Construction of the tangent space in GR

[Alan Grayson]

One way to do it, say at point P on the spacetime manifold, is to imagine test particles of all velocities less than c, passing through point P. Then, presumably, these vectors define the tangent space at P. But for these vectors to form a vector space, on which the metric tensor is defined, we have to include vectors with velocities greater than c. So, the original set of velocity vectors are NOT closed under usual addition. We must expand the original set of velocity vectors moving at velocities less than c, to presumably close that set in order to construct the vector space, without which the metric tensor is undefined. 

What bothers me about this procedure is that applying it to the closed interval, say [0.1], and asking whether it forms a vector space under addition (with the real numbers forming the scalar field), it's claimed this interval is NOT a vector space since it isn't CLOSED under addition. Well, if we can add elements to the tangent vectors with velocities less than c, with those having velocities greater than c, doesn't the original set of vectors also fail the closure test, and therefore fails to make the total set a vector space, necessary for defining the metric tensor? 

TY, AG

[Alan Grayson]

A related question is this; if the manifold is spacetime, and at some point P we consider all test particles moving at velocities less than c, containing point P, is the set of these tangent vectors sufficient to define the tangent plane at P? IOW, can we omit all test particles moving at velocities greater than c moving through point P, and still have a well defined tangent plane at point P?  If so, the problem I am struggling with is whether omitting such test particles on the GR manifold, as we must, will the result be a tangent plane whose tangent vectors at P do NOT form a vector space, since it isn't closed, and therefore have a problem with defining the metric tensor at P which is defined on a vector space. TIA, AG

[Alan Grayson]

If calculating the metric tensor is ill-defined, it seem impossible to use Einstein's Field Equations to get any useful prediction, but yet that's the claim.  Baffling, to say the least. AG

[smitra]

You only need 4 linearly independent vectors; orthogonality is not
needed. Once you have this linear space you can construct an orthogonal
basis, but you can't directly probe spacelike vectors.

Saibal

On 24-08-2024 04:32, Alan Grayson wrote:
> If calculating the metric tensor is ill-defined, it seem impossible to
> use Einstein's Field Equations to get any useful prediction, but yet
> that's the claim. Baffling, to say the least. AG
>
> On Thursday, August 22, 2024 at 2:32:51 AM UTC-6 Alan Grayson wrote:
>
>> A related question is this; if the manifold is spacetime, and at
>> some point P we consider all test particles moving at velocities

>> LESS THAN C, containing point P, is the set of these tangent vectors
>> sufficient to define the tangent plane at P? IOW, can we OMIT all
>> test particles moving at velocities GREATER THAN C moving through

>> point P, and still have a well defined tangent plane at point P? If
>> so, the problem I am struggling with is whether omitting such test
>> particles on the GR manifold, as we must, will the result be a
>> tangent plane whose tangent vectors at P do NOT form a vector space,
>> since it isn't closed, and therefore have a problem with defining
>> the metric tensor at P which is defined on a vector space. TIA, AG
>>
>> On Tuesday, August 20, 2024 at 2:26:16 AM UTC-6 Alan Grayson
>> wrote:
>>
>>> One way to do it, say at point P on the spacetime manifold, is to

>>> imagine test particles of all velocities LESS THAN C, passing

>>> through point P. Then, presumably, these vectors define the
>>> tangent space at P. But for these vectors to form a vector space,

>>> ON WHICH THE METRIC TENSOR IS DEFINED, we have to include vectors
>>> with velocities GREATER THAN C. So, the original set of velocity

>>> vectors are NOT closed under usual addition. We must expand the
>>> original set of velocity vectors moving at velocities less than c,
>>> to presumably close that set in order to construct the vector
>>> space, without which the metric tensor is undefined.
>>>
>>> What bothers me about this procedure is that applying it to the
>>> closed interval, say [0.1], and asking whether it forms a vector
>>> space under addition (with the real numbers forming the scalar
>>> field), it's claimed this interval is NOT a vector space since it
>>> isn't CLOSED under addition. Well, if we can add elements to the
>>> tangent vectors with velocities less than c, with those having
>>> velocities greater than c, doesn't the original set of vectors

>>> ALSO fail the closure test, and therefore fails to make the total
>>> set a VECTOR SPACE, necessary for defining the metric tensor?
>>>
>>> TY, AG
>
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[Alan Grayson]

TY, but if you use time-like velocity vectors on spacetime as the manifold to construct a tangent plane, that set won't form a vector space (on the tangent plane) on which the metric tensor is allegedly defined, because a vector space is defined using ordinary addition of vectors, and the only way to preserve time-likeness is to use the Lorentz transformation, but a frame transformation is not the same as ordinary addition. AG

[Brent Meeker]

The difference of two time-like vectors can be space-like.  That's why the CMB at opposite points on the celestial globe have not interacted.

Brent

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[Alan Grayson]

The vectors you refer to; are they the velocity vectors on a tangent space of a manifold? 

I'm confused about the definition of a tangent space on a manifold, say at point P. If the manifold is spacetime and we must exclude all test particles moving at velocities > c. Does the set of tangent vectors at P form a vector space? I don't see how it can, since under ordinary addition parallel vectors can sum to velocities > c, negating the closure property for a vector space. If they don't form a vector space, this compromises the definition of the metric tensor which is defined on a vector space on the tangent space at each point P. TY, AG

[Brent Meeker]

On 8/24/2024 8:54 PM, Alan Grayson wrote:

The vectors you refer to; are they the velocity vectors on a tangent space of a manifold? 

I'm confused about the definition of a tangent space on a manifold, say at point P. If the manifold is spacetime and we must exclude all test particles moving at velocities > c.

