# Thread: Quaternion Physics and Redshift Revealed

1. Originally Posted by yawyaw
could the 1.75' be gravitational lensing deflection?
It is

2. Originally Posted by zerocold
It is
Thanks! I still don't know anything about lensing but it is nice to not get it confused with redshift deflection.

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## Quaternions and de Broglie oscillation

Yawyaw. I would like to ask you how works the de Broglie oscillation in the quaternion theory?.
There is a repulsive energy according to Pauli exclusion principle and gravitational attractive warping of the space. That way the matter interacts repulsive and attractive.

4. Originally Posted by yawyaw
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.
Are you saying that the quaternion and the four-vectors formulations are equivalent?
If so, that was my point, and you refute you claim that "The mathematics was compromised to satisfy physics dogma".

Originally Posted by yawyaw
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!
I am still waiting for you to show us that imaginary part in the space-time interval...

5. Originally Posted by papageno
Are you saying that the quaternion and the four-vectors formulations are equivalent?
If so, that was my point, and you refute you claim that "The mathematics was compromised to satisfy physics dogma".

I am still waiting for you to show us that imaginary part in the space-time interval...
Quaternions and 4-vectors NOT the same, see 1.,2.,3.

see point 3. for imaginary: point p and p^2 for 4-vector -(ct) + x^2 + y^2 + z^2

6. Originally Posted by czeslaw
Yawyaw. I would like to ask you how works the de Broglie oscillation in the quaternion theory?.
There is a repulsive energy according to Pauli exclusion principle and gravitational attractive warping of the space. That way the matter interacts repulsive and attractive.
I am not familiar with de Broglie oscillations. I looked it up and here is a link that may help.

http://www.calphysics.org/mass.html

My work relates Planck's Constant h to the fine structure constant thru z, the free space impedance. z=W/C =375 Ohms and h=WC where W= 500 atto Webers (volt second) and C=4/3 E-18 (atto) Coulombs. The Ether Vacuum has quantum and magnetic charges, C and W. h=zC^2.

The Fine Structure Constant alpha=1/2 (e/C)^2/n . This may help.

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Originally Posted by yawyaw
Thanks! I still don't know anything about lensing but it is nice to not get it confused with redshift deflection.
What is "redshift deflection"?

What we have been talking about is the deflection of a photon as it passes by the Sun.

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Originally Posted by yawyaw
The energy-momentum 4-vector is a identical to the quaternion gravity energy, with the E=-mu/R and mcv=pc.
Is it your claim that gravity is described by a classical Newtonian potential?

Does that potential propagate with infinite velocity, or at some other speed?

9. 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
Off Topic.

My undergraduate physical chemisty professor used to include a bonus question on the final exam to reduce the failure rate or to entertain himself.

He'd post a picture and ask: Who is this?

It was J. Willard Gibbs, the inventor of Physical Chemistry.
Last edited by John Jones; 2008-Apr-30 at 02:23 AM.

10. Originally Posted by John Jones
Off Topic.

My undergraduate physical chemist professor used to include a bonus question on the final exam to reduce the failure rate or to entertain himself.

He'd post a picture and ask: Who is this?

It was J. Willard Gibbs, the inventor of Physical Chemistry.
Thanks for the comment. A nice diversion. Gibbs was a great physicist and educator, with many contributions. I looked him up after your post.

11. ## Quaternion Gravity Wave propagation

Originally Posted by Fortis
Is it your claim that gravity is described by a classical Newtonian potential?

Does that potential propagate with infinite velocity, or at some other speed?
I claim that Gravity is described by a classical quaternion energy E= -mu/R + mcv. This energy is quaternion containing the classical Newtonian potential(real) energy and the vector energy of motion mcv. The Quaternion second derivative is the wave equation. The first derivative is X=(d/dR + Del)= (d/cdt + Del). The second derivative is the square of the first derivative:
X^2 = ((d^2/c^2dt^2 - Del^2) + 2d/dR Del) The wave equation for Gravity is a quaternion and consists of a real part and a vector part: a real longitudinal wave equation and a vector Transverse wave equation. The

X^2E= ((d^2/c^2dt^2 - Del^2)(-mu/R) -2mcd/dR Del.v) + ((d^2/c^2dt^2 - Del^2)mcv + 2d/dR(mcDelxv + Del(-mu/R))

x^2E = ((d^2/c^2dt^2 - Del^2)(-mu/R) + 2mcvcos(g)/R^2) + ((d^2/c^2dt^2 - Del^2)mcv + 2mv(csin(g)/R^2 + v r/R3))

Here you see the Quaternion wave equations, a real longitudinal wave and the vector transverse wave. At the equilibrium condition, 0=XE, the waves are simplified.

