Bondi ##k##-factor concept from first principles

[robphy]

I tried to reformat the \LaTeX in the earlier post.
If one right-clicks on an equation (or, or say, this \LaTeX ), a pop-up option shows the typesetting commands in \LaTeX or MathML. In \LaTeX, use square-brackets with opening and closing tags: [ itex ] and [ /itex ] .

 

[robphy]

"A(OQ→)=(U→×V→)=−2(UtVx−UxVt)=−2((Qt+Qx2)(−Qt−Qx2)−(Qt+Qx2)(Qt−Qx2))=Qt2−Qx2=ΔsQ→2."
The second line was clipped in the post. I see it now after pasting it. There is still a problem, however. Trying to refer to the paper in the link, and it says I don't have access unless I buy it.

Here's a screenshot from that closing section.

 

[tom.capizzi]

I've been thinking about this information all week, and I believe there is a misunderstanding. When I used the transform pair, it was in the context of invariance of the magnitude in transforming from a vector to a bivector hyperbolic coordinate system. In point of fact, the standard Lorentz matrix has a pair of real eigenvalues and a pair of real eigenvectors. You recognized the transform as being the transform to light-cone coordinates, and assumed that's why I use them. The elaborate derivation above is, of course, mathematically correct. It's just that, even though it is the same transform, in Minkowski spacetime, the results are complex. And while the math is true, it is not the math of real eigenvectors. The two real eigenvectors are perpendicular, and define a flat plane. So, it would appear that to work with the eigenvectors, I must use Euclidean units, since the symmetry vectors are composites of both of the original axes. For the eigenvectors to be real, both contributions have to be real. So, I might have a problem with general relativity, but I'm only trying to model special relativity. The advantage is that the dot product of the eigenvectors is 0, and the cross-product becomes s², independent of rapidity. We don't need the imaginary signature to get the difference of the squares. It's an inherent property of the product of eigenvectors, and the associated hyperbolic trig. The x dot x term is not -1, it is +1. So, u dot u is 1 and v dot v is 1 and u dot v = 0. This is only the case for real eigenvectors, not light-cone vectors. At least, do you agree these are different cases? And have I overlooked anything?

 

[vanhees71]

Lightlike vectors are real vectors. If you consider only (1+1)-dimensional spacetime (the "Minkowski plane") then the two lightlike eigenvectors of a proper orthochronous Lorentz transformation form a basis but of course not a Minkowski-orthonormal basis (see the discussion above that two light-like Minkowski orthogonal vectors are in fact parallel to each other, so there cannot be a Minkowski-orthogonal basis of light-like vectors). Indeed the two eigenvectors are not Minkowski-orthonormal:
$$(1,1) \cdot (1,-1)=1 \times 1 - 1 \times (-1)=2.$$

 

[tom.capizzi]

But my point was that the light-cone axes are not the same as the eigenvector axes. For one thing, the surface of the light-cone encloses a volume, so that any arbitrary point not on the surface requires 3 coordinates. The eigenvector plane has no depth. I understand your argument that the light-cone axes are not Minkowski-orthonormal. The question is, are they Euclidean-orthonormal? Because, if they are, then, is it even necessary to use Minkowski spacetime? As I heard it, the great improvement of Minkowski was to insert a factor of "i" into the 4-coordinates, producing the formula for the invariant. With Euclidean eigenvectors, the minus sign that defines the invariant is a feature of the real eigenvectors, themselves. Perhaps Minkowski coordinates can describe special relativity, but aren't required to accomplish that, if flat, real eigenvectors are used? Implying that there is something fundamentally different about special relativity compared to general relativity, despite seeming to be a limiting case. After all, the short version of general relativity is matter tells space how to curve and space tells matter how to move. More formally, general relativity defines the relationship between the 2nd derivative with respect to time and the second derivative with respect to space. In this context, special relativity defines the relationship between the first derivative with respect to time and the first derivative with respect to space, v/c = tanh(w) = sinh(w)/cosh(w) = Δr/Δct, in an inertial frame. So, in the absence of gravity, only special relativity effects are evident, but mathematically 1st and 2nd order derivatives are distinct entities. Worst case scenario, as I see it, is that the eigenvector plane is an isomorphism of Minkowski spacetime. It happens to linearize the relationships that are distorted in Minkowski coordinates, but it does not fit general relativity. But if the goal is to create a more understandable model of special relativity, and it happens to be Euclidean, wouldn't an isomorphism be acceptable? Is there a contradiction I've overlooked?

