| Topic: |
Science > Physics |
| User: |
"Koobee Wublee" |
| Date: |
03 May 2007 05:54:18 PM |
| Object: |
Re: Gravitational Radiation |
On May 3, 11:46 am, wrote:
Wikipedia's page on gravitational waves lists the following formula
for calculating the energy loss of a two-body system over time from
gravitational radiation as:
E = -(32/pi) * (G^4/c^5) * [(M1*M2)^2 * (M1 + M2)] / r^5
Can anyone give reference on how this is derived? Thanks.
In the meantime, the observed speed-up in the orbit of the pulsar
binary system can easily be explained through the geodesic equations.
In particular, the Euler-Lagrange equation associated with the
longitudinal displacement variable. If the curvature in spacetime is
independent of the longitude, conservation of angular momentum is
observed. If not, we would not observe a conservation of angular
momentum. This does not mean energy is radiated away.
In the model where the geodesics follow the path of maximum spacetime
(or proper time), energy does not have to be conserved. However,
light would never propagate through space. Since light does propagate
through space, this traditional model of geodesics must be invalid.
The other model of choice is the one where the geodesics follow the
path of minimum elapsed time (principle of least time as in the
Snell's law). In this model, conservation of energy becomes a basic,
fundamental principle. There is no room for gravitational radiation.
Thus, a change in angular momentum does not mean energy is carried
off. In conjunction, the orbit can shrink and still obey the
conservation of energy.
What would cause the curvature of spacetime to be a function of
longitudinal displacement as well? The answer is the speed of
gravity. I don't think it has to be the speed of light. However, it
cannot exceed the speed of light like Dr. Van Flandern is suggesting.
The gravitational pull of one star is not going to come from the other
star directly passing through the center-of-mass of the binary system
if the speed of gravity is finite. Because of a finite speed of
gravity, the gravitational effect is going to come at an angle away
from the center-of-mass. At this angle, the radial gravitational
force is going to be a little stronger than if the speed of gravity is
infinite, and the tangential force is what is causing the increase in
orbital speed. If the speed of gravity is infinite, this tangential
force would be zero.
Although the earth-sun system does manifest this increase in angular
momentum of the earth, its effect is very small because the center-of-
mass is very close to the center of the sun. However, Mercury's
orbital advance is totally different from this phenomenon of finite
speed in gravity.
.
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| User: "Eric Gisse" |
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| Title: Re: Gravitational Radiation |
03 May 2007 09:36:06 PM |
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On May 3, 3:54 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On May 3, 11:46 am, wrote:
Wikipedia's page on gravitational waves lists the following formula
for calculating the energy loss of a two-body system over time from
gravitational radiation as:
E = -(32/pi) * (G^4/c^5) * [(M1*M2)^2 * (M1 + M2)] / r^5
Can anyone give reference on how this is derived? Thanks.
Why ask when it is based on mathematics that you claim to be faulty
from the ground up?
In the meantime, the observed speed-up in the orbit of the pulsar
binary system can easily be explained through the geodesic equations.
In particular, the Euler-Lagrange equation associated with the
longitudinal displacement variable. If the curvature in spacetime is
independent of the longitude, conservation of angular momentum is
observed. If not, we would not observe a conservation of angular
momentum. This does not mean energy is radiated away.
The quantity known as curvature does not enter into the geodesic
equations.
In the model where the geodesics follow the path of maximum spacetime
(or proper time), energy does not have to be conserved. However,
light would never propagate through space. Since light does propagate
through space, this traditional model of geodesics must be invalid.
As you have been told _many times_ before, ds^2 = 0 for null paths and
thus the proper time parameterization can not be used. That does not,
in any way, mean that the 'traditional model' of geodesics are
invalid.
The other model of choice is the one where the geodesics follow the
path of minimum elapsed time (principle of least time as in the
Snell's law). In this model, conservation of energy becomes a basic,
fundamental principle. There is no room for gravitational radiation.
*laughs*
The same "argument" applies to Maxwell's equations. So, by the same
logic, there is no such thing as electromagnetic radiation.
Thus, a change in angular momentum does not mean energy is carried
off. In conjunction, the orbit can shrink and still obey the
conservation of energy.
Oh, neat! That means I can move arbitrarily close to and from the
Earth without an energy cost. Oh wait...that's fucking stupid.
Read the phrase "Thus, a change in angular momentum does not mean
energy is carried off. In conjunction, the orbit can shrink and still
obey the conservation of energy." and consider carefully whether you
want to stand by that statement.
What would cause the curvature of spacetime to be a function of
longitudinal displacement as well? The answer is the speed of
gravity. I don't think it has to be the speed of light. However, it
cannot exceed the speed of light like Dr. Van Flandern is suggesting.
I can't tell - do you _really_ think curvature enters into the
equations of motion, or are you simply using curvature to mean
something else despite its' established meaning?
The gravitational pull of one star is not going to come from the other
star directly passing through the center-of-mass of the binary system
if the speed of gravity is finite. Because of a finite speed of
gravity, the gravitational effect is going to come at an angle away
from the center-of-mass. At this angle, the radial gravitational
force is going to be a little stronger than if the speed of gravity is
infinite, and the tangential force is what is causing the increase in
orbital speed. If the speed of gravity is infinite, this tangential
force would be zero.
....and KW discovers the concept of retarded fields.
Although the earth-sun system does manifest this increase in angular
momentum of the earth, its effect is very small because the center-of-
mass is very close to the center of the sun. However, Mercury's
orbital advance is totally different from this phenomenon of finite
speed in gravity.
.
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| User: "Koobee Wublee" |
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| Title: Re: Gravitational Radiation |
04 May 2007 01:10:05 AM |
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On May 3, 7:36 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 3, 3:54 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On May 3, 11:46 am, wrote:
Wikipedia's page on gravitational waves lists the following formula
for calculating the energy loss of a two-body system over time from
gravitational radiation as:
E = -(32/pi) * (G^4/c^5) * [(M1*M2)^2 * (M1 + M2)] / r^5
Can anyone give reference on how this is derived? Thanks.
Why ask when it is based on mathematics that you claim to be faulty
from the ground up?
Why not?
In the meantime, the observed speed-up in the orbit of the pulsar
binary system can easily be explained through the geodesic equations.
In particular, the Euler-Lagrange equation associated with the
longitudinal displacement variable. If the curvature in spacetime is
independent of the longitude, conservation of angular momentum is
observed. If not, we would not observe a conservation of angular
momentum. This does not mean energy is radiated away.
The quantity known as curvature does not enter into the geodesic
equations.
You are so wrong as always. <shrug>
In the model where the geodesics follow the path of maximum spacetime
(or proper time), energy does not have to be conserved. However,
light would never propagate through space. Since light does propagate
through space, this traditional model of geodesics must be invalid.
As you have been told _many times_ before, ds^2 = 0 for null paths and
thus the proper time parameterization can not be used. That does not,
in any way, mean that the 'traditional model' of geodesics are
invalid.
This is total BS. You cannot make a special case for photons.
<Shrug>
The other model of choice is the one where the geodesics follow the
path of minimum elapsed time (principle of least time as in the
Snell's law). In this model, conservation of energy becomes a basic,
fundamental principle. There is no room for gravitational radiation.
*laughs*
This is not the Improv.
The same "argument" applies to Maxwell's equations. So, by the same
logic, there is no such thing as electromagnetic radiation.
Gravity is not electromagnetism. <shrug>
Thus, a change in angular momentum does not mean energy is carried
off. In conjunction, the orbit can shrink and still obey the
conservation of energy.
Oh, neat! That means I can move arbitrarily close to and from the
Earth without an energy cost. Oh wait...that's fucking stupid.
You are off in the left field once again. This is about the case of
conservation of energy, not energy.
Read the phrase "Thus, a change in angular momentum does not mean
energy is carried off. In conjunction, the orbit can shrink and still
obey the conservation of energy." and consider carefully whether you
want to stand by that statement.
That is correct. This is according to the Euler-Lagrange equations
which in this special case are known as the geodesic equations that
follow the principle of least time and not that absurd make-believe
maximum spacetime BS. Time is reality. Spacetime is a combination of
space and time. Get over with it.
What would cause the curvature of spacetime to be a function of
longitudinal displacement as well? The answer is the speed of
gravity. I don't think it has to be the speed of light. However, it
cannot exceed the speed of light like Dr. Van Flandern is suggesting.
I can't tell - do you _really_ think curvature enters into the
equations of motion, or are you simply using curvature to mean
something else despite its' established meaning?
The curvature enters the geodesic equations. <shrug>
The gravitational pull of one star is not going to come from the other
star directly passing through the center-of-mass of the binary system
if the speed of gravity is finite. Because of a finite speed of
gravity, the gravitational effect is going to come at an angle away
from the center-of-mass. At this angle, the radial gravitational
force is going to be a little stronger than if the speed of gravity is
infinite, and the tangential force is what is causing the increase in
orbital speed. If the speed of gravity is infinite, this tangential
force would be zero.
