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Old 26-April-2008, 04:24 PM
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Default Quaternion Physics and Redshift Revealed

ATM is a much needed forum. Most other places on the web are opposed to new ideas. The establishment physics police insist on conformity to the current dogma. The idea that The Big bang and Expanding Universe is incorrect is anathema.

Enough on why ATM is a good thing. There is a short coming on ATM. It is probably hard to explain a really revolutionary idea in a page or two. Another shortcoming is that the Thread Approach privatizes ideas, where physics is a public subject. However, ATM is the best available in that people here are questioning and generally well informed.

Redshift is the key concept underlying the idea of the Big Bang and the Expanding Universe. A little background is necessary. Newton created modern physics with his theory of gravity, the idea that every bit of matter attracts every other bit of matter. On the face of it this would lead one to forecast the coming together of all of matter.

Einstein, when he developed Relativity Theory, recognized this gravitational collapse. He also so no evidence of the collapse, so here the theory called for collapse, the observations didn't show it. Einstein created the "cosmological constant", to counter the collapse. He later dropped the cosmological constant.

Later, observational physicists, discovered the redshift and came to the conclusion, that not only was there no gravitational collapse, there was Expansion! Edwin Hubble corelated the redshift data and distance data and the expansion became the current dogma. The dogma of expansion has become the test of whether one is a scientist or not. The redshift is seen in light spectra and the shift is the pattern of an atomic spectra toward the red end of the spectrum. The redshift z, is defined as the change in wavelength divided by the unshifted wavelength, z=delta wavelength/wavelength=v/c.

Surprisingly, Edwin Hubble did not believe in the expansion idea. He said the redshift was a "hitherto unknown principle of nature." Observational physicists could measure the redshift with ease and they ran with the ball and the expansion dogma. Why could that be? The problem was that the theorists had no good derivation of the redshift. Thats where it now stands.

My interest came about not from a fascination with redshift. I was interested in Maxwell's Equations and electromagnetism. I researched the origins of the equations and discovered something. The mathematics of physics is out of date or defective. To cut to the chase, Einstein in his Relativity Theory introduced the idea that the universe is four dimensional. Minkowski and others jumped on this idea and introduced the imaginary into mainstream physics. As an electrical engineer, the imaginary was a standard part of our field. While it is still not clear the imaginary shows up in Einstein's Interval:

Interval I = x^2 + y^2 + z^2 - (ct)^2 OR I=(ct)^2 - (x^2 + y^2 + z^2)
as spacetime is space oriented or time oriented.

In searching Maxwell's Equations, I ran across William Rowan Hamilton's Quaternions. Quaternions consist of a real number and three vector numbers, thus a four dimensional space and mathematical division algebra. Why was this not being used in physics? The answer is instructive and maybe important.

Maxwell was introduced to quaternions as a way to mathematize Faraday's "Lines of Force". These forces had directions and Hamilton's Vectors were directional. This made them very attractive for the new field of electricity with their directional fields. There was a problem however. The problem was Hamilton's Rules for vectors i,j,k: i^2 = j^2 =k^2 = - 1. This minus one was disliked by maxwell and other physicists. They said, if I drop a ball there is a displacement in the direction of the force (gravity) there should be a positive energy, yet Hamilton's quaternions gives a negative energy! It is still the rule in physics that a displacement in the direction of the force is still called energy. The reason is that around 1900, Oliver Heaviside and J, Willard Gibbs listened to the physicists and developed vector Analysis where the Rule is: i^2 = j^2 = k^2 = +1. Here is a case where The mathematics was compromised to satisfy physics dogma. What Hamilton's mathematics was saying is that when a displacement is in the direction of the force, there is EXERGY not ENERGY. This is easily seen with thought. Exergy is the root for exercise, ex-ergs = out ergs! Energy is en-erg = in ergs! Physically, Hamilton's mathematics accurately describes the physics of the situation. The minus sign in the vector product indicates explosion and the plus sign indicates implosion. In 1900, Gibbs Vectors displaced quaternions in text books and quaternions were forgotten effectively. Gibbs Vectors are the standard in today's physics with i^2=1.

When I found quaternions I looked for areas where they would be used and found few, mostly in video games for rotations. Space orbit operations also use them. Medical uses in artificial limbs also use them, but by and large they are forgotten. I finally found a case where they are the difference, Gravity Theory!

Newton,s Theory of Gravity is based on real numbers E= -mMG/R=-mu/R. Einstein in his Relativity Theory followed Newton in keeping a real Energy.
I noticed that there was no vector energy. I added a vector energy to convert Gravity Energy E to a quaternion energy E= -mu/R + mcv and created a Quaternion Theory of Gravity.

Hamilton when he invented quaternions, created the idea of vectors and the idea of a vector derivative, Del = id/dx + jd/dy = kd/dz. I noticed this and decided to create a quaternion derivative X=d/dR + Del=d/cdt + Del.

The Quaternion Theory of Gravity needs two things, quaternion numbers and quaternion derivatives. This we have and can develop the Quaternion Theory of Gravity and clear up the issue redshift.

Before we start, lets do a thought experiment to understand the mathematics of redshift. Consider yourself standing near a railroad track and a train is coming from afar on your left. The train is far off and the angle between you and the train is a very small angle. As the train approaches you on the left the frequency you hear is higher than the train whistle frequency. As the train gets closer, the frequency you hear is closer to the train whistle frequency, and the angle gets larger. When the train is right in front of you, the frequency you hear is identical to the train whistle frequency and the angle is 90 degrees. As the train passes you the angle is still large but less than 90 degrees and the frequency you hear is lower than the frequency of the train whistle frequency. As the train gets far away on the right the frequency gets lower and the angle gets smaller again. This is the physical background for the Quaternion mathematics.

