Mass causes space time curvature

[shounakbhatta]

If I take it by literally meaning: Mass causes space time to curve. A rubber sheet where the mass is there, it causes the dent, the curvature.

So it means the greater the momentum, the greater the curve or the dent.

Now if we have a very big mass, I mean to say big in terms of size, the greater will be the curve and smaller the size, smaller the mass.

Is that so?

In that case where are the areas that would have a bigger curve on space time rather than smaller one?

-- Shounak

 

[HomogenousCow]

Space time curvature is measured by a tensor.
More qualitatively, the curvature at a given event in space time is measured by how much a vector changes if you take it around an infitesimal loop at that event
It is difficult to answer your question, the stress energy tensor causes curvature.
The most well known solution is that of a non rotating non charged sphere, spacetime deviates more and more from the flat minkowski space time the closer you approach the sphere, if you want a picture to wrap your mind around, http://en.wikipedia.org/wiki/File:Flamm.jpg this might help.

 

[shounakbhatta]

Yes, thank you for the reply.

I went through the stress-energy tensor and understood the components. Here energy-momentum tensor the energy and momentum is conserved right?

My question is:

The bigger the <size> of the mass, the bigger the dent(curve) in the space time.

Sun, moon etc. if we imagine them as mass spheres, does the curvature is caused due to the volume of the planets?

May be it is a very wrong question but still.....

-- Shounak

 

[Fredster1765]

I'm probably not qualified to answer your question technically (I don't study physics), but I would think that the geometric size of the object is irrelevant to the curvature it causes. Using the rubber-membrane analogy, an inflated balloon would cause much less curvature than a small iron ball as it has less mass. I'd argue the same thing applies to the fabric of space-time, as black holes are tiny objects with incredible mass that still manage to warp space-time to a degree that even light can't escape it on the inside of the event-horizon.

 

[Nugatory]

I'm probably not qualified to answer your question technically (I don't study physics), but I would think that the geometric size of the object is irrelevant to the curvature it causes. Using the rubber-membrane analogy, an inflated balloon would cause much less curvature than a small iron ball as it has less mass. I'd argue the same thing applies to the fabric of space-time, as black holes are tiny objects with incredible mass that still manage to warp space-time to a degree that even light can't escape it on the inside of the event-horizon.

Argument from analogy (especially an analogy as flawed and misleading as the rubber-membrane one) is always suspect, but in this case it works just fine. The only thing I'd add is that the it's not just the mass that matters but how concentrated it is. Compare the effect of a kilogram of wood (maybe 2 cm by 20 by 20, lying flat) and a kilogram sphere of lead. Far away from these objects the curvature of the membrane will be the same, but nearby the curvature will be much greater for the lead sphere.

 

[Bill_K]

if you want a picture to wrap your mind around, http://en.wikipedia.org/wiki/File:Flamm.jpg this might help.

Flamm's paraboloid is an unfortunate analogy, misleading in almost every respect. For one thing it appears to sanctify the rubber sheet and the picture of a gravitational field as a dent in a membrane. Questions it typically leads to: Does this mean that spacetime is embedded in a higher dimensional space? What is the vertical coordinate? And why are objects attracted downwards?

The image it conjures is that of a planet as a little ball rolling on the membrane, and it's the inward force of the sloped surface that causes the planet to go around and around in an orbit. It implies, moreover, that the curvature of the surface represents the curvature of spacetime (which it doesn't), and that this in turn represents gravitational attraction (which it doesn't).

Flamm's paraboloid is a slice through the spacetime at constant t, whereas a planetary orbit is not at constant t; so the properties of this surface have little to do with planetary motion. Orbital motion, in fact, is caused by "curvature in time", i.e. the g₀₀ component of the metric tensor. In linearized theory, g₀₀ = 1 - 2V where V is the Newtonian potential, and it's the effect of a varying V that enters the geodesic equation and causes them to spiral around the origin.