...and KW discovers the concept of retarded fields.
No, Koobee Wublee applies the retarded fields to the binary system for
the first time in the history of science.
Although the earth-sun system does manifest this increase in angular
momentum of the earth, its effect is very small because the center-of-
mass is very close to the center of the sun. However, Mercury's
orbital advance is totally different from this phenomenon of finite
speed in gravity.
Speechless after so many blunders.
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| User: "Eric Gisse" |
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| Title: Re: Gravitational Radiation |
04 May 2007 02:41:43 AM |
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On May 3, 11:10 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
[...]
The quantity known as curvature does not enter into the geodesic
equations.
You are so wrong as always. <shrug>
Don't blame me for not knowing what _you_ mean by curvature.
The only thing that enters into the geodesic equations are the
Christoffel symbols. What is commonly referred to in the literature
[do you read _anything_?] as curvature is one of the following: A rank
4 tensor, a rank 2 tensor, or a scalar. None of those enter into the
geodesic equations.
If you disagree, give a literature or web citation that agrees with
you or otherwise prove it.
In the model where the geodesics follow the path of maximum spacetime
(or proper time), energy does not have to be conserved. However,
light would never propagate through space. Since light does propagate
through space, this traditional model of geodesics must be invalid.
As you have been told _many times_ before, ds^2 = 0 for null paths and
thus the proper time parameterization can not be used. That does not,
in any way, mean that the 'traditional model' of geodesics are
invalid.
This is total BS. You cannot make a special case for photons.
<Shrug>
Draw a space-time diagram. Draw the path a photon takes. Look at what
ds^2 that would have to correspond to for that path to be taken. .
Realize your mistake, and move the ***** on. Your constant obsessing
over what would cover, at most, a lecture's worth of time is quite
irritating.
The other model of choice is the one where the geodesics follow the
path of minimum elapsed time (principle of least time as in the
Snell's law). In this model, conservation of energy becomes a basic,
fundamental principle. There is no room for gravitational radiation.
*laughs*
This is not the Improv.
The same "argument" applies to Maxwell's equations. So, by the same
logic, there is no such thing as electromagnetic radiation.
Gravity is not electromagnetism. <shrug>
Explain how your argument applies to gravitation and not
electromagnetism. Free particles travel along geodesics in classical
mechanics too - just simpler ones.
Do you understand how radiation is created in electrodynamics and
gravitation?
Thus, a change in angular momentum does not mean energy is carried
off. In conjunction, the orbit can shrink and still obey the
conservation of energy.
Oh, neat! That means I can move arbitrarily close to and from the
Earth without an energy cost. Oh wait...that's fucking stupid.
You are off in the left field once again. This is about the case of
conservation of energy, not energy.
So where does the energy go?
Just like with electrodynamics - power that goes off to infinity is
_defined_ to be radiation. Admittedly it is slightly more difficult
with GR, but the concept stands.
Read the phrase "Thus, a change in angular momentum does not mean
energy is carried off. In conjunction, the orbit can shrink and still
obey the conservation of energy." and consider carefully whether you
want to stand by that statement.
That is correct. This is according to the Euler-Lagrange equations
which in this special case are known as the geodesic equations that
follow the principle of least time and not that absurd make-believe
maximum spacetime BS. Time is reality. Spacetime is a combination of
space and time. Get over with it.
The geodesic equations _ARE_ the Euler-Lagrange equations - there is
no special case ***** involved. They are the result of extremizing
a function - this is proven in any reasonable calculus of variations
and/or classical mechanics textbook.
The "maximum spacetime" stuff _IS_ BS - nobody calls it that but you.
The proper terminology is, once again, maximizing proper time. It is a
concept that is well-motivated - if you are a particle just traveling
without the action of external forces, what path are you going to
take? It will be a geodesic - which happens to be the case of maximal
proper time for particles on timelike paths.
What would cause the curvature of spacetime to be a function of
longitudinal displacement as well? The answer is the speed of
gravity. I don't think it has to be the speed of light. However, it
cannot exceed the speed of light like Dr. Van Flandern is suggesting.
I can't tell - do you _really_ think curvature enters into the
equations of motion, or are you simply using curvature to mean
something else despite its' established meaning?
The curvature enters the geodesic equations. <shrug>
http://en.wikipedia.org/wiki/Geodesic_(general_relativity)
Point them out to me. If they are not in there to your satisfaction,
write them here.
Merely asserting that I am wrong carries no weight with me. You have
to actually try to support your arguments.
The gravitational pull of one star is not going to come from the other
star directly passing through the center-of-mass of the binary system
if the speed of gravity is finite. Because of a finite speed of
gravity, the gravitational effect is going to come at an angle away
from the center-of-mass. At this angle, the radial gravitational
force is going to be a little stronger than if the speed of gravity is
infinite, and the tangential force is what is causing the increase in
orbital speed. If the speed of gravity is infinite, this tangential
force would be zero.
...and KW discovers the concept of retarded fields.
No, Koobee Wublee applies the retarded fields to the binary system for
the first time in the history of science.
Be sure to let us know when you have the paper written.
[---]
.
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| User: "" |
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| Title: Re: Gravitational Radiation |
04 May 2007 11:26:58 AM |
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On May 4, 12:41 am, Eric Gisse <jowr...@gmail.com> wrote:
Be sure to let us know when you have the paper written.
[---]
Way to go, force these assholes to write down the math and see them
run away or, even better write some stupid formulas, easy to debunk!
Don't let them write prose!
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| User: "Koobee Wublee" |
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| Title: Re: Gravitational Radiation |
04 May 2007 12:57:26 PM |
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On May 4, 12:41 am, Eric Gisse <jowr...@gmail.com> wrote:
On May 3, 11:10 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
You are so wrong as always. <shrug>
Don't blame me for not knowing what _you_ mean by curvature.
OK.
The only thing that enters into the geodesic equations are the
Christoffel symbols. What is commonly referred to in the literature
[do you read _anything_?] as curvature is one of the following: A rank
4 tensor, a rank 2 tensor, or a scalar. None of those enter into the
geodesic equations.
So, you know what I mean by curvature. Thus, you lied.
If you disagree, give a literature or web citation that agrees with
you or otherwise prove it.
There are many ways to express a point. <shrug>
Draw a space-time diagram. Draw the path a photon takes. Look at what
ds^2 that would have to correspond to for that path to be taken. .
Photons propagate through space and not through spacetime. <shrug>
Realize your mistake, and move the ***** on. Your constant obsessing
over what would cover, at most, a lecture's worth of time is quite
irritating.
I did not make these mistakes. You were fed with these mistakes.
<shrug>
Gravity is not electromagnetism. <shrug>
Explain how your argument applies to gravitation and not
electromagnetism.
Sure, but you are not ready for it. The understanding of
electromagnetism is also flawed since the middle of the 19th century.
This will be another chapter of rigorous discussion. In the meantime,
let's just concentrate on GR and SR.
Free particles travel along geodesics in classical
mechanics too - just simpler ones.
This is too vague.
Do you understand how radiation is created in electrodynamics and
gravitation?
Yes.
You are off in the left field once again. This is about the case of
conservation of energy, not energy.
So where does the energy go?
The energy is conserved. If you are not brain dead, you should have
figured it out.
Just like with electrodynamics - power that goes off to infinity is
_defined_ to be radiation. Admittedly it is slightly more difficult
with GR, but the concept stands.
The mathematics of the Euler-Lagrange equations which are the geodesic
equations proves no such a case. These equations prove the energy is
conserved but not the angular momentum. I told you to learn your
calculus of variations, didn't I?
That is correct. This is according to the Euler-Lagrange equations
which in this special case are known as the geodesic equations that
follow the principle of least time and not that absurd make-believe
maximum spacetime BS. Time is reality. Spacetime is a combination of
space and time. Get over with it.
The geodesic equations _ARE_ the Euler-Lagrange equations - there is
no special case ***** involved.
What I meant special is the binary star system.
They are the result of extremizing
a function - this is proven in any reasonable calculus of variations
and/or classical mechanics textbook.
Minimizing or maximizing what?
The "maximum spacetime" stuff _IS_ BS - nobody calls it that but you.
You call it as proper time which is spacetime in disguise. <shrug>
The proper terminology is, once again, maximizing proper time. It is a
concept that is well-motivated - if you are a particle just traveling
without the action of external forces, what path are you going to
take? It will be a geodesic - which happens to be the case of maximal
proper time for particles on timelike paths.
Geodesics maximizing spacetime is total BS because it does not allow a
photon to propagate. <shrug>
The curvature enters the geodesic equations. <shrug>
http://en.wikipedia.org/wiki/Geodesic_(general_relativity)
Yes, the curvature enters the geodesic equations. <shrug>
Point them out to me. If they are not in there to your satisfaction,
write them here.
