- #561
Mentz114
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Bon voyage, Doug.
Lut
Lut
The geometric product makes sense if you fear the unknown Lorentz invariant interval,
sweetser said:Hello:
Quaternions as events in spacetime physics are not the Clifford algebra CL(0, 2). Let me first explain what that bit of jargon means.
Clifford algebras were an attempt to generalize the math started by Hamilton, passing through Grassman. These algebras are independent of coordinates, and can be written in arbitrary dimensions. As algebras, it is not important that they are invertible, although that can happen.
The Clifford algebra CL(0, 2) is a multivector, meaning it has two parts. First there is the scalar, the 0 part, a pure number. The bivector is the second part. They use a wedge product, [itex]\wedge[/itex], a generalized cross product to form a 3-term bivector.
[tex]q(t, x_1, x_2, x_3) = t ~+~ x_1 e_2 \wedge e_3 ~+~ x_2 e_3 \wedge e_1 ~+~ x_3 e_1 \wedge e_2 \quad eq~1[/tex]
Why use a wedge product? The reason is that wedge products are axial vectors, those that switch handedness in a mirror. The other sort is a polar vector. If one looks at an event in a mirror, this makes physical sense.
It is not worth the time to argue with math types who think they can define whatever they want, however they see fit. I have the perspective of a mathematical physicists, where any math definition can be cow-roped by physical meaning (I am a crude mathematical physics who uses rodeo analogies).
The unspoken assumption behind CL(0, 2) is that one should use a mirror on this event. That is like taking (t, x1, x2, x3) to (t, -x1, -x2, -x3). That sort of transformation does have the handedness needed for an axial vector. Yet it is easy enough to think of counter examples. What about a time reflection, where an event (t, x1, x2, x3) goes to (-t, x1, x2, x3)? This transformation does not change handedness, it is represented by a polar vector. Compare the two functions
mirror reflection: q -> q' = q*
time reflection: q -> q' = -q*
Since these are so close to each other as functions, I don't think one should take precedence over the other, as asserting one must use a bivector for the spatial part of a quaternion does.
The fundamental currency of the Universe is a bare event which stands alone in a vacuum, not in front of a mirror waving a left hand. General relativity has a message about basis vectors. Should we have a toy Universe with two events, the interval between these two events is found by taking the dot product of the difference between the two events:
[tex](c dt, dx_1, dx_2, dx_3).(c dt, dx_1, dx_2, dx_3) = c^2 dt^2 ~+~ dx_1^2 (e_2 \wedge e_3).(e_2 \wedge e_3) ~+~ dx_2^2 (e_3 \wedge e_1).(e_3 \wedge e_1) ~+~ dx_3^2 (e_1 \wedge e_2).(e_1 \wedge e_2)[/tex]
[tex]= c^2 dt^2 ~-~ dx_2^2 ~-~ dx_2^2 ~-~ dx_2^2 \quad eq~2[/tex]
So far, so good. Now repeat the measurement in a toy Universe where nothing has been altered for the generators of the two events, but one has added a mass. A metric theory of gravity does not alter events, but does change the measure of events. In this setup, the differences are the same. The only thing that could change are the sizes of the basis vectors themselves. The problem here is that there is no basis vector associated with the dt2 term.
The deep message of special relativity is to treat time as we do space. In the geometric algebra approach, time is a scalar, whereas the space parts use two basis vectors as a bivector. Contrast this with the way I define a quaternion:
[tex]q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3[/tex]
This is not a Clifford algebra because the basis vector e_0 commutes with the others. In the definition of a Clifford algebra, all the basis elements anti-commute. Physicists would call this a 4-vector, because one can add them or multiply it by a scalar, and it transforms like a 4-vector. It is this last phrase, on how it transforms under a Lorentz boost, that holds the magic. The goal of a great definition is to remove the magic and let the math speak for itself. Square the difference between two events using the quaternion definition as written in equation 3:
[tex](c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)
\quad eq~4[/tex]
To be consistent with special relativity, we make the following map:
[tex]e_0^2 = +1[/tex]
[tex]e_1^2 = -1[/tex]
[tex]e_2^2 = -1[/tex]
[tex]e_3^2 = -1 \quad eq~5-8[/tex]
To be consistent with general relativity to first order tests of weak gravity fields for non-rotating, spherically symmetric sources, all we need is:
[tex]e_0 = exp(-G M/c^2 R)[/tex]
[tex]1/e_0 = e_1 = e_2 = e_3 \quad eq~9-10[/tex]
so according to the GEM proposal,
[tex]e_0^2 = exp(-2 G M/c^2 R)[/tex]
[tex]e_1^2 = -exp(2 G M/c^2 R)[/tex]
[tex]e_2^2 = -exp(2 G M/c^2 R)[/tex]
[tex]e_3^2 = -exp(2 G M/c^2 R) \quad eq~11-15[/tex]
I forget who said it, but one can only hope to find an important law in physics if one also finds an invariance principle. When there is a gravitational source, the terms in the 3-vector part of the square (2 dt dx e_0 e_1, 2 dt dy e_0 e_2, 2 dt dz e_0 e_3) remain invariant under the presence of such a gravitational field.
