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Comment by cynicalkane | Hacker News Reader

Really annoying to see false expertise pop up again and again.

> conservation of momentum, which is just a subset of conservation of energy

This is false. Momentum and energy are two different things, even with mass-energy equivalence. In particular, momentum is a vector while energy is a scalar, and their mass-energy equivalent units (impulse vs energy) are also different.

Also: pair production doesn't necessarily require some amazing quantum device to keep the pair separate.

> There is even some research into using a vibrating "quantum mirror" that forces virtual photons to become real using only energy from quantum fluctuations but that's even more complicated and difficult to engineer.

This is true if you consider conservation of energy to be an engineering problem. Under no scientific theory can you pluck energy from the vacuum.

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> Momentum and energy are two different things, even with mass-energy equivalence. In particular, momentum is a vector while energy is a scalar, and their mass-energy equivalent units (impulse vs energy) are also different.

None of this is correct when you take relativity into account. In relativity, momentum and energy are both parts of a 4-vector, and that 4-vector is conserved. And in natural units, in which c = 1, energy and momentum have the same units. Particle physicists use these units all the time.

Right, most people think of mass energy equivalence which is E=mc^2 but this only describes energy for a mass at rest, with a momentum of zero. The full equation relating energy to intrinsic mass and momentum is E^2=(pc)^2+(mc^2)^2 which is why smashing particles in a collider works. The momentum of two particles colliding at relativistic speeds is "conserved" in the classical sense but it is really converted to photons and massive particles, among other things.

What I meant in my original post was that intuition regarding conservation of momentum in classical mechanics may not apply here because as long as the total energy of the system is conserved, the EM drive is free to convert energy to particles and motion because momentum will still be conserved between the drive and the new particles (which is how regular chemical propulsion works). The system experiences a net change in momentum of zero but what matters is that the total energy of the EM drive slowly falls as energy is converted to particles that provide propulsion.

> the EM drive is free to convert energy to particles and motion because momentum will still be conserved between the drive and the new particles

I agree that this would conserve momentum if if were what was happening in the EmDrive. But all of the papers I have read insist that it is not what is happening; no particles are being ejected from the test device; there is no "rocket exhaust".

How do they measure the exhaust? With the EM field in the cavity I would imagine precisely measuring the photon emissions would be very difficult and the SNR might not be high enough to draw a conclusion. If the force is generated by new massive particles it would be headline news in experimental physics as the most efficient energy-mass converter ever made but if it's somehow generated by high energy photons, what are the chances they have adequate imaging to measure UV, X, and gamma rays, especially if the emissions have some sort of complex interference pattern? (although the theoretical maximum for force/power input is much lower with photon pressure than the results in this paper so it's still unlikely)

> How do they measure the exhaust?

I don't know that they are specifically looking for any exhaust other than RF emissions (which they say are not observed). If there were in fact an exhaust at some other frequency I don't know that it would be detected. (If such were present it would still be rather unexpected, in my view; there is nowhere near enough energy being pumped into the apparatus to start creating particle-antiparticle pairs, or to trigger any kind of process that would produce UV or X or gamma rays.)

Yes, I know how the 4-dimensional momentum vector in relativity works.

I'm not aware of a physical view for which energy and momentum are the same for the purposes of something like the EmDrive. In particular, you cannot completely turn one into the other by a choice of coordinates, and you cannot mix them by any physical process. And I could be wrong, but I'm 99% sure you need both energy and momentum conservation to get 4-momentum conservation.

No matter your view, conserving the spacelike components of 4-momentum is not "a subset of the conservation of energy". And the truth/falsehood of this view has nothing to do with the "non quantum world". Relativity was an extension of classical electrodynamics before it was anything else.

> And I could be wrong, but I'm 99% sure you need both energy and momentum conservation to get 4-momentum conservation.

I'm almost certain you're wrong. The most mathematically precise definition I remember for conservation of 4-momentum only requires that the interaction is invariant to spacetime translations. I don't see how that requires you to conserve both separately from the whole.

