Does general relativity give curvature a magnitude?

[zeromodz]

I see that general relativity uses tensors to calculate curvature. How exactly does relativity calculate actual curvature. Are the units of curvature m^-1, like regular curvature units?

For example, using SI units

Ruv - 1/2guvR = (8πG/c^4)Tuv
R00 - 1/2g00R = (8πG/c^4)T00
R00 + R/2 = 8πGρ/c^2

Since it is just the energy density component, the c^4 = c^2 in this case and T00 equals rho. guv also equals negative one at the time time component (g00). So how would I go from here to get an actual answer for curvature. Say the density is 5 kg / m^3. What is the energy density?

 

[bcrowell]

I see that general relativity uses tensors to calculate curvature. How exactly does relativity calculate actual curvature.

What do you mean by "actual?"

Are the units of curvature m^-1, like regular curvature units?

Different people have different ways of thinking about the units of tensors. Personally my way of thinking about it is that coordinates can have any units, e.g., your coordinates could be (t,r,theta,phi), with units of (seconds,seconds,unitless,unitless). In that case, you can't make any general statements about the units of a tensor like the Riemann tensor.

Say the density is 5 kg / m^3. What is the energy density?

In SI units? I guess you'd multiply the density by c^2. But please, for the love of god, don't try to do GR in SI units :-)

 

[PeterDonis]

So how would I go from here to get an actual answer for curvature. Say the density is 5 kg / m^3. What is the energy density?

Energy density (in conventional units) is just c^2 times mass density. 5 kg/m^3 is about 4.5 x 10^17 Joules/m^3. However, as bcrowell hinted, it is much, much easier to do GR in "natural" units, where c = G = 1. In those units, energy density has units of inverse length squared--energy and mass have the same units, length, so energy density is just length divided by length cubed, or inverse length squared. Since energy density is equated to curvature in the Einstein Field Equation, curvature also has units of inverse length squared.

For the case you gave, it's easier to convert the mass to length units; the conversion factor is G / c^2, or about 7 x 10^-28 m / kg. So 5 kg/m^3 equates to about 3.5 x 10^-27 m^-2 in curvature units. That's why it's not normal in everyday life to think of mass/energy density as curvature--the curvature corresponding to ordinary densities is way too small.

 

[zeromodz]

Energy density (in conventional units) is just c^2 times mass density. 5 kg/m^3 is about 4.5 x 10^17 Joules/m^3. However, as bcrowell hinted, it is much, much easier to do GR in "natural" units, where c = G = 1. In those units, energy density has units of inverse length squared--energy and mass have the same units, length, so energy density is just length divided by length cubed, or inverse length squared. Since energy density is equated to curvature in the Einstein Field Equation, curvature also has units of inverse length squared.

For the case you gave, it's easier to convert the mass to length units; the conversion factor is G / c^2, or about 7 x 10^-28 m / kg. So 5 kg/m^3 equates to about 3.5 x 10^-27 m^-2 in curvature units. That's why it's not normal in everyday life to think of mass/energy density as curvature--the curvature corresponding to ordinary densities is way too small.

Sorry, I didn't mean to say how do I calculate the energy density. I want to know how to calculate the curvature.

 

[zeromodz]

What do you mean by "actual?"

By "actual", I mean a quantitative answer. Like a numerical magnitude.

 

[Mentz114]

Here are the non-zero components of R_abcd for the Schwarzschild spacetime.

[tex]
\begin{align*}
R_{1100}&=-\frac{2\,{c}^{2}\,m}{{r}^{3}}\\
R_{2200}&=\frac{{c}^{2}\,m\,\left( r-2\,m\right) }{{r}^{2}}\\
R_{2211}&=-\frac{m}{r-2\,m}\\
R_{3300}&=\frac{{c}^{2}\,m\,\left( r-2\,m\right) \,{sin\left( \theta\right) }^{2}}{{r}^{2}}\\
R_{3311}&=-\frac{m\,{sin\left( \theta\right) }^{2}}{r-2\,m}\\
R_{3322}&=2\,m\,r\,{sin\left( \theta\right) }^{2}
\end{align*}
[/tex]

and the Riemann scalar is [itex]R=48\,{m}^{2}/r^6[/itex]. Tensor indexes are for coordinates [itex]x^0=t,\ x^1=r,\ x^2=\theta,\ x^3=\phi[/itex].

 

[yuiop]

If I look at a point in space and say "wow, that has a lot of curvature!" what is the physical significance of that? Does it means that two nearby clocks that are at rest wrt each other will be running at significantly different rates or does it means that two nearby inertial test masses will accelerate rapidly away from each other or does it simply mean a great distortion of Euclidean geometry?

 

[Mentz114]

If I look at a point in space and say "wow, that has a lot of curvature!" what is the physical significance of that? Does it means that two nearby clocks that are at rest wrt each other will be running at significantly different rates or does it means that two nearby inertial test masses will accelerate rapidly away from each other or does it simply mean a great distortion of Euclidean geometry?

I think you know the answer to that.

The Riemann scalar is useful in this case because it tells us there is a physical curvature singularity at r=0, R being invariant.

 

[PAllen]

If I look at a point in space and say "wow, that has a lot of curvature!" what is the physical significance of that? Does it means that two nearby clocks that are at rest wrt each other will be running at significantly different rates or does it means that two nearby inertial test masses will accelerate rapidly away from each other or does it simply mean a great distortion of Euclidean geometry?

