Is the flat METRIC still FLAT ?

[George Hammond]

If I multiply the ordinary flat space metric
by an arbitrary function of time: a(t) ,
will it remain a "flat space-time", IOW
will the curvature still be zero?

I t say YES but without plugging it into
Mathematica and calculating the
Ricci scalar curvature (R) and checking
that it is still zero – R=0., I'm not sure?

But someone must know the answer
off the top of their head without going to
all that trouble.

George Hammond MS physics

[Stan Fultoni]

On Tuesday, May 17, 2022 at 12:04:46 AM UTC-7, ghamm...@gmail.com wrote:
> If I multiply the ordinary flat space metric by an arbitrary function of time: a(t) ,
> will it remain a "flat space-time", IOW will the curvature still be zero?

No, not generally.

> I say YES but without plugging it into Mathematica and calculating the

> Ricci scalar curvature (R) and checking that it is still zero – R=0., I'm not sure?

The vanishing of the Ricci scalar, and even the Ricci tensor, does not imply the vanishing of the Riemann curvature tensor, which is the actual measure of curvature. This should be obvious, because the vacuum field equations are just the vanishing of the Ricci tensor, which certainly doesn't imply that there is no curvature... if it did, there could be no gravity.

[George Hammond]

[GE Hammond MS Physics]
Do you mean that if I plug the following metric
into Mathematica:

|a 0 0 0 |
|0 a 0 0 | = the given spacetime metric
|0 0 a 0 |
|0 0 0 -a|

where: a = a(t) (a is defined as a simple
well behaved function of time)

You say the Ricci-scalar could actually turn out to
be NON-ZERO ? And possibly Riemann also ?

Do you believe that ?

I certainly need to confirm that ?

I'll even PAY someone to check it on Mathematica-
whatever they ask (within reason) to find out
since I don't have access to Mathematica and
am not familiar with it. MAXIMA says it is
NON-zero but I don't trust Maxima !

It is of VITAL importance to me involving other
research ! Money is no object, if I can get
the CORRECT answer !

I'm absolutely desperate to know what
Mathematica says !

George Hammond (80 yrs old btw)

I can be contacted by email

[mitchr...@gmail.com]

If gravity is a curved metric how can it be flat?
but why would a parabolic metric create elliptical
orbits?

[Stan Fultoni]

On Tuesday, May 17, 2022 at 8:07:49 PM UTC-7, ghamm...@gmail.com wrote:

> > > If I multiply the ordinary flat space metric by an arbitrary function of time: a(t) ,
> > > will it remain a "flat space-time", IOW will the curvature still be zero?
> >
> > No, not generally.
> >
> > > I say YES but without plugging it into Mathematica and calculating the
> > > Ricci scalar curvature (R) and checking that it is still zero – R=0., I'm not sure?
> >
> > The vanishing of the Ricci scalar, and even the Ricci tensor, does not imply the vanishing of the Riemann curvature tensor, which is the actual measure of curvature. This should be obvious, because the vacuum field equations are just the vanishing of the Ricci tensor, which certainly doesn't imply that there is no curvature... if it did, there could be no gravity.
>

> Do you mean that if I plug the following metric into Mathematica:
> |a 0 0 0 |
> |0 a 0 0 | = the given spacetime metric
> |0 0 a 0 |
> |0 0 0 -a|

> where: a = a(t) (a is defined as a simple well behaved function of time) you

> say the Ricci-scalar could actually turn out to be NON-ZERO ? And possibly
> Riemann also ?

Your first question was whether the curvature was zero, and the answer is No, not generally. Applying a variable scale factor, such as a(t) to the metric line element generally results in a curved manifold.

Then you started talking about the Ricci scalar, and I pointed out that you're mixing up different things, because the vanishing of the Ricci scalar and even the Ricci tensor does not imply zero curvature. For example, the Ricci tensor and the Ricci scalar both vanish throughout Schwarzschild spacetime, but the curvature is definitely not zero.

> Do you believe that ?

That the curvature is generally non-zero? Of course. Yes, you can verify this for yourself, just be careful not to confuse curvature with the Ricci scalar.

> I'll even PAY someone to check it on Mathematica - whatever they ask

> (within reason) to find out since I don't have access to Mathematica and
> am not familiar with it. MAXIMA says it is NON-zero but I don't trust Maxima !

First I think you should try to decide what you are seeking. Are you trying to find out if the Riemann curvature is zero (in general, it isn't), or are you trying to find out if the Ricci scalar or Ricci tensor are zero? Those are very different things.