The abstract of The Quaternion Theory of Gravity is that there is Conservation of Gravitational Energy. As a consequence the sum of the central forces is zero and the sum of the directional forces is zero. Here is this mathematically. Conservation means the first derivative of energy is zero.

Force is the first derivative energy:
Force F=XE = (d/dR + Del)(-mu/R + mcv)

F = m(u/R^2 - cDel.v) + m(dcv/dR + cDelxv + Del(-u/R))

F = m(g - cDel.v) + m(cdv/dR + cv/R sin(rv) + g r/R)

F= m(v^2/R - cv/R cos(rv)) + m( dcv/cdt + cv/R sin(rv) + gr/R)

This is the quaternion derivative of the Energy. Lets discuss each term.

The first term, v^2/R = GM/R^2 is the gravitational centripetal(center seeking) acceleration.

The second term, -cDel.v= -cV/R cos(rv) is the centrifugal acceleration caused by the gravitational velocity v=sqrt(GM/R). Note that v is gravity generated. This centrifugal acceleration or force is in effect the cosmological "constant", that balances the gravitational centripetal forces. The negative sign is the result of the direction of the velocity. The divergence, Del.v = -v/Rcos(rv) is negative when the direction is outward, indicating a redshift. The divergence is positive when the velocity is inward indicating a blue shift. The divergence is zero when the velocity is perpendicular, not inward or outward, not approaching or receding. Now you see the importance of Hamilton's Rules, in indicating direction. If we used Gibbs vector rules redshift would be blue shift.

The sum of the centripetal acceleration and the centrifugal (center fleeing) acceleration gives zero acceleration at equilibrium. This equilibrium condition is the same as the boundary condition or the limit condition or the continuity condition all of these terms indicate the same thing, the extremum condition. Here we refine the redshift in the real equation of equilibrium:
(The sum of the vector forces is also zero and is the basis of Newton's For every action thee is an equal and opossite reaction. However, the vector equilibrium condition is not a player in redshift and will not be discussed here further).

0=g - cDel.v

g=cDel.v

v^2/R = cv/R cos(rv)

v/c=cos(rv)

Here you see that v/c the, the cosine of the angle(rv) between the radius vector and the velocity v; redshift z=v/c is specifically the cos(rv).

Quaternion Physics provides the derivation of the redshift and explains its importance in cosmology. Notice that at v=c, the redshift is 1=v/c=c/c. At redshift equal 1, the angle is zero, meaning that the direction of the velocity is radial to the center of gravity. Notice that when the redshift is zero, the angle is 90 degrees or the cosine is zero. This is not possible except at velocity equal zero. This requires v=sqrt(GM/R)=0 is only true at zero mass or infinite separation.

The other point is that the redshift indicates equilibrium or conservation of energy. The redshift relationship v/c=z is not true unless g=cDel.v.

Quantum redshift is the same quaternion story except the real energy is electrical rather than gravitational: E= -e^2/4pi eR + mcv

F=XE = (d/dR + Del)(-e^2/2piR + mcv)
F= (e^2/2pi eR^2 - cv/R cos(rv)) + (dmcv/dR + mcDelxv + Del (-e^2/4pi eR))

F=(e^2zc/4pi R^2 - mcv/r cos(rv)) + (mcdv/dR + mccv/R sin(rv) + e^2zcr/4pi R^2)

Again setting the force to zero at equilibrium gives for the real part:
e^2zc/4pi R^2 = mcv/R cos(rv) This gives the following

e^2zc/4pi R = nhc cos(rv)/2pi R
e^2z/2nh =cos(rv)

This gives Fine structure Constant alpha=e^2z/2nh=cos(rv) !

The Fine Structure Constant is seen to be e^2z/2nh=7.2E-3/n =cos(rv).
The quantized redshift is the same cos(rv) and is related to the Fine Structure Constant. The electrical redshift is larger than the gravitational redshift at the sun at n under about 5; alpha= 7.2E-3 > sqrt(GM/Rc^2)=sqrt(2.1164E-6) at the sun.

In my work on Maxwell's Equations, I have discovered that Planck''s Constant is h=WC where W is the Quantum magnetic Charge and C is the quantum electric charge . W = 500 atto Webers(volt seconds) and C= 4/3 atto Coulombs. Maxwell's Ether is the frees space impedance z=W/C = 375 Ohms. Planck's Constant h=zC^2 = W/C C^2=WC= 2/3 k atto atto Joules seconds. The Fine Structure constant is e^2z/2zC^2n=(e/C)^2/2n

This is a long post and there is more to say about it. The Universe is very interesting and there is lots to find. I think Quaternion Physics is the new frontier.

Again it a good thing to have a place to challenge to mainstream physics.

Quaternion Gravity Theory has made known, Hubble's "hitherto unknown principle", redshift !


For ATM is the best.

