1. ## Quaternion Relativity Theory

Einstein's Relativity has been around for 100 years and Physicists have got to know and like it. But is it not time to really look at alternatives.

Quaternion Relativity differs from Einstein's Relativity, Here is the Quaternion Formulation in general:

Here is a basic operation in Relativity.
Transform (Rotate) Point 2 around Point 1
(order counts in Quaternions, this is called non-commutativity, AB is not BA).

P1P2=(ct1 + ix1 + jy1 + kz1)(ct2 + ix2 + jy2 + kz2)=
(c^2t1t2 - (x1x2 + y1y2 + z1z2) ) + ct1(ix2 + jy2 + kz2) + ct2(ix1 + jy1 + kz2) + (ix1 + jy1 + kz1)X(ix2 + jy2 + kz2)

(where X is the cross product of vectors that I will not multiply out here.)

Now we see the the first parentheses is like the 4-vector Interval:
(c^2t1t2 - (x1x2 + y1y2 + z1z2))

Note that there are other terms in Quaternion Relativity. Note that the cross product term can be zero if the vectors are parallel! Thus Einstein's Relativity to be for parallel vectors only. However, the points ct1(ix2 + jy2 + kz2) and ct2(ix1 + jy1 + kz1) are still missing from Einstein's Relativity Theory.

What happens if the the vectors are perpendicular? Then x1x2 + y1y2 + z1z2 =0 and the Interval reduces to "c^2t1t2" and the cross product term that is missing in Einstein's Relativity comes back. The cross product term is non-commutative like Pauli's Spinors and introduces non-commutativity into Relativity.

Finally the Interval can reduce to zero if c^2t1t2 = x1x2 + y1y2 + z1z2, however, ct1(ix2 + jy2 + kz2) and ct2(ix1 + jy1 + kz1) and the cross product non-commutative term is not zero. I hold as a point of physics that the Interval is never negative, namely that the speed of light is the limit and c^2>=(x1x2 + y1y2 + z1z2)/t1t2= v1v2.

Now you see the difference between Quaternion Relativity and Einstein's Relativity.

Here is another view of the Quaternion Relativity that might be more familiar:

(cosA + asinA)(cosB + bsinB) = cos(A + B) + csin(A+ B) =
(cosAcosB - sinAsinBcos(ab)) + asinAcosB + bsinBcosA + sinAsinBsin(ab)axb

This last formulation of Quaternion Relativity is quite familiar to those familiar with "Complex Number Theory". However, the last cross product term"sinAsinB sin(ab)axb" is not in complex numbers because complex numbers are commutative while quaternions are non-commutative. Non-commutativity is key to Quantum Theory and is also important in physics in general, whirlpool, handedness, etc.

Relativity is about Transformations aka Rotations in four space. Quaternions are the premier mathematical tool for rotations. There are differences in Quaternion Relativity and Einstein's Relativity Theory.

Einstein among his many achievements brought the idea of 4 dimensions into physics. This was revolutionary. However, there is only one 4 dimensional Associative Division Algebra, Quaternions. Associative means A(BC)=(AB)C.
In fact quaternions is uniquely the only Associative Division Algebra. The two other associative division algebras are Real Number Algebra and Complex Number Algebra, and they are commutative subsets of Quaternions.

The only other Division Algebra is the non-Associative Octonions, a 8 Dimensional Division Algebra.

There is more to Quaternions but this should give one reason to re-examine Einstein's Relativity Theory, there are things missing. Einstein's Relativity is related to situations where the angles A+B are multiples of pi (180 degrees) where the result is a real number. When angle A+B=multiples of 90 degrees, the result is vectors.
Last edited by yawyaw; 2008-May-03 at 12:13 PM.

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Originally Posted by yawyaw
Here is a basic operation in Relativity.
Transform (Rotate) Point 2 around Point 1
(order counts in Quaternions, this is called non-commutativity, AB is not BA).

P1P2=(ct1 + ix1 + jy1 + kz1)(ct2 + ix2 + jy2 + kz2)=
(c^2t1t2 - (x1x2 + y1y2 + z1z2) ) + ct1(ix2 + jy2 + kz2) + ct2(ix1 + jy1 + kz2) + (ix1 + jy1 + kz1)X(ix2 + jy2 + kz2)
Seems a little vague. A rotation usually has an angle associated with it. What is it in the example that you have given?