If you are not brain dead, you should have no problem seeking the
solution yourself. <shrug>
Merely asserting that I am wrong carries no weight with me. You have
to actually try to support your arguments.
I don't really care if you understand anything at all.
No, Koobee Wublee applies the retarded fields to the binary system for
the first time in the history of science.
Be sure to let us know when you have the paper written.
OK, don't hold your breath.
.
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| User: "Eric Gisse" |
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| Title: Re: Gravitational Radiation |
04 May 2007 05:08:50 PM |
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On May 4, 10:57 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On May 4, 12:41 am, Eric Gisse <jowr...@gmail.com> wrote:
On May 3, 11:10 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
You are so wrong as always. <shrug>
Don't blame me for not knowing what _you_ mean by curvature.
OK.
The only thing that enters into the geodesic equations are the
Christoffel symbols. What is commonly referred to in the literature
[do you read _anything_?] as curvature is one of the following: A rank
4 tensor, a rank 2 tensor, or a scalar. None of those enter into the
geodesic equations.
So, you know what I mean by curvature. Thus, you lied.
You have a proven history of using words with an established meaning
to mean something unique to you. None of the accepted definitions for
curvature are used in the geodesic equation.
If you disagree, give a literature or web citation that agrees with
you or otherwise prove it.
There are many ways to express a point. <shrug>
What's the matter, you don't have a reference to support your
argument?
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
Draw a space-time diagram. Draw the path a photon takes. Look at what
ds^2 that would have to correspond to for that path to be taken. .
Photons propagate through space and not through spacetime. <shrug>
You _are_ aware that photons travel at a finite speed, right?
Realize your mistake, and move the ***** on. Your constant obsessing
over what would cover, at most, a lecture's worth of time is quite
irritating.
I did not make these mistakes. You were fed with these mistakes.
<shrug>
Your misunderstandings are not mistakes on my part.
Gravity is not electromagnetism. <shrug>
Explain how your argument applies to gravitation and not
electromagnetism.
Sure, but you are not ready for it. The understanding of
electromagnetism is also flawed since the middle of the 19th century.
Really? Based upon what understanding do you make _this_ claim?
This will be another chapter of rigorous discussion. In the meantime,
let's just concentrate on GR and SR.
Rigorous discussion it is! Do you think that when I scroll down I will
see some actual mathematics, or at least some references to them? Or
will I just read more denial formed into prose?
Free particles travel along geodesics in classical
mechanics too - just simpler ones.
This is too vague.
Is it?
A free particle is a particle that is not under the influence of any
external forces. It will travel along a geodesic, which is defined as
a straight path. It is simpler in classical mechanics because
classical mechanics occurs on an Euclidean metric, of which all
Christoffel components are zero [Only in rectilinear coordinates, at
least].
Have you studied classical mechanics? Do you think it is equally
crippled?
Do you understand how radiation is created in electrodynamics and
gravitation?
Yes.
We'll see about that.
You are off in the left field once again. This is about the case of
conservation of energy, not energy.
So where does the energy go?
The energy is conserved. If you are not brain dead, you should have
figured it out.
Yes, and conservation of energy is half the motivation _OF_ radiation.
We see there is power going off to infinity, and we want conservation
of energy. Motivated by that, some approximations are made and we find
electromagnetic radiation. The story is more complicated than that,
but that's pretty much it.
Just like with electrodynamics - power that goes off to infinity is
_defined_ to be radiation. Admittedly it is slightly more difficult
with GR, but the concept stands.
The mathematics of the Euler-Lagrange equations which are the geodesic
equations proves no such a case. These equations prove the energy is
conserved but not the angular momentum. I told you to learn your
calculus of variations, didn't I?
Think about the physics of the geodesic equations. They explain how a
free particle [you can attach a force, but not in this specific case]
moves on a manifold. They do not, nor should they, explain dynamical
properties of the manifold - that is the job of the field equations.
http://pancake.uchicago.edu/~carroll/notes/
Give #6 a bit of studying.
Think back to classical mechanics, which I seriously question your
understanding of, where Lagrangians and such were introduced. Every
quantity that the Lagrangian is independent of is associated with a
conserved quantity.
The situation is analogous in GR - with the extremizing of proper
time, the square root of the metric under some parameterization is
analogous to the Lagrangian. They obey the same equations [Euler-
Lagrange] and obey the same conditions [extremizing over a path].
However in general, there are no conserved quantities in GR - even
energy and angular momentum.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
That is correct. This is according to the Euler-Lagrange equations
which in this special case are known as the geodesic equations that
follow the principle of least time and not that absurd make-believe
maximum spacetime BS. Time is reality. Spacetime is a combination of
space and time. Get over with it.
The geodesic equations _ARE_ the Euler-Lagrange equations - there is
no special case ***** involved.
What I meant special is the binary star system.
They are the result of extremizing
a function - this is proven in any reasonable calculus of variations
and/or classical mechanics textbook.
Minimizing or maximizing what?
The function under some parameterization.
You have studied classical mechanics, right? If you actually studied
the calculus of variations, you would understand that any function
that is being extremized over a path obeys the Euler-Lagrange
equations.
You wouldn't have been doing something as silly as telling me to study
something which you don't understand, now would you?
The "maximum spacetime" stuff _IS_ BS - nobody calls it that but you.
You call it as proper time which is spacetime in disguise. <shrug>
I, along with pretty much every physicist, calls it proper time.
Spacetime is _NOT_ proper time, no matter how much you abuse
terminology.
The proper terminology is, once again, maximizing proper time. It is a
concept that is well-motivated - if you are a particle just traveling
without the action of external forces, what path are you going to
take? It will be a geodesic - which happens to be the case of maximal
proper time for particles on timelike paths.
Geodesics maximizing spacetime is total BS because it does not allow a
photon to propagate. <shrug>
Really? What makes you say that?
The Euler-Lagrange equations are defined with respect to _some_
[affine] parameterization - it doesn't have to be proper time. Just
pick something that you can parameterize light with - perhaps
wavelength.
Consider ye olde Minkowski space in one space and time dimension as a
simple example.
ds^2 = 0 = dt^2 - dx^2
0 = [dt/da]^2 - [dx/da]^2
dt/da = dx/da
1 = dx/dt [=c if I retained c, of course]
A photon travels along, in God's units, equal amounts of space and
time. Isn't that interesting - I seem to recall mentioning space-time
diagrams and photons.
The curvature enters the geodesic equations. <shrug>
http://en.wikipedia.org/wiki/Geodesic_(general_relativity)
Yes, the curvature enters the geodesic equations. <shrug>
Point them out to me. If they are not in there to your satisfaction,
write them here.
If you are not brain dead, you should have no problem seeking the
solution yourself. <shrug>
Point out the specific quantity you are calling curvature, since the
quantity _I_ - along with the rest of the physics community - calls
curvature does not enter into the geodesic equations.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
Merely asserting that I am wrong carries no weight with me. You have
to actually try to support your arguments.
I don't really care if you understand anything at all.
No, Koobee Wublee applies the retarded fields to the binary system for
the first time in the history of science.
Be sure to let us know when you have the paper written.
OK, don't hold your breath.
Wow, you are doing something for the _first time in the history of
science_, and you aren't even going to bother writing a paper on it?
Must not be that interesting if you can't be bothered to write a paper
about it.
.
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| User: "Koobee Wublee" |
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| Title: Re: Gravitational Radiation |
05 May 2007 12:46:02 AM |
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On May 4, 3:08 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 4, 10:57 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
You have a proven history of using words with an established meaning
to mean something unique to you. None of the accepted definitions for
curvature are used in the geodesic equation.
You have a proven history of using vocabularies to describe physics.
To be more point blank, that is total BS.
What's the matter, you don't have a reference to support your
argument?
My arguments are sound by themselves in accordance with sound
mathematics and logic. Unlike you, I don't need BS to decorate by
arguments.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
I notice your references are total BS. <shrug>
Photons propagate through space and not through spacetime. <shrug>
You _are_ aware that photons travel at a finite speed, right?
Yes.
I did not make these mistakes. You were fed with these mistakes.
<shrug>
Your misunderstandings are not mistakes on my part.
Your lack of not be able to identify the mistakes fed to you is none
of my concern nor my mistake. <shrug>
Sure, but you are not ready for it. The understanding of
electromagnetism is also flawed since the middle of the 19th century.
Really? Based upon what understanding do you make _this_ claim?
Stay tuned and fine out.
This will be another chapter of rigorous discussion. In the meantime,
let's just concentrate on GR and SR.
Rigorous discussion it is! Do you think that when I scroll down I will
see some actual mathematics, or at least some references to them? Or
will I just read more denial formed into prose?
If the discussion is fruitful, you will see some mathematics. If not,
you will see me pointing out the matheMagics fed to you.
Free particles travel along geodesics in classical
mechanics too - just simpler ones.
This is too vague.