Last edited by yawyaw; 29-April-2008 at 01:29 AM..
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Old 26-April-2008, 05:37 PM
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Einstein, when he developed Relativity Theory, recognized this gravitational collapse. He also so no evidence of the collapse, so here the theory called for collapse, the observations didn't show it. Einstein created the "cosmological constant", to counter the collapse. He later dropped the cosmological constant.
It wasn't collapse that worried him. It was the absence of a static solution, i.e. the universe must be contracting or expanding, that he found to be problematic. This was particularly worrying because the mainstream view at the time was that the universe was in a steady state. If you consider the history in a little more depth, you will find that Friedmann derived the general properties of a homogeneous, isotropic universe in the early 1920s. It was noted that in the absence of a cosmological constant, the solution was not static. In 1929 Edwin Hubble formulated Hubble's Law, relating redshift to distance from the Earth. Needless to say that this relationship is exactly what was predicted by what later became known as the FLRW solution to the equations of GR.
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Old 26-April-2008, 05:40 PM
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Observational physicists could measure the redshift with ease and they ran with the ball and the expansion dogma. Why could that be? The problem was that the theorists had no good derivation of the redshift.
Not true. It is quite consistent with the FLRW solution, though there now appears to be evidence of a non-zero cosmological constant.
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Old 26-April-2008, 05:59 PM
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Last August there was a thread by Doug Sweetser; "Causality and the Quaternion Derivative":

Causality and the Quaternion Derivative

It was a fairly lively thread and worth reading.

Jim
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Old 26-April-2008, 08:58 PM
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Originally Posted by yawyaw
Quaternion Physics provides the derivation of the redshift and explains its importance in cosmology. Notice that at v=c, the redshift is 1=v/c=c/c. At redshift equal 1, the angle is zero, meaning that the direction of the velocity is radial to the center of gravity. Notice that when the redshift is zero, the angle is 90 degrees or the cosine is zero. This is not possible except at velocity equal zero. This requires v=sqrt(GM/R)=0 is only true at zero mass or infinite separation.
I'm not sure I totally understand, but it would appear that at some point, past zero, gravity should become repulsive rather than attractive, if it is to follow Maxwell's equations. Is this not so? And if it is so, and we know gravity is never repulsive, how does The Quaternion Theory of Gravity overcome this problem? Just curious is all.
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Old 26-April-2008, 09:15 PM
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i2 = -1
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Old 26-April-2008, 10:15 PM
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Originally Posted by trinitree88 View Post
i2 = -1
Yup, I saw that in yawyaw's OP here:
Quote:
Maxwell was introduced to quaternions as a way to mathematize Faraday's "Lines of Force". These forces had directions and Hamilton's Vectors were directional. This made them very attractive for the new field of electricity with their directional fields. There was a problem however. The problem was Hamilton's Rules for vectors i,j,k: i^2 = j^2 =k^2 = - 1. This minus one was disliked by maxwell and other physicists. They said, if I drop a ball there is a displacement in the direction of the force (gravity) there should be a positive energy, yet Hamilton's quaternions gives a negative energy! It is still the rule in physics that a displacement in the direction of the force is still called energy. The reason is that around 1900, Oliver Heaviside and J, Willard Gibbs listned to the physicists and developed vector Analysis where the Rule is: i^2 = j^2 = k^2 = +1. ... The minus sign in the vector product indicates explosion and the plus sign indicates implosion. In 1900, Gibbs Vectors displaced quaternions in text books and quaternions were forgotten effectively. Gibbs Vectors are the standard in today's physics with i^2=1.
So gravity must become 'explosion' per this minus sign? Doesn't work right, unless 'infalling' for gravitational force is replaced with 'push gravity'. But there is no real justification for this, as had been ATM'ed in the past. Push gravity runs into trouble eventually, if I remember all those posts. Pull gravity is what we actually experience, and our physics on that is pretty well proven with space probes making it to their destinations, not to mention space-time curvature of GR.

I don't think distant red shift at cosmological distances can be explained this easily, though I am personally of the opinion such red shift may be from other than Doppler space-expansion, so am open to suggestion.
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Old 26-April-2008, 11:41 PM
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Quote:
Originally Posted by yawyaw View Post
Newton,s Theory of Gravity is based on real numbers E= -mMG/R=-mu/R. Einstein in his Relativity Theory followed Newton in keeping a real Energy.
I noticed that there was no vector energy. I added a vector energy to convert Gravity Energy E to a quaternion energy E= -mu/R + mcv and created a Quaternion Theory of Gravity.
How does your "vector energy" compare to the energy-momentum 4-vector that is regularly used in SR & GR?

And now for an obvious question. It is not sufficient for a new theory to explain a single phenomena, such as the cosmological red-shift. If you have a new theory of gravitation then it should also explain:
1) The behaviour of the perihelion of Mercury,
2) Gravitational red-shift
3) Gravitational lensing, and the gravitational deflection of light by the Sun,

as well as the more usual gravitational effects.

Does your model do this?
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Old 27-April-2008, 01:31 AM
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Quote:
Originally Posted by yawyaw View Post
ATM is a much needed forum. Most other places on the web are opposed to new ideas. The establishment physics police insist on conformity to the current dogma. The idea that The Big bang and Expanding Universe is incorrect is anathema.
Please, spare us the usual excuse ATM proponents use to justify their failures.


Quote:
Originally Posted by yawyaw View Post
Enough on why ATM is a good thing. There is a short coming on ATM. It is probably hard to explain a really revolutionary idea in a page or two.
You should try and read some scientific papers.


Quote:
Originally Posted by yawyaw View Post
Another shortcoming is that the Thread Approach privatizes ideas, where physics is a public subject. However, ATM is the best available in that people here are questioning and generally well informed.
The ATM forum is also publicly visible.


Quote:
Originally Posted by yawyaw View Post
Redshift is the key concept underlying the idea of the Big Bang and the Expanding Universe.
The redshift-distance relation (Hubble's law) is one of the observations.


Quote:
Originally Posted by yawyaw View Post
A little background is necessary. Newton created modern physics with his theory of gravity,... [SNIP!]
Galileo Galilei described the method at the basis of modern science. The method is what distinguishes science from pseudo-science.