This should be basic stuff if you are using quaternion notation to work with rotations. I don't think that the operation that you are describing does rotate point 2 around point 1, but I'd be more than happy to be proven wrong.

3. ## The rotation angel in Relativity

Originally Posted by Fortis
Seems a little vague. A rotation usually has an angle associated with it. What is it in the example that you have given?

This should be basic stuff if you are using quaternion notation to work with rotations. I don't think that the operation that you are describing does rotate point 2 around point 1, but I'd be more than happy to be proven wrong.
You have to work to see it, nobody can prove it to you. You have to prove it to yourself, yes or no. |P1|= sqrt((Ct1)^2 + x1^2 + y1^2 + z1^2)
|P1|(ct1/|P1| +( ix1 + jy1 + kz1)/|P1|)=|P1|(cos A + a sinA). The cos A = ct1/|P1|.

You can take it from there. it is the same as complex numbers, just more complex, 4 versus 2 dimensions.

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Originally Posted by yawyaw
You have to work to see it, nobody can prove it to you. You have to prove it to yourself, yes or no. |P1|= sqrt((Ct1)^2 + x1^2 + y1^2 + z1^2)
|P1|(ct1/|P1| +( ix1 + jy1 + kz1)/|P1|)=|P1|(cos A + a sinA). The cos A = ct1/|P1|.

You can take it from there. it is the same as complex numbers, just more complex, 4 versus 2 dimensions.
Let's take the complex numbers as you suggest.

z1=r1exp(i.theta1)

and

z2=r2exp(i.theta2)

If we multiply them together we get

z1.z2 = r1.r2exp(i.(theta2+theta2))

If z2 is of unit magnitude, then this gives a rotation of the point given by z1 about the origin, through an angle given by theta2.

This is not a rotation about the point z2.

Am I missing something?

5. I think yawyaw sort of described the "dot product" in quaternions, where the cos A is just the cosine of the angle between the two "vectors".

I think yawyaw should start from scratch, first define what a quaternion is (I guess I could go to Wiki or search my books at work) and yawyaw should explain a little better (maybe by using the "sup" and "sub" comments on the board to get more readable equations), and then maybe we get to see something that (s)he thinks is important.

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Originally Posted by yawyaw
Here is a basic operation in Relativity.
Transform (Rotate) Point 2 around Point 1
(order counts in Quaternions, this is called non-commutativity, AB is not BA).

P1P2=(ct1 + ix1 + jy1 + kz1)(ct2 + ix2 + jy2 + kz2)=
(c^2t1t2 - (x1x2 + y1y2 + z1z2) ) + ct1(ix2 + jy2 + kz2) + ct2(ix1 + jy1 + kz2) + (ix1 + jy1 + kz1)X(ix2 + jy2 + kz2)

(where X is the cross product of vectors that I will not multiply out here.)
If point 1 and point 2 are the same, then the result of rotating a point about itself should be the same as the point you first thought of. What happens if we try it using your expression, which is just the regular quaternion product.

P2=(c2t2-(x2+y2+z2)) + 2.c.t.(i.x.+j.y+k.z)

Obviously the cross product term disappears.

Anyway, it is apparent that P2 does not equal P, and hence it is not the result of rotating the point P about itself.

Did you really mean what you said in italics?

7. Originally Posted by Fortis
If point 1 and point 2 are the same, then the result of rotating a point about itself should be the same as the point you first thought of. What happens if we try it using your expression, which is just the regular quaternion product.

P2=(c2t2-(x2+y2+z2)) + 2.c.t.(i.x.+j.y+k.z)

Obviously the cross product term disappears.

Anyway, it is apparent that P2 does not equal P, and hence it is not the result of rotating the point P about itself.

Did you really mean what you said in italics?
"If point 1 and point 2 are the same, then the result of rotating a point about itself should be the same as the point you first thought of."

Your statement is not true. There is one rotation that fits your statement. If you figure it out you will gain insight into rotations.

I really meant what I said.

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Originally Posted by yawyaw
"If point 1 and point 2 are the same, then the result of rotating a point about itself should be the same as the point you first thought of."

Your statement is not true. There is one rotation that fits your statement. If you figure it out you will gain insight into rotations.

I really meant what I said.
What is that "one rotation"?

Can you give an example where rotating a point about itself does not lead to the original point being recovered?
Last edited by Fortis; 2008-May-04 at 11:01 AM. Reason: To add the last question for clarification

9. Originally Posted by Fortis
What is that "one rotation"?

Can you give an example where rotating a point about itself does not lead to the original point being recovered?
Yes. The point you just used in your example, did not lead to itself. I am waiting for you to find the point that does lead to itself.