Is it?
Yes.
A free particle is a particle that is not under the influence of any
external forces. It will travel along a geodesic, which is defined as
a straight path. It is simpler in classical mechanics because
classical mechanics occurs on an Euclidean metric, of which all
Christoffel components are zero [Only in rectilinear coordinates, at
least].
You knew the answer before. To be more thorough, you need to show so
through the calculus of variations after identifying a valid
Lagrangian that satisfy as a density to a meaningful action that needs
to be extremized (usually minimized).
Have you studied classical mechanics? Do you think it is equally
crippled?
Classical mechanics appears to be crippled in cause if you have not
stay tuned to experiments performed in the past a hundred years or
so. MMX is a fine example. <shrug>
The energy is conserved. If you are not brain dead, you should have
figured it out.
Yes, and conservation of energy is half the motivation _OF_ radiation.
We see there is power going off to infinity, and we want conservation
of energy.
We don't see any energy lost. Energy is conserved according to the
very mathematics of Noether's theorem. <shrug>
Motivated by that, some approximations are made and we find
electromagnetic radiation. The story is more complicated than that,
but that's pretty much it.
Again, electromagnetism is not gravity. Current model is flawed right
from the start. It is another long chapter of discussion. Let's get
gravity out of the way first before we talk about that.
The mathematics of the Euler-Lagrange equations which are the geodesic
equations proves no such a case. These equations prove the energy is
conserved but not the angular momentum. I told you to learn your
calculus of variations, didn't I?
Think about the physics of the geodesic equations. They explain how a
free particle [you can attach a force, but not in this specific case]
moves on a manifold. They do not, nor should they, explain dynamical
properties of the manifold - that is the job of the field equations.
Yes.
http://pancake.uchicago.edu/~carroll/notes/
Give #6 a bit of studying.
It is in postscript. How do I convert it to pdf?
Think back to classical mechanics, which I seriously question your
understanding of, where Lagrangians and such were introduced. Every
quantity that the Lagrangian is independent of is associated with a
conserved quantity.
This is not so. Your understanding of the calculus of variations is
still faulty. In fact, it is still non-existent.
The situation is analogous in GR - with the extremizing of proper
time, the square root of the metric under some parameterization is
analogous to the Lagrangian.
As I said many times already, the proper time or the quantity known as
spacetime is total BS, and I have explained why it is so. <shrug>
They obey the same equations [Euler-
Lagrange] and obey the same conditions [extremizing over a path].
However in general, there are no conserved quantities in GR - even
energy and angular momentum.
There are many ways to interpret the geodesic motions, and this point
is more general than GR. Maximizing the spacetime or the proper time
leads to the absurdity of not allowing photons to propagate. Only
minimizing the elapsed time in accordance to Fermat's principle of
least time, one will truly find a set of geodesic equations.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
I have no way of evaluating your reference because it is presented in
an archaic form of document. I am very certain that I am correct.
Thus, if your reference is not presented in a modern presentation, I
certainly would not bend over my back to decipher it. <shrug>
Minimizing or maximizing what?
The function under some parameterization.
No, not the function. Try again.
You have studied classical mechanics, right? If you actually studied
the calculus of variations, you would understand that any function
that is being extremized over a path obeys the Euler-Lagrange
equations.
No, the Euler-Lagrange equations are the result of extremizing if and
only if a valid Lagrangian is identified.
You wouldn't have been doing something as silly as telling me to study
something which you don't understand, now would you?
Oh, no.
You call it as proper time which is spacetime in disguise. <shrug>
I, along with pretty much every physicist, calls it proper time.
Spacetime is _NOT_ proper time, no matter how much you abuse
terminology.
Proper time = Spacetime divided by the speed of light
Do you not agree that the speed of light is constant?
Geodesics maximizing spacetime is total BS because it does not allow a
photon to propagate. <shrug>
Really? What makes you say that?
Simple. The spacetime for a photon is already zero. Trying to
maximizing spacetime for a photon is thus so absurd beyond words.
When [hanson] understands that, he will laugh his head off.
The Euler-Lagrange equations are defined with respect to _some_
[affine] parameterization - it doesn't have to be proper time.
Yes.
Just
pick something that you can parameterize light with - perhaps
wavelength.
It makes no justification to parameterize light with wavelength.
Geodesics does not depend on a photon's wavelength.
You have been fed crap.
Consider ye olde Minkowski space in one space and time dimension as a
simple example.
ds^2 = 0 = dt^2 - dx^2
0 = [dt/da]^2 - [dx/da]^2
dt/da = dx/da
1 = dx/dt [=c if I retained c, of course]
A photon travels along, in God's units,
So, you are a disciple of the great reverend Hammond. I should have
known.
equal amounts of space and
time. Isn't that interesting - I seem to recall mentioning space-time
diagrams and photons.
I vaguely remember that BS. Well, what next?
Point out the specific quantity you are calling curvature, since the
quantity _I_ - along with the rest of the physics community - calls
curvature does not enter into the geodesic equations.
The metric.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
No, I did not say 'smart'. I said 'brain dead'. <shrug>
Wow, you are doing something for the _first time in the history of
science_, and you aren't even going to bother writing a paper on it?
I never said I am not.
Must not be that interesting if you can't be bothered to write a paper
about it.
Are you encouraging me to write a paper about this simple finding?
.
|
|
|
| User: "Eric Gisse" |
|
| Title: Re: Gravitational Radiation |
05 May 2007 01:21:57 AM |
|
|
On May 4, 10:46 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On May 4, 3:08 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 4, 10:57 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
You have a proven history of using words with an established meaning
to mean something unique to you. None of the accepted definitions for
curvature are used in the geodesic equation.
You have a proven history of using vocabularies to describe physics.
One of the many tyrannies of education.
To be more point blank, that is total BS.
What's the matter, you don't have a reference to support your
argument?
My arguments are sound by themselves in accordance with sound
mathematics and logic. Unlike you, I don't need BS to decorate by
arguments.
Why is it that you resist all requests for the mathematics, then?
[...]
A free particle is a particle that is not under the influence of any
external forces. It will travel along a geodesic, which is defined as
a straight path. It is simpler in classical mechanics because
classical mechanics occurs on an Euclidean metric, of which all
Christoffel components are zero [Only in rectilinear coordinates, at
least].
You knew the answer before. To be more thorough, you need to show so
through the calculus of variations after identifying a valid
Lagrangian that satisfy as a density to a meaningful action that needs
to be extremized (usually minimized).
.....but curvature does _not_ enter into that. It is just Christoffel
symbols - they are not curvature, though they and curvature are
related.
Have you studied classical mechanics? Do you think it is equally
crippled?
Classical mechanics appears to be crippled in cause if you have not
stay tuned to experiments performed in the past a hundred years or
so. MMX is a fine example. <shrug>
It remains exceptionally useful. Though the truly useful stuff is in
the quantum and/or relativistic domain, a lot of techniques from
classical mechanics are applicable.
The energy is conserved. If you are not brain dead, you should have
figured it out.
Yes, and conservation of energy is half the motivation _OF_ radiation.
We see there is power going off to infinity, and we want conservation
of energy.
We don't see any energy lost. Energy is conserved according to the
very mathematics of Noether's theorem. <shrug>
Motivated by that, some approximations are made and we find
electromagnetic radiation. The story is more complicated than that,
but that's pretty much it.
Again, electromagnetism is not gravity. Current model is flawed right
from the start. It is another long chapter of discussion. Let's get
gravity out of the way first before we talk about that.
The mathematics of the Euler-Lagrange equations which are the geodesic
equations proves no such a case. These equations prove the energy is
conserved but not the angular momentum. I told you to learn your
calculus of variations, didn't I?
Think about the physics of the geodesic equations. They explain how a
free particle [you can attach a force, but not in this specific case]
moves on a manifold. They do not, nor should they, explain dynamical
properties of the manifold - that is the job of the field equations.
Yes.
http://pancake.uchicago.edu/~carroll/notes/
Give #6 a bit of studying.
It is in postscript. How do I convert it to pdf?
Install this:
http://www.cs.wisc.edu/~ghost/
Then this:
http://www.ghostgum.com.au/
But if you are really hard up and want it converted...
http://www.ps2pdf.com/convert/convert.htm
Think back to classical mechanics, which I seriously question your
understanding of, where Lagrangians and such were introduced. Every
quantity that the Lagrangian is independent of is associated with a
conserved quantity.
This is not so. Your understanding of the calculus of variations is
still faulty. In fact, it is still non-existent.
The situation is analogous in GR - with the extremizing of proper
time, the square root of the metric under some parameterization is
analogous to the Lagrangian.
As I said many times already, the proper time or the quantity known as
spacetime is total BS, and I have explained why it is so. <shrug>
Proper time is simply the amount of time an observer would read on his
clock.
The proper time parameterization allows one, for timelike paths, to
use quantities that are measured in the observer's frame.