Quote:
Originally Posted by yawyaw View Post
Edwin Hubble correlated the redshift data and distance data and the expansion became the current dogma.
Hubble's law is an established observation, not a dogma.


Quote:
Originally Posted by yawyaw View Post
The dogma of expansion has become the test of whether one is a scientist or not. [SNIP!]
No, the method is the test whether one is a scientist or a crackpot.


Quote:
Originally Posted by yawyaw View Post
My interest came about not from a fascination with redshift. I was interested in Maxwell's Equations and electromagnetism. I researched the origins of the equations and discovered something. The mathematics of physics is out of date or defective.
I am curious to know what mathematics of physics you have studied....


Quote:
Originally Posted by yawyaw View Post
To cut to the chase, Einstein in his Relativity Theory introduced the idea that the universe is four dimensional. Minkowski and others jumped on this idea and introduced the imaginary into mainstream physics. As an electrical engineer, the imaginary was a standard part of our field. While it is still not clear the imaginary shows up in Einstein's Interval:

Interval I = x^2 + y^2 + z^2 - (ct)^2 OR I=(ct)^2 - (x^2 + y^2 + z^2)
as spacetime is space oriented or time oriented.
There is no imaginary in that interval, only squares of real numbers. Maybe you are just imagining it...


Quote:
Originally Posted by yawyaw View Post
In searching Maxwell's Equations, I ran across William Rowan Hamilton's Quaternions. Quaternions consist of a real number and three vector numbers, thus a four dimensional space and mathematical division algebra. Why was this not being used in physics? The answer is instructive and maybe important.
The answer is that four-vectors are more intuitive and give just as much information as quaternions. And they fit nicely into the whole formalism developed for theoretical physics.


Quote:
Originally Posted by yawyaw View Post
Maxwell was introduced to quaternions as a way to mathematize Faraday's "Lines of Force". These forces had directions and Hamilton's Vectors were directional. This made them very attractive for the new field of electricity with their directional fields. There was a problem however. The problem was Hamilton's Rules for vectors i,j,k: i^2 = j^2 =k^2 = - 1. This minus one was disliked by maxwell and other physicists. They said, if I drop a ball there is a displacement in the direction of the force (gravity) there should be a positive energy, yet Hamilton's quaternions gives a negative energy! It is still the rule in physics that a displacement in the direction of the force is still called energy. The reason is that around 1900, Oliver Heaviside and J, Willard Gibbs listned to the physicists and developed vector Analysis where the Rule is: i^2 = j^2 = k^2 = +1. Here is a case where The mathematics was compromised to satisfy physics dogma.
Provide the evidence that something has been compromised.


Quote:
Originally Posted by yawyaw View Post
What Hamilton's mathematics was saying is that when a displacement is in the direction of the force, there is EXERGY not ENERGY. This is easily seen with thought. Exergy is the root for exercise, ex-ergs = out ergs! Energy is en-erg = in ergs! Physically, Hmailtn's mathematics accurately describes the physics of the situation. The minus sign in the vector product indicates explosion and the plus sign indicates implosion. In 1900, Gibbs Vectors displaced quaternions in text books and quaternions were forgotten effectively. Gibbs Vectors are the standard in today's physics with i^2=1.

[SNIP!]
I always ask quaternion fans to show us that quaternions work better than four-vectors.
Considering that you see an imaginary in the very real Einstein interval you quoted, I won't hold my breath.
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Old 27-April-2008, 02:54 AM
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Quote:
Originally Posted by Fortis View Post
How does your "vector energy" compare to the energy-momentum 4-vector that is regularly used in SR & GR?

And now for an obvious question. It is not sufficient for a new theory to explain a single phenomena, such as the cosmological red-shift. If you have a new theory of gravitation then it should also explain:
1) The behaviour of the perihelion of Mercury,
2) Gravitational red-shift
3) Gravitational lensing, and the gravitational deflection of light by the Sun,

as well as the more usual gravitational effects.

Does your model do this?
Thanks for your questions. I don't know how far this Theory goes.

1) I have developed the orbit equation and get r=P(1 +ct/R)/(1 + ecos v)
where P=R(sin(rv))^2. This may or may not answer to Mercury's perihelion.

The vector equilibrium equation is 0=dcv/cdt + cDelxv t + ur/R3.

2 and 3) I think the Quaternion Theory handles the gravitational redshift and I am uninformed about gravitational lensing. On gravitational deflection by the sun, this Theory challenges Einstein's deflection. This theory clearly gives the deflection as cos(rv) = v/c=sqrt(GM/c^2R) = 1.457859E-3 giving a angle of 89degrees 54' 59.93". This gives a deflection angle of 5 arc minutes and 0.07 arc seconds. This is much larger than Einstein's 1.75' arc seconds.

This difference raises issues. The computation here is the redshift deflection caused by gravity changing the angle of the velocity away from the sun. Is Einstein's deflection changing the angle away or towards the sun, a blue shift? Einstein computes the angle by the tangent. The cosine is the redshift angle, so is Einstein's angle a redshift deflection?

Looking at Einstein's computation, the tangent is computed using the time across the diameter of the sun and the ratio of the deflection to the radius or half the denominator: Tan v =y/R=1/2gt^2/R should be the Tan. Einstein drops the 1/2 and doubles the time t=2R/c giving him a factor of 4 over what I think it should be and this accounts for the 1.75
instead of 0.44'. This gives Einstein a Tan= y/R=4GM/Rc^2=8.4656E-6. I think a case can be made for Tan=y/2R=GM/Rc^2=2.1164E-6 and angle 0.44'.