10. why have yougot 2 threads going? shouldn't you concentrate on the first one before you jump to this as well?

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Originally Posted by yawyaw
Yes. The point you just used in your example, did not lead to itself. I am waiting for you to find the point that does lead to itself.
Let's go back to the 2-d case, with points on the complex plane, i.e.

z = r.exp(i.theta)

A rotation about the origin can be generated by multiplying z by exp(i.phi), where phi is the angle of rotation.

This leads to a new value of z, let's call it z' which is given by

z' = r.exp(i.(theta + phi))

As all these rotations are about the origin, and we want to examine rotations of a point about itself we can, without loss of generality, apply a rotation to the origin. As r for the origin is equal to zero, i.e

z = 0.exp(i.theta)

then rotating that through an angle phi gives

z' = 0.exp(i.(theta + phi))

It is trivial to see that in this case of rotating a point about itself, z'=z.

It is also trivial to see that this is also true if you rotate any point about itself.

Can you show that I am wrong?

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Originally Posted by yawyaw
Now we see the the first parentheses is like the 4-vector Interval:
(c^2t1t2 - (x1x2 + y1y2 + z1z2))

Note that there are other terms in Quaternion Relativity. Note that the cross product term can be zero if the vectors are parallel! Thus Einstein's Relativity to be for parallel vectors only. However, the points ct1(ix2 + jy2 + kz2) and ct2(ix1 + jy1 + kz1) are still missing from Einstein's Relativity Theory.
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Finally the Interval can reduce to zero if c^2t1t2 = x1x2 + y1y2 + z1z2, however, ct1(ix2 + jy2 + kz2) and ct2(ix1 + jy1 + kz1) and the cross product non-commutative term is not zero.
The key point about the interval in SR is that it is frame independent, i.e. if you carry out a transformation between two different inertial frames, the interval remains the same.

Do you believe this to be the case for your quaternion interval?

13. Originally Posted by Fortis
Let's go back to the 2-d case, with points on the complex plane, i.e.

z = r.exp(i.theta)

A rotation about the origin can be generated by multiplying z by exp(i.phi), where phi is the angle of rotation.

This leads to a new value of z, let's call it z' which is given by

z' = r.exp(i.(theta + phi))

As all these rotations are about the origin, and we want to examine rotations of a point about itself we can, without loss of generality, apply a rotation to the origin. As r for the origin is equal to zero, i.e

z = 0.exp(i.theta)

then rotating that through an angle phi gives

z' = 0.exp(i.(theta + phi))

It is trivial to see that in this case of rotating a point about itself, z'=z.

It is also trivial to see that this is also true if you rotate any point about itself.

Can you show that I am wrong?
Nice try and I'll accept that, the zero transformation rotation is itself for all angles. This brings up the point that the Transformation has two parts the magnitude |P|, which can be zero, and the rotation Angle A. The general; transformation is a rotation with expansion of contraction dependent on the magnitude greater than 1 or less than 1. The sign is controlled by the rotation, A=pi. A pure rotation has magnitude = 1. Here is the answer.

The rotation that returns itself is simple and shows much about transformation. Lets set up your condition
p^2 = p: this is
(cosA +asinA)^2 = cos(2A) + a sin(2A)=cosA + asinA
cos2A = cosA and sin2A=sinA the angle that solves this is A=n2pi=360n-integer degrees, including n=0:

cos(2.0)=cos(0) = 1
cos(4pi)=cos(2pi)=1

sin(2.0) = sin(0)= 0
sin(4pi) = sin(2pi)=0

thus the p=1 + 0 = 1. This is the rotation around angle multiples of 2pi(360 degrees) and returns to itself.

Now Einstein/Lorentz Transformation doesn't have any vector terms in it. This means that sin(A+B)=0 or the sum of the angles A+B= n pi or multiples of pi. We could call this the Relativity of transformations whose angles sum to pi. To get the vectors we could have A+B= npi/2 and get just vector terms and no real terms. This is the Relativity of the speed of light,
cos(A + B)= c^2t1t2 - (x1x2 + y1y2 + z1z2)=0 thus the real term goes to zero, the sum of angles A+B= pi/2=90 degrees. If A=B then we have the square of the transformation and 2A=90 degrees and A= 45 degrees.