> No matter your view, conserving the spacelike components of 4-momentum is not "a subset of the conservation of energy". And the truth/falsehood of this view has nothing to do with the "non quantum world". Relativity was an extension of classical electrodynamics before it was anything else.

That was a poor choice of words on my part. I meant that an understanding of conservation of momentum based in classical mechanics is rather meaningless when talking about a drive that might use energy to create particles. What matters is that the total energy of the system is conserved and I stand by my choice of "subset" because momentum is not independently conserved. Calling conservation of momentum a subset of conservation of energy is not even close to calling energy the same thing as momentum and I have absolutely no clue where you got that idea.

> The most mathematically precise definition I remember for conservation of 4-momentum only requires that the interaction is invariant to spacetime translations. I don't see how that requires you to conserve both separately from the whole.

"Conservation" of a 4-vector means its components in a given inertial frame do not change. That entails that, from the viewpoint of a given inertial frame, the total energy and each individual component of the momentum are all conserved separately.

There's a confusion here which you still don't get. Momentum is not 4-momentum. Conservation of latter implies the former but the reverse is not necessarily true. And nothing in the above is a "subset of conservation of energy"--that's absurd.

Actually, I'm not 99% sure, I'm 100% sure. It's very easy to imagine a relativistic system where momentum is conserved but energy isn't. Imagine if the universe was made of inelastic billiard balls, with no internal "heat", so every collision just lost energy. But momentum is still conserved.

> It's very easy to imagine a relativistic system where momentum is conserved but energy isn't. Imagine if the universe was made of inelastic billiard balls, with no internal "heat", so every collision just lost energy.

This won't work. Let's suppose we have just two billiard balls, and that 3-momentum is conserved in the center of mass frame (in which it is by definition zero), but energy in this frame decreases at each collision. (Ignore the fact that this obviously implies that the 4-momentum of the system is not conserved, even though it is undergoing no external interactions.) Now transform into any other inertial frame. You will find that 3-momentum is not conserved either; it has a different nonzero value after a collision than before.

This is a bizarre comment. If we forget about energy, this is equivalent to saying 3-momentum is not conserved in inelastic collisions in real life. Which is false.

> this is equivalent to saying 3-momentum is not conserved in inelastic collisions in real life.

No, it's not. In real life, energy is conserved, so the energy that is lost from kinetic energy in inelastic collisions needs to go somewhere. And whatever receives that energy will also have momentum (even if it doesn't in the center of mass frame, it will in other frames), so it has to be included in the accounting of momentum conservation as well as energy conservation. A relativistic theory that conserves energy as well as 3-momentum (i.e., standard relativity theory) will still conserve both energy and 3-momentum when you change frames, even in inelastic collisions. (In the simplest case, the lost kinetic energy goes into increasing the rest mass of the two billiard balls, by increasing their temperature.)

What you hypothesized was something different: a hypothetical relativistic theory (which obviously does not match our actual world) in which energy is not conserved but 3-momentum is. In such a hypothetical theory, inelastic collisions could take place without the lost kinetic energy going anywhere: it simply disappears. I am simply explaining why such a relativistic theory is not possible: if energy is not conserved, 3-momentum cannot be conserved either (except in one particular frame, the center of mass frame, but that violates the principle of relativity so it is not allowed in a relativistic theory).

No, the 4-momentum is conserved as a vector given a reference frame, which means all 4 parts are conserved, which means 3-momentum is conserved. It doesn't "go somewhere".

You seem to be obsessed about conservation of the magnitude of 4-momentum under Lorentz transformations. It's the whole vector I'm talking about, not the magnitude. and change-of-frame is not a necessary part of conservation laws.

The claim that you can observe conversation in one frame but not another is absurd. The laws are affine; they hold in any frame or none.

> the 4-momentum is conserved

In standard relativity, yes. But not in your hypothetical "relativistic system where momentum is conserved but energy isn't". You said such a system was easy to imagine. I simply pointed out that, easy to imagine or not, such a system doesn't work. Nothing that I said about 3-momentum being conserved in one frame but not another applies to standard relativity where the 4-momentum is conserved. It only applies to the hypothetical system that you claimed was easy to imagine. I don't understand why you keep talking as though my comments about 3-momentum being conserved in one frame but not another referred to standard relativity.