Asserting that it is BS does not make it so.
They obey the same equations [Euler-
Lagrange] and obey the same conditions [extremizing over a path].
However in general, there are no conserved quantities in GR - even
energy and angular momentum.
There are many ways to interpret the geodesic motions, and this point
is more general than GR. Maximizing the spacetime or the proper time
leads to the absurdity of not allowing photons to propagate. Only
minimizing the elapsed time in accordance to Fermat's principle of
least time, one will truly find a set of geodesic equations.
Look - as I have been trying to explain, it _doesn't have to be proper
time_! The parameterization is completely arbitrary - it just has to
be an affine parameter.
You are right though - it is more general than GR. Which is what I
have been trying to point out all along - this is not a concept
restricted to GR.
Fermat's principle of least time simply doesn't work in curved
spacetime in the way it works in classical mechanics. Who decides what
'least time' is? In relativity, t' =/= t. Something else has to be in
play.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
I have no way of evaluating your reference because it is presented in
an archaic form of document. I am very certain that I am correct.
Thus, if your reference is not presented in a modern presentation, I
certainly would not bend over my back to decipher it. <shrug>
You think postscript is _archaic_ ?
Postscript is one of the two major documents that the arXiv uses, and
you have never heard of it and couldn't spend 30 seconds on google
looking on how to view it? Why didn't you spend that 30 seconds
looking at the rest of the page where links to both the HTML and PDF
versions were given?
Minimizing or maximizing what?
The function under some parameterization.
No, not the function. Try again.
You have studied classical mechanics, right? If you actually studied
the calculus of variations, you would understand that any function
that is being extremized over a path obeys the Euler-Lagrange
equations.
No, the Euler-Lagrange equations are the result of extremizing if and
only if a valid Lagrangian is identified.
No.
It is more general than that. Any function that is extremized along a
given path will obey the Euler-Lagrange equations.
You wouldn't have been doing something as silly as telling me to study
something which you don't understand, now would you?
Oh, no.
You call it as proper time which is spacetime in disguise. <shrug>
I, along with pretty much every physicist, calls it proper time.
Spacetime is _NOT_ proper time, no matter how much you abuse
terminology.
Proper time = Spacetime divided by the speed of light
Not quite.
dT^2 = - ds^2 / c.
The minus sign because we want proper time to be positive.
Do you not agree that the speed of light is constant?
Geodesics maximizing spacetime is total BS because it does not allow a
photon to propagate. <shrug>
Really? What makes you say that?
Simple. The spacetime for a photon is already zero. Trying to
maximizing spacetime for a photon is thus so absurd beyond words.
What a coincidence - nobody has ever said that is what is done. There
is no extremizing done for photons because they _already_ travel along
an extremal path of zero proper time.
For a photon, ds^2 = 0. Period.
Think about the chain rule for a moment - dx = [dx/dq]*dq. I can do
the same thing with the line element, and parameterize the photon with
a specified parameterization and integrate.
When [hanson] understands that, he will laugh his head off.
Hanson will understand something of moderate mathematical complexity
around the time I jump the grand canyon while riding my Unicorn with
my Phoenix keeping the airspace clear.
The Euler-Lagrange equations are defined with respect to _some_
[affine] parameterization - it doesn't have to be proper time.
Yes.
Just
pick something that you can parameterize light with - perhaps
wavelength.
It makes no justification to parameterize light with wavelength.
Geodesics does not depend on a photon's wavelength.
Why not?
The geodesic equation is with respect to an _arbitrary_
parameterization - I could parameterize a photon's path with candy
corn so long as it is an affine parameter.
You have been fed crap.
Consider ye olde Minkowski space in one space and time dimension as a
simple example.
ds^2 = 0 = dt^2 - dx^2
0 = [dt/da]^2 - [dx/da]^2
dt/da = dx/da
1 = dx/dt [=c if I retained c, of course]
A photon travels along, in God's units,
So, you are a disciple of the great reverend Hammond. I should have
known.
Try to distinguish between my personal eccentricities and absolute
fruitloops like [Hammond].
equal amounts of space and
time. Isn't that interesting - I seem to recall mentioning space-time
diagrams and photons.
I vaguely remember that BS. Well, what next?
The recollection that photons travel along a 45 degree line
delineating the light cone. Which is only possible when ds^2 = 0.
Consult an introductory relativity text for further details, or ask a
specific question.
Point out the specific quantity you are calling curvature, since the
quantity _I_ - along with the rest of the physics community - calls
curvature does not enter into the geodesic equations.
The metric.
Well, what you call curvature _obviously_ enters into the geodesic
equations via the Christoffel symbols. It is just that nobody but you
calls the metric "curvature".
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
No, I did not say 'smart'. I said 'brain dead'. <shrug>
Wow, you are doing something for the _first time in the history of
science_, and you aren't even going to bother writing a paper on it?
I never said I am not.
Must not be that interesting if you can't be bothered to write a paper
about it.
Are you encouraging me to write a paper about this simple finding?
I am encouraging you to write a reasoned argument with literature
references that doesn't insult the reader.
.
|
|
|
| User: "hanson" |
|
| Title: Re: Gravitational Radiation |
06 May 2007 02:03:16 PM |
|
|
My young friend Eric Gisse, the Jowr, wrote below
dT^2 = - ds^2 / c.
[hanson]
Assuming that you are not a blind Einstein Dingleberry
who does see & imagine things that are in reality not
existing, or that you are not a Nazi parrot, like Yehiel Porat
loves to call you, can you tell me how & why in your equat.
the dimension of time [T^2] became a negative dimension
of L^2/(L/T) = -[LT], which generates a term reduction to
dimensions of [T] = - [L]
I will not buy that this is a typo, not from you, because you
are an aspiring physicist (with a BSc coming soon, you
hope) and you always insist that you are correct in your
important statements (which are of course merely parroted
portions out of some text book, from the establishment,
that makes you believe that because its printed it must
be "true fact") --- So, Eric, kindly justify your operation of
how time became a length, and being negative to boot.
Don't use any "untimely backpedaling to cover any length"
That will not count, Eric... ahahahaha... ahahahaha.....
Thanks for the laughs... ahahaha... ahahahahanson
"Eric Gisse" <jowr.pi@gmail.com> wrote in message
news:1178346117.685625.15800@e65g2000hsc.googlegroups.com...
On May 4, 10:46 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On May 4, 3:08 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 4, 10:57 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
You have a proven history of using words with an established meaning
to mean something unique to you. None of the accepted definitions for
curvature are used in the geodesic equation.
You have a proven history of using vocabularies to describe physics.
One of the many tyrannies of education.
To be more point blank, that is total BS.
What's the matter, you don't have a reference to support your
argument?
My arguments are sound by themselves in accordance with sound
mathematics and logic. Unlike you, I don't need BS to decorate by
arguments.
Why is it that you resist all requests for the mathematics, then?
[...]
A free particle is a particle that is not under the influence of any
external forces. It will travel along a geodesic, which is defined as
a straight path. It is simpler in classical mechanics because
classical mechanics occurs on an Euclidean metric, of which all
Christoffel components are zero [Only in rectilinear coordinates, at
least].
You knew the answer before. To be more thorough, you need to show so
through the calculus of variations after identifying a valid
Lagrangian that satisfy as a density to a meaningful action that needs
to be extremized (usually minimized).
....but curvature does _not_ enter into that. It is just Christoffel
symbols - they are not curvature, though they and curvature are
related.
Have you studied classical mechanics? Do you think it is equally
crippled?
Classical mechanics appears to be crippled in cause if you have not
stay tuned to experiments performed in the past a hundred years or
so. MMX is a fine example. <shrug>
It remains exceptionally useful. Though the truly useful stuff is in
the quantum and/or relativistic domain, a lot of techniques from
classical mechanics are applicable.
The energy is conserved. If you are not brain dead, you should have
figured it out.
Yes, and conservation of energy is half the motivation _OF_ radiation.
We see there is power going off to infinity, and we want conservation
of energy.
We don't see any energy lost. Energy is conserved according to the
very mathematics of Noether's theorem. <shrug>
Motivated by that, some approximations are made and we find
electromagnetic radiation. The story is more complicated than that,
but that's pretty much it.
Again, electromagnetism is not gravity. Current model is flawed right
from the start. It is another long chapter of discussion. Let's get
gravity out of the way first before we talk about that.
The mathematics of the Euler-Lagrange equations which are the
geodesic
equations proves no such a case. These equations prove the energy is
conserved but not the angular momentum. I told you to learn your
calculus of variations, didn't I?
Think about the physics of the geodesic equations. They explain how a
free particle [you can attach a force, but not in this specific case]
moves on a manifold. They do not, nor should they, explain dynamical
properties of the manifold - that is the job of the field equations.
Yes.
http://pancake.uchicago.edu/~carroll/notes/
Give #6 a bit of studying.
It is in postscript. How do I convert it to pdf?