Quaternion Relativity handles all these cases from just reals to just vectors to reals and vectors.

A quaternion is the sum of the real and the vector using Hamilton's Rules i^2 =j^2=k^2=ijk=-1. (sorry I don't know about sup and sub)

The Interval I=(ct)^2 - (x^2 + y^2 + z^2) is quaternion, the Interval I=(x^2 + y^2 + z^2) -(ct)^2 is not quaternion it is Minkowskian and reflects Minkowski's use of imaginary for the spacetime 4th dimension. I use R=ct for the 4th dimension, the answer to another of your questions. I sometimes use r=ct in p=r + ix + jy + kz
other times I use w=ct, so that p= w +ix + jy + kz, in all cases the real spacetime dimension is ct. Minkowski uses ict, where i is a vector like j and k but the 3 space dimensions are all real. So that Minkowski space is real and time is a vector. The real dimension in spacetime is R=r=w=ct and space dimensions are vectors (ix + jy + kz). The space dimensions are vectors because of the i, j and k. These are the vectors, not the x, y and z. x,y and z are real numbers.

Thanks for your questions. I hope you have a better feel for quaternions. Have a nice day.
Last edited by yawyaw; 2008-May-04 at 06:01 PM.

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Originally Posted by yawyaw
Nice try and I'll accept that, the zero transformation rotation is itself for all angles. This brings up the point that the Transformation has two parts the magnitude |P|, which can be zero, and the rotation Angle A. The general; transformation is a rotation with expansion of contraction dependent on the magnitude greater than 1 or less than 1. The sign is controlled by the rotation, A=pi. A pure rotation has magnitude = 1. Here is the answer.
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We may be misunderstanding each other.

Let's take the point P. I now carry out a rotation about the point P. Has the point P moved to a different location?

Or

I take a painting. I now start it spinning about a point in the painting (a star perhaps?). As I look at the painting, which point remains in the same location?

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Originally Posted by yawyaw
The Interval I=(ct)^2 - (x^2 + y^2 + z^2) is quaternion, the Interval I=(x^2 + y^2 + z^2) -(ct)^2 is not quaternion it is Minkowskian and reflects Minkowski's use of imaginary for the spacetime 4th dimension.
In SR the signature of the metric tensor is pretty arbitrary. Some people choose to define the interval as

ds2 = (c2t2)-(x2+y2+z2)

and others choose

ds2 = -(c2t2)+(x2+y2+z2)

It is just a matter of convention.

16. Originally Posted by Fortis
In SR the signature of the metric tensor is pretty arbitrary. Some people choose to define the interval as

ds2 = (c2t2)-(x2+y2+z2)

and others choose

ds2 = -(c2t2)+(x2+y2+z2)

It is just a matter of convention.
I don't consider that physics should have this level of arbitrariness. I think version of the Interval with -(ct)2 unsound mathematics (Non-division algebra) and physics ( 4th dimension is real, R=ct, not imaginary R= ict). mathematically only Quaternions constitute an Associative Division Algebra. It is hard to have sound physics based on unsound mathematics, my opinion!

(thanks for the showing me the [sup] notation)

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Originally Posted by yawyaw
I don't consider that physics should have this level of arbitrariness.
Why not?

The key here is that the interval is invariant. It doesn't matter if it is positive or negative. It still leads to spacetime possessing the same geometry.
I think version of the Interval with -(ct)2 unsound mathematics (Non-division algebra)
Do you mean that division does not exist for complex numbers?
and physics ( 4th dimension is real, R=ct, not imaginary R= ict).
The use of ict is really a form of short hand. It is fine if you want to constrain yourself to a flat spacetime, but really it is just obscuring the underlying geometry.

Are you at all familiar with differential geometry, and tensor calculus? It is all very clear when you work with metric tensors. No imaginary numbers that might alarm people. Time is still real-valued, as is space.
mathematically only Quaternions constitute an Associative Division Algebra. It is hard to have sound physics based on unsound mathematics, my opinion!
Again, are you at all familiar with tensor alculus and differential geometry? I would imagine that you would at least have a passing acquaintance with them if you are criticising GR.
(thanks for the showing me the [sup] notation)
No problem.

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By the way, I'm still very interested to know if you believe that the "quaternion interval" is an invariant in all inertial reference frames. Do you believe that?

