Suppose you have a system that can decompose matter to energy - and it can do so while the matter is moving through the system - what becomes of the momentum of that matter? Does energy have momentum or is the described hypothetical device a reactionless drive?
I think what you’re really asking is, “Does the law of conservation of matter & energy apply to momentum?”
Example: We have a sealed but transparent glass box which includes some matter to be used as fuel, and a device to convert that fuel to light with zero byproducts. The light will escape the box. Photons have no mass and no momentum. But the mass and momentum will get lower and lower as the fuel is used up. The box will continue to travel at the same velocity (if there’s nothing to slow it down) but it will be easier stop as time goes on.
Photons do have momentum: “Even though photons (the particle aspect of light) have no mass, they still carry momentum. This leads to applications such as the solar sail.” So the momentum of the object will decrease, but it does not disappear, it is “carried away” by the photons.
Momentum–whether in the form of inertial mass in motion or some nebulous kind of “energy”–is always conserved within a closed system. Converting mass into, say, photons would simply result in the requisite distribution of photons which would equal the original momentum to the tune of:p[sub]system[/sub] = Σp[sub]i[/sub] = Σ(h·ν/c)[sub]i[/sub]
where h is the Planck constant, ν is the frequency, and c is the speed of light in a vaccum.
Stranger
oops. my bad! :smack:
Well, as I’ve said many times before: Mass cannot be converted to energy.
Mass is not a thing, it’s a property of a system that has energy that can’t be transformed away.
Therefore for a closed system its mass, energy and momentum are conserved
Ring, you may have said it many times before, but I didn’t read it. If mass cannot be converted to energy, what are we to make of E=mc^2?
It means that for a system at rest the energy of the system is equal to its mass times the speed of light squared. In other words this is a system that has energy of E and a mass of m.
BTW the complete equation is
E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]2[sup]2[/sup]
And doesn’t nuclear energy derive from converting mass to energy? Didn’t the Hiroshima bomb convert some grams of fissile material to energy?
No. The electromagnetic and kinetic energy generated by a fission reaction comes from the reduction of the potential energy of the nucleus.
In other words potential energy is converted to electromagnetic and kinetic energy—one form of energy is just converted to another form.
The mass of the original system remains the same. The local mass defect is caused by the reduction of its potential energy.
I’m off to bed. Goodnight
Ring, I think you mean well, but you’re being a little cryptic here.
A closed system has some total energy E[sub]t[/sub] (kinetic energy, potential rest energy, and otherwise) and total momentum p[sub]t[/sub]. No matter what happens inside this closed system, its energy and momentum will always be the same. We can define the “mass” of this closed system by m[sup]2[/sup]c[sup]4[/sup] = E[sub]t[/sub][sup]2[/sup] - p[sub]t[/sub][sup]2[/sup] c[sup]2[/sup], and since the total energy and momentum of the system can’t change, then the mass defined thusly will always be the same as well.
However, what’s not true is that the mass of a “system” consisting of two objects equals the sum of the masses of each object considered separately. So a uranium-235 nucleus at rest has a given mass; when it fissions into a krypton nucleus and a barium nucleus, the momenta and energies of the two daughter nuclei will add up in such a way that the mass of the system (as defined above) will be the same as that of the original uranium nucleus. What’s not true, however, is that the mass of the krypton nucleus (as defined above) plus the mass of the barium nucleus (as defined above) will equal the original mass of the uranium nucleus. In this sense, mass not conserved.
How about this.
Mass and energy are exactly the same things. They just manifest in different forms. Think of an analogy to ice and water vapor. Although popular usage loosely says that ice can be “converted” to water vapor by adding heat, water molecules are always identical water molecules under any circumstances. The two phases are simply systems bound together in different ways.
Similarly, mass and energy are both composed of particles, which - despite appearances - are identical in their essences. One is simply bound up, “frozen,” more than the other. We loosely say that mass is “converted” to energy or vice versa (as in particle accelerators), but measurements will always show that the starting and ending products are equivalent. These are phases of the same thing.
Ring, you may be technically correct that the *popular * usage of the word “convert” is wrong or even misleading, but your explanations don’t convey that nuance. This is a matter of language rather than physics.
As is common in science, the more we discover, the less clear our earlier definitions become. At a quantum level, “mass” doesn’t really have much meaning (hence physicists refer to the energy of particles such as protons and electrons rather than their mass) since there are mass to energy “conversions” going on constantly.