Install this:
http://www.cs.wisc.edu/~ghost/
Then this:
http://www.ghostgum.com.au/
But if you are really hard up and want it converted...
http://www.ps2pdf.com/convert/convert.htm
Think back to classical mechanics, which I seriously question your
understanding of, where Lagrangians and such were introduced. Every
quantity that the Lagrangian is independent of is associated with a
conserved quantity.
This is not so. Your understanding of the calculus of variations is
still faulty. In fact, it is still non-existent.
The situation is analogous in GR - with the extremizing of proper
time, the square root of the metric under some parameterization is
analogous to the Lagrangian.
As I said many times already, the proper time or the quantity known as
spacetime is total BS, and I have explained why it is so. <shrug>
Proper time is simply the amount of time an observer would read on his
clock.
The proper time parameterization allows one, for timelike paths, to
use quantities that are measured in the observer's frame.
Asserting that it is BS does not make it so.
They obey the same equations [Euler-
Lagrange] and obey the same conditions [extremizing over a path].
However in general, there are no conserved quantities in GR - even
energy and angular momentum.
There are many ways to interpret the geodesic motions, and this point
is more general than GR. Maximizing the spacetime or the proper time
leads to the absurdity of not allowing photons to propagate. Only
minimizing the elapsed time in accordance to Fermat's principle of
least time, one will truly find a set of geodesic equations.
Look - as I have been trying to explain, it _doesn't have to be proper
time_! The parameterization is completely arbitrary - it just has to
be an affine parameter.
You are right though - it is more general than GR. Which is what I
have been trying to point out all along - this is not a concept
restricted to GR.
Fermat's principle of least time simply doesn't work in curved
spacetime in the way it works in classical mechanics. Who decides what
'least time' is? In relativity, t' =/= t. Something else has to be in
play.
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
I have no way of evaluating your reference because it is presented in
an archaic form of document. I am very certain that I am correct.
Thus, if your reference is not presented in a modern presentation, I
certainly would not bend over my back to decipher it. <shrug>
You think postscript is _archaic_ ?
Postscript is one of the two major documents that the arXiv uses, and
you have never heard of it and couldn't spend 30 seconds on google
looking on how to view it? Why didn't you spend that 30 seconds
looking at the rest of the page where links to both the HTML and PDF
versions were given?
Minimizing or maximizing what?
The function under some parameterization.
No, not the function. Try again.
You have studied classical mechanics, right? If you actually studied
the calculus of variations, you would understand that any function
that is being extremized over a path obeys the Euler-Lagrange
equations.
No, the Euler-Lagrange equations are the result of extremizing if and
only if a valid Lagrangian is identified.
No.
It is more general than that. Any function that is extremized along a
given path will obey the Euler-Lagrange equations.
You wouldn't have been doing something as silly as telling me to study
something which you don't understand, now would you?
Oh, no.
You call it as proper time which is spacetime in disguise. <shrug>
I, along with pretty much every physicist, calls it proper time.
Spacetime is _NOT_ proper time, no matter how much you abuse
terminology.
Proper time = Spacetime divided by the speed of light
Not quite.
dT^2 = - ds^2 / c.
The minus sign because we want proper time to be positive.
Do you not agree that the speed of light is constant?
Geodesics maximizing spacetime is total BS because it does not allow
a
photon to propagate. <shrug>
Really? What makes you say that?
Simple. The spacetime for a photon is already zero. Trying to
maximizing spacetime for a photon is thus so absurd beyond words.
What a coincidence - nobody has ever said that is what is done. There
is no extremizing done for photons because they _already_ travel along
an extremal path of zero proper time.
For a photon, ds^2 = 0. Period.
Think about the chain rule for a moment - dx = [dx/dq]*dq. I can do
the same thing with the line element, and parameterize the photon with
a specified parameterization and integrate.
When [hanson] understands that, he will laugh his head off.
Hanson will understand something of moderate mathematical complexity
around the time I jump the grand canyon while riding my Unicorn with
my Phoenix keeping the airspace clear.
The Euler-Lagrange equations are defined with respect to _some_
[affine] parameterization - it doesn't have to be proper time.
Yes.
Just
pick something that you can parameterize light with - perhaps
wavelength.
It makes no justification to parameterize light with wavelength.
Geodesics does not depend on a photon's wavelength.
Why not?
The geodesic equation is with respect to an _arbitrary_
parameterization - I could parameterize a photon's path with candy
corn so long as it is an affine parameter.
You have been fed crap.
Consider ye olde Minkowski space in one space and time dimension as a
simple example.
ds^2 = 0 = dt^2 - dx^2
0 = [dt/da]^2 - [dx/da]^2
dt/da = dx/da
1 = dx/dt [=c if I retained c, of course]
A photon travels along, in God's units,
So, you are a disciple of the great reverend Hammond. I should have
known.
Try to distinguish between my personal eccentricities and absolute
fruitloops like [Hammond].
equal amounts of space and
time. Isn't that interesting - I seem to recall mentioning space-time
diagrams and photons.
I vaguely remember that BS. Well, what next?
The recollection that photons travel along a 45 degree line
delineating the light cone. Which is only possible when ds^2 = 0.
Consult an introductory relativity text for further details, or ask a
specific question.
Point out the specific quantity you are calling curvature, since the
quantity _I_ - along with the rest of the physics community - calls
curvature does not enter into the geodesic equations.
The metric.
Well, what you call curvature _obviously_ enters into the geodesic
equations via the Christoffel symbols. It is just that nobody but you
calls the metric "curvature".
Notice how I actually give references and some motivational
explanation rather than the usual "If you are smart you can find it
yourself" ***** you feed me?
No, I did not say 'smart'. I said 'brain dead'. <shrug>
Wow, you are doing something for the _first time in the history of
science_, and you aren't even going to bother writing a paper on it?
I never said I am not.
Must not be that interesting if you can't be bothered to write a paper
about it.
Are you encouraging me to write a paper about this simple finding?
I am encouraging you to write a reasoned argument with literature
references that doesn't insult the reader.
.
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| User: "Androcles" |
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| Title: Re: Gravitational Radiation |
06 May 2007 02:56:07 PM |
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"hanson" <hanson@quick.net> wrote in message =
news:UDp%h.4221$Q96.3729@trnddc04...
My young friend Eric Gisse, the Jowr, wrote below
dT^2 =3D - ds^2 / c.
He must be an only-one-speed-of-light kook who doesn't accept the =
principle of relativity.
http://www.androcles01.pwp.blueyonder.co.uk/PoR/PoR.htm
Amazing how these Dingleberries believe what they are told to =
hallucinate
when the empirical evidence proves them wrong.=20
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac.htm
And he's your friend?
You can't teach kooks, even if they are friends.
[hanson]
Assuming that you are not a blind Einstein Dingleberry
Oh, do come on... how much evidence do you need?
Of course he's a Blind Einstein Dingleberry (BED), it's empirically =
evident!
"In agreement with experience" "it is clear" "we further assume" that
"we establish by definition that the ``time'' required by a Dingleberry =
to=20
travel from A to B equals the ``time'' it requires to travel from B to =
A."=20
Assume 'm'akes an '*****' out of 'u' and 'E'.
.
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| User: "hanson" |
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| Title: Re: Gravitational Radiation |
06 May 2007 03:58:25 PM |
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AHAHAHA....ahahaha... AHAHAHAHA... ahahahaha...
Andro, don't be so hard on them, BED's are still evolving...
Remember at their age we were exactly like them, young
enthusiastic and ... gullible... We both posted about it.
You just recently posted that you once wrote an "ode" to
Einstein and further back we both posted how we defended
SR/GR and Einstein.. and you even told us that you did
physically beat the ***** out of a fellow student who had
uttered doubts about Einstein...
Eventually all Einstein Dingleberries will fall off/from Albert's
sphincter.. once they have their fill of that Rela-smell... that
got them nowhere in its and their darkness... ahahaha...
**All Einstein Dingleberries are destined to biome EARs**
(Einstein-Anti-Relativists)... ahahahaha... ahahaha....
[Andro]
And he's your friend?
You can't teach kooks, even if they are friends.
[hanson]
Of course, Eric & all other E-Dingleberries are my friends.
Most of them are sad to fanatical results/cases who fell
victim to a gigantic 100+ year old jerk-off as/when they (like
you and me) got indoctrinated & brainwashed at a tender
and impressionable age... Bad scene, sad case, but most
of the will out-grow it... Nature does a better job then Albert.
I am not here to teach them nor do I regard them as kooks
because as long as Einsteinism is their beloved FAITH I will
let'em suffer in it. (See their own, decade old, daily fights of
Dingleberry1 against Dingleberry2 in NGs sp, spr, sp-res.)