> It doesn't "go somewhere".

By "go somewhere" I simply meant energy can get transferred from one part of the system to another. I didn't mean that total energy wasn't conserved (in standard relativity).

> change-of-frame is not a necessary part of conservation laws.

It is in a relativistic theory, since the principle of relativity requires that the laws of physics, including conservation laws, must hold in all frames.

> The claim that you can observe conversation in one frame but not another is absurd.

For standard relativity, of course it is. But not for your hypothetical system.

> The laws are affine; they hold in any frame or none.

This seems to contradict what you said earlier in the same post, that "change of frame is not a necessary part of conservation laws".

You seem confused by how conservation works, and intent on misunderstanding it.

All I can say is that momentum conservation is Lorentz invariant in my system, and hence cannot depend on choice of reference frame. The conserved quantity is not invariant, and conserved quantities do not need to be Lorentz invariant, and fact almost always are not. If this confuses you, sorry, but I cannot follow your reasoning anymore.

> All I can say is that momentum conservation is Lorentz invariant in my system

If by "my system" you mean your hypothetical theory where energy is not conserved but 3-momentum is, I strongly suspect you have not done the math. See below.

> The conserved quantity is not invariant, and conserved quantities do not need to buslye Lorentz invariant, and fact almost always are not. If this confuses you, sorry, but I cannot follow your reasoning anymore.

I understand what you are saying here, but it does not refute what I was saying. You are saying that, for example, in standard relativity, energy is conserved--in a given frame, it stays the same through any series of events--but it is not frame invariant; changing frames changes the energy. I agree with that.

What I am saying is something different: in your hypothetical relativistic theory in which energy is not conserved, then 3-momentum cannot be conserved in any frame other than one particular one (which in my example was the center of mass frame). I am not saying the numerical value of the 3-momentum has to be the same in all frames. I am saying that in all frames but one, the numerical value of the 3-momentum, as evaluated in that frame, is different after an inelastic collision than before, i.e., 3-momentum is not conserved in that frame.

Here is the math backing up that assertion. Suppose 3-momentum is conserved in the center of mass frame. In that frame the 3-momentum is zero. We assume that our system consists of just two billiard balls, each with the same rest mass m. Before the collision, one ball has speed v and the other has speed -v (we can restrict ourselves to one spatial direction). After the collision, one ball has speed w and the other has speed -w, where w < v (because the collision is inelastic). So 3-momentum is zero before and after the collision in this frame, and energy is not conserved--it is smaller after the collision.

Now transform to any other frame. Call the relative velocity of the Lorentz transformation u. Then, in this frame, the speeds of the two balls before the collision are (I am using units where c = 1)

v+' = (u + v)/(1 + uv)

v-' = (u - v)/(1 - uv)

and the speeds after the collision are

w+' = (u + w)/(1 + uw)

w-' = (u - w)/(1 - uw)

Now we evaluate the 3-momentum in this frame. Before the collision, it is

m ( v+'/sqrt(1 - (v+')^2) + v-'/sqrt(1 - (v-')^2) )

After the collision, it is

m ( w+'/sqrt(1 - (w+')^2) + w-'/sqrt(1 - (w-')^2) )

Substituting and straightforward algebra simplifies these to

before = 2mu/sqrt[(1-u^2)(1-v^2)]

after = 2mu/sqrt[(1-u^2)(1-w^2)]

These are obviously not equal, hence 3-momentum is not conserved in this frame. (In standard relativity, where energy is conserved, 3-momentum would be conserved as well; as I mentioned before, the simplest way for that to happen would be for the two billiard balls to heat up, increasing their rest mass. More complicated ways would involve other particles, or the billiard balls emitting radiation, or something like that.)