There also is the question of the direction of the deflection is it a redshift or blueshift, away from the sun or towards the sun. I welcome your comments about the tangent and whether it is a red or blue shift. My inclination is to go with Quaternion Gravity. Could there also be a defect in the observation of Einstein's deflection and does the observation show redshift or blueshift away or towards the sun.

Thanks again for your comments.
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Old 27-April-2008, 04:58 AM
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Quote:
Originally Posted by Fortis View Post
How does your "vector energy" compare to the energy-momentum 4-vector that is regularly used in SR & GR?

And now for an obvious question. It is not sufficient for a new theory to explain a single phenomena, such as the cosmological red-shift. If you have a new theory of gravitation then it should also explain:
1) The behaviour of the perihelion of Mercury,
2) Gravitational red-shift
3) Gravitational lensing, and the gravitational deflection of light by the Sun,

as well as the more usual gravitational effects.

Does your model do this?
Your question on 4-vectors.
I am not familiar with 4 vectors. From wiki 4-vector:
inner product of u v=u0v0 -u1v1 -u2v2-u3v3 seems to be the same as the scalar product of quaternions.
In quaternions uv=u0v0 -u1v1 -u2v2 -u3v3 +u0(v1 + v2 + v3) + v0(u1 + u2 =u3) + (u1 + u2 +u3)x(v1 +v2 +v3) .

The rule for multiplying quaternions is
AB=(Ar +Av)(Br + Bv)=(ArBr -Av.Bv) + (ArBv + AvBr + AvxBv)

Where AvxBv = -BvxAv

For spacetime position p=r + ix + jy + kz where r=ct
p^2 = (r^2 -(x^2 + y^2 +z^2)) + 2r(ix + jy +kz)

pp* =r^2 + x^2 + y^2 + z^2 where p* = r-(ix + jy + kz)

Quaternions and 4-vectors seem similar but do 4-vector use Hamilton's
rules for i^2=j^2=k^2=ijk=-1?
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Old 27-April-2008, 06:30 AM
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2 and 3) I think the Quaternion Theory handles the gravitational redshift and I am uninformed about gravitational lensing. On gravitational deflection by the sun, this Theory challenges Einstein's deflection. This theory clearly gives the deflection as cos(rv) = v/c=sqrt(GM/c^2R) = 1.457859E-3 giving a angle of 89degrees 54' 59.93". This gives a deflection angle of 5 arc minutes and 0.07 arc seconds. This is much larger than Einstein's 1.75' arc seconds.

Current measurements have verified GR to within about 2 percent of the 1.75 arc seconds. Your theory's prediction is outside of that uncertainty and hence, does not make an accurate prediction of the bending of light around the sun.

I was going to ask how exactly does your theory tie into SR and how exactly does your theory allow covariance in physical laws. But, since it's prediction is so far outside the actual observations, on the bending of light around the sun, it's pretty much useless anyway.
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Old 27-April-2008, 05:49 PM
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Originally Posted by nutant gene 71 View Post
I'm not sure I totally understand, but it would appear that at some point, past zero, gravity should become repulsive rather than attractive, if it is to follow Maxwell's equations. Is this not so? And if it is so, and we know gravity is never repulsive, how does The Quaternion Theory of Gravity overcome this problem? Just curious is all.
I am proposing that gravitational energy E= -mu/R + mcv. The real energy is centripetal (center seeking) the vector energy mcv is centrifugal (center fleeing). Your comment makes the point that gravity is both attractive and repulsive. Newton and Einstein's real energy only model is only attractive and has no repulsive feature. However the redshift is evidence of the centrifugal force. The Universe is in equilibrium because of the balance of the centripetal and centrifugal forces of gravity. Gravity is the source of the velocity in mcv.

This is not a gravity problem. Gravity has two aspects the real and the vector. The real is attractive and the vector is 'repulsive".

Thanks for your comment.

Last edited by yawyaw; 27-April-2008 at 05:51 PM.. Reason: Forgot comment
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Old 27-April-2008, 06:05 PM
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Thanks for your questions. I don't know how far this Theory goes.

1) I have developed the orbit equation and get r=P(1 +ct/R)/(1 + ecos v)
where P=R(sin(rv))^2. This may or may not answer to Mercury's perihelion.

The vector equilibrium equation is 0=dcv/cdt + cDelxv t + ur/R3.
General Relativity gives the correct rate for the anomalous shift of the perihelion of Mercury. If you have what you believe to be a theory of gravity, then you should try to calculate the value as predicted by your theory.
Quote:
2 and 3) I think the Quaternion Theory handles the gravitational redshift and I am uninformed about gravitational lensing. On gravitational deflection by the sun, this Theory challenges Einstein's deflection. This theory clearly gives the deflection as cos(rv) = v/c=sqrt(GM/c^2R) = 1.457859E-3 giving a angle of 89degrees 54' 59.93". This gives a deflection angle of 5 arc minutes and 0.07 arc seconds. This is much larger than Einstein's 1.75' arc seconds.
Given that your model does not fit observation, you have a bit of a problem. It probably means that your model is wrong.
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Old 27-April-2008, 08:03 PM
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Originally Posted by yawyaw View Post
Your question on 4-vectors.
I am not familiar with 4 vectors. From wiki 4-vector:
inner product of u v=u0v0 -u1v1 -u2v2-u3v3 seems to be the same as the scalar product of quaternions.
In quaternions uv=u0v0 -u1v1 -u2v2 -u3v3 +u0(v1 + v2 + v3) + v0(u1 + u2 =u3) + (u1 + u2 +u3)x(v1 +v2 +v3) .