19. Originally Posted by Fortis
By the way, I'm still very interested to know if you believe that the "quaternion interval" is an invariant in all inertial reference frames. Do you believe that?
Complex numbers have two dimensions, the Interval has 4 dimensions, thus I don't consider the Interval with the complex number 'i' in it a complex number. It is a 4 dimensional something that is not a associative division algebra.

I have heard of Tensors but I don't use them.

I believe that the quaternion Interval is invariant in all inertial frames being identical to the Interval I= (ct)^2 - (x^2 + y^2 + z^2).

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Originally Posted by yawyaw
I believe that the quaternion Interval is invariant in all inertial frames being identical to the Interval I= (ct)^2 - (x^2 + y^2 + z^2).
I just want to be clear here. The quaternion product that you gave, which includes the traditional SR 4-vector interval, is

p2 = (c2t2-(x2+y2+z2)) + 2.c.t.(i.x.+j.y+k.z)

Is it the whole of this that you regard as being invariant, or is it only the first term?

21. Originally Posted by Fortis
I just want to be clear here. The quaternion product that you gave, which includes the traditional SR 4-vector interval, is

p2 = (c2t2-(x2+y2+z2)) + 2.c.t.(i.x.+j.y+k.z)

Is it the whole of this that you regard as being invariant, or is it only the first term?
It appears that Einstein was looking for an invariant distance in 4 D that is equivalent to the 3D distance. In this sense, I think both terms are invariant in the sense that a vector's characteristics should remain invariant when described in different reference frames. For example two vectors parallel in a frame will be parallel in the new frame, or perpendicular in one frame should be perpendicular in the new frame. If the angles sum to 90 degrees in one frame (the light condition) then the new angles sum to 90 degrees.

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Originally Posted by yawyaw
It appears that Einstein was looking for an invariant distance in 4 D that is equivalent to the 3D distance. In this sense, I think both terms are invariant in the sense that a vector's characteristics should remain invariant when described in different reference frames. For example two vectors parallel in a frame will be parallel in the new frame, or perpendicular in one frame should be perpendicular in the new frame. If the angles sum to 90 degrees in one frame (the light condition) then the new angles sum to 90 degrees.
Just double checking, because I want to be sure about what it is that you are claiming.

If I carry out a coordinate transformation from one inertial reference frame to another,

p2 = (c2t2-(x2+y2+z2)) + 2.c.t.(i.x.+j.y+k.z)

will be the same in the transformed coordinates. Is that correct?

Or are you just suggesting that it is invariant only in some qualitative sense, not in a quantiative sense?

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Originally Posted by yawyaw
ds2 = (c2t2)-(x2+y2+z2)

and others choose

ds2 = -(c2t2)+(x2+y2+z2)
I don't consider that physics should have this level of arbitrariness. I think version of the Interval with -(ct)2 unsound mathematics (Non-division algebra) and physics ( 4th dimension is real, R=ct, not imaginary R= ict). mathematically only Quaternions constitute an Associative Division Algebra. It is hard to have sound physics based on unsound mathematics, my opinion!
You can't get rid of the implicit imaginary unit. If you write

I = ds2 = -(ct)2+(x2+y2+z2)

then that's the same as

I = ds2 = (ict)2+(x2+y2+z2)

and if you write

I' = ds2 = (ct)2-(x2+y2+z2)

that's the same as

I' = ds2 = (ct)2+[(ix)2+(iy)2+(iz)2]

The two versions are symmetrical, I' = -I. Basically, you can't force Minkowski spacetime into a Euclidean metric.

24. Originally Posted by Disinfo Agent
You can't get rid of the implicit imaginary unit. If you write

I = ds2 = -(ct)2+(x2+y2+z2)

then that's the same as

I = ds2 = (ict)2+(x2+y2+z2)

and if you write

I' = ds2 = (ct)2-(x2+y2+z2)

that's the same as

I' = ds2 = (ct)2+[(ix)2+(iy)2+(iz)2]

The two versions are symmetrical, I' = -I. Basically, you can't force Minkowski spacetime into a Euclidean metric.
Quaternions give the I' = ds2 = (ct)2+[(ix)2+(jy)2+(kz)2]

The but ct + ix + iy + iz= ct + i(x + y +z) is a 2D complex number as opposed to q= ct + ix + jy + kz which is a 4d quaternion number. Minkowski has time imaginary and space real. Space is a vector domain where direction, parallel and perpendicular are involved. Real Number Domain has no concept of parallel or perpendicular, 5.4 is not 0 similarly 5.5 is not 0. Quaternion vectors are all imaginary in the sense of i^2=j^2=k^2=ijk=-1.

