Hello,

I really hope I could get an answer here, because I couldnt find it on
the internet and cant contact my prof atm.

I have to calculate the curvature scalar for spherical coordinates
(theta, phi). So the metric tensor is:

g_00=1 , g_11=sin^2(theta), g_01=g_10=0

From my calculations I get:

R=-1/tan^2(theta)

Can anyone verify if that's correct? Also, can anyone explain why this
theta dependence is to be expected (if correct)?

Your help is really appreciated.

Thanks

"Ikke" wrote in message oups.com...
Hello,

I really hope I could get an answer here, because I couldnt find it on
the internet and cant contact my prof atm.

I have to calculate the curvature scalar for spherical coordinates
(theta, phi). So the metric tensor is:

g_00=1 , g_11=sin^2(theta), g_01=g_10=0

From my calculations I get:

R=-1/tan^2(theta)

The Ricci scalar is -2.
You probably made a little mistake somewhere.

Dirk Vdm


Can anyone verify if that's correct? Also, can anyone explain why this
theta dependence is to be expected (if correct)?

Your help is really appreciated.

Thanks

Ikke ha scritto:
I have to calculate the curvature scalar for spherical coordinates
(theta, phi). So the metric tensor is:

g_00=1 , g_11=sin^2(theta), g_01=g_10=0

From my calculations I get:

R=-1/tan^2(theta)

So, Dirk Van de moortel answered you the correct result.
Anyway, you could have been able to see that your result was not good
precisely because there is a dependence on theta, which is not to be
expected, since you are calculating a scalar.

There are only three things which are invariant in GR: the Kronecker
delta, the zero tensors and the scalars; in each and everyone of them,
if you calculate it and you realize the presence of coordinates which
shouldn't be there for symmetry reasons, you can always conclude that
the result is wrong, even if you can not know what is the good result.

Also, I do not know if there is a problem of convention, but you are
calculating the curvature of a sphere, hence, it is at least strange
that you can get a negative result.

Cheers.

"Dirk Van de moortel" wrote
in message ...

[anip]

xi, x'?

Androcles

Dirk Van de moortel wrote:
"Ikke" wrote in message oups.com...
Hello,

I really hope I could get an answer here, because I couldnt find it on
the internet and cant contact my prof atm.

I have to calculate the curvature scalar for spherical coordinates
(theta, phi). So the metric tensor is:

g_00=1 , g_11=sin^2(theta), g_01=g_10=0

From my calculations I get:

R=-1/tan^2(theta)

The Ricci scalar is -2.
You probably made a little mistake somewhere.

Ricci Tensor is zero in free space even under the curvature of
spacetime. Thus, you definitely made a mistake somewhere. Ricci
scalar is ZERO for flat spacetime (in this case) with the following
metric.

** g_00 = g_11 = 1
** g_22 = r^2 cos^2(phi)
** g_33 = r^2
** Others = 0

Where

** phi = latitude

Ricci scalar is a very poor gauge to measure the curvature of
spacetime. Hilbert's Lagrangian that leads to Einstein Field
Equations is totally unqualified and unjustified.

Hello,

I probably made a mistake (I corrected one since my previous post), but
somehow I can't get a good answer (still have theta dependence which is
illogical). I first calculated that Ricci tensor and calculated the
curvature scalar by contracting the Ricci tensor.

My Ricci tensor looks like this:

R_11=1 ; R_22=1 ; R_12=R_21=0

where _ denotes subscript.

From this I now get the curvature scalar:

R = G^ij*R_ij = g^11*R_11 + g^22*R_22 = 1 + 1/sin^2(theta)

My calculated Christoffel symbols a

F^1_22 = -sin(theta)cos(theta)
F^2_12 = cos(theta)/sin(theta) = F^2_21 (symmetry in lower index)

where ^ denotes upper index.

I can totally understand if you don't want to check all this for
yourself, but maybe you see a mistake at first sight. Also, doesn't
your answer seem a bit strange as well? You get a negative number for
the curvature (if Ricci scalar is the same as curvature scalar; I'm
assuming so), but we are considering a sphere, and doesn't a sphere
have positive curvature?

Thanks.



The Ricci scalar is -2.
You probably made a little mistake somewhere.

Dirk Vdm

Yes, I realized this, but even after going over my calculations a few
times I couldn't (and still can't) find the mistake. So, to make sure
that I understood this well I posted it here.

But since I still can't find my mistake, I might be making some
fundamental mistake elsewhere. Maybe I should take a look at this again
tomorrow ;-)

Thanks for your help.


Luca schreef:

Ikke ha scritto:
I have to calculate the curvature scalar for spherical coordinates
(theta, phi). So the metric tensor is:

g_00=1 , g_11=sin^2(theta), g_01=g_10=0

From my calculations I get:

R=-1/tan^2(theta)

So, Dirk Van de moortel answered you the correct result.
Anyway, you could have been able to see that your result was not good
precisely because there is a dependence on theta, which is not to be
expected, since you are calculating a scalar.

There are only three things which are invariant in GR: the Kronecker
delta, the zero tensors and the scalars; in each and everyone of them,
if you calculate it and you realize the presence of coordinates which
shouldn't be there for symmetry reasons, you can always conclude that
the result is wrong, even if you can not know what is the good result.

Also, I do not know if there is a problem of convention, but you are
calculating the curvature of a sphere, hence, it is at least strange
that you can get a negative result.

Cheers.

"Koobee Wublee" wrote in message ups.com...

Dirk Van de moortel wrote:
"Ikke" wrote in message oups.com...
Hello,

I really hope I could get an answer here, because I couldnt find it on
the internet and cant contact my prof atm.

I have to calculate the curvature scalar for spherical coordinates
(theta, phi). So the metric tensor is:

g_00=1 , g_11=sin^2(theta), g_01=g_10=0

From my calculations I get:

R=-1/tan^2(theta)

The Ricci scalar is -2.
You probably made a little mistake somewhere.

Ricci Tensor is zero in free space even under the curvature of
spacetime. Thus, you definitely made a mistake somewhere. Ricci
scalar is ZERO for flat spacetime (in this case) with the following
metric.

No one is talking about spacetime. The question was about
spherical coordinates (theta, phi).

In two dimensions with the metric
g_00=1 , g_11=sin^2(theta), g_01=g_10=0
i.o.w. with
ds^2 = (d\theta)^2 + sin^2(theta) (d\phi)^2
the curvature is not zero.
Go ahead and calculate it.

And yes, in 3 dimensions with spherical coordinates (r,theta,phi) with
ds^2 = (dr)^2 + r^2 (d\theta)^2 + r^2 sin^2(\theta) (d\phi)^2
the Ricci tensor and scalar are zero.

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...

[blink]

xi, x' ?

Androcles

"Ikke" wrote in message
ups.com...
| Yes, I realized this, but even after going over my calculations a few
| times I couldn't (and still can't) find the mistake. So, to make sure
| that I understood this well I posted it here.
|
| But since I still can't find my mistake, I might be making some
| fundamental mistake elsewhere. Maybe I should take a look at this again
| tomorrow ;-)


Of course you are, and you should.
The fundamental mistake is division by zero, elsewhere.
Dig deep.

Androcles.