The rule for multiplying quaternions is
AB=(Ar +Av)(Br + Bv)=(ArBr -Av.Bv) + (ArBv + AvBr + AvxBv)

Where AvxBv = -BvxAv

For spacetime position p=r + ix + jy + kz where r=ct
p^2 = (r^2 -(x^2 + y^2 +z^2)) + 2r(ix + jy +kz)

pp* =r^2 + x^2 + y^2 + z^2 where p* = r-(ix + jy + kz)

Quaternions and 4-vectors seem similar but do 4-vector use Hamilton's
rules for i^2=j^2=k^2=ijk=-1?
The energy momentum 4-vector is given by (E/c, p).
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Old 27-April-2008, 10:04 PM
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Default Energy momentum 4-vector is similar to the Gravity energy

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Originally Posted by Fortis View Post
The energy momentum 4-vector is given by (E/c, p).
The energy momentum 4-vector is similar to the gravity energy quaternion:
E= -mu/R + mcv = (-E/c + p)c if the vectors are i^2=j^2=k^2=ijk=-1.

If 4-vectors are Minkowski space then they are not the same as quaternions. Minkowski space has -(e0)^2 = (e1)^2 =(e2)^2 =(e3)^2=1.

Hamilton's quaternions are a associative division algebra and is unique. Real numbers and complex numbers are subsets.

The energy momentum 4-vector is a quaternion momentum if it used quaternion rules. This momentum 4-vector would also define the redshift. The key point is that gravity is only partly defined by the real energy, the momentum vector energy is also an aspect of gravity. This was left out and thus the need for dark energy. The physics of the situation call for the momentum energy and the math of the situation call for quaternions.

The ubiquitous redshift is an indicator of the gravitational equilibrium condition and a non expanding universe. To the extent Relativity demands expansion, there is problem with relativity Theory. Redshift is not an indicator of expansion, to the contrary redshift is an indicator of equilibrium..
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Old 27-April-2008, 11:09 PM
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Yawyaw, as you get the wrong value for the deflection of light by the Sun, the rest of this seems pretty academic.
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Old 27-April-2008, 11:52 PM
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Default Gravity has a "repulsive part", mcv the vector energy

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Originally Posted by nutant gene 71 View Post
I'm not sure I totally understand, but it would appear that at some point, past zero, gravity should become repulsive rather than attractive, if it is to follow Maxwell's equations. Is this not so? And if it is so, and we know gravity is never repulsive, how does The Quaternion Theory of Gravity overcome this problem? Just curious is all.
The real gravity energy, -mu/R is attractive, the vector gravity energy, mcv is repulsive, -mcDel.v. Gravity has two aspects the real and the vector. The vector part mcv, was left out. it is the source of what some are caling "dark energy". the v in mcv is v=sqrt(GM/R), it is created by gravity.
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Old 28-April-2008, 03:42 AM
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Originally Posted by yawyaw View Post
To the extent Relativity demands expansion, there is problem with relativity Theory.
Why don't you get this straight. Relativity theory, whey applied to cosmological problems, does not demand expansion. It only prohibits static solutions (it could be shrinking). Given you can't get this straight, not to mention, you weren't even aware of gravitational lensing or four vectors, your claims of relativity theory being wrong seem very weak. I'm even wondering if you are aware of the tensor notation used in GR.

As far as your quaternion idea goes for gravity, as I said, and Fortis again noted, it doesn't get the right value for the deflection of light around the sun, so any other ideas you think it may work for are pretty much an exercise in futility.
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Old 28-April-2008, 09:19 AM
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Originally Posted by yawyaw View Post
The real gravity energy, -mu/R is attractive, the vector gravity energy, mcv is repulsive, -mcDel.v. Gravity has two aspects the real and the vector. The vector part mcv, was left out. it is the source of what some are caling "dark energy". the v in mcv is v=sqrt(GM/R), it is created by gravity.
You wrote something very important.
The attractive potential gravitational energy and repulsive kinetic energy are connected together. I have an idea of the space related to the problem. http://www.charge.glt.pl/ Quaternion describes rotation operations http://en.wikipedia.org/wiki/Quatern...atial_rotation

Every particle oscillate http://en.wikipedia.org/wiki/De_Broglie_hypothesis
Due to half-integer spin of the fermion, as an observer circles a fermion (or as the fermion rotates 360° about its axis) the wavefunction of the fermion changes sign. A related phenomenon is called an antisymmetric wavefunction behavior of a fermion. Fermions obey Fermi-Dirac statistics, meaning that when one swaps two fermions, the wavefunction of the system changes sign. http://en.wikipedia.org/wiki/Fermion

My assumption is that matter has a wave structure in the space. Something like Milo Wolff's idea - http://members.tripod.com/mwolff/
The wave goes in and out of the particle. There is a phase transition in the point where the probability function is most possible (wave function collapse ?).

There is a Qouantum Loop lattice created and the virtual particle goes forth in the loop and virtual antiparticle goes back.
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Old 28-April-2008, 01:33 PM
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Quote:
Originally Posted by Fortis View Post
Yawyaw, as you get the wrong value for the deflection of light by the Sun, the rest of this seems pretty academic.
I computed the redshift deflection. Maybe the 1.75' deflection is not the redshift deflection at the sun, but some other deflection. I provided my formula. What formula gives the 1.75' deflection?
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Old 28-April-2008, 02:31 PM
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I computed the redshift deflection. Maybe the 1.75' deflection is not the redshift deflection at the sun, but some other deflection. I provided my formula. What formula gives the 1.75' deflection?
Why, the formula is well-known to anyone versed in General Relativity. It is (in SI units) 4*G*M/(r*c2). Here G is the Newtonian gravitational constant, M is the mass of the Sun, r is the radius of the Sun, and c is the speed of light.