J. Willard Gibbs Vectors have I^2=J^2=K^2=+1 not -1. They are not assoicaitive because of the +1. I(IJ)= -J but (II)J= +J. The signs matter.
Quaternions are assoiciative, i(ij)=-j and (ii)j=-j.

25. Originally Posted by Fortis
Just double checking, because I want to be sure about what it is that you are claiming.

If I carry out a coordinate transformation from one inertial reference frame to another,

p2 = (c2t2-(x2+y2+z2)) + 2.c.t.(i.x.+j.y+k.z)

will be the same in the transformed coordinates. Is that correct?

Or are you just suggesting that it is invariant only in some qualitative sense, not in a quantitative sense?
I think the invariance is quantitative.

26. Originally Posted by yawyaw
I think the invariance is quantitative.
I should hope so!

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Originally Posted by yawyaw
J. Willard Gibbs Vectors have I^2=J^2=K^2=+1 not -1. They are not assoicaitive because of the +1. I(IJ)= -J but (II)J= +J. The signs matter.
It's true that the cross product is not associative, but you have an error, there: i x i=0, of course, so (i x i) x j =0.

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Originally Posted by yawyaw
Quaternions give the I' = ds2 = (ct)2+[(ix)2+(jy)2+(kz)2]

The but ct + ix + iy + iz= ct + i(x + y +z) is a 2D complex number as opposed to q= ct + ix + jy + kz which is a 4d quaternion number.
However,

ds = ct.e0 + i.(x.e1 + y.e2 +z.e3)

is a 4-d vector (with some imaginary components), where I have used bold to indicate a vector quantity, and ei would be the ith basis vector in an orthonormal basis set.

The dot product,
ds.ds = c2t2-(x2+y2+z2)

which is just the usual interval of SR.
Minkowski has time imaginary and space real.
There is nothing really stopping you from making space imaginary and time real. As I have said before, the key thing is that the interval is invariant under a coordinate transformation from one inertial frame to another. At times (real ones ) I wonder if you are constructing a strawman version of SR.
Space is a vector domain where direction, parallel and perpendicular are involved. Real Number Domain has no concept of parallel or perpendicular, 5.4 is not 0 similarly 5.5 is not 0.
See the above. If I stick an "i" in front of the spatial 3-vector, instead of the time scalar, in the 4-vector, it all works equally fine.
Quaternion vectors are all imaginary in the sense of i^2=j^2=k^2=ijk=-1.
So in that sense you don't have a problem with imaginary space, though you seem to have a problem with imaginary time. This doesn't seem to be very consistent.
J. Willard Gibbs Vectors have I^2=J^2=K^2=+1 not -1. They are not assoicaitive because of the +1. I(IJ)= -J but (II)J= +J.
What is the operation II, or IJ?
Is it the dot product, the vector product, or the dydic product?
The signs matter.
Quaternions are assoiciative, i(ij)=-j and (ii)j=-j.
Why does the coordinatization of space time have to be associative? This seems to be a big issue for you, but I have never felt the need to multiply Alice's location by Bob's location, by Charlie's location.
Last edited by Fortis; 2008-May-05 at 08:18 PM. Reason: To fix tags and subscripts

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Originally Posted by yawyaw
I think the invariance is quantitative.
As the quaternion invariant is made up of one component which is"real" and 3 which sit in front of a "i", a "j", and a "k" respectively, then each of these components must be independently invariant, i.e.

(c2t2-(x2+y2+z2)) is invariant

and

2.c.t.i.x, or t.x, is invariant

and

2.c.t.j.y, or t.y is invariant

and

2.c.t.k.z, or t.z is invariant.

All of these must be simultaneously invariant.

Do you agree with this so far?

30. Originally Posted by Fortis
As the quaternion invariant is made up of one component which is"real" and 3 which sit in front of a "i", a "j", and a "k" respectively, then each of these components must be independently invariant, i.e.

(c2t2-(x2+y2+z2)) is invariant

and

2.c.t.i.x, or t.x, is invariant

and

2.c.t.j.y, or t.y is invariant

and

2.c.t.k.z, or t.z is invariant.

All of these must be simultaneously invariant.

Do you agree with this so far?
No. What is the criteria with respect to vectors that are not in the Interval, they all seem to be zero.
Last edited by yawyaw; 2008-May-06 at 02:30 AM.

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