Just to clean up one item: I did assume in the above that the rest mass of the billiard balls was unchanged in the inelastic collision. But we can drop that assumption and still prove that, if energy is not conserved in the center of mass frame, 3-momentum cannot be conserved in any frame other than the center of mass frame. Suppose the rest mass of the balls after the collision is M (instead of m). Then we have for the 3-momentum before (P0') and after (P1') the collision, in a frame with velocity u relative to the center of mass frame:

P0' = 2mu/sqrt[(1-u^2)(1-v^2)]

P1' = 2Mu/sqrt[(1-u^2)(1-w^2)]

But we can simplify this by writing down the total energy before (E0) and after (E1) the collision, in the center of mass frame:

E0 = 2m/sqrt(1-v^2)

E1 = 2M/sqrt(1-w^2)

So we can see that

P0' = E0 (u/sqrt(1-u^2))

P1' = E1 (u/sqrt(1-u^2))

Hence, if E0 > E1, we must also have P0' > P1. In other words, the only reason we happen to have P0 = P1 in the center of mass frame is that being in that frame is equivalent to having u = 0 in the above formulas.

(It is also straightforward to show that if energy is not conserved in the center of mass frame, it is not conserved in any frame. So the Lorentz invariance of the two conservation laws, energy and momentum, cannot be separated--they are inseparably linked.)

"Under no scientific theory can you pluck energy from the vacuum."

You can if you're very patient, have a lot of material, and lucky aim, thanks to the metric expansion of space.

From within a galaxy cluster shoot one end a very very long wire into the central black hole of a galaxy in a distant galaxy to anchor the far parts of the wire, and wind some near part of the wire around the pole of an electrical generator in your galactic cluster.

You won't get much power for a long time by human standards, but you'll get an awful lot of energy (and eventually substantial power give or take divergence in the stress-strain tangent modulus and related quantities) even after the gravitationally anchored far end has receded outside your generator's Hubble volume.

Mining \Lambda dark energy for fun and profit, brought to you by $nabla_\mu T^{\mu\nu} = 0$, where $nabla_mu$ is the covariant derivative. It only takes a large up-front investment that you'll wish you had made billions of years ago.

Did you see my response to the parent, where I give a sketch of how conservation of momentum follows directly from conservation of energy in all frames of reference? What is your objection to that argument?

Remove either conservation law and work out the equations for an elastic collision, and you'll see a new degree of freedom pop up. In classical mechanics they are logically independent, and since the space is affine, frames of reference don't help you.

Your sketch takes as a premise, something that is actually a consequence of conservation laws: the way quantities change under frame-of-reference changes.

You can get both by looking at the least action principle under frame of reference changes and using Noether's theorem, but that's not what your sketch is about.

Is there some objective sense in which conservation of momentum is more fundamental than conservation of energy in a different reference frame? Since either can be derived from the other, how do you determine which is "actually a consequence" of which? (I ask out of genuine curiosity)

Well, what is fundamental isn't the point. If you imagine a world where different quantities are conserved, there would no doubt be different change-of-frame equations describing how they are. The point is you can imagine a world where energy is conserved but momentum is not. Actually it's possible to imagine physical systems, with affine frames of reference, where nothing is conserved at all.

Fundementality is a different thing, but it's more natural to assume conservation laws and take frame-of-reference equations are a consequence. You can do it the other way (I'm pretty sure, but not 100% sure because my physics math is rusty) but you're still making assumptions, and you still don't have one conservation law following from the other.

You've got it backwards: in an elastic collision, kinetic energy is conserved by definition. Removing one conservation law makes the whole exercise physically meaningless because elastic collisions require as a given that kinetic energy will not be converted to other forms of energy. This premise guarantees that the momentum is treated as a quantity separate from the total energy of the system (which, as a reply to your first post pointed out, when expressed in relativistic units, includes not just the energy-mass equivalency but energy-momentum as well).

In reality, momentum in almost every collision will be converted into other forms of energy such as when intramolecular bonds are broken and materials are deformed in a car accident or when momentum is converted into mass in particle colliders. The elastic collision equations are only useful as a rough approximation precisely because they go out of their way to simplify the energy-momentum equivalence to conservation of kinetic energy just like Newtonian mechanics is a simpler but useful rough approximation of relativity at low energies.

You can't use elastic collisions to reason about energy-momentum equivalence because their most basic assumption is that there is no energy-momentum equivalence.

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