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Old 28-April-2008, 06:50 PM
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The key point is that gravity is only partly defined by the real energy, the momentum vector energy is also an aspect of gravity.

I missed this the first time around. You are obviously not aware that in GR, the energy and the momentum are accounted for together, by the stress-energy tensor. That includes both the density and flux of the energy and momentum.

Quote:
Originally Posted by yawyaw View Post
This was left out and thus the need for dark energy.
Dark energy has nothing to do with it. Why don't you do a bit more reading about GR, before making wrong pronouncements about what GR does and does not require. It's becoming quite obvious that what you know about GR has come from misunderstanding some popular science articles about it.

Quote:
Originally Posted by yawyaw View Post
The physics of the situation call for the momentum energy and the math of the situation call for quaternions.
The momentum energy is included in the physics of the situation and math of the situation call for differential geometry (or as it used to be known, tensor calculus, or even futher back absolute differential calculus), not quaternions.
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Old 28-April-2008, 07:41 PM
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Why, the formula is well-known to anyone versed in General Relativity. It is (in SI units) 4*G*M/(r*c2). Here G is the Newtonian gravitational constant, M is the mass of the Sun, r is the radius of the Sun, and c is the speed of light.

Get thee to a library!
Thanks for your comment. I have that formula in my post#10. Have you ever questioned that formula?


Looking at Einstein's computation, the tangent is computed using the time across the diameter of the sun and the ratio of the deflection to the radius or half the denominator: Tan v =y/R=1/2gt^2/R should be the Tan. Einstein drops the 1/2 and doubles the time t=2R/c giving him a factor of 4 over what I think it should be and this accounts for the 1.75
instead of 0.44'. This gives Einstein a Tan= y/R=4GM/Rc^2=8.4656E-6. I think a case can be made for Tan=y/2R=GM/Rc^2=2.1164E-6 and angle 0.44'.

There also is the question of the direction of the deflection is it a redshift or blueshift, away from the sun or towards the sun. I welcome your comments about the tangent and whether it is a red or blue shift.

I would welcome an explanation why the angle is the ratio of y/R but the time is over 2R?
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Old 28-April-2008, 08:36 PM
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Quote:
Originally Posted by yawyaw View Post
Thanks for your comment. I have that formula in my post#10. Have you ever questioned that formula?


Looking at Einstein's computation, the tangent is computed using the time across the diameter of the sun and the ratio of the deflection to the radius or half the denominator: Tan v =y/R=1/2gt^2/R should be the Tan. Einstein drops the 1/2 and doubles the time t=2R/c giving him a factor of 4 over what I think it should be and this accounts for the 1.75
instead of 0.44'. This gives Einstein a Tan= y/R=4GM/Rc^2=8.4656E-6. I think a case can be made for Tan=y/2R=GM/Rc^2=2.1164E-6 and angle 0.44'.

There also is the question of the direction of the deflection is it a redshift or blueshift, away from the sun or towards the sun. I welcome your comments about the tangent and whether it is a red or blue shift.

I would welcome an explanation why the angle is the ratio of y/R but the time is over 2R?
You can start off with the solution to the field equations for a static, spherically symetric, distribution of mass. You can then derive the equation describing null-geodesics (i.e. the world-lines of photons with zero rest mass) in this spacetime. When you integrate the differential equation that you have just derived, you end off with a deflection, for a photon grazing the Sun, of 1.75''.

I recommend that you do a bit of background reading on GR, (Shutz, or Misner Thorne and Wheeler.)

As to being able to make a case for o.44'', that is not what is observed experimentally, and hence it is wrong.
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Old 28-April-2008, 10:15 PM
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GR also has a very intuitive explanation for the equivalence of gravitational mass and inertial mass. Is there an equivalent explanation within the framework of your model?
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Old 28-April-2008, 10:21 PM
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Default Is Einstein's 1.75' deflection the REDSHIFT Deflection?

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You can start off with the solution to the field equations for a static, spherically symetric, distribution of mass. You can then derive the equation describing null-geodesics (i.e. the world-lines of photons with zero rest mass) in this spacetime. When you integrate the differential equation that you have just derived, you end off with a deflection, for a photon grazing the Sun, of 1.75''.

I recommend that you do a bit of background reading on GR, (Shutz, or Misner Thorne and Wheeler.)

As to being able to make a case for o.44'', that is not what is observed experimentally, and hence it is wrong.
Thanks for your comment and I agree with your last statement.

My question is, " Is the 1.75' deflection a redshift deflection"? My Quaternion Gravity Theory is discussing redshift deflections. I claim the the redshift deflection is 5" away from the sun, z=cos(89 degrees, 55 ").

Is the 1.75' deflection toward or away from the sun and
the 1.75' IS THE REDSHIFT DEFLECTION at the sun. There may be other deflections that Einstein was considering. I have not heard the 1.75' deflection referred to as a redshift deflection. it could be we are not discussing the same deflection type, could the 1.75' be gravitational lensing deflection?
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Old 29-April-2008, 03:23 AM
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Default Quaternion Physics and Redshift Revealed in Relativity Theory.

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Originally Posted by papageno View Post
Please, spare us the usual excuse ATM proponents use to justify their failures.



You should try and read some scientific papers.



The ATM forum is also publicly visible.



The redshift-distance relation (Hubble's law) is one of the observations.



Galileo Galilei described the method at the basis of modern science. The method is what distinguishes science from pseudo-science.



Hubble's law is an established observation, not a dogma.