In summary, I see two flaws to the assertion that quaternions are the Clifford algebra CL(0, 2). First, while some quaternions are axial vectors, others are polar vectors. Second, the scalar needs to have a basis vector associated with it to work with metric theories of gravity. As far as I can tell, everyone working with geometric algebra in physics works with the assumption that the quaternions are faithfully represented by CL(0, 2).
Doug
sweetser said:Hello Ghost_of_PL:
Great question! I had not looked at all the details, and your post got me to think about them.
There is hope that the algebra represented in equations 11-15 will be a division algebra since as M approaches zero, the algebra approaches the standard quaternion division algebra of equations 5-8.
To solve this problem, I fired up Mathematica and defined the quaternion using a 4x4 real matrix. This definition is identical to the standard one except for factors of Exp(G M/c2 (x2 + y2 + z2)1/2 (a minus factor on the t term).
In the limit of M->0, the inverse matrix is exactly the same as a Hamilton's quaternions. Those are defined for all possible values of t, x, y, and z because the divisor is the norm, t + x2 + y2 + z2, which is positive definite. The inverse is only undefined if t=x=y=0.
In the general case, the inverse looks like so:
[tex]\frac{1}{q(t,x,y,z)} = (t exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -x ~ exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -y ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -z ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm) \quad eq ~1[/tex]
[tex]where: ~norm = t^2 ~+~ (x^2 ~+~ y^2 ~+~ z^2) exp(\frac{4 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}) \quad eq ~1[/tex]
Again the norm is positive definite. With Hamilton's quaternions, the point set that must be excluded is zero for all four numbers. With the curved by mass quaternions, one must also exclude where x=y=z=0, which is the entire real line. This calls for a refinement of the definition of the algebra in equation 11-15, that should R=0, then the algebra becomes the standard Hamilton case, equations 5-8.
To answer your question specifically, the inverse in spacetime curved by mass of (1, 1, 0, 0) is:
[tex]\frac{1}{(1, 1, 0, 0)} = (\frac{e^{\frac{G M}{c^2}}}{1 ~+~ e^{\frac{4 G M}{c^2}}},-\frac{e^{\frac{3 G M}{c^2}}}{1~+~ e^{\frac{4 G M}{c^2}}}, 0, 0) \quad eq ~ 2[/tex]
Let's do a sanity check at least for the first term. If we hope to get a 1 out of multiplying (1, 1, 0, 0) by its inverse, we have to get out a [itex]1~+~ e^{\frac{4 G M}{c^2}}[/itex], which shows up in the denominator of the inverse. For the time terms, there will be a positive and a negative exponent that cancel each other out, yielding a 1. From the x terms, we get a positive exponent plus an exponent times 3 which combine for the exponent to the fourth power. Looks good. If anyone is interested, the notebook is here.
By refining the definition to cover the case where R=0, I hope I have shown the curved basis quaternions are a division algebra isomorphic to the flat spacetime quaternions defined by Hamilton.
Thanks for the question,
Doug
sweetser said:Hello Ghost_of_PL:
Good old fashioned algebra rocks! The only way I really learn something is with paper and pencil (so I can erase a lot). I did not understand what your issue was, but now I have a better grasp of it. I did define what I meant by a quaternion in this expression:
[tex]q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3[/tex]
Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.
The reason I needed Mathematica was to answer a different problem because I misunderstood your question, namely finding the inverse of a quaternion with the basis vectors in equations 5-8. That was a fun calculation anyway :-)
Doug
sweetser said:Hello Ghost_of_PL:
There are only 4 slots in a quaternion, something that is obvious from the 4x4 real matrix representation used in the Mathematica notebook. Looking back, I not only used a variation on a common definition (which would normally have i, j, k for e_1, e_2, e_3, no e_0):
[tex]q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3[/tex]
sweetser said:Sometimes people have complained about the meaning of the plus sign since this is not like the sum of real numbers. I also wrote out explicitly what is meant by squaring a quaternion:
[tex](c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)\quad eq~4[/tex]
The first term of eq.4 is a pure number without a factor of e_0 and violates "that every term has a basis vector, no exceptions" since c^2dt^2e_0^2 +dx1^2e_1^2 +dx2^2e_2^2 +dx3^2 = c^2dt^2- dx1^2- dx2^2 -dx3^2 . The right hand side of this last equation has no factor of e_0. Your matrix representation can fare no better since it also admits stand alone real numbers.sweetser said:Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.