Science Question

[BrianW.]

Hello,

If a Nuclear Bomb was detonated inside of a box that could withstand the blast what would happen to the weight of the box?

Would it decrease because some of the mass was converted to energy?

[Fnord]

Assuming that you mean the total mass of the box and contents, then yes, the mass would indeed decrease by an amount equal to E/(c*c), where "E" is the energy released in ergs, where the mass is measured in kilograms, and where "c" is the speed of light.

[Jackalgirl]

Assuming that you mean the total mass of the box and contents, then yes, the mass would indeed decrease by an amount equal to E/(c*c), where "E" is the energy released in ergs, where the mass is measured in kilograms, and where "c" is the speed of light.

Would that also be the case if the box completely contained the blast (i.e., all of the energy released by the blast)?

[Lynx2174]

Would that also be the case if the box completely contained the blast (i.e., all of the energy released by the blast)?

Yes, but the box would now be very much warmer than it was before; the amount of thermal energy equal to the value of the mass converted.

[BrianW.]

Assuming that you mean the total mass of the box and contents, then yes, the mass would indeed decrease by an amount equal to E/(c*c), where "E" is the energy released in ergs, where the mass is measured in kilograms, and where "c" is the speed of light.

Yes that is what I mean,

But instead of the decrease is it possible that mass is converted to photons and since photons have mass the weight of the box remains unchanged?

[mhaze]

But instead of the decrease is it possible that mass is converted to photons and since photons have mass the weight of the box remains unchanged?

No. The heat came from mass that no longer exists. I suspect also that the photons have much more energy than their measurable mass; eg if there was no kinetic energy released, only light and gamma photons, mass should be less.

[boooeee]

Yes, but the box would now be very much warmer than it was before; the amount of thermal energy equal to the value of the mass converted.

I think it may depend on how we're defining "mass" and "weight".

If we define "mass" as the combined rest mass of the particles in the box, then the answer is "yes", the mass would be less after the explosion.

However, if we're thinking of "weight", then I think the weight of the box would be the same, assuming the explosion was entirely contained within the box. So, if we had our box on a scale, the needle on the scale wouln't move.

Also, if we're thinking of "inertia", in terms of how hard it is to push the box around, I also don't think that would change either.

So, for most intents and purposes, the mass wouldn't change. I think.

[Madalch]

Assuming that you mean the total mass of the box and contents, then yes, the mass would indeed decrease by an amount equal to E/(c*c), where "E" is the energy released in ergs, where the mass is measured in kilograms, and where "c" is the speed of light.
If you're measuring the speed of light in m/s, then shouldn't energy be measured in joules? One J = 1 kgm^2/s^2.

[theturtlemoves]

Yes that is what I mean,

But instead of the decrease is it possible that mass is converted to photons and since photons have mass the weight of the box remains unchanged?

The mass is converted to photons, but photons are massless. They have *exactly* 0 mass.

[Cuddles]

Yes that is what I mean,

But instead of the decrease is it possible that mass is converted to photons and since photons have mass the weight of the box remains unchanged?

Yes. In fact it won't be photons because they will mostly be absorbed by the box. Most of the energy will be thermal energy in the box molecules, and all energy is equivalent to mass. The box will gradually lose weight as it cools down.

No. The heat came from mass that no longer exists. I suspect also that the photons have much more energy than their measurable mass; eg if there was no kinetic energy released, only light and gamma photons, mass should be less.

Who says that the heat no longer exists? Since energy cannot be created or destroyed it must exist somewhere, and if it is specified that the box keeps everything inside then the energy must still be inside somewhere.

The mass is converted to photons, but photons are massless. They have *exactly* 0 mass.

No, photon have zero rest mass. Relativistic mass, which is what is relevant for this question, is very much non-zero.

[Thabiguy]

So, to sum up:

If the box perfectly contained all of the explosion - as in, you could sit right on the box and never notice that there was just a nuclear explosion inside it - then the weight of the box would not change at all.

Of course, in practice, you would find it practically impossible to perfectly isolate the extreme pressure and temperature inside the box and prevent them from seeping outside. And any energy leaking out - in the form of heat, light, radiation, vibrations, etc. - will carry away corresponding mass with it.

At leak rates that you commonly encounter in life, this mass descrease is practically unmeasurable, but if you allowed to leak out enormous amounts of energy such as those liberated in a nuclear explosion, you could measure it; up to some 0.1% of mass could be lost.

[BrianW.]

Thank you all for your replies.

[Fnord]

If you're measuring the speed of light in m/s, then shouldn't energy be measured in joules? One J = 1 kgm^2/s^2.

D'OH! :bwall

[Ladewig]

I think it may depend on how we're defining "mass" and "weight".

If we define "mass" as the combined rest mass of the particles in the box, then the answer is "yes", the mass would be less after the explosion.

However, if we're thinking of "weight", then I think the weight of the box would be the same, assuming the explosion was entirely contained within the box. So, if we had our box on a scale, the needle on the scale wouln't move.


Why would the mass be less and the weight be the same?

[boooeee]

Why would the mass be less and the weight be the same?


