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Physical model of measurement for affine geometry, dual 
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#1
Feb913, 08:38 PM

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We recently had a long thread http://physicsforums.com/showthread.php?t=666861 about cases where raising and lowering indices isn't completely natural, i.e., where a vector "naturally" wants to be upperindex or lowerindex.
If you have a metric, then it's pretty clear to me what realworld measurement processes correspond to the use of the metric. In a Riemannian space, they're basically the ones referred to in Euclid: compass and straightedge or, in some weaker sense, a marked straightedge. In a semiRiemannian space, you need the universe to contain at least one clock. (A ruler will also work.) Geroch has a very nice popularlevel book, Relativity from A to B, that systematically works out this interpretation of GR. What is the right model of measurement for a space where you don't have a metric but you do have parallelism, e.g., the tangent space on a manifold where no metric exists? In affine geometry, you have a measure of area and you have a measure of length along a line and lines parallel to it. You don't have angular measure, a notion of perpendicularity, congruence, or comparability of vectors that are not parallel. You can take products of vectors with dual vectors, giving real numbers. The best metaphor I've been able to come up with is the ghost of a clock. What I mean by this is that you have a clock that cruises like a ghost through space, moving inertially. You can't interact with it, so you can't, e.g., force it to travel along a curved worldline and measure its proper time. It can move at c or at speeds higher than c. Unlike a physically manipulable clock, it doesn't suffice to establish a frame of reference. You can make copies of it moving along parallel worldlines, but you can't synchronize the copies (because synchronization is perpendicularity). Basically if you take the product of a dual vector with a vector, [itex]\omega\cdot v[/itex], you can think of it as comparing the rate of one ghostclock against the rate of another ghostclock. Is this a correct model? Is there a simpler, more accurate, or more natural model, or one stated in terms of more realistic measuring devices? 


#2
Feb913, 08:57 PM

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How about the geometry of null geodesics? That's what I usually think of. But there isn't any real notion of "length" , until you specify a frame of reference. So I'm not sure if it's what you are actually looking for.



#3
Feb913, 10:13 PM

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#4
Feb1013, 02:45 AM

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Physical model of measurement for affine geometry, dual
I wouldn't get rid of the metric tensor to define a null geodesic. But if you have an affinely parameterized curve which is a null geodesic, the length of any segment of the curve will be equal to zero. What I should perhaps have said to be more precise, is that the length (found via the metric tensor) is always zero.
If we used this notion of length, the one from the metric tensor, we'd say that all segments of a null curve "had the same length", because zero = zero. Which is trivally true, and more or less useless. However, some parameterizations of the null geodesic are affine parameterizations and some are not affine. If we have an affinely parameterized geodesic, we can mark off "equal intervals" along the geodesic according to equal spacings of the affine parameter. I.e. if the affine parameter is [itex]\lambda[/itex], the interval between [itex]\lambda=0[/itex] and [itex]\lambda=1[/itex] is the same as the interval between [itex]\lambda=1[/itex] and [itex]\lambda=2[/itex]. This notion of 'equal length' is more restrictive enough than the trivial notion of having the same length according to the metric tensor, which is always satisfied and not useful because it's trivial. Thus we can use Schild's ladder, for instance, to do parallel transport along null geodesics, because we have this notion of "equal length". So my interpretation of the underlying geometry (the geometry we use to good effect in Schild's ladder, for instance) is that there is a frame independent way to compare lengths (by which I mean the purely spatial part of the distance between two points along a null geodesic), but that the value of said lengths are totally dependent on one's choice of frame and uncertain up to a multiplicative factor. 


#5
Feb1013, 07:01 AM

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So if objects are constrained to only rotate in the horizontal direction, then there is no good way to compare distances in the vertical direction with distances in the horizontal direction. The most natural geometry for this situation would have separate metrics for the horizontal and vertical directions, and there would be no nondegenerate 3D metric. 


#6
Feb1113, 02:35 PM

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Let me expand on my original answer in a way that I hope is a bit clearer.
(I hope Schild's ladde is familiar? One can find a reference to it in MTW if it's not. It's not really a terribly popular construct, but I personally like it a lot. I have a forum post on it  somewhere or other.) You have a metric  but it's trivial. The length of any null vector is zero, the dot product of any two null vectors is zero. You don't have a Lorentz invariant definition of angle. You can somewhat sensibly talk about the "angle" between two light beams (You basically project the null vectors into nonnull vectors in some particular frame to get the angle) but the result depends on what frame you use. Or in other words if you construct "orthogonal" light beams in one frame (by making their spatial projectios orthogonal in that frame) they won't be orthogonal in some moving frame. So there isn't any true notion of angle in this space. Now to repeat, the question was: And the proposed answer is, in this space, the nontrivial notion of distance is given, up to a scale factor, not by the metric, but by the affine parameterization. This is all an "intuitive leap" sort of observation, its not something that I've made rigorous. We can physically interpret this affine parameteriztion. If you have an EM plane wave, you can mark regular intervals along it in some given frame by taking locations where the Efield is zero in that frame. All frames will agree that the points are "evenly spaced" (I'm not sure they'll all agree that the Efield is zero, but that isn't the claim, the only claim is that they'll agree the points are evenly spaced). And this is the basis of marking out "equal distances" in our affine geometry, rather than using the metric (which always gives zero). The EM wave is just to provide some physical insight by providing a physical example. Mathematically, all you need is the affine parameterization. 


#7
Feb1113, 04:03 PM

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Pervect, that's food for thought. Here's the Wiki article on Schild's Ladder, for those who ( like me ) need it it,
http://en.wikipedia.org/wiki/Schild%27s_ladder 


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