 

[robphy]

A metric (or dot-product) where the eigenvector axes (along the future-forward and future-backward lightcone) are orthogonal \vec u \cdot \vec v=0 can't capture the causal character of the vectors with a Minkowski metric.

\vec u +\vec v is a timelike vector in Minkowski spacetime, and
\vec u -\vec v is a spacelike vector in Minkowski spacetime. (Think radar measurements.)

But
(\vec u+\vec v)\cdot(\vec u+\vec v)=(\vec u \cdot \vec u)+(\vec v \cdot \vec v) +2 (\vec u \cdot \vec v)=<br /> (\vec u \cdot \vec u)+(\vec v \cdot \vec v)<br />
and
(\vec u-\vec v)\cdot(\vec u-\vec v)=(\vec u \cdot \vec u)+(\vec v \cdot \vec v) -2 (\vec u \cdot \vec v)=<br /> (\vec u \cdot \vec u)+(\vec v \cdot \vec v)<br />
means that this metric (dot-product) doesn't distinguish timelike from spacelike.

From my "rotated graph paper" approach, the Minkowski metric for (1+1)-Minkowski spacetime seems to be captured by the signed area of the diamond for the vector along the diagonal of that diamond. This area (on this plane) does not rely on the dot-product of either Minkowski spacetime or Euclidean space.UPDATE:
One could also consider the eigenvectors of the Galilean transformation in (1+1)-Galilean spacetime…. the PHY 101 position-v-time graph.

Again, there are no timelike eigenvectors (in accord with the Relativity Principle).
There is an eigenvector which Galilean-null and Galilean-spacelike [absolute time] (using the degenerate temporal-metric \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 0 \end{array} \right) )
and its eigenvalue is 1 ["absolute length"].

As in the Minkowski case, I don’t think Euclidean geometry can capture the Galilean structure.

Spacetime geometry is a non-Euclidean geometry.

 

[tom.capizzi]

I'm trying to follow, but I'm unclear what timelike eigenvectors have to do with the Relativity Principle. Could you elaborate?

 

[robphy]

Inertial observers have future timelike 4-velocities. A timelike eigenvector would imply that one of those 4-velocities was special, violating the Relativity Principle. In special relativity, the lightlike vectors are eigenvectors, corresponding to an invariant speed, the speed of light.

 

[tom.capizzi]

I see your point, but I wonder, isn't that the whole point of Minkowski spacetime, that time is different from space? In any case, the discussion is focused on those light-like vectors. In the context of this being an isomorphism, the dot product is not the only way to determine causal character. The formula s² = c²t²-r² still distinguishes time-like and space-like from light-like. It equals (ct+r)(ct-r) = s/e^w u x s*e^w v = s² u x v = s². In this isomorphism, it is the cross-product which determines causal character, as well as the directed area of the invariant bivector, s² u x v. So, is the issue that there is no distinction using the dot product, or that the distinction is determined by the cross-product, instead?

 

[robphy]

I see your point, but I wonder, isn't that the whole point of Minkowski spacetime, that time is different from space? In any case, the discussion is focused on those light-like vectors. In the context of this being an isomorphism, the dot product is not the only way to determine causal character. The formula s² = c²t²-r² still distinguishes time-like and space-like from light-like. It equals (ct+r)(ct-r) = s/e^w u x s*e^w v = s² u x v = s². In this isomorphism, it is the cross-product which determines causal character, as well as the directed area of the invariant bivector, s² u x v. So, is the issue that there is no distinction using the dot product, or that the distinction is determined by the cross-product, instead?

The point of Minkowski spacetime is to explain and predict the results of experiments… not to force a structure that appeals to us.

My rotated graph paper exploits a property of areas in (1+1)-Minkowski spacetime and is in tune with radar methods and the Bondi k-calculus. In (2+1) and higher, the role of a metric now becomes important.

(In a 1-dim vector space one can compare vectors without a metric… but in 2dim, one needs a metric. Areas on a plane are like a 1dim vector space… to describes areas in space would require a metric.)

For a higher dimensions of Minkowski spacetime, I don’t have a formulation… so I don’t have anything to declare about the role of the cross-product in general.

I’m not sure where this conversation started from or where it’s going. But I think I have made my point about being aware of the eigenvectors of the boost transformation in the k-calculus, the meaning of no timelike eigenvectors as a statement of the relativity principle, and that the Euclidean metric is not sufficient to handle spacetime geometry. I make no other claims other than my geometrical construction with the diamonds and methods of counting calculation seem to capture many features of (1+1)-Minkowski spacetime using the Minkowski metric.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

An off topic subthread has been deleted. Thread reopened.