# Momentum is Fundamental?

1. Nov 9, 2013

### VernonNemitz

To the best of my knowledge, General Relativity considers Momentum to be as fundamental a thing as Energy. However, in day-to-day descriptions of Things in Physics, Momentum always seems to be dependent upon something else (such as Mass in motion). It seems to me that if Momentum was truly a Fundamental thing, it should be able to exist independently of anything else (like a piece of Energy can exist independently of other things, in the form of a photon).

Has any work been done along the lines of attempting to figure out whether or not Momentum can exist in some independent form? Or, to rephrase that Question, has any work been done to prove that Momentum must always exist as a dependency of something else (like Mass in motion)?

I only know of one place in Physics where formally-published papers have had a chance of touching upon an Answer to that Question, and those papers discuss a hypothetical thing called "negative mass". Basically, consider this simple hypothetical interaction between two particles:

(m)(v)---> (poof!) <---(-m)(-v)

If the masses and velocities are equal and opposite, then Zero mass will be left over after the event (the process was called "nullification" by Dr. Robert L. Forward), and there also will be Zero kinetic energy left over. However, *all* the Momentum will be left over ...! What *form* could that Momentum possess? Unfortunately, so far as I know, none of the papers on negative mass ever got around to considering that question.

What can you folks tell me?

2. Nov 9, 2013

### Staff: Mentor

A photon also has momentum and spin and wavelength and frequency, so I don't know why you would think it is an example of energy existing independently of other things.

3. Nov 9, 2013

### dauto

You don't have to go to GR for that. Momentum is as fundamental as energy in SR and in Newtonian physics as well. It is just as fundamental as energy in every single physical theory out there.

4. Nov 9, 2013

### WannabeNewton

Energy doesn't have any more of an independent existence than momentum. Classically, energy and momentum are carried by both particles and fields (and can be exchanged between them) so it is clear that neither has any more of an independent existence than the other.

For example in the case of the classical electromagnetic field, the 3-momentum density of the electromagnetic field is given by the Poynting vector $\vec{S} = \frac{c}{4\pi}\vec{E}\times \vec{B}$ and the energy density is given by $\varepsilon = \frac{1}{8\pi}(E^2 + B^2)$. Furthermore we can use Maxwell's equations to derive a conservation of energy-momentum equation involving $\vec{S}$ and ${\varepsilon}$: $\vec{\nabla}\cdot \vec{S} + \partial_t \varepsilon = -c\vec{j}\cdot \vec{E}$.

5. Nov 9, 2013

### Staff: Mentor

Also, I have never heard of nullification. We cannot discuss speculative topics here. Can you provide a legitimate reference for that?

6. Nov 9, 2013

### Mentz114

Energy and momentum are the Noether conserved quantities under time-translation symmetry and spatial-translation symmetry respectively.

Last edited: Nov 9, 2013
7. Nov 9, 2013

### dauto

I can tell you that negative mass doesn't really exist.

8. Nov 9, 2013

### WannabeNewton

You can't really take these as the fundamental definitions. If a massive particle with 4-momentum $P$ passes by an observer with 4-velocity $U$ at some event $p$ on both their worldlines then the observer will measure its energy at $p$ to be $E = -g_p (P,U)$. This relation holds in any space-time $(M,g)$, regardless of if a time-like killing field is present or not. Even when we have a time-like killing field $\xi$, for massive particles we can only define a conserved energy $E = -g(P,\xi)$ along time-like geodesic worldlines with 4-velocity $U$ because only for these do we have $\nabla_U g(P,\xi) = 0$ on account of $\nabla_U P = 0$ and $\nabla \xi^{\flat} = -\frac{1}{2}d\xi^{\flat}$. If the space-time is also asymptotically flat then we can interpret $E = -g(P,\xi)$ as the energy as measured by a stationary observer at infinity so it is clear that the conserved energy has its limitations in usage.

9. Nov 9, 2013

### bcrowell

Staff Emeritus
A photon that has energy also has momentum. A photon that has momentum also has energy.

Energy and momentum are unified in relativity as different parts of the energy-momentum four-vector, so their logical status is exactly the same.

10. Nov 9, 2013

### K^2

The actual fundamental quantity is Energy-stress tensor. Energy and momentum, by themselves, are frame-dependent. Observer in one frame of reference will see one energy and momentum, observer in another will see something else. And how these transform depends on exactly what sort of system we are talking about.

Energy-stress tensor is what brings it all together. It is a conserved quantity in all reference frames. Indeed, it is the conserved charge of General Relativity and its local symmetries. Furthermore, given Energy-stress tensor and a reference frame, one can compute energy and momentum.

11. Nov 9, 2013

### WannabeNewton

Well it's fundamental insofar as it has the ability to characterize the energy-momentum density of (in this context classical) fields that aren't the gravitational field. For the energy-momentum of the gravitational field, we are relegated to local quantities such as the Landau-Lifshitz pseudotensor or directly global quantities like, for example, the Komar mass of a stationary asymptotically flat space-time with time-like killing field $\xi$: $M = -\frac{1}{8\pi}\int _{S^2_{\infty}}\star d\xi^{\flat}$.

12. Nov 10, 2013

### pervect

Staff Emeritus
I don't know what you mean by "exist in some independent form". I can say that massless objects, like photons, can and do have momentum, so your concept of momentum as "mass in motion" is probably flawed.

I can also say that momentum isn't a tensor, and that energy isn't a tensor, but the combination of the two in the form of an energy-momentum 4-vector is - but unless you share my belief that tensors are in some sense "fundamental objects", I don't know how much use this observation will be to you.

I can also add that we currently have multiple defintions of mass in General relativity, NONE of which can be applied in all situations, so that "mass" is probably a lot less "fundamental" than you think it is.

13. Nov 10, 2013

### stevendaryl

Staff Emeritus
The fundamental equation relating mass, energy and momentum for a particle is:

$E^2 - p^2 c^2 = m^2 c^4$

According to this equation, it is impossible to have $|E| < |p| c$, because that would make $m^2$ negative, which is impossible (unless there is some way to make sense of imaginary mass). So it's not possible to have a particle with zero energy and nonzero momentum.