[Tom Roberts]

On 5/17/22 3:04 AM, George Hammond wrote:
> If I multiply the ordinary flat space metric by an arbitrary
> function of time: a(t) , will it remain a "flat space-time", IOW
> will the curvature still be zero?

[a(t) cannot be completely arbitrary, it must be both
finite and non-zero everywhere; otherwise the product
is not a valid metric. That implies a(t) > 0 everywhere,
and it is often notated λ^2.]

[Also: "space metric" => "spacetime metric"]

Not in general. This is an instance of "conformally flat" -- there
exists a function (here 1/a(t)) such that the metric tensor times the
function yields a flat metric. The function can be an arbitrary
positive-definite non-singular function on the manifold, it need not be
simply a function of t (presumably some time coordinate).

> I t say YES but without plugging it into Mathematica and calculating
> the Ricci scalar curvature (R) and checking that it is still zero –
> R=0., I'm not sure?

"Flat" means the Riemann curvature tensor is zero. The Ricci scalar can
be zero while Riemann is nonzero (= manifold is not flat). Indeed that
is the case throughout Schwarzschild spacetime -- the Ricci tensor and
scalar are both zero, but the Weyl tensor and the Riemann tensor are
nonzero.

[The Einstein field equation of GR directly implies that
in any vacuum region (i.e. T=0) the Ricci tensor is
zero, which implies the Ricci scalar is also zero. But
there can still be gravity, as Weyl and Riemann need not
be zero.]

Tom Roberts

[George Hammond]

May 18th 2022

Hi Tom Roberts –
I know very well who you are from sci.physics.relativity.

Perhaps as long as 7 years ago I talked to you there about
this metric:

|a 0 0 0 |
|0 a 0 0 | = the given spacetime metric
|0 0 a 0 |
|0 0 0 -a|

where: a = a(t) = "the scale factor" is a
simple well behaved function of time.
(Well behaved in the sense of the FRW
scale factor).

And I told you that plugging this metric into MAXIMA
I got the following result for the Ricci scalar:

6 a (a..) - 3 (a.)^2
_______________ = R = Ricci scalar curvature

2 a^3

(a..) = 2nd derivative of a w.r.t. tme
(a.) = 1st derivative of a w.r.t. time

And I asked you if you would plug the metric into
Mathematica and tell me if you got the same answer?

And you said that you did, and that "the answer was correct".

Notice that the answer MAXIMA gave, is very similar to
the well known Ricci scalar for the FRW metric !

And I assumed the difference was that I was not using
"conformer time".
That and also the fact that I used "a" for the scale factor
and not "(a^2)" !

At any rate, is there any chance that you could
enter that metric again in Mathematica and confirm
that the above result is actually true?

Some people have argued that the curvature is actually ZERO

It is a matter of considerable and urgent importance to
research (unrelated to gravity) in another academic field
which I will not mention (but related to psychology).


Thanks in advance, absolutely desperate,
Kurvature (George Hammond)

[JanPB]

Yes, it's correct. It's a bit simpler of you use a^2 instead of a.
The result is then:

R = 6a../a^3

Ricci components (still *using a^2 instead of a*). Let's denote:

A = (a.. a - (a.)^2)/a^4
B = (a.. a + (a.)^2)/a^4

then:

R_00 = -3A
R_11 = R_22 = R_33 = B
R_ij = 0 otherwise.

Riemann components:

R^0_101 = R^0_202 = R^0_303 = A
R^1_212 = R^1_313 = R^2_323 = (a.)^2/a^4
R^a_bcd = 0 otherwise.

--
Jan

[JanPB]

Forgot to add: Ricci and Riemann components are with
respect to the normalised frame:

a dt, a dx, a dy, a dz

(as opposed to the coordinate dt, dx, dy, dz).

--
Jan

[Ross A. Finlayson]

On Tuesday, May 17, 2022 at 12:04:46 AM UTC-7, ghamm...@gmail.com wrote:

It's contracting.

[George Hammond]

Hi JanPB -
Mathematica apparently uses (0,1,2,3) while
Maxima uses (1,2,3,4) , so I am guessing
that you're using Mathematica
while I'm using Maxima.

I have rerun the problem on Maxima using
"a^2" instead of "a". And I got exactly the
same result you got for R –

R = 6a../a^3

Also it returned the identical results you got
for the 4 nonzero components of the Ricci
tensor.