I wish I could find an online source for this, but J. L. Synge, in his 1960 book, has a substantial discussion of this question. He demonstrates that measurements at a 5 or more points are needed to distinguish actual curvature of spacetime from other phenomena (e.g. non-inertial motion, including rotation). Of course it is still your choice whether to interpret such measurements as curvature or the impact of gravity on measuring devices, but that is a separate philosophical issue.

 

[pervect]

As far as physical significance goes, I think its more relevant to talk about what the Riemann curvature tensor is in a frame-field, rather than in a coordinate basis.

For the Schwarzschild metric, we can set up a coframe basis

w1 = sqrt(1-r_s/r) dt
w2 = dr/sqrt(1-r_s/r)
w3 = [itex]r d\theta[/itex]
w4 = [itex]r \sin \theta d\phi[/itex]

The metric then becomes diagonal, equal to

ds^2 = -c^2 w1^2 + w2^2 + w3^2 + w4^2

With this metric, the dual of w1 is a unit vector in the t direction, usually written as [itex]\hat{t}[/itex], a "t" with a "hat" over it (the latex seems blurry).

Similarly the dual of w2 is a unit vector in the r direction, the dual of w3 a unit vector in the theta direction, and the dual of w4 a unit vector in the phi direction.

Then we can write
[tex]R^{2}{}_{121} = -\frac{r_s}{c^2\, r^3}[/tex]
[tex]R^{3}{}_{131} = R^{4}{}_{141} = \frac{r_s}{2\, c^2\, r^3}[/tex]

I've chosen the initial superscript to tie-in with the geodesic deviation equation, but I'm running out of time to explain the details thereof.

If we take the first step towards geometric units by making c=1, we can see that the units of the Riemann in the frame field are 1 / length^2, where "length" could be a space interval (say feet), or a time interval (say nanoseconds) as long as you choose units where c=1 (1 foot/ nanosecond works fairly well, though people don't use feet much in the literature for distance units! - and 1 foot per nanosecond is only approximately correct).

Note that the Riemann curvature has the same units as gaussian curvature, which can be defined as the the products of the two principal curvatures for a 3-d space

http://en.wikipedia.org/w/index.php?title=Gaussian_curvature&oldid=388108115

however, the gaussian curvature is a measure of curvature that's independent of any specific embedding , unlike the "principal curvature".

Sectional curvature is another interesting topic to read on that is closely related - unfortunately I don't use it very much. I'm pretty sure there is a nice way to explain how gaussian curvature, sectional curvature, and the Riemann curvature tensor all tie in together, but I only actually use the Riemann curvature personaly.

 

[zeromodz]

Here are the non-zero components of R_abcd for the Schwarzschild spacetime.

[tex]
\begin{align*}
R_{1100}&=-\frac{2\,{c}^{2}\,m}{{r}^{3}}\\
R_{2200}&=\frac{{c}^{2}\,m\,\left( r-2\,m\right) }{{r}^{2}}\\
R_{2211}&=-\frac{m}{r-2\,m}\\
R_{3300}&=\frac{{c}^{2}\,m\,\left( r-2\,m\right) \,{sin\left( \theta\right) }^{2}}{{r}^{2}}\\
R_{3311}&=-\frac{m\,{sin\left( \theta\right) }^{2}}{r-2\,m}\\
R_{3322}&=2\,m\,r\,{sin\left( \theta\right) }^{2}
\end{align*}
[/tex]

and the Riemann scalar is [itex]R=48\,{m}^{2}/r^6[/itex]. Tensor indexes are for coordinates [itex]x^0=t,\ x^1=r,\ x^2=\theta,\ x^3=\phi[/itex].

Thank you, this is exactly what I was looking for. Could you tell me where you found these and how to derive them if possible. I especially am interested in the Scalar curvature formula and how all that works out. Thanks again.

 

[PeterDonis]

Sorry, I didn't mean to say how do I calculate the energy density. I want to know how to calculate the curvature.

The Einstein Field Equation that you wrote down equates the two, so in a sense calculating one *is* calculating the other. However, if you want a definition of the Ricci tensor or Einstein tensor that you wrote down, it's just a contraction of the Riemann tensor that bcrowell wrote down:

[itex]R_{ab} = R^{c}_{acb}[/itex]

and the Ricci scalar R is a contraction of that:

[itex]R = R^{a}_{a} = g^{ab}R_{ab} = g^{ab}R^{c}_{acb}[/itex]

The 0-0 component of the field equation then reads (all units are curvature units, so the T term on the right is in the units I gave in an earlier post):

[itex]R_{00} - \frac{1}{2}g_{00} R = T_{00}[/itex]

You can use the Riemann tensor components and the known components of g in the Schwarzschild metric to fill in the above (note that you'll have to use some of the symmetries of the Riemann tensor to get the component indices into the form I've written above).

 

[Mentz114]

Thank you, this is exactly what I was looking for. Could you tell me where you found these and how to derive them if possible. I especially am interested in the Scalar curvature formula and how all that works out. Thanks again.

They were calculated for me by Maxima using this formula, (notation: the ',' means partial differentiation so g_ab,k is dg_ab/dx^k ).

[tex]
{R^r}_{mqs}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}
[/tex]

where

[tex]
{\Gamma^m}_{ab}=\frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})
[/tex]

For the derivations, see any textbook on GR. This is R calculated in the coordinate basis. Pervect has done the calculation for a frame field in the Schwarzschild spacetime, hence the difference.

 

[George Yuill]

I follow the algebra leading to your Rrmqs but get in a muddle in applying this.Must there be local and general coordinates.Or is it usable with just one set of coordinates.how does one apply contraction to this;what is the physical meaning of the contracted version.many thanks
maths amateur drowning