er tensors in Einstein's Field Equation because I am doubtful
that it matters, since any tangent space on which the metric tensor is
defined, will always contain an infinite set of vectors in the vector space
defining that tangent space, so the same problem arises; namely, which pair
of v
er tensors in Einstein's Field Equation because I am doubtful
that it matters, since any tangent space on which the metric tensor is
defined, will always contain an infinite set of vectors in the vector space
defining that tangent space, so the same problem arises; namely, which pair
of v
that it matters, since any tangent space on which the metric tensor is
defined, will always contain an infinite set of vectors in the vector space
defining that tangent space, so the same problem arises; namely, which pair
of vectors must one chose, to calculate the metric tensor? TY, AG
--
Yo
Wiki states it's a linear function of two variables whose domain is the
vector space of velocity vectors on the tangent plane at some point P, say,
on a spacetime manifold. But assuming we know the function, given by its
4x4 matrix representation defined on some coordinate system, which pair of
2:13 AM (1 minute ago)
>
>
>
> to Everything List
> I've posed this question to several physicists, but so far none has been
> able to solve it. Namely, since the metric tensor is defined on *vector
> spaces on tangent planes on the spacetime* *manifold*, and it'
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Alan Grayson
2:13 AM (1 minute ago)
to Everything List
I've posed this question to several physicists, but so far none has been
able to solve it. Namely, since the metric tensor is defined on *vector
spaces on tangent planes on the
I've posed this question to several physicists, but so far none has been
able to solve it. Namely, if the metric tensor is defined on tangent spaces
on the spacetime manifold, and it's a bilinear map to the real numbers, how
do we know which pair of vectors to use to perform the c
Wiki says the METRIC tensor is a bilinear function which maps to the real
numbers. And the METRIC tensor FIELD is a scalar field, thus a real number
at every point of spacetime. AG
On Saturday, August 17, 2024 at 6:55:37 PM UTC-6 Brent Meeker wrote:
You can't "resolve" a tenso
You can't "resolve" a tensor into a number.
Brent
On 8/17/2024 3:59 PM, Alan Grayson wrote:
Since the metric tensor is defined on the vector space on the tangent
plane, it's defined on pairs (u, v), some of which can be tachyons.
But how do we choose which pairs, with or
Since the metric tensor is defined on the vector space on the tangent
plane, it's defined on pairs (u, v), some of which can be tachyons. But how
do we choose which pairs, with or without tachyons, to resolve the metric
tensor into some real number? AG
On Saturday, August 17, 2024 at
the Metric Tensor,
denoted as g (without its two subscripts), is a bilinear function of
vectors *u *and*v,* in the *_vector_ _space_* resident in the tangent
space, say at point P, of an underlying manifold, which maps to the
real numbers. There’s also the concept of the Metric Tensor *Field*
According to Wikipedia and other reliable sources, the Metric Tensor,
denoted as g (without its two subscripts), is a bilinear function of
vectors *u *and* v,* in the *vector space* resident in the tangent space,
say at point P, of an underlying manifold, which maps to the real numbers
According to Wikipedia and other reliable sources, the Metric Tensor,
denoted as g (without its two subscripts), is a bilinear function of
vectors *u *and* v,* in the *vector space* resident in the tangent space,
say at point P, of an underlying manifold, which maps to the real numbers
On Tuesday, May 17, 2022 at 11:42:37 AM UTC-6 meeke...@gmail.com wrote:
> Of course it depends on the energy distribution. That's what's on the
> right side of Einstein's equation and if it's given you can solve for the
> metric on the left side. The proper time between two time-like events
Of course it depends on the energy distribution. That's what's on the
right side of Einstein's equation and if it's given you can solve for
the metric on the left side. The proper time between two time-like
events is the metric measure along a geodesic between them, which is
force-free path.
I'm getting it. Defined abstractly, without reference to any particular
spacetime, the inner products from which the metric tensor derives, must be
abstract. It's only by applying the EFE's that we can SOLVE for the
specific metric tensor associated with a particular spacetime.
Looks like I'm spinning my wheels and getting nowhere, since the Metric
Tensor has nothing to do any particular spacetime one is assuming and its
energy distribution, when writing and trying to solve EFE's, since the
inner products on the basis vectors in V, are completely arbitr
It appears in Einstein's Field Equations and presumably allows us to
calculate the Metric of spacetime, which is a bilinear map from a pair of
spacetime coordinates to a real number. If the foregoing is correct, what
is the definition of distance between any pair of spacetime points, and
does i
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