I was using a very specific definition of mass. The definition I specified is "the combined rest mass of the particles". However, this ignores relativistic mass (e.g. the mass of a photon - its rest mass is zero). Since weight is a function of the relativistic mass, that is why the combined rest mass of the particles can be less, but the weight stay the same.

I think that this definition of mass (combined rest mass) has very little physical significance. A point I tried to make (poorly) in my post.

[Dilb]

So, to sum up:

If the box perfectly contained all of the explosion - as in, you could sit right on the box and never notice that there was just a nuclear explosion inside it - then the weight of the box would not change at all.

Precisely. That's conservation of mass. Without it you could build a perpetual motion machine by dropping, the box, then exploding it, then raising the lighter box, then un-exploding it. The fact that this is utterly impossible in practice doesn't matter.

Of course, in practice, you would find it practically impossible to perfectly isolate the extreme pressure and temperature inside the box and prevent them from seeping outside. And any energy leaking out - in the form of heat, light, radiation, vibrations, etc. - will carry away corresponding mass with it.

The biggest challenge would be containing any neutrinos. You'd need a box of solid lead a few light years thick to hope to catch most of the neutrinos, and while neutrinos have very little rest mass they can carry away a lot of kinetic energy.

[Ladewig]

Without it you could build a perpetual motion machine by dropping, the box, then exploding it, then raising the lighter box, then un-exploding it. The fact that this is utterly impossible in practice doesn't matter.


I'm having a devil of a time understanding your hypothetical. What does un-exploding mean?

[Dilb]

I'm having a devil of a time understanding your hypothetical. What does un-exploding mean?

Undoing whatever change an explosion created. If we simplify it down to one uranium atom in a box:

Split uranium atom, get a few photons/whatever (and, not actually true, less mass).
Lift up lighter box, doing some work.
Fuse uranium atom back together with the energy released from the splitting. (get heavier box)
Lower the heavier box, extracting more work than was used to lift it.
Repeat as needed for a perpetual motion machine.

Fusing a uranium atom back together with the same energy is impossible, but there are more realistic systems that you could use to try and do this. The point being that lifting or accelerating 'energy' requires just as much work as lifting/accelerating matter, otherwise we could get free energy.

[Yllanes]

I was using a very specific definition of mass. The definition I specified is "the combined rest mass of the particles". However, this ignores relativistic mass (e.g. the mass of a photon - its rest mass is zero). Since weight is a function of the relativistic mass, that is why the combined rest mass of the particles can be less, but the weight stay the same.

I think that this definition of mass (combined rest mass) has very little physical significance. A point I tried to make (poorly) in my post.

The 'rest mass' (from now on just mass) of a system of two particles is not the sum of their rest masses, even if tehy don't interact at all. Mass is best defined as the magnitude of 4-momentum. \footnotesize
$p^\mu = (E/c,\ \vec p)$, where the first component E is the energy and the other 3-momentum. From now on c = 1. Mass is defined as

\foonotesize\[
-m^2 =p_\mu p^\mu= |\vec p\,|^2-E^2\]


For example, photons have E = pc, so their mass is 0. Examples of 4-momentum for photons are p1=(E,0,0,E) or p2=(E,E,0,0). Now assume we have a system consisting of those two photons, the total momentum is P=p1+p2=(2E,E,0,E). Now the mass of the system is

\footnotesize
\[-M^2 = \bigl|\vec P\bigr|^2 - (2E)^2=(E^2+E^2)- 4E^2 = - 2E^2\]


So a system consisting entirely of massless particles can have mass.

[Ladewig]

Fuse uranium atom back together with the energy released from the splitting. (get heavier box)

Doesn't returning the uranium to its original state require more energy than was released?

As for the wide range of comments, I am leary of believing in some of these arguments. I'd like to see either a science website used as a citation or a list of credentials of the posters making claims.

[RecoveringYuppy]

Under realistic conditions returning uranium to it's original state requires more energy than you could recover from the splitting of the uranium.

But in a hypothetical where you can imagine boxes that can retain absolutely everything you are imagining that you can recover all the energy from the splitting. Since mass and energy can't be destroyed, you should be able to reconstruct the uranium under such circumstances.

In realistic scenarios you are not going to be able to recover all the energy. The energy is going to spread out and it's unlikely to ever return to as compact a state as it existed in the uranium atom ever again.

[Dilb]

Right, in practice you can't fuse uranium back together with anything approaching a similar amount of energy, although in principle the conservation of energy means that it should take the exact same amount as was produced. You could use some other, slightly less impossible system. For example, you can trap light by putting it in a cavity of a photonic crystal. If you put also quantum dot in the cavity, you can get a periodic transition between the photon stuck in the cavity, and the photon being absorbed by the quantum dot. You can imagine that, if light somehow didn't add to the mass of our box, that the box would periodically become lighter and heavier. If we then put the box on a spring with the same natural frequency, we'd have a perpetually bouncing box, and getting free energy would be a simple matter of attaching a magnet.

The point being that you can't change the mass of a system by purely internal changes, otherwise it would be possible to violate conservation of energy. It shouldn't be too difficult to see that it would also allow you to violate conservation of momentum.