> if mass can be converted into energy, then you would have more energy and less mass, so E=mc^2 would no longer hold
I'm confused why this wouldn't hold? IANAP so I don't know if E=mc2 comes into play in a "non-nuclear" mass -> force conversion. Say a rocket with x kg of fuel mass lifting off with y J of energy spent. Naively I would say that before liftoff there was x kg of fuel which is equivalent to xc^2 energy and no(?) energy y = 0. At a later time (b) When the rocket has spent half of the fuel, the energy yb spent this far could be calculated as (x/2)c^2 and energy left in mass (xb) would (x/2)*c^2. So that the total amount of energy/mass is the same at start and at time b.
Hmm, this got really confused, so... I just hit "reply".
The conversion you are talking about is true only if you annihilate your fuel (which creates photons), and then you convert photons into speed in some way with 100% efficiency.
But really this never happens, rockets work by ejecting mass at a tremendous speed. Thus not all the mass of the fuel is not converted into energy.
But otherwise, yeah, converting mass to energy completely works. This is what happens in stars and nuclear bombs, actually.
> The conversion you are talking about is true only if you annihilate your fuel (which creates photons), and then you convert photons into speed in some way with 100% efficiency.
Maybe it's hard to explain from a real world example like a rocket but if mass isn't converted to energy in that case, in the E=mc2 sense, where does the energy come from?
Well, in case of a rocket, there are two phenomenon resulting into thrust.
The first one is the action-reaction principle: fuel goes in one direction, rocket goes in the opposite one.
The second one is indeed the E=mc^2 equation: as the fuel burns, some of its mass (a negligible part, but still results in some decent amount of energy) is converted into energy and this energy can be used to propel the rocket. The rest of the mass is ejected, see first point.
So if you compare the mass of the rocket + fuel before the take-off and the mass of the rocket + burned fuel after the take-off, you will indeed see a delta, which comes from a partial conversion of mass into energy of the fuel during combustion.
Actually, as pointed out in an other comment, every reaction (or most of) leads to a decrease of the energy of the objects: an empty battery has less energy than a charged battery (yes I know I'm a genius for pointing that out). The difference of energy results, according to m = E/c², to a loss of mass. In the case of a battery, the delta is waaaay to small to be detected, but it exists.
While it might be true that the temperatures of rocket fuel burning are enough to create plasma and some reactions that lead to mass turning into energy, I think Jabavu is right in that the energy comes from the chemical reaction, not from E=mc2.
Well, yes, but it's the same thing, really. It's what my last paragraph above is about.
E=mc² is not about nuclear reactions, it's about every reaction. A chemical reaction of combustion transforms some fuel molecules into new molecules. If you compute the mass of the products of the combustion and compare it to the mass of the fuel before burning, you will notice a sliiiiight difference (probably negligible, barely detectable). The disappearing mass has been converted to kinetic energy (and heat, but who cares).
From chemical reactions (breaking chemical bonds), as opposed to nuclear reactions. There's a certain amount of energy required to shove a bunch of stable molecules together into a (more) unstable molecule (fuel, explosive). When that unstable molecule (fuel) is burned, it breaks apart into more stable products and releases the energy that was required to combine them in the first place.
If I heat 100 molecules of water ice to melting, the 100 molecules of liquid water aren't more massive. The energy went largely into kinetic energy, not changing the atomic number of the atoms. Ok, that's a phase change, not a chemical reaction. I've changed the momentum, not the mass. Are you saying this would be encompassed in the full form E^2=m^2 c^4 + p^2 c^2 ?
Indeed, in the m term, not in the p, which would be the total momentum of the water system in your frame.
For more mind-boggling numbers see for example [1]
Holy sh*t! Is this effect typically taught / introduced in a GR course? Why did I not know this!?
EDIT> Is there any way to derive this outside of GR?
EDIT2> I notice that you were careful to say "heavier", not more "massive". Does this point to the effect being part of the gravitational interaction specifically?
I'm confused why this wouldn't hold? IANAP so I don't know if E=mc2 comes into play in a "non-nuclear" mass -> force conversion. Say a rocket with x kg of fuel mass lifting off with y J of energy spent. Naively I would say that before liftoff there was x kg of fuel which is equivalent to xc^2 energy and no(?) energy y = 0. At a later time (b) When the rocket has spent half of the fuel, the energy yb spent this far could be calculated as (x/2)c^2 and energy left in mass (xb) would (x/2)*c^2. So that the total amount of energy/mass is the same at start and at time b.
Hmm, this got really confused, so... I just hit "reply".