“Rest mass” is synonymous with “mass”. Any other meaning of the word “mass” is now deprecated in physics.
Thanks.
Far be it for me to dispute a physicist (I’m pretty sure Chronos is a physicist), but I thought mass = energy, and is conserved absolutely. So a positron and an electron may annihilate each other in an explosion of “massless” gamma rays, but the mass inherent in the energy of the resultant photons equals the mass of the original two particles plus whatever energy they started out with. Which sounds like it agrees with your first post.
What throws me is the line about “rest mass” being synonymous with “mass”. Doesn’t that directly contradict your first post, Chronos? What am I missing?
If I’m understanding him correctly, the mass of the system which includes both gamma rays is equal to the mass of the electron and positron; it’s only when you look at the individual gamma rays, each of which has a mass of zero, that you’d get confused, because by treating the gamma rays separately, you’re ignoring some of the energy/mass.
I was mistaken when I mentioned rest mass, because that’s the only kind of mass Chronos was talking about.
But I could be wrong - it’s been a long day.
Yeah, but if rest mass is the only kind of mass, how does it follow that mass is conserved only for dynamic systems and not for each particle observed from a state of rest? Can a dynamic system be said to be at rest?
Whether mass is conserved (in special relativity) just depends on how you look at it. Mass is the norm (i.e. the ‘length’) of the four momentum vector P. For an isolated system P[sub]system[/sub] and hence it’s norm |P[sub]system[/sub]| is constant, i.e. the rest mass of the system considered as a whole is conserved
Let’s say the system is made up of a number of n particles with momenta P[sub]1[/sub], P[sub]2[/sub], …, P[sub]n[/sub], then
P[sub]system[/sub] = P[sub]1[/sub] + P[sub]2[/sub] + … + P[sub]n[/sub]
however the sum (which I just call M for convenience,not to imply that it’s more proper to think of it as the mass of the system)
M[sub]system[/sub] = |P[sub]1[/sub]| + |P[sub]2[/sub]| + … + |P[sub]n[/sub]|
(where P[sub]1[/sub], etc, are the rest masses of the individual particles) will not neccesarily be equal to |P[sub]system[/sub]| and is not necessarily constant, i.e. it is not conserved.
In many ways most prefer to talk about the conservation of|P[sub]system[/sub]| in terms of the conservation of energy (it’s just the statement that the system’s energy is conserved in it’s rest frame) rather than describe at as the conservation of mass. On the other hand it’s a much of a muchness: |P[sub]system[/sub]| is conserved and M isn’t conserved so if you choose to talk about the conservation of mass in terms of the former then mass is conserved and if you choose the latter it isn’t conserved.
Implicitly, each individual particle in the isolated system does not in itself constitute an isolated system and for this reason the statements we make about quantities that are conserved for isolated systems do not apply to the constituent particles.
A dynamic system has a center of mass frame of reference - and in the case of the electron and positron converted to gamma rays, the same frame of reference that has the electron and positron stationary initially is the center of mass frame of reference for the gamma rays. If you’re calculating energy of the system before and after the annihilation, and you keep the same frame of reference before and after, you’ll see energy/mass conserved - but if you switch reference frames (by, for example, considering each gamma ray photon individually) you won’t.
Also note that if you change frames like that, you won’t find energy to be conserved in any form. This just underscores the importance of being consistent in your definitions of things like that.
I thought that was true, but was afraid I was missing some obvious counterexample where changing frames wouldn’t matter…
Well, you could very carefully pick two frames which happened to have the same energy (i.e., two frames which are both moving at the same speed, but in different directions, relative to the zero-momentum frame), but in general, it’ll be different.
This is only true for very small-scale chemical reactions. In fact, on a joule-to-joule comparison, chemical reactions and nuclear reactions exhibit the same E=mC[sup]2[/sup] energy-to-mass equivalency.
Case in point:
The space shuttle launches with massive quantities of hydrogen and oxygen on board. When the main engines finish their two-minute burn during the initial phase of the ascent, the resulting water weighs about two grams less than the total mass of hydrogen and oxygen they had on board. Those missing two grams are manifested as kinetic and potential energy of the space shuttle and as thermal energy (waste heat) in the hot steam being fired out the back of the main engines.
If you extracted the same total amount of energy from a nuclear reaction, you would expect to see the same two-gram mass change.
Which also means that with a good enough fusion reactor, you could launch a shuttle into orbit using just 2 grams of water.
Only if you could somehow convert the entire two grams of water into pure energy. If you’re talking about extracting the hydrogen from 2 grams of water and fusing it into helium, the energy you will get out of that is quite a bit less.
Hydrogen comprises on 11% of the mass of water. After that, you smash the hydrogen together into helium, and the resulting energy is equivalent to the difference in mass between the hydrogen fuel and the helium product.
In your fusion reactor, you would need to fuse enough hydrogen into helium to create a 2-gram mass difference - just like the SSME’s do with their chemical reaction.
On the other hand, though, most of the weight the Shuttle’s engines are lifting is more fuel for the engines. Cut out almost all the fuel, and you could launch using much less energy.
And 2 grams is still undetectable, out of all of the many tons of fuel that went into the process. Especially considering that the water produced is scattered hither, thither and yon.
Yes, but the fuel is also the reaction mass for the rocket engines. You can replace the fuel with something without a lot of chemical potential energy, replacing that with the energy extracted from the total energy conversion of that two grams of water, but you still need to have a lot of spare mass to throw out the back to make the rocket go.
Unless, of course, you’re planning to launch the shuttle from some kind of slingshot. 
Hm - speaking of slingshots… I always wonder why the rockets are launched from standstill. Wouldn’t it make sense to give the rocket the initial boost from something like a giant railgun?
Doing it for a vertical takeoff would probably be impossible, but an angled one could be done. Wouldn’t that save a significant amount of reaction mass and increase payloads?
An angled launch trajectory means you have to go through more atmosphere, which is bad news. You could straighten out to near vertical right at the end of your launching rail, but that would also cost you. And you’ve also got a lot of complicated infrastructure that would cost money to set up, and would be potentially subject to a variety of failures. I’m sure that NASA considered it, but rejected it in favor of less expensive methods.
EDIT: Oh, and as for reaction mass, that won’t be the same for all launch methods. You could get away with less reaction mass, if you throw it out the back quicker. I don’t know exactly by how much, but I’m sure you could get a significantly more efficient launch overall if you had some fuel that would annihilate completely, even if you did still need reaction mass on top of that.
I still don’t get it. How is the mass of the system of two gamma photons still be the “rest mass”? Isn’t the whole point that you can’t measure the photons at rest?
And, anyways, what does no longer referring to rest mass accomplish? In other words, where is the rest mass concept deficient to the point where it has to be replaced? Is it just that the mass isn’t always at “rest”?
BTW, it sure wasn’t obsolete six years ago. What changed?
Orbital velocity is somewhere in excess of 17,400 MPH. It would take a very large machine - and destructively large accelerations - to impart any meaningful initial velocity before the onboard rocket engines take over. Rather than build a long, complicated slingshot, it’s easier just to put more fuel in the vehicle and let the engines run a bit longer.
However, there is a free boost that’s available without any slingshot required. Ever wondered why NASA usually launches its rockets from Florida instead of Maine? 