Vdm
Dimensional analysis:
http://groups.google.com/group/sci.physics.relativity/msg/6fe7633a0e8130f8
Dono
<juanREM...@canonicalscience.com> wrote:
> AlbertShito wrote on Mon, 19 May 2008 10:56:33 -0700:
>
Juan R.
Dirk, if you do not know what are the units for V/c^2 ask to Eric :-)
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.html
Albertito
<juanREM...@canonicalscience.com> wrote:
> Dirk Van de moortel wrote on Tue, 20 May 2008 16:13:27 +0200:
>
>
>
> > message
> > pan.2008.05.20.06.33...@canonicalscience.com
>
> > [snip]
>
> >>> And you're claiming you have studied physics? V/c^2 is dimensionless,
> >>> fuckhead. V is a Newtonian gravitational potential, which has
> >>> dimensions of a velocity squared.
>
> >> That has an easy explanation Albertito.
>
> >> The 'master' is 'studying' in the only University of the world (and
> >> rest of the Cosmos) where *dimensional analysis* is believed to be some
> >> Chinese food.
>
> > Juanita R. González-Álvarez follows guidelines:
> > http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
>
> > Dirk Vdm
>
> Dirk, if you do not know what are the units for V/c^2 ask to Eric :-)
>
> --http://canonicalscience.org/en/miscellaneouszone/guidelines.html
Why are all those fanatics of Einstein's relativity so braindead?
:-)
Albertito
combination of addition and Doppler shift like this,
u ln(z +1) = U(w) + U(v),
where
w and v are velocities of two bodies in the same frame of
reference,
U(w) and U(v) are their respective vector rapidities,
z is a Doppler shift of a photon emitted by one of those
bodies and observed by the other.
u is a unit vector pointing in the direction of the vector
rapidity sum.
A transverse Doppler shift is implicit in the above equation.
In the case v and w formed an angle phi=pi/2, then
ln(z +1) = sqrt(|U(w)|^2 + |U(v)|^2),
z = Exp( - sqrt(|U(w)|^2 + |U(v)|^2) ) - 1
is the transverse Doppler shift.
From this combination, we can easily see, if we make
w = c velocity of light, and perform the subtraction
u ln(z +1) = U(c) - U(v),
then
c' = U'(u ln(z +1)) = U'(U(c) - U(v))
is a vector velocity of light, with norm |c'| = |c|.
Thus, the angle between c' and v is the
aberration angle of light.
Under Galilean relativity, the unary operator U is
defined as the identity U(v) = v, so v = U'(v).
Then, velocity c' reduces to c' = u ln(z +1) = c - v.
Under SR, it is slightly more complicated. The subtraction
U(c') = U(c) - U(v) can't be performed because of
U(c) = (c/|c|) arctanh(|c|/|c|), diverges in magnitude,
arctanh(1) = oo. And, that's the reason why a relativistic
aberration formula is nonsense, as I already have pointed
out in this thread
http://groups.google.com/group/sci.physics.relativity/msg/e3b8a5258fb50227
Juan R.
>> Dirk, if you do not know what are the units for V/c^2 ask to Eric :-)
>>
>> --http://canonicalscience.org/en/miscellaneouszone/guidelines.html
>
> Why are all those fanatics of Einstein's relativity so braindead? :-)
La wikipedia es tan sabia
http://en.wikipedia.org/wiki/Fanaticism
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org
Dono
Juanshito,
You need to hire your pal, Albertshito to work at your "Center for
Canonical Science" :-)
You guys are a perfect fit.You may even be able to swap meds.
Androcles
| Albertito wrote on Thu, 22 May 2008 03:57:03 -0700:
|
| >> Dirk, if you do not know what are the units for V/c^2 ask to Eric :-)
| >>
| >> --http://canonicalscience.org/en/miscellaneouszone/guidelines.html
| >
| > Why are all those fanatics of Einstein's relativity so braindead? :-)
|
| La wikipedia es tan sabia
|
| http://en.wikipedia.org/wiki/Fanaticism
|
|
associated with a position or opinion which is so far from the norm as to
appear ludicrous and/or provably wrong"
Thus the question should be:
"Why are all those cranks of Einstein's relativity so braindead? "
--
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
Androcles
http://www.androcles01.pwp.blueyonder.co.uk/
Albertito
> "Juan R. González-Álvarez" <juanREM...@canonicalscience.com> wrote in
Well, then we should invent a hybrid term for describing
them as half a fanatic and half a crank of Einstein relativity,
what about a 'fancrankeinstein' ? :-)
Juan R.
I buy this one :-)
Anyone to write a wiki article?
I recommend this image for the article
http://img2.freeimagehosting.net/uploads/d5814688f0.jpg