Is energy-momentum invariant and/or conserved?

[greypilgrim]

Hi.

I'm reading an introductory text that somehow seems to confuse if $E^2-(cp)^2=const$ means that the left side is invariant (under Lorentz transformations) or conserved (doesn't change in time). As far as I understand it, they only prove Lorentz invariance.

Are they both true? If so, isn't conservation trivial by conservation of energy and conservation of momentum? If not, I assume there should be a corresponding symmetry according to Noether's theorem, which would be what?

 

[Orodruin]

Are they both true? If so, isn't conservation trivial by conservation of energy and conservation of momentum?

Yes and yes. Mixing up invariance with conservation is not an unusual mistake.

 

[Andrew Mason]

Hi.

I'm reading an introductory text that somehow seems to confuse if $E^2-(cp)^2=const$ means that the left side is invariant (under Lorentz transformations) or conserved (doesn't change in time). As far as I understand it, they only prove Lorentz invariance.

Are they both true? If so, isn't conservation trivial by conservation of energy and conservation of momentum? If not, I assume there should be a corresponding symmetry according to Noether's theorem, which would be what?

$E^2-(cp)^2=(m_0c^2)^2$ and since m₀ is a property that is the same for all observers, so it is Lorentz invariant and conserved. But energy and momentum are frame dependent so they vary depending on the observer.

AM

 

[Orodruin]

$E^2-(cp)^2=(m_0c^2)^2$ and since m₀ is a property that is the same for all observers, so it is Lorentz invariant and conserved. But energy and momentum are frame dependent so they vary depending on the observer.

AM

The mass is the same for all observers because it is constructed as an invariant, not the other way around. Being the same for all observers also in no way guarantees conservation.

 

[greypilgrim]

Yes and yes.

I've often seen this equation being used to analyze particle collisions. I think one needs the conservation property there. In Newtonian physics, we need both conservation of energy and of momentum. Why does the energy-momentum relation seem to be enough in the relativistic case? Only $E^2-(cp)^2=const$ seems to impose less restrictions than $E=const$ and $\vec{p}=const$.

 

[Orodruin]

Why does the energy-momentum relation seem to be enough in the relativistic case?

It is not enough. You need to conserve 4-momentum. However, you can make a lot of useful inferences by combining the 4-momentum (of the entire system or of individual particles) into invariants.

Can you give an example of a Newtonian situation where you need to use both energy and momentum conservation where in the rlativistic case you only need the system’s invariant mass to get the corresponding information? It may be easier to discuss a concrete example.

 

[Dale]

If not, I assume there should be a corresponding symmetry according to Noether's theorem, which would be what?

The conservation of four-momentum follows from the spacetime translation invariance of the action, just as in the nonrelativistic case with the conservation of three momentum and invariance with respect to spatial translations.

 

[pervect]

One relatively advanced point is worth pointing out. $E^2 - (cp)^2$ is an invariant only for an isolated system.

To take a specfic example, if one has a "box of light", and one ignores the walls of the box (so that one only considers the light trapped inside the box), $E^2 - (cp)^2$ is not the same for all observers. It is not invariant for this non-isolated system. The light isn't isolated because it's interacting with the walls of the box. If one considers the entire system, including the walls, then the system is isolated and then $E^2 - (cp)^2$ is invariant.

Moving onto conservation laws from invariance, the conservation law that applies to the box of light can be formulated in terms of the vanishing of the divergence of the stress energy tensor. This is somewhat similar to conservation laws in fluid mechanics, the form of the law involves partial differential equations (PDE) , sometimes called continuity equations, that are always satisfied. The tensor version of this conservation principle is at graduate level, because that is where tensors are introduced. Potentially, the basic principles can be understood without tensors by considering the PDE's of fluid flow, though.

See for instance https://en.wikipedia.org/w/index.php?title=Continuity_equation&oldid=856772993

In the wiki

$$\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho u) = 0 $$

would be the motivational idea I mentioned. The actual relevant relativistic formula for energy momentum conservation would be the one that says $\partial_\nu T^{\mu\nu} = 0$, which one can see mentioned briefly in the wiki article.

To fully appreciate this formulation of the conservation of energy-momentum, though, one needs to know about the stress-energy tensor, which implies that one knows what tensors are, which implies some graduate level knowledge.