No, the method is the test whether one is a scientist or a crackpot.



I am curious to know what mathematics of physics you have studied....



There is no imaginary in that interval, only squares of real numbers. Maybe you are just imagining it...



The answer is that four-vectors are more intuitive and give just as much information as quaternions. And they fit nicely into the whole formalism developed for theoretical physics.



Provide the evidence that something has been compromised.



I always ask quaternion fans to show us that quaternions work better than four-vectors.
Considering that you see an imaginary in the very real Einstein interval you quoted, I won't hold my breath.
1. The Group property of Associativity is lost with 4-vectors.
Associativity 4-vectors: (II)J=+J but I(IJ)=IK= -J
Associativity Quaternions: (ii)j=-j and i(ij)=ik = -j
2. Identity: What 4-vector is the Identity element, v1=1v=v
quaternions 1, i, j, k, 1 is the identity element. No identity element , no inverse, no division algebra.
3. point in four space:
Quaternion: p=ct + ix + jy + kz; p^2= ((ct)^2 -(x^2 + y^2 + z^2)) + 2(ct)(ix + jy + kz)

4-vector: p= (ct)e0 + xe1 + ye2 ze3; p^1 = (-1(ct)^2 + x^2 + y^2 + z^2) + 2(ct)e0(xe1 + ye2 + ze3)

It appears that for Relativity Theory, the 4-vector is a QUATERNION, and not the Minkowski 4-vector elsewhere.

4. It also appears that GR's Beta=v/c = z= cos(g) the redshift and the Lorentz Transformation 1/Gamma= sqrt(1-(v/c)^2) = sin(g). It appears the Quaternions and redshift is the essence of GR!
see:
ttp://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c2
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Old 29-April-2008, 03:34 AM
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Default Quaternion Physics and Redshift Revealed in Relativity Theory.

Quote:
Originally Posted by papageno View Post
Please, spare us the usual excuse ATM proponents use to justify their failures.



You should try and read some scientific papers.



The ATM forum is also publicly visible.



The redshift-distance relation (Hubble's law) is one of the observations.



Galileo Galilei described the method at the basis of modern science. The method is what distinguishes science from pseudo-science.



Hubble's law is an established observation, not a dogma.



No, the method is the test whether one is a scientist or a crackpot.



I am curious to know what mathematics of physics you have studied....



There is no imaginary in that interval, only squares of real numbers. Maybe you are just imagining it...



The answer is that four-vectors are more intuitive and give just as much information as quaternions. And they fit nicely into the whole formalism developed for theoretical physics.

Quaternion point: p=ct + ix + jy + kz; p^2= ((ct)^2 -(x^2 + y^2 + z^2)) + 2ct(ix + jy +kz)

It appears that for Relativity Theory, the 4-vector is a QUATERNION, and not the Minkowski 4-vector elsewhere.

4-vector point: p= e0(ct) + e1x + e2y +e3z; p^2=(-(ct)^2 +x^2 + y^2 + z^2) + 2cte0(e1x + e2y + e3z)

It also appears that GR's Beta=v/c = z= cos(g) is the redshift and the Lorentz Transformation 1/Gamma= sqrt(1-(v/c)^2) = sin(g). It appears the Quaternions and redshift is the essence of GR!
see:
ttp://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html#c2


Provide the evidence that something has been compromised.
1. The Group property of Associativity is lost with 4-vectors.
Associativity 4-vectors: (II)J=+J but I(IJ)=IK= -J

2. Identity: What 4-vector is the Identity element, v1=1v=v
quaternions 1, i, j, k, 1 is the identity element. No identity element , no inverse, no division algebra.


I always ask quaternion fans to show us that quaternions work better than four-vectors.
Considering that you see an imaginary in the very real Einstein interval you quoted, I won't hold my breath.
1. The Group property of Associativity is lost with 4-vectors.
Associativity 4-vectors: (II)J=+J but I(IJ)=IK= -J
Associativity Quaternions: (ii)j=-j and i(ij)=ik = -j

2. Identity: What 4-vector is the Identity element, v1=1v=v
quaternions 1, i, j, k, 1 is the identity element. No identity element , no inverse, no division algebra.

It also appears that GR's Beta=v/c = z= cos(g) the redshift and the Lorentz Transformation 1/Gamma= sqrt(1-(v/c)^2) = sin(g). It appears the Quaternions and redshift is the essence of GR!
see:
http://hyperphysics.phy-astr.gsu.edu...v/vec4.html#c2
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Old 29-April-2008, 04:01 AM
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Default The energy-momentum 4-vector is a quaternion energy!

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Originally Posted by Fortis View Post
How does your "vector energy" compare to the energy-momentum 4-vector that is regularly used in SR & GR?

And now for an obvious question. It is not sufficient for a new theory to explain a single phenomena, such as the cosmological red-shift. If you have a new theory of gravitation then it should also explain:
1) The behaviour of the perihelion of Mercury,
2) Gravitational red-shift
3) Gravitational lensing, and the gravitational deflection of light by the Sun,

as well as the more usual gravitational effects.

Does your model do this?
The energy-momentum 4-vector is a identical to the quaternion gravity energy, with the E=-mu/R and mcv=pc. GR uses the same rules as quaternions for the inner vector product is negative.
http://hyperphysics.phy-astr.gsu.edu...v/vec4.html#c2

The same redshift conclusion is derivable from the conservation of the energy-momentum 4 vector! In the link above, the redshift=v/c=cos(g) is the Beta of GR and the sin(g)=sqrt(1-(v/c)^2) = 1/Gamma of the Lorentz Transformation.
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