I fact I do do thank then of letting me hire them in cyberspace
for a few minutes & treat them like the bumpersticker says:
***** "Hire the handicapped. They are fun to watch ******
Thanks for the laughs, Andro... ahahaha... ahahanson
-------- above is top posted ---------
"Androcles" <Engineer@hogwarts.physics.co.uk> wrote in
message news:rpq%h.149183$Zb2.8006@fe2.news.blueyonder.co.uk...
"hanson" <hanson@quick.net> wrote in message
news:UDp%h.4221$Q96.3729@trnddc04... or in
http://groups.google.com/group/sci.physics/msg/b89f85df7957b341
My young friend Eric Gisse, the Jowr, wrote below
dT^2 = - ds^2 / c.
[Andro]
He must be an only-one-speed-of-light kook who doesn't accept the
principle of relativity.
http://www.androcles01.pwp.blueyonder.co.uk/PoR/PoR.htm
Amazing how these Dingleberries believe what they are told to hallucinate
when the empirical evidence proves them wrong.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac.htm
And he's your friend?
You can't teach kooks, even if they are friends.
[hanson]
Assuming that you, Eric, are not a blind Einstein Dingleberry
[Andro]
Oh, do come on... how much evidence do you need?
Of course he's a Blind Einstein Dingleberry (BED), it's empirically evident!
"In agreement with experience" "it is clear" "we further assume" that
"we establish by definition that the ``time'' required by a Dingleberry to
travel from A to B equals the ``time'' it requires to travel from B to A."
Assume 'm'akes an '*****' out of 'u' and 'E'.
.
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| User: "Androcles" |
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| Title: Re: Gravitational Radiation |
06 May 2007 06:00:58 PM |
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"hanson" <hanson@quick.net> wrote in message =
news:Rjr%h.938$LJ3.489@trnddc02...
AHAHAHA....ahahaha... AHAHAHAHA... ahahahaha...
Andro, don't be so hard on them, BED's are still evolving...
I love being hard on beds. Is there a better place?
Oh... wait... you mean BEDs. Nah, they'll never evolve, they still think =
the Earth is=20
flat and doesn't go anywhere.=20
http://www.androcles01.pwp.blueyonder.co.uk/Clockgain.PNG
Einstein's Flat Earth Society is alive and kicking.
It's been that way since before Nickel-arse Copper-knickers ***** his =
pants and=20
discovered two dissimilar metals in an organic compound produced an =
electric cell,
and then that wap Volta got the credit for inventing voltaic =
haemmorhoids...=20
oops... I mean voltaic piles, and all he did was stack 'em up.=20
http://en.wikipedia.org/wiki/Alessandro_Volta
Shocking, I say.=20
Copernicus would have been booted out of the Church if he'd told anyone, =
'cos he also discovered relativity, poor bugger. Well... not =
"discovered",
but used the PoR to determine the Earth moved and not the Sun.
http://en.wikipedia.org/wiki/Nicolaus_Copernicus
Remember at their age we were exactly like them, young
enthusiastic and ... gullible... We both posted about it.
I'd rather forget that. I was sure Einstein was right until I grew up,
and then a fellow student asked for it to be proven. He was smarter
than me and I didn't listen. Today I bear that shame.
I used to know everything, but the older I get the more certain I am
that I know next to nothing. At least I made my own discovery and told
the world about it while still alive. =20
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
This is conformation, but again, only those with insight will see it.
http://www.androcles01.pwp.blueyonder.co.uk/Analemmae/Analemmae.htm
Only by delving into the data with an open mind (and a hint of =
intution)
can anyone make any sense of it.
The one it matters to is me, but I really can rank myself amongst=20
the great scientists. Someone else can claim the credit, Wilson is =
trying=20
hard and Kennaugh likes his term "BaTh" more than he likes me=20
saying "*****". Oh well... I'll go down in history as "Androcles",=20
but my work will be recognised after I'm dead, as was Copernicus.
No man gets more than one discovery of Nature's secrets to his credit.
I own ALL variable stars because I can explain them all:
http://www.britastro.org/vss/baalc.html=20
including this:=20
http://www.androcles01.pwp.blueyonder.co.uk/Copernicus/image021.jpg
You just recently posted that you once wrote an "ode" to
Einstein and further back we both posted how we defended
SR/GR and Einstein.. and you even told us that you did
physically beat the ***** out of a fellow student who had
uttered doubts about Einstein...
Yeah, I'm guilty as charged. I was a fuckhead who didn't listen.
Eventually all Einstein Dingleberries will fall off/from Albert's
sphincter.. once they have their fill of that Rela-smell... that
got them nowhere in its and their darkness... ahahaha...
**All Einstein Dingleberries are destined to biome EARs**
(Einstein-Anti-Relativists)... ahahahaha... ahahaha....
But you see, I'm not an anti-relativist. I'm an =
anti-Only-one-speed-of-lighter.
The Principle of Relativity is sacrosanct.=20
It is an axiom.=20
One-speed-of-lighters (OSoLs) are kooks of the first magnitude.=20
This includes aetherialists such as O'Barr and kenseto, along with all
Einstein dingleberries.
[Andro]
And he's your friend?
You can't teach kooks, even if they are friends.
[hanson]
Of course, Eric & all other E-Dingleberries are my friends.
I'd say acquaintances because they are quaint....=20
Most of them are sad to fanatical results/cases who fell
victim to a gigantic 100+ year old jerk-off as/when they (like
you and me) got indoctrinated & brainwashed at a tender
and impressionable age...=20
Exactly. But I'd like to move on, there are quasars and Seifert galaxies
out there and NOBODY has solved them. But I have nobody to talk
to about them and inspire me. I'm not interested in dingleberry =
solutions,=20
they are WRONG. I'm not much closer now than I was 10 years ago.
I shall probably die not knowing.
From Google: Results 1 - 10 of about 48,700 for Seifert galaxy --- none =
are=20
explanatory.
Bad scene, sad case, but most
of the will out-grow it... Nature does a better job then Albert.
Albert was useless from the get-go.
I am not here to teach them nor do I regard them as kooks
because as long as Einsteinism is their beloved FAITH I will
let'em suffer in it. (See their own, decade old, daily fights of
Dingleberry1 against Dingleberry2 in NGs sp, spr, sp-res.)
Ok, so why ARE you here?
I'm here because I want to discover all I can about Nature.=20
I'm fairly well educated in human nature, and generally all I see
is homo neanderthalensis, summarised by "***** and laugh".
That's the sum total of their intelligence.
If I teach, then who will learn? Yet I try.
I fact I do do thank then of letting me hire them in cyberspace
for a few minutes & treat them like the bumpersticker says:
***** "Hire the handicapped. They are fun to watch ******
Thanks for the laughs, Andro... ahahaha... ahahanson
Laughs are easy. Fucks are easy. Finding real intelligence in
another is a haemmorhoid you'd easily confuse with a dingleberry.
My American sweetheart sent me this:
One day, when a seamstress was sewing while sitting close to a=20
river, her thimble fell into the river. When she cried out, the Lord=20
appeared and asked, "My dear child, why are you crying?"
=20
The seamstress replied that her thimble had fallen into the water=20
and that she needed it to help her husband in making a living for their =
family.
=20
The Lord dipped His hand into the water and pulled up a golden=20
thimble set with pearls.
=20
"Is this your thimble?" the Lord asked.
=20
The seamstress replied, "No."
=20
The Lord again dipped into the river.=20
He held out a silver thimble ringed with sapphires.
=20
"Is this your thimble?" the Lord asked.
Again, the seamstress replied, "No".
=20
The Lord reached down again and came up with a leather thimble. "Is=20
this your thimble?" the Lord asked.
=20
The seamstress replied, "Yes."
=20
The Lord was pleased with the woman's honesty and gave her all=20
three thimbles to keep, and the seamstress went home happy.
Some years later, the seamstress was walking with her husband along=20
the riverbank, and her husband fell into the river and disappeared under =
the water.
=20
When she cried out, the Lord again appeared and asked her,=20
"Why are you crying?"
=20
"Oh Lord, my husband has fallen into the river!"
=20
The Lord went down into the water and came up with George Clooney.
=20
"Is this your husband?" the Lord asked.
=20
"Yes!" cried the seamstress.
=20
The Lord was furious. "You lied! That is an untruth!"
=20
The seamstress replied, "Oh, forgive me, my Lord. It is a =
misunderstanding. =20
You see, if I had said 'no' to George Clooney,=20
you would have come up with Brad Pitt. Then if I said 'no' to him, you=20
would have come up with my husband. Had I then said 'yes,' you would =
have
given me all three. Lord, I'm not in the best of health and would=20
not be able to take care of all three husbands, so THAT'S why I said =
yes' to George Clooney.
=20
And so the Lord let her keep him.
=20
The moral of this story is: Whenever a woman lies, it's for a good=20
and honorable reason, and in the best interest of others.
=20
That's our story, and we're sticking to it.
=20
.