 

[HomogenousCow]

It does provide a rare chance to visualize something in a theory as abstract as GR.
However, the radial component of the metric tensor in the schwarschild metric makes a non negligible contribution to the r^-3 correction term for orbital trajectories.

 

[Bill_K]

the radial component of the metric tensor in the schwarschild metric makes a non negligible contribution to the r^-3 correction term for orbital trajectories.

How often is it advertised that way? How many people, seeing the Flamm paraboloid for the first time, are told that it is not what it looks like, and not what they naturally think it is.

 

[GarageDweller]

The general public generally regards physics as magic, does it matter?

 

[nitsuj]

The general public generally regards physics as magic, does it matter?

No, because that's not supported. If it was than perhaps.

 

[pervect]

I think Flamm's paraboloid depicts spatial curvature, and not space-time curvature. If it was going to represent space-time curvature, one of the embedded coordinates would have to be timelike. But both embedded coordinates are space-like, at least in the exterior region.

 

[shounakbhatta]

Well, lot of thanks to Bill. At least we come to know that Flamn's paraboloid is not what is shown in the equations. Well, I would like to extend my special thanks to Bill to point out what it actually means. Yes, this is the problem with popular science and try to depict anything on a common way. But indeed for understanding General Relativity and what is actually a curvature of space time, we need an image in order to visualize what POSSIBLY can be.

I would request Bill to enlighten us kindly in a simple language (which I know it is difficult) what is actually a space time curvature.

I also have one question to Nugatory . As you have told :The only thing I'd add is that the it's not just the mass that matters but how concentrated it is."

In the stress energy tensor component there are components like 'energy density' & 'momentum density'. Does that actually mean "HOW MUCH ENERGY" & "HOW MUCH MOMENTUM" it shows? I mean to say what you told is the 'concentration'?

 

[Naty1]

yes to the last two questions above. extended, low density mass, can't cause all that much curvature of spacetime...analogous to the siutation where you pass thru a hollow cylinder to the center of the Earth where gravitational attraction diminishes...

But visualizing how gravitational potential curves spacetime is beyond me.

have no notes with me but I may have a spacetime curvature illustration reference
and will post next week if i can remember...
This is not spacetime curvature caused by earth: http://simple.wikipedia.org/wiki/Spacetime
despite the illustration explanation...could not find an illustration ...

Explaining 'curvature' is not so easy...you can start via the Riemann curvature tensor
and you'll find further references like Ricci...etc,etc...andthere is extrinsic (embedded)
curvature and intrinsic curvature of the Riemannian manifold which Einstein utilized.

If you are moving at constant speed but experiencing acceleration you are in curved space or time or both...

 

[shounakbhatta]

Thanks Naty. So the intensity of momentum and energy determines HOW much the curve. But it is really interesting that determining curvature is so difficult. But I would really like to know just a little bit about intrinsic curvature. What does it mean?

One more question, sun,moon, other planets are also mass which causes the curvature in the spacetime manifold. So, is it bigger the mass (bigger the size of the planet) the bigger the curvature in that area of space?

Thanks,

-- Shounak

 

[Naty1]

the standard interpretation of curvature takes Ricci curvature as a represntation of
a volume element change and the Weyl component as a measure of tidal forces...Exactly why those have been so widely adopted is 'above my paygrade'...but likely it is because they offer valid physical insight.

For insights into curvature, try here
http://en.wikipedia.org/wiki/Riemann_curvature_tensor

and

http://en.wikipedia.org/wiki/Riemann_curvature_tensor#Geometrical_meaningHow the different measures of curvature compare and the advantages and disadvantages of different formulations is something i have not gotten into...
but apparently Einstein's utilization of existing mathematics from Riemann's
work has pretty well stood the test of time. There have been discussions in these
forums of some 'upgrades' that have been investigated but I did not follow
those closely. Riemann's curvature tensor is but one way to express curvature of a
Riemann manifold: http://en.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds

 

[Nugatory]

I also have one question to Nugatory . As you have told :The only thing I'd add is that the it's not just the mass that matters but how concentrated it is."