X^2E = -(d^2/c^2dt^2 + Del^2)(-mu/R + mcv).
Last edited by yawyaw; 2008-Apr-30 at 04:38 AM.

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Originally Posted by yawyaw
I claim that Gravity is described by a classical quaternion energy E= -mu/R + mcv.
What is "R"?

13. Originally Posted by yawyaw
Quaternions and 4-vectors NOT the same, see 1.,2.,3.
I never said they are.

But you said: "It appears that for Relativity Theory, the 4-vector is a QUATERNION... "
So I asked whether - according to you - the formulation of Relativity in terms of quaternions is equivalent to the formulation in terms of four-vectors.

Of course, you are utterly wrong when you finish the sentence with "and not the Minkowski 4-vector elsewhere", since the "Einstein" space-time interval you quoted is at the basis of the definition of Minkowskian vector-space.

Originally Posted by yawyaw
see point 3. for imaginary: point p and p^2 for 4-vector -(ct) + x^2 + y^2 + z^2
No, you claimed that there is an imaginary in this space-time interval: I = x^2 + y^2 + z^2 - (ct)^2.
So, where is it? I see only squares of real numbers.

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Originally Posted by papageno
No, you claimed that there is an imaginary in this space-time interval: I = x^2 + y^2 + z^2 - (ct)^2.
So, where is it? I see only squares of real numbers.
Ah. The joy of metrics (or pseudo-metrics...)

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Originally Posted by papageno
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)
There is no imaginary in that interval, only squares of real numbers. Maybe you are just imagining it...
No, x2 + y2 + z2 - (ct)2 can be written as x2 + y2 + z2 + (ict)2.
Not that I'm advocating the replacement of vectors with quaternions, or anything.

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Originally Posted by Disinfo Agent
No, x2 + y2 + z2 - (ct)2 can be written as x2 + y2 + z2 + (ict)2.
Not that I'm advocating the replacement of vectors with quaternions, or anything.
SR is frequently taught in that fashion, but I think that the forcing of SR onto a Euclidean geometry, via the use of i, masks the underlying geometry of Minkowski spacetime. The use of a metric (or pseudo-metric) tensor allows you to move forward in a more natural fashion, and is pretty much essential when working with GR. Mind you, that is a personal opinion.

17. ## Quaternion Relativity Theory

Originally Posted by papageno
I never said they are.

But you said: "It appears that for Relativity Theory, the 4-vector is a QUATERNION... "
So I asked whether - according to you - the formulation of Relativity in terms of quaternions is equivalent to the formulation in terms of four-vectors.

Of course, you are utterly wrong when you finish the sentence with "and not the Minkowski 4-vector elsewhere", since the "Einstein" space-time interval you quoted is at the basis of the definition of Minkowskian vector-space.

No, you claimed that there is an imaginary in this space-time interval: I = x^2 + y^2 + z^2 - (ct)^2.
So, where is it? I see only squares of real numbers.
"So I asked whether - according to you - the formulation of Relativity in terms of quaternions is equivalent to the formulation in terms of four-vectors."

See my new ATM post on Quaternion Relativity!

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Please can you tell me what "R" is in
Originally Posted by yawyaw
I claim that Gravity is described by a classical quaternion energy E= -mu/R + mcv.

19. Originally Posted by Fortis
Please can you tell me what "R" is in
"R" is the separation distance between m and M, in the gravitational Real or Potential energy E= mu/R=mGM/R.

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Originally Posted by yawyaw
"R" is the separation distance between m and M, in the gravitational Real or Potential energy E= mu/R=mGM/R.
If "R" is a purely spatial coordinate, then how can you simply replace "d/dR" with "d/c.dt" in
The first derivative is X=(d/dR + Del)= (d/cdt + Del).
?

21. Originally Posted by Fortis
If "R" is a purely spatial coordinate, then how can you simply replace "d/dR" with "d/c.dt" in
?
Think distance R=ct; dR= cdt.

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Originally Posted by yawyaw
Think distance R=ct; dR= cdt.
The second part may follow from the first, but it is the first part that needs explaining.

Why is "R=ct"?

