# Does space warp reduce the space volume?

1. Mar 22, 2009

### Gerinski

Take as example a neutron star. Its mass warps the space(time) which the neutron star itself occupies (the space inside its perimeter). Does this effect make the neutron star smaller? If we could put a neutron star in unwarped space, would it occupy a bigger volume?

I mean, does it look smaller to an external observer.

Last edited: Mar 22, 2009
2. Mar 22, 2009

### A.T.

Depends how it is warped. In GR the spatial volume is increased by mass. There is more space around a mass, than an outside observer would assume judging by the circumference of that area.

3. Mar 22, 2009

### Mentz114

Gerinski ,
your question can't be answered as it stands. It is not possible to have a neutron star in flat space. Try to rephrase your question in terms of observers with clocks, lasers, radar sets, rulers etc.

4. Mar 24, 2009

### Gerinski

Thanks, I'l try to rephrase the question.
Let's start by an observer observing a distant planet of normal density. He observes the trajectories of asteroids approaching it and being deflected by its gravitational field, if they get close enough they will turn around it into an orbit. These asteroids trajectories are geodesics along the warped space surrounding the planet. From the observation of many such asteroids the observer can infer a sort of imaginary grid of the space surrounding the planet, or a set of concentrical spheres surrounding it forming a reference of the "shape of the space" around the body, and he can tell that an asteroid approaching from a certain direction at a certain velocity will follow a certain path along these imaginary grids.

Now, he goes on to observe another body of the same apparent size but much bigger density (I know a neutron start will not be so big as a planet, so either we just assume it could or you take something else of the same size as the planet but much bigger mass).
He now sees that the paths followed by asteroids coming exactly the same as before, turn out to get much closer to the object and to produce tighter orbits. The geodesic path originating from a same direction and velocity has been pulled inwards towards the massive body. The geodesic path may be again considered as "the shape of space".

From these observations it seems reasonable that he would deduct that the increased gravity has "sucked in" the space fabric towards the center of the body. The volume occupied by a certain set of points in the grid or reference spheres has shrinked, the same points (which were defined by the result paths taken by certain approaching trajectories) are now closer to the body.

The original apparent volume of a certain portion of the grid seems to have shrinked by the higher gravity. This would seem to mean that gravity has sucked in or shrinked the apparent volume of a certain piece of space.

Extrapolating the effect to the space in the interior of the body itself, one might as well deduct that the massive body is apparently smaller than it would be if the object wasn't there. The points marked on the grid are now the surface of the body, and following the previous logic they are closer to the center that they would be if gravity was smaller. One might therefore believe that the apparent small size of the body is partly due to the space which it occupies being "sucked in" or shrinked.

5. Mar 26, 2009

### stevebd1

I think what Gerinski is asking is if the coordinate radius (or reduced circumference) of the neutron star matches the proper radius of the neutron star. If we consider the Schwarzschild vacuum solution, the proper change in distance up to the neutron star surface is-

$$dr'=dr\left(1-\frac{2M}{r}\right)^{-1/2}$$

considering the Schwarzschild interior solution, the proper change in distance within the neutron star might be-

$$dr'=dr\left(1-\frac{2Mr^2}{R^3}\right)^{-1/2}$$

where R is the r coordinate at the surface of the NS and M=Gm/c2

Based on the above and assuming a coordinate radius of 10 km (which is smaller than QM predicts) at the surface of a 2 sol mass neutron star, we can see that when r=10 km, the two solutions match (dr'=1.563dr). As we progress down through the NS, using the interior solution, the difference between dr and dr' reduces until dr=dr' at the centre. Based on dr' calculated at various interior coordinate radii we get an average of dr'=1.2dr from surface to centre which provides an approx. proper radius of 12 km for the NS which is closer to the radius predicted for a 2 sol mass NS by QM. So in answer to your question, I'm going to say yes, a static neutron star would appear smaller due to the warping of space (i.e. a neutron star with a predicted radius of 12 km might appear to an observer to have a radius of something in the region of 10 km) but it would occupy the same volume as predicted by QM.

Interior metric- https://www.physicsforums.com/showpost.php?p=1543402&postcount=8

Last edited: Mar 27, 2009
6. Mar 26, 2009

### Gerinski

Thanks Steve, as a layman I can't fully grasp all of your math but you seem to have understood my question and your reply seems to confirm my intuition that the space warp caused by a very massive object in the very space it occupies contributes to a smaller apparent size of the object to an external observer.

Just to get rid of a couple of doubts:
1. your last sentence that "the NS would appear smaller than its predicted radius, while it would occupy the same volume" stems from the fact that the measuring units in that region shrink as well (from the external observer's viewpoint) so the locally measured volume is always the same. Right ?

2. AT's reply simply reconfirms the same: an observed volume of space occupied by a massive body actually contains more space than the size it looks. The same but said the other way around.

7. Mar 27, 2009

### stevebd1

Hi Gerinski

1) Yes, volume would be based on the proper radius (12 km in this case)

2) Yes

Last edited: Mar 27, 2009
8. Mar 27, 2009

### Phrak

Nicely done steve, though I think you have a sign error in

$$dr'=dr\left(1-\frac{2M}{r}\right)^{-1/2}$$

where it should be

$$dr'=dr\left(1+\frac{2M}{r}\right)^{-1/2} ,$$

shouldn't it? M, the Schwarzschild radius, is much smaller than r. Would this change anything?

Last edited: Mar 27, 2009
9. Mar 27, 2009

### stevebd1

Hi Phrak

Click on the link and you'll see where the second equation originates from.

Last edited: Mar 28, 2009
10. Mar 29, 2009

### Phrak

OK. It looks good. I made a sign error.