# Light in the box gravity - question.

1. Oct 26, 2009

### Dmitry67

I have a massless and perfectly reflecting sphere or a box. Inside I have 0.5kg of matter + 0.5kg of antimatter.

Far away from this sphere I have a small test body. I detects a gravity from mass of 1kg.

Now matter and antimatter annihilate. As nothing happens outside, the test body is still detecting the same gravity. But there is no more 'rest' mass left.

My question is: the gravitation which comes from the sphere or a box - what components affect it and how much:
1 pressure of light
2 tension of the walls of the box (because they resist the pressure)
If 1 and 2 are oppoiste and equal what is left?
3 Stress energy tensor INSIDE the box? (from a light trapped inside?)

Thank you

2. Oct 26, 2009

### Ich

The energy of the photons, which is the same as the energy of the matter before annihilation.

3. Oct 26, 2009

### hamster143

1 & 2 aren't opposite. Gravitation comes from the stress energy tensor of radiation plus the stress energy tensor of the box. Stress energy tensor of radiation is $$diag(\rho,p,p,p)$$ where $$p= \rho / 3$$. But spatial components don't really have any effect on the static Schwarzschild metric. So the test body should not be able to notice the annihilation.

4. Oct 26, 2009

### Dmitry67

Thank you.
Another question: if light beam passes by, does it create a frame dragging effect?

5. Oct 26, 2009

### hamster143

I don't see a rotating object here ...

6. Oct 26, 2009

### Dmitry67

whats about a system of 4 (or more) ideally reflecting mirrors?
So light goes in circles?

7. Oct 26, 2009

### pervect

Staff Emeritus
I think I have to mildly disagree with some of the responses here. The answer is that all of the effects listed contribute to the gravitational field outside the box.

Just to be clear, I assume we are assuming a static system (and not trying to model any transient effects of the explosion such as emitted gravity waves which are assumed to be negligible), and measuring the gravitational field with an accelerometer, which is stationary with respect to the box. The notion of "stationary" is defined because the system is static.

However, you'll see no change in the gravity outside the box when the anti-matter is combined with the matter, which is what the original poster asked.

But - if you place a gravity probe just inside the surface of the box, you _will_ see a change in gravity. It will increase. ((I assume a spherical box for simplicity)). It'll have to be pretty tough probe not to melt, of course :-).

Assuming that the box is under tension but has no mass is equivalent to assuming it is made out of exotic matter. Normal matter wouldn't be able to decrease the gravitational field in the matter that this "thought experiment" indicates (the field must decrease going from inside to outside the box). The assumption that the box has no mass but is under tension is only possible for exotic matter, however, not normal matter.

8. Oct 26, 2009

### lightarrow

If you mean proper = invariant mass, why there is 'no more'? I would say mass is exactly the same as before: mass is a system's energy divided by c2 if the system is stationary, as in this case.

9. Oct 26, 2009

### Dmitry67

no, I meant 'invariant mass'.
I understand that relativistic mass is conserved.

10. Oct 26, 2009

### DrGreg

To expand on what lightarrow said, the invariant mass of a system of particles is not necessarily equal to the sum of the invariant masses of each particle -- it can be higher.

In the system's "rest frame" -- i.e. the frame in which the sum of the particle momenta is zero, the "centre of momentum frame" -- the invariant mass is the sum of the particle energies divided by c2.

Assuming no energy escapes, the invariant mass of the box remains 1 kg throughout.

11. Oct 26, 2009

### pervect

Staff Emeritus
Just a quick note - calculating the invariant mass of the box doesn't necessarily tell us much about the gravitational field.

The concept of mass that's most applicable to this problem is the GR notion of Komar mass - it's the concept of mass associated with the time translation symmetry of the system.

If we have a static system (or more generally, a stationary system), as we do in this case when we ignore any transients associated with the explosion, the product of the force-at-infinity times the area of the enclosing sphere (or, more generally, the normal component of the force-at-infinity times the area of any enclosing body) is a constant which is proportional to the Komar mass.

See for instance http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=284528862, or, preferably, the discussion in Wald that's referenced by the wiki link (though the Wald discussion is rather technical).

12. Oct 27, 2009

### Dmitry67

Another question. Proton is much heavier then u and d quarks, and most of the mass comes from the relativistic motion of quarks.

So, if we look at proton at whole we get an invariant mass, if we look at quarks, we see small relativistic mass bug big relativistic one.

Hm... what is a right question... while technically the description above works, but are there more... consistent description of mass? The value we use for mass depends on the point of view... of our knowledge of the system.