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| User: "Eric Gisse" |
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| Title: Re: Gravitational Radiation |
06 May 2007 03:27:16 PM |
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On May 6, 12:03 pm, "hanson" <han...@quick.net> wrote:
My young friend Eric Gisse, the Jowr, wrote below
dT^2 = - ds^2 / c.
[hanson]
Assuming that you are not a blind Einstein Dingleberry
who does see & imagine things that are in reality not
existing, or that you are not a Nazi parrot, like Yehiel Porat
loves to call you, can you tell me how & why in your equat.
the dimension of time [T^2] became a negative dimension
of L^2/(L/T) = -[LT], which generates a term reduction to
dimensions of [T] = - [L]
That's actually a good catch. It should be - ds^2 / c^2. I usually
deal with c = 1 units so I don't have to track them, that way proper
time is simply parameterized by - ds^2.
I will not buy that this is a typo, not from you, because you
are an aspiring physicist (with a BSc coming soon, you
hope) and you always insist that you are correct in your
important statements (which are of course merely parroted
portions out of some text book, from the establishment,
that makes you believe that because its printed it must
be "true fact") --- So, Eric, kindly justify your operation of
how time became a length, and being negative to boot.
Don't use any "untimely backpedaling to cover any length"
That will not count, Eric... ahahahaha... ahahahaha.....
Thanks for the laughs... ahahaha... ahahahahanson
No, it means I made a mistake. It happens.
It does not mean relativity is wrong, or that the entire foundation of
physics is going to collapse.
Relativities standing in physics is completely uncoupled from
inquisition and ranting by people on USENET. I use USENET, in a real
bass-ackwards way, to indirectly solidify my understanding of
relativity.
[snip]
.
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| User: "hanson" |
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| Title: Re: Gravitational Radiation |
06 May 2007 04:44:55 PM |
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"Eric Gisse" <jowr.pi@gmail.com> wrote in message
news:1178483236.561224.142740@h2g2000hsg.googlegroups.com...
On May 6, 12:03 pm, "hanson" <han...@quick.net> wrote:
My young friend Eric Gisse, the Jowr, wrote below
dT^2 = - ds^2 / c.
[hanson]
Assuming that you are not a blind Einstein Dingleberry
who does see & imagine things that are in reality not
existing, or that you are not a Nazi parrot, like Yehiel Porat
loves to call you, can you tell me how & why in your equat.
the dimension of time [T^2] became a negative dimension
of L^2/(L/T) = -[LT], which generates a term reduction to
dimensions of [T] = - [L]
[Eric]
That's actually a good catch. It should be - ds^2 / c^2. I usually
deal with c = 1 units so I don't have to track them, that way proper
time is simply parameterized by - ds^2.
[hanson]
.... ahahaha.. well, your escape attempt is a bit of a weak
weaseling attempt.... but l let you get away with it... ahahaha...
ahahaha... Now go make amends with Koobee Wooblee....
[hanson]
I will not buy that this is a typo, not from you, because you
are an aspiring physicist (with a BSc coming soon, you
hope) and you always insist that you are correct in your
important statements (which are of course merely parroted
portions out of some text book, from the establishment,
that makes you believe that because its printed it must
be "true fact") --- So, Eric, kindly justify your operation of
how time became a length, and being negative to boot.
Don't use any "untimely backpedaling to cover any length"
That will not count, Eric... ahahahaha... ahahahaha.....
Thanks for the laughs... ahahaha... ahahahahanson
[Eric]
No, it means I made a mistake. It happens.
It does not mean relativity is wrong, or that the entire foundation of
physics is going to collapse.
Relativities standing in physics is completely uncoupled from
inquisition and ranting by people on USENET. I use USENET, in a real
bass-ackwards way, to indirectly solidify my understanding of
relativity.
[hanson]
For that you will get kudos, Eric.
Not for your "mistake" admission though.... because out
in the real world and in biz where you'd like to work in one
day, such "mistakes" do cost lives and billions of dollars.
Only teachers, theoreticians, and useless Einstein Dingle-
berries can afford to make mistakes... because the worst thing
that can happen is that they ***** up a youngster's life for a few years.
But kudos you do get, Eric, for your MO, to sharpen and exercise
what you have to learn for your upcoming BSc. Congrats, dude!
hanson
.
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| User: "Koobee Wublee" |
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| Title: Re: Gravitational Radiation |
06 May 2007 01:09:02 AM |
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On May 4, 11:21 pm, Eric Gisse <jowr...@gmail.com> wrote:
On May 4, 10:46 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
You have a proven history of using vocabularies to describe physics.
One of the many tyrannies of education.
No, it shows your weakness in the subject that require rigorous math.
<shrug>
My arguments are sound by themselves in accordance with sound
mathematics and logic. Unlike you, I don't need BS to decorate by
arguments.
Why is it that you resist all requests for the mathematics, then?
I did not resist. I just cannot provide you with voodoo math of your
liking. <shrug>
You knew the answer before. To be more thorough, you need to show so
through the calculus of variations after identifying a valid
Lagrangian that satisfy as a density to a meaningful action that needs
to be extremized (usually minimized).
....but curvature does _not_ enter into that. It is just Christoffel
symbols - they are not curvature, though they and curvature are
related.
Metric = description of curvature if the coordinate system is decided
already.
Classical mechanics appears to be crippled in cause if you have not
stay tuned to experiments performed in the past a hundred years or
so. MMX is a fine example. <shrug>
It remains exceptionally useful. Though the truly useful stuff is in
the quantum and/or relativistic domain, a lot of techniques from
classical mechanics are applicable.
Yes, to the first order. <shrug>
But if you are really hard up and want it converted...
http://www.ps2pdf.com/convert/convert.htm
Thanks.
As I said many times already, the proper time or the quantity known as
spacetime is total BS, and I have explained why it is so. <shrug>
Proper time is simply the amount of time an observer would read on his
clock.
Wrong!
The proper time parameterization allows one, for timelike paths, to
use quantities that are measured in the observer's frame.
Asserting that it is BS does not make it so.
Denial does not make it right.
There are many ways to interpret the geodesic motions, and this point
is more general than GR. Maximizing the spacetime or the proper time
leads to the absurdity of not allowing photons to propagate. Only
minimizing the elapsed time in accordance to Fermat's principle of
least time, one will truly find a set of geodesic equations.
Look - as I have been trying to explain, it _doesn't have to be proper
time_! The parameterization is completely arbitrary - it just has to
be an affine parameter.
It must be a parameter that must be extremized.
You are right though - it is more general than GR. Which is what I
have been trying to point out all along - this is not a concept
restricted to GR.
Fermat's principle of least time simply doesn't work in curved
spacetime in the way it works in classical mechanics.
I find to be the exact opposite of your silly claim.
Who decides what 'least time' is?
Mathematics of the calculus of variations does. <shrug>
In relativity, t' =/= t. Something else has to be in
play.
Since geodesic equations are based on one single observer, his time
measurement counts.
I have no way of evaluating your reference because it is presented in
an archaic form of document. I am very certain that I am correct.
Thus, if your reference is not presented in a modern presentation, I
certainly would not bend over my back to decipher it. <shrug>
You think postscript is _archaic_ ?
Yes.
Postscript is one of the two major documents that the arXiv uses, and
you have never heard of it and couldn't spend 30 seconds on google
looking on how to view it? Why didn't you spend that 30 seconds
looking at the rest of the page where links to both the HTML and PDF
versions were given?
I always see pdf format.
No, the Euler-Lagrange equations are the result of extremizing if and
only if a valid Lagrangian is identified.
No.
It is more general than that. Any function that is extremized along a
given path will obey the Euler-Lagrange equations.
I have to say we are both correct at this point.
Proper time = Spacetime divided by the speed of light
Not quite.
dT^2 = - ds^2 / c.
The minus sign because we want proper time to be positive.
You are getting too lunatic talking about minus time.
Simple. The spacetime for a photon is already zero. Trying to
maximizing spacetime for a photon is thus so absurd beyond words.
What a coincidence - nobody has ever said that is what is done. There
is no extremizing done for photons because they _already_ travel along
an extremal path of zero proper time.
For a photon, ds^2 = 0. Period.
Yes, that means all the paths a photon can take is already accumulated
with a spacetime of zero. There is no such unique path or geodesic
path for such a photon to travel through space. Thus, this point is
invalid.
It makes no justification to parameterize light with wavelength.
Geodesics does not depend on a photon's wavelength.
Why not?
Geodesics does not depend on a photon's wavelength.
The geodesic equation is with respect to an _arbitrary_
parameterization - I could parameterize a photon's path with candy
corn so long as it is an affine parameter.
I have been telling you the proper parameter is time, or the
observer's own time measurement. Now, do you get it?
So, you are a disciple of the great reverend Hammond. I should have
known.
Try to distinguish between my personal eccentricities and absolute
fruitloops like [Hammond].
I have failed to distinguish that, and I still see you as a disciple
of the great reverend.
I | | | | | | | | |