In the stress energy tensor component there are components like 'energy density' & 'momentum density'. Does that actually mean "HOW MUCH ENERGY" & "HOW MUCH MOMENTUM" it shows? I mean to say what you told is the 'concentration'?

They pretty much tell you both density and how much is there in total - density is how much is at a point, and if you look at a bunch of neighboring points you'll get the total amount in that area.

BUT - This is a vague and handwavy sort of description of a clear and precise mathematical concept. If you want real understanding, you have to move beyond these vague analogies (remember what I said about the rubber membrane in the post you quoted from) and look at the math.

 

[WannabeNewton]

the standard interpretation of curvature takes Ricci curvature as a represntation of
a volume element change...Exactly why those have been so widely adopted is 'above my paygrade'...but likely it is because they offer valid physical insight.

It can be shown mathematically that such interpretations are valid of course. For example one such relation: given a riemannian manifold M and a point p in M, letting S^{n-1} be the sphere in TpM and A(S^{n-1}) the area of said sphere, you can express the scalar curvature at p, K(p) , in terms of the standard ricci curvature Ric_{p}(x), by K(p) = \frac{1}{A(S^{n-1})}\int_{S^{n-1}}Ric_{p}(x)dS^{n-1}. There are various other interpretations of ricci and scalar curvatures I would presume (Lee gives an ostensibly different interpretation of the two in his riemannian manifolds text). A standard introductory text in differential geometry seems to treat the geometric interpretation of Riemann curvature in great detail but not so much the ricci and scalar curvatures. Maybe if someone knows an advanced text that does indeed treat the geometric interpretations of the latter two in much greater detail that would be quite a blessing indeed.

 

[Naty1]

Here is a nice discussion of relativity...
and some comments on curvature here:
http://math.ucr.edu/home/baez/einstein/node2.html

 

[K^2]

The bigger the <size> of the mass, the bigger the dent(curve) in the space time.

No. The metric for a non-rotating spherical body is a function of mass ONLY. If you go the same distance from center of the object with a fixed mass, you will find exactly the same curvature of space-time regardless of how dense the object is.

If the Sun was to be replaced by black hole of equal mass right now, all of the planets would keep on moving in exactly the same trajectories. In terms of gravity, we wouldn't notice the difference.

Like people have said, you shouldn't take the analogy with object on an elastic surface too literally.

 

[Nugatory]

No. The metric for a non-rotating spherical body is a function of mass ONLY. If you go the same distance from center of the object with a fixed mass, you will find exactly the same curvature of space-time regardless of how dense the object is.

Although if the object is denser, you can move closer to the center of mass while while still remaining outside the surface of the object where the vacuum solution is valid. Thus you can experience stronger gravitational and tidal forces and more curvature; this increase is probably what shounakbhatta was getting at when he spoke of a "bigger dent".

Which just goes to show...

Like people have said, you shouldn't take the analogy with object on an elastic surface too literally.

 

[K^2]

Thus you can experience stronger gravitational and tidal forces and more curvature; this increase is probably what shounakbhatta was getting at when he spoke of a "bigger dent".

If that's what he was looking for, then yes. But it sounded like he expected same curvature at the surface, but further out as object size increases, which isn't correct.

 

[bahamagreen]

Is the interior of a hollow sphere Ric=0?

 

[K^2]

Is the interior of a hollow sphere Ric=0?

Yeah, vacuum is always Ricci-flat.

 

[bahamagreen]

Would we say the interior of a hollow sphere is a case where mass causes space time non-curvature... in so far as additional regional curvatures will still be present within the sphere, but the sphere's regional curvature will contain a "flat spot"?

 

[K^2]

It's actually completely flat. The only spherically symmetric solution that is Ricci-flat and doesn't have mass at the center singularity will be flat. The interior will be time-dilated compared to exterior, but otherwise, identical to Minkowski space-time.