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.
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.)
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.