For example if I convert some mass into energy (using nuclear fusion or fission or whatever…) will gravity caused by that mass disappear? Or will it stay the same?
Stays the same.
This comes up as a range of cute puzzles. For instance a wound clock weighs more than an unwound one - because there is energy stored in the spring. (The amount it weighs more is however essentially impossible to measure.)
Yep - go check out stress-energy tensoron Wikipedia. It’s typically written out as T[sup]uv[/sup] and you’ll find it in Einstein’s field equations.
It relates energy, momentum, energy flux and momentum flux in spacetime. With GR it gives the curvature of spacetime as a function of the tensor.
Yes. The object that is most sensible to call the “source” of gravity is the Stress-Energy tensor, and one component (usually the most significant in everyday situations) is the energy density.
For most everyday objects, the bulk of the energy density is locked up as the mass of the object, so saying gravity is caused by mass instead of by energy is a good approximation.
So how much more does my phone battery weigh when charged vs. dead?
I understand the difference is imperceptible to us in use, but are there scales sensitive enough to detect such a change?
This also explains why my 10 year old son weighs more than he looks.
There’s a lot of dualities in fundamental physics. Space and time. Electricity and magnitism. Likewise, matter and energy are the same thing. We know in the presence of more and more matter, the stronger the gravitational field. The same is true of pure energy (E=mc[sup]2[/sup])
That said, while there are plenty of theories, which is part of all the hullabaloo over finding the Higgs Boson, no one is quite sure exactly what causes gravity.
e=mc^2.
My cell phone came with a 5.99 watt-hour battery. That translates to 2.4*10^-13 kg. You need to measure the weight of the battery to 12-digit accuracy to measure this difference. I doubt there is any scale that can do that.
I just bought a really hip looking Planck scale from Target for $12.99. That oughta do it.
I believe E=mc^2 doesn’t apply here, because there is no mass-energy conversion involved. In fact mass difference would be a lot lot smaller.
Why?
There is never any mass-energy conversion involved in anything, because you can’t convert mass to energy. Mass is already, always, a form of energy.
scr4’s calculation is correct.
Can you convert a form of energy that is not mass *into *mass?
There is pair production, where you have a single photon that splits into a particle and it’s antiparticle.
As far as I know, a single photon system has zero mass, but after pair production, the system would have non-zero mass.
The total amount of energy before and after production would be the same though.
I’d wait for chronos to explain though, I am probably wrong.
An isolated photon cannot convert into an electron/positron pair, as a single particle going to two particles can never conserve both energy and momentum. Pair production only occurs in the vicinity of charged particles that can provide or absorb energy and momentum. Another way of looking at it is that pair production only occurs when the original photon has a virtual photon to interact with.
In the end, the mass, as talked about in this thread (a.k.a. the “invariant mass”) is, well, invariant.
You can’t do pair production with a single photon. You need at least two, and even though individual photons are massless, a pair of photons can have mass. In fact, the pair must have at least as much mass as the resulting electron and positron in order for pair production to happen.
Or another real photon; that works too.
Haha that didn’t take long. Thanks for clearing it up! I never knew that pair production required more than one photon.