# How little does a hydrogen atom curve space?

1. Dec 1, 2009

### Spinnor

I assume general relativity allows us in principle to calculate volumes of some region of space?

If so, if I have a large volume of space far from other mass and energy does it make sense to ask how much a large volume of space changes depending on whether or not it contains the mass of single hydrogen atom? For simplicity assume a point mass. I know any such change will be very small but in principle can this question be answered? Does a large box of space with matter in it have a greater volume then a similar sized empty box?

Thanks for any help!

Last edited: Dec 1, 2009
2. Dec 1, 2009

### pliu123123

Ya,if we compare the two systems :
1. infinitisimal volume nearby a hydrogen atom
2. infinitsimal volume nearby a vacuum

we may find that the volumes are different for describing physical laws
but with suitable change of coordinates(change of speed,etc), we can have the same description to the physical law,i.e. the so called"volume" of the two cases

Last edited: Dec 1, 2009
3. Dec 2, 2009

### Spinnor

If loop quantum gravity is on the right track then I am under the impression that volume is quantized. If mass or energy changes spacetime I wondered if the smallest masses (elementary particles) or the smallest energies ( say a radio wave photon) correspond somehow to changes in quantized volumes?

It seems there is no lower limit to photon energy but according to loop quantum gravity there is a minimum change in spacetime? I guess I'm missing something %^)

4. Dec 2, 2009

### atyy

If empty space contains nothing, then we will have Minkowski spacetime.

If "empty" space contains a point mass, then you can get the Schwarzschild solution, which is a black hole.

Suppose there are "test particles" whose mass is so much smaller than even the point mass in the Schwarzschild solution, so that they themselves do not curve spacetime. We can use the motion of such test particles to determine whether spacetime is flat Minkowski or curved Schwarzschild.

If you define "similar sized" empty box as "same volume", then the large box will have the same volume in both cases.

Try asking this question in the Beyond the Standard Model forum, and Marcus will probably give you a good lead.

5. Dec 2, 2009

### Spinnor

Instead of boxes, use spherical regions. If we could both measure the circumference and radius of a large spherical region with and without a point mass, I was under the impression that the ratios of these two numbers was a measure of the curvature of space? I thought this would give rise to two different volume measurements for spherical regions of the same circumference depending weather or not a point mass were present?

Similar question posted in Beyond the Standard Model forum.

6. Dec 3, 2009

### clamtrox

The volume of a region of spacetime is

$$V = \int_V d^3x \sqrt{|g|}$$

where g is the determinant of the 3-metric. Schwartzschild solution metric reads

$$ds^2 = (1-r_s/r) dt^2 - \frac{dr^2}{1-r_s/r} - r^2(d \theta^2 + \sin^2 \theta d\phi^2)$$

Notice however, that you cannot just integrate over the entire Schwarszchild solution -- the radial coordinate does not behave well inside the event horizon so you may need to integrate from r_s to the edge of the box instead.

7. Dec 3, 2009

### Spinnor

Thank you!

So for flat space we know:

$$dV = r^2 sin\theta dr d\theta d\phi$$

with your help we guess a point mass will change dV to:

$$dV_m = \frac{ r^2 sin\theta dr d\theta d\phi}{\sqrt{1-\frac{r_s}{r}}}$$

so for

$$\frac{r_s}{r}}$$

much less then 1

$${\sqrt{1-\frac{r_s}{r}}} = 1 + \frac{r_s}{2r}}$$

Edit: That should be 1 over the square root.

r_s for the mass of a hydrogen atom is tiny ( r_s = 2Gm/c^2 = 2.5E-54m). Let us only integrate down to 100r_s

$$V_m - V = 2\pi r_s R^2$$

Where R is the size of the box. So if this is close then a box the size of our Universe will have its volume change by about 1 m^3 when a single hydrogen atom is present?