In a nuclear reaction, what we typically call “mass” is being converted into “energy”, though what’s really happening is atomic nuclei are being converted into photons. They’re both energy, and hence they’re both also mass, but they’re different forms and behave differently, so it’s easier to split them into arbitrary categories we’re familiar with to make it easier to visualize.
OK, I may have been a little sloppy in my terminology - let’s forget ‘mass’ and talk about matter instead - as was my original intent.
An object in space is capable of repeatedly accelerating matched particles of matter and antimatter down a tube, where they annihilate each other - presumably being converted (or transformed, if converted is not the right word) into energy.
Accelerating the particles in one direction caused a reaction in the other direction - so the object propelling the particles has been pushed in one direction, but what happens to the momentum that was previously owned by the now-annihilated particles? If it’s inherited by the photons forming the released energy, is the vector preserved?
If so, my next question is: What happens to a mirror when a photon bounces off it? - does the photon impart momentum to the mirror?
Yup. The total (vector) momentum of the particles before annihilation equals that of the photons after.
Yup. Make a big enough mirror and you’ve got a solar sail.
Fair enough. Now with the mirror thing though…
The photon has momentum - it bounces off the mirror, imparting momentum and zooms off in the opposite direction - doesn’t the photon now have less momentum of its own? What does that mean in practical terms? Does it get redshifted as a result of this transaction, or is the momentum of photons completely separate from their other properties?
You need to view the photon and the mirror as one system in this case – before the collision, the system has the momentum p[sub]phot,in[/sub], the momentum of the photon. After the collision, conservation of momentum requires the system to still have momentum p[sub]phot,in[/sub] – but the photon is now flying in the opposite direction, and remember momentum is a vector, so the whole system must satisfy the equation
p[sub]phot,in[/sub] = p[sub]phot,out[/sub] + p[sub]mir[/sub].
Roughly speaking, p[sub]phot,out[/sub] ~ -p[sub]phot,in[/sub], the photon will have nearly the same momentum, but in the opposite direction. Using this in the equation for the conservation of momentum yields p[sub]mir[/sub] ~ 2p[sub]phot,in[/sub].
However, obviously, the mirror is now moving, and thus has kinetic energy, and energy is not a vector quantity, so the photon must have imparted some of its energy onto the mirror in order to not violate energy conservation:
E[sub]phot,in[/sub] = E[sub]phot,out[/sub] + E[sub]kin,mir[/sub].
The energy of a photon is E = hv, where v is its frequency and h Planck’s constant. Thus, the photon does experience a change in frequency:
hv[sub]in[/sub] = hv[sub]out[/sub] + p²[sub]mir[/sub]/2m, using E[sub]kin,mir[/sub] = p²[sub]mir[/sub]/2m (since E = mv²/2 and p = mv, v here being the speed of the mirror), which yields for the frequency shift
Δv = v[sub]in[/sub] - v[sub]out[/sub] = p²[sub]mir[/sub]/2mh;
now, the momentum of a photon is also connected to its frequency, via p = hv/c, c being the vacuum speed of light. Using this in the equation for momentum conservation (in the form hv[sub]in[/sub]/c = -hv[sub]out[/sub]/c + p[sub]mir[/sub], the ‘-’ being necessary because the momentum vector of the outgoing photon now has the opposite direction), it’s not too hard to figure out an equation for the frequency shift depending on the frequency of the incoming photon, the mass of the mirror, c and h, it just looks a bit messy due to the quadratic equation involved.
Anyway, yes, the photon does get a bit redshifted, but it’ll be a really tiny amount since the photon doesn’t impart terribly much kinetic energy to the mirror.
Thanks for all that. Sounds like its back to the drawing board for my reactionless interstellar drive…
I don’t disagree at all but that helps me ralize how confused I am about this.
First, I am a bit confused as to the relationship of Kinetic Energy (.5Mvv) and Momentum (Mv) ; the fact that they both depend on Mass and velocity makes me wonder if Momentum is just some re-statement of Kinetic energy.
Actual calculations are further compounded by rotation; A spinning top, (or a flywheel) has rotational momentum and kinetic energy although the whole object’s center of gravity/mass has zero velocity.
Also; does heat represent a form of enery? Given two items with equal mass, velocity, and altitude ( same Ke and Pe and Pt, presumably) if one is 'hot" and the other is “cold”, does one have more total energy than the other?
It would have been just like my Physics professor to ask for a calculation of the total energy of a red-hot yo-yo taken to the top of a mountain, spun, and then thrown sideways.
I’m not sure what question I’m even asking here but I am seeing there is a lot to the subject of measuing the energy of a system.
Apropos of this discussion, xkcd discusses solar sails (warning, some NSFW language).