You don't exclude vectors.  You just realize that some vectors don't represent velocities.

Brent

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[Alan Grayson]

So if one wants to define the tangent plane on a manifold, and that manifold is spacetime, one includes vectors corresponding to v > c even though they're physically disallowed? This seems wrong since the tangent space would be defined by violating a key postulate of relativity. What about the "vectors" on the light cone you referred to earlier? What exactly are they? They seem different from those used to define the tangent plane. TY, AG

[Brent Meeker]

On 8/25/2024 3:17 AM, Alan Grayson wrote:

So if one wants to define the tangent plane on a manifold, and that manifold is spacetime, one includes vectors corresponding to v > c even though they're physically disallowed? This seems wrong since the tangent space would be defined by violating a key postulate of relativity.

Vectors are mathematics, like coordinate systems.  Relativity is about physics.  You use numbers to represent positions; that doesn't mean number are positions.

Brent

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[Alan Grayson]

FWIW, this sounds like a totally ridiculous way of defining a tangent plane on a manifold when the manifold is spacetime. On the manifold you have velocity vectors moving at v < c, and other vectors representing nothing which can be conceptualized, and the combination being a vector space. Not in the slightest rigorous IMO, but as we know my opinion doesn't matter.

Let's go on to the more difficult problem; namely, how do we calculate the metric tensor when it's a bilinear function of two vectors on the manifold, and there exist an uncountable set of pairs of vectors which can be used as arguments of the metric tensor?

AG

[Jesse Mazer]

"On the manifold you have velocity vectors moving at v < c, and other vectors representing nothing which can be conceptualized"

I already mentioned how spacelike intervals/vectors can be conceptualized as ruler distances, do you have any reason for ruling out this interpretation?

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[Jesse Mazer]

We can measure spacelike intervals in special relativity--this is the "proper distance" which just corresponds to the distance between two events measured by a rigid ruler that's at rest in the inertial frame where the events are simultaneous. So like I said earlier, I'd think spacelike vectors in the tangent space could correspond physically to the same notion, where we're talking about short free-falling rulers at rest in the local inertial frame where the two ends of the vector are simultaneous (see https://www.einstein-online.info/en/spotlight/equivalence_principle/ on notion of local inertial frames in curved spacetime which approach the physics of special relativity in the limit as size of local region goes to zero...someone who knows more about the math of GR can correct me if I'm wrong, but maybe one could think of the tangent plane conceptually as what you'd get if you took this infinitesimal local inertial frame and imagined blowing it up to a finite region of spacetime, preserving the local proportionality between coordinate increments and proper times/proper distances, also preserving angles between 4-vectors which are used in the dot products).

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[Jesse Mazer]

The tangent space in relativity is not limited to timelike vectors (those connecting points on worldlines of particles moving slower than than c), it also includes null vectors (vectors between points on the worldline of a particle moving at c) and spacelike vectors (which have no physical interpretation in terms of separation between points on any worldline, although there are other physical interpretations like pairs of events that are simultaneous in some local inertial frame). 

I mentioned before that the metric is used to construct the line element in a particular spacetime coordinate system, you integrate the line element along a path through spacetime to get proper time along an extended timelike worldline if you know its coordinate representation. One can equally well integrate the line element to get proper distance along *spacelike* paths through the spacetime, these don't represent wordlines of physical particles but they can be thought of as measurements along a chain of short rulers whose ends coincide at the moment of measurement, where each ruler's length is measured in its local inertial rest frame (see https://en.wikipedia.org/wiki/Comoving_and_proper_distances#Uses_of_the_proper_distance ) and its surface of simultaneity is tangent to the spacelike path at the moment of measurement. For this use of the metric to measure proper distance along spacelike paths, you need those spacelike vectors in the tangent space.

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[Jesse Mazer]

If you have a global coordinate system, can't a freefalling observer in a small neighborhood of spacetime check how a small increment of a spacelike coordinate relates to proper distance along a free-falling ruler, one which is arranged along that coordinate axis? Is that different from what you mean by probing spacelike vectors?

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[Alan Grayson]

To Jesse Mazer: TY for those comments. They helped me solve, to some extent, my issue with tangent planes in GR. I think I had the wrong perspective. Although at some point P on the spacetime manifold, there exists an uncountable set of vectors which form a vector space if the non-physical vectors are included, on which the metric tensor is defined, we're NOT interested in calculating the metric tensor between any two of THOSE vectors, but of vectors along a geodesic to calculate the proper time as measured by an observer moving along it. And since the path of the geodesic is presumably known and likely to be curved, we need to use calculus and partition the increments along the geodesic, so when going to the limit of an infinite number of increments, we can calculate the proper elapsed time, from the start point to the end of the geodesic. Hence, the two points on which the metric tensor is defined, is a series of pairs of points along the geodesic, which summed via calculus, yields the total elapsed proper time from the beginning to the end of the path. AG

[Alan Grayson]

As for spacelike vectors, they're necessary for the tangent vectors at the point of contact of the tangent plane with the manifold to form a vector space. But why is it necessary to define the metric tensor on a vector space? IOW, why can't the unphysical vectors be dispensed with, and still have a viable metric tensor? AG

[Alan Grayson]

There's an evident error this analysis, although I think it's on the right track; namely, no mention of the Riemann Curvature Tensor, without which, presumably, we can't define a geodesic path. (Does Lawrence Crowell post here any more?).  AG

On Tuesday, August 27, 2024 at 10:46:44 PM UTC-6 Alan Grayson wrote:

[smitra]

Yes, but you can only see the effects of actions at different positions
when they move into each other's lightcones. But this is not a problem.

Saibal

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