23. as always it has to be to make it work.

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Originally Posted by Fortis
The second part may follow from the first, but it is the first part that needs explaining.

Why is "R=ct"?
In the other thread you said
I use R=ct for the 4th dimension, the answer to another of your questions.
So does this mean that where you said
E= -mu/R + mcv
you actually could have written

E= -mu/ct + mcv

and that these two statements are identical?

25. Originally Posted by Fortis
In the other thread you said

So does this mean that where you said

you actually could have written

E= -mu/ct + mcv

and that these two statements are identical?
Yep!

for example d/dR(-mu/ct) = d/cdt (-mu/ct)= -mu/c2d/dt 1/t=mu/(ct)2 = mu/R2

The derivative would be the time derivative rather than the space derivative. Spacetime has space units thus ct=meters. Space and Time came before spacetime. Much of physics is still in that mode. For example electromagnetism uses space and time in maxwell's Equations: Faraday's law: 0= dBv/dt + DelxE Del is a space derivative. My version is 0= dEv/dR + DelxEv= dEv/cdt + DelxEv. This shows that Bv = Ev/c.

This is like saying the distance to the sun from earth is 150 Giga meters or 500 light seconds(ct=c500).

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Originally Posted by yawyaw
Yep!

for example d/dR(-mu/ct) = d/cdt (-mu/ct)= -mu/c2d/dt 1/t=mu/(ct)2 = mu/R2

The derivative would be the time derivative rather than the space derivative. Spacetime has space units thus ct=meters. Space and Time came before spacetime. Much of physics is still in that mode. For example electromagnetism uses space and time in maxwell's Equations: Faraday's law: 0= dBv/dt + DelxE Del is a space derivative. My version is 0= dEv/dR + DelxEv= dEv/cdt + DelxEv. This shows that Bv = Ev/c.

This is like saying the distance to the sun from earth is 150 Giga meters or 500 light seconds(ct=c500).
So if

E = -mu/R + mcv

and

E = -mu/ct + mcv

then if I sit at some fixed location relative to a gravitating body and wait, then that would be equivalent to me moving further away from it. Doesn't the 1/t dependence imply that gravitational fields weaken with time?

27. Originally Posted by Fortis
So if

E = -mu/R + mcv

and

E = -mu/ct + mcv

then if I sit at some fixed location relative to a gravitating body and wait, then that would be equivalent to me moving further away from it. Doesn't the 1/t dependence imply that gravitational fields weaken with time?
Nope. spacetime has units of space, meters R is the metric. R=ct gives a time based radar like way to measure the meters. You can measure the space with a yardstick or a radar gun.
Last edited by yawyaw; 2008-May-05 at 03:54 AM.

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Originally Posted by yawyaw
Nope. spacetime has units of space, meters R is the metric. R=ct gives a time based radar like way to measure the meters. You can measure the space with a yardstick or a radar gun.
If R=ct, then R gets bigger the longer you wait. This means that the "potential" term mu/R gets smaller the longer you wait. If this is not the case the R does not equal ct.

Please can you explain the error in this reasoning?

29. Originally Posted by Fortis
If R=ct, then R gets bigger the longer you wait. This means that the "potential" term mu/R gets smaller the longer you wait. If this is not the case the R does not equal ct.

Please can you explain the error in this reasoning?
The error in the reasoning is to consider "t" as an indicator of "flowing time" rather than a "date", like May 5. Radar time gives the best reasoning model, or light year. distance. I generally use R because it reflects that I am working with distances, in calculus. Think of R as 500 seconds out as a distance. I think of the sun as 8 1/3 minutes away.

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Originally Posted by yawyaw
The error in the reasoning is to consider "t" as an indicator of "flowing time" rather than a "date", like May 5.
So it would take a different value on 4th July? How is that different to "flowing time"?
Radar time gives the best reasoning model, or light year. distance. I generally use R because it reflects that I am working with distances, in calculus. Think of R as 500 seconds out as a distance. I think of the sun as 8 1/3 minutes away.
But that is not time, that is space. I can talk about distances in terms of light seconds, and in essence that is what we do if we use the convention where we set c=1. That still doesn't mean that in a particular inertial frame we can just replace spatial derivatives by time derivatives. Do you really believe that for a function f, you can write

df/dR = (1/c)df/dt ?

The left hand side is the rate of change of f with respect to a spatial displacement, whereas the righthand side is a rate of change with respect to time.

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