And by the way, thank you immensely for
checking the Riemannian components
You got 6 nonzero components while Maxima
returns 12 (6 of them identical to yours) – but
don't worry about that – the only thing I am
concerned with is this question –

IS RIEMANN NON-ZERO, YES OR NO ?

And apparently, according to both Mathematica
and Maxima –

RIEMANN IS DEFINATELY NON-ZERO !

So the space-time is definitely CURVED !

That's all I really need to know for the research
I am currently engaged in.

Once more, I can't thank you enough for taking the time
to run the metric on Mathematica – I am deeply indebted to
you for being so considerate, and very appreciative for your
responding to a frantic call for help.

And above all else, I am greatly relieved to discover
that the metric is definitely curved.

George Hammond

[George Hammond]

> It's contracting.

[George]
Yes – this is not a "gravitational" problem, it actually
comes from an unrelated academic field of research,
where yes, the space is actually contracting.

[JanPB]

On Friday, May 20, 2022 at 12:58:08 PM UTC-7, ghamm...@gmail.com wrote:
> Hi JanPB -
> Mathematica apparently uses (0,1,2,3) while
> Maxima uses (1,2,3,4) , so I am guessing
> that you're using Mathematica
> while I'm using Maxima.

No, I used paper and pen :-) using Cartan's calculus which
is much faster than Christoffel symbols.

The best pen to use is a Pilot pen which for some reason is
not sold in the US, you need to buy it in Asian imports stores
or on eBay (or abroad): it's called Pilot Ballpoint Hi-Tec-C (it's NOT
the gel one). For me the blue 0.4mm is the best, Riemann trensors
come out in no time:
https://www.ebay.com/itm/222274319068?hash=item33c094f2dc:g:kaUAAOSwmfhX6jCB

> I have rerun the problem on Maxima using
> "a^2" instead of "a". And I got exactly the
> same result you got for R –
>
> R = 6a../a^3
>
> Also it returned the identical results you got
> for the 4 nonzero components of the Ricci
> tensor.
>
> And by the way, thank you immensely for
> checking the Riemannian components
> You got 6 nonzero components while Maxima
> returns 12 (6 of them identical to yours) – but
> don't worry about that – the only thing I am
> concerned with is this question –

Ah, you mean the index flips:

R^0_110 = -R^0_101
R^0_220 = -R^0_202

...etc.

also:

R^1_001 = R^0_101

...etc. etc. I didn't bother writing them down, sorry!

I automatically calculate only R^a_bcd with a < b and c < d, and then
flip as needed:

R^a_bcd = -R^a_bdc

and:

R^a_bcd = - g^aa g_bb R^b_acd [NO summation convention!]

...where all g^ij and g_ij are +/-1 always in Cartan's method (one
reason this method is so much faster than Christoffel's).

> IS RIEMANN NON-ZERO, YES OR NO ?
>
> And apparently, according to both Mathematica
> and Maxima –
>
> RIEMANN IS DEFINATELY NON-ZERO !

Right.

> So the space-time is definitely CURVED !
>
> That's all I really need to know for the research
> I am currently engaged in.

There is an entire sub-industry of differential geometry called
"conformally flat manifolds" which is what you are looking at.

--
Jan

[George Hammond]

To: Ross A. Finlayson:

Dear Sir:
In the FLRW metric the scale factor "a(t) is a positive number
greater than 1 which increases with time.

And the FLRW metric is known to "expand" with time – and
since my metric is similar – what makes you think that it' is
"contracting"?

George

[Ross A. Finlayson]

How is that flat?

To have that and a flat metric at the time seems at odds,
why is seems that "the metric" is "almost completely flat",
with curvature though "infinity and one over infinity: greater
than zero", not "infinity and one over infinity: zero".

Then I think FLRW metric is over time part of inflationary
theory, it basically goes to zero, any it's "non flat-ness".

Einstein's cosmological constant is an infinitesimal,
besides whether it's zero or finite.

Then, for "how MOND" - if you'll excuse me, whether any
two objects share a frame like "this all existed since the
Big Bang, in theory" or "this all exists since the LaniaKea
Jet, which science thought was the Big Bang, in theory",
makes for most all observations that "most all around here
is part of the LaniaKea Jet and the Big Bang, in theory".

There's still of course for the ephemeris that "since the
LaniaKea Jet, things have been pretty quiet", .... That is,
"FLRW works out the red-shift differences of the graininess".

I.e. FLRW (theory) was kind of built on top of the LaniaKea Jet
which it thought in theory was the Big Bang.