# Intrinsic momentum

1. Sep 1, 2010

### wdlang

is there intrinsic momentum for some particle?

there are intrinsic angular momentum (spin) and intrinsic parity, so why not intrinsic momentum?

2. Sep 2, 2010

### fscman

Even if a particle did have intrinsic momentum, it would be undetectable, since we can always choose an inertial reference frame where this equals zero. Intrinsic angular momentum is detectable since in every inertial frame we agree that the particle possesses angular momentum.

3. Sep 2, 2010

### wdlang

but the particle does not move at all

it is surely a surprise that a static particle has momentum

4. Sep 2, 2010

### granpa

Last edited: Sep 2, 2010
5. Sep 2, 2010

### Bob S

If you include the "intrinsic" momentum of a decay particle in a two-body decay, then the decay of a charged pion at rest into a muon and a neutrino qualifies. The measured muon "intrinsic" momentum in pion decay is ~29.7877±0.0014 MeV/c. See

http://prd.aps.org/abstract/PRD/v20/i11/p2692_1

Bob S

6. Sep 2, 2010

### fscman

How do we measure momentum? As far as I know, there is no simple way of measuring momentum by itself; we need to measure the velocity of the particle with respect to a certain frame, and multiply that by m (or by $$\gamma m$$). This will give you the total momentum of the particle, not just the intrinsic or extrinsic parts, since we cannot break the 4-velocity into "intrinsic" and "extrinsic" parts. Then, how do we define the intrinsic parts of the momentum?

Another way of looking at it is, the product $$\gamma m u$$ where u is the 4-velocity cannot just represent the extrinsic momentum. If it does, then, how does the intrinsic momentum manifest itself?

To be sure, an intrinsic momentum could be defined, but it isn't measurable and is not physically meaningful.

As for the cited experiment, I do not believe that the muon was at rest after it was emitted. The "at rest" specifies the pion.

Hope that helps to explain my view,
fscman

7. Sep 2, 2010

### Dr Lots-o'watts

Doesn't any photon have an intrinsic linear momentum p=hk ?

No dependency on inertial frame for a photon...

8. Sep 2, 2010

### fscman

The photon momentum do have a dependency on an inertial frame.
According to relativity, E=pc. We also know that E=hf, so
p=hf/c.
The thing to know is, f is dependent on an inertial frame. In fact, f is the temporal part of a 4-vector with the wave vector k being the spatial parts.

So, the momentum does depend on a inertial frame. The question is, does that count as "intrinsic" momentum?

9. Sep 2, 2010

### Staff: Mentor

10. Sep 3, 2010

### Dr Lots-o'watts

Ah yes... Doppler.

11. Oct 16, 2010

### Xia Ligang

An idea bursts into my brain. Could intrinsic momentum be the "mass"? You know, they are nearly the same from the view that E square is the sum of m square and p square. Maybe by introducing the intrinsic momentum, we could find some relations among the masses of all particles.

12. Oct 16, 2010

### Bob S

Momentum and intrinsic momentum are frame dependent. Intrinsic (or invariant) mass, given by

(Mc2)2 = [E2 - (pc)2]½, is frame independent.

Bob S

13. Oct 16, 2010

### Xia Ligang

As we can change the orientation of an intrinsic spin, but cannot change its magnitude, it is likely that the magnitude of "intrinsic" moemtum is invariant, but its direction could be altered. If "intrinsic" momentum possesses this property, we can extend the perception of mass. A scalar "mass" becomes a vector "intrinsic" mementum. "mass" has directions. But it is just an alternative. It might not be better than the original mass. Anyway, everything needs to be tested in the experiments.