The zero momentum frame of reference

[skotos]

According to my understanding of physics, there should be a frame of reference, such that the net momentum of the observable universe is zero (or as close to zero as possible, given quantum mechanics). Google has been unhelpful in this regard, so I pose the following question to the denizens of SDN:

1) Does such a frame of reference exist?

2) What is the Earth's velocity relative to it?

[Master of Cards]

What you are looking for is called a Privlaged Frame of Refcence which does not exist

[darthbob88]

According to my understanding of physics, there should be a frame of reference, such that the net momentum of the observable universe is zero (or as close to zero as possible, given quantum mechanics). Google has been unhelpful in this regard, so I pose the following question to the denizens of SDN:

1) Does such a frame of reference exist?

2) What is the Earth's velocity relative to it?

1) No, by relativity. If I understand it correctly, the simple answer is that there is no such thing as a fixed frame of reference. See wikipedia on the subject.
2) See 1)

[Kuroneko]

Most likely yes, to a fairly reasonable approximation--approximation because "total momentum of the universe" is not a definable concept in general relativity unless some various simplifying assumptions are undertaken. In the standard Friedmann-Robertson-Walker family of models, for example, there is indeed a "privileged frame" [1] in which the "cosmic dust" is maximally isotropic. This frame also experiences the most proper time, which is the resolution to an interesting version of twin paradox for closed universes [2]. For our universe in particular, this translates to the frame in which the cosmic background radiation is maximally isotropic. Since measurements of this radiation are indeed very isotropic from Earth, one can conclude that Earth's velocity in this frame is "very low" in relativistic terms. I don't have the data to make a more precise estimate.

To see why the total momentum might not be meaningful, take the simplest case of a spherical universe, suppress one spatial dimension, and suppose the whole universe consists of only two particles, one at the north pole of the sphere and one at, say, equator and 0° longitude. The particles may be moving any which way, and thus have any momenta. However, the momentum vector will be a tangent vector at the particle's location. One can visualize this universe as an ordinary sphere embedded in Euclidean space, so the tangent vector in a plane tangent to the sphere at that point. (Differential geometry defines tangent vectors independently of any embedding, but let's not worry about that now.)

One problem should become immediately obvious: if the momentum vectors of the two particles "live" in completely tangent spaces (here, visualized as tangent planes), how can we meaningfully add them? We could "transport" the momentum vectors from one point to the other, keeping the vector always tangent to the sphere. Unfortunately, this depends on the path taken. To see this, say the north particle's vector is along the prime meridian (two possible directions). Transporting it along the prime meridian to the second particle's location gives a momentum vector directed toward either north or south. Transporting it first along the 90° meridian to the equator and then along the equator to the second particle's location, however, gives a vector that's directed toward either east or west. In other words, momentum "here" vs. momentum "there" is not comparable.

Now, it so happens that the sphere is nice enough for us to pick a way to compare the vectors that's mathematically consistent and physically reasonable. However, a general solution to the Einstein field equation might not be nearly so serendipitous.

[1] Not in regards to the laws of physics themselves! It may be useful to use "preferred" for that sort of phenomenon and reserve "privileged" for frames in which symmetry occurs.
[2] Suppose that the universe is closed, and hence finite; let's say the Einstein static universe in which expansion/contraction is not a factor. Then one twin may go around the universe without and meet up with the other twin without changing reference frames at all (thus getting rid of the standard resolution in STR). Each one considers the other to be temporally dilated. Which one is older?

[darthbob88]

(After reading above)Or, of course, I could be wrong. As in most things, you can get close enough, but I maintain that an exact frame of reference at the center of the universe cannot be found.

[CmdrWilkens]

The problem lies if we have, as by all appearences we do, a non-static universe. With all spatial dimensions currently spreading themselves and expanding all particles are in motion relative to one another thus while one could have a very minimal momentum relative to everything else it is quite impossible to be perfectly at rest relative to everything else. Still I suppose one could try to spin relativity on its head and try to function as if everything is expanding away from your point of reference and the point itself being not in motion (as a function of being a point it can't spread). That said I have no where near the level of understanding it would take to judge whether it would be possible to do so mathmatically or observationally.

[Kuroneko]

The spatial expansion of the universe does not contribute any momentum to any object. The FRW family of models do, in fact, have global momentum and angular momentum conservation, but not that of energy (at least, not in general). And since the FRW family are symmetrical about only one reference frame, there is indeed a "privileged frame" in which the total momentum (and angular momentum) is zero. How seriously this is to be taken in regards to our own universe depends on how appropriate an FRW metric is to our universe.

[drachefly]

Actually, the CMBR does have substantial momentum in respect to us. Not exactly relativistic, but the anisotropy due to the doppler shift is large compared to all of its other anisotropies.

Note, physics works the same in this frame as in any other.

[Kuroneko]

Actually, the CMBR does have substantial momentum in respect to us. Not exactly relativistic, but the anisotropy due to the doppler shift is large compared to all of its other anisotropies.

Ah; neat. I couldn't recall of any quantitative measurement of this... you wouldn't happen to have one somewhere, would you?

[drachefly]

That search didn't, irritatingly enough. You'd think a FAQ would have that.

Googling "CMBR dipole velocity" yields a figure of around 620 kilometers per second for not us in particular but pretty much all the galaxies in our general area. I guess that would be a more interesting figure for cosmologists.

By comparison, our galactical orbital speed is 16.5 km/s; our solar system orbital speed is 30 km/s; the speed of LEO is 8 km/s.

[drachefly]

Also for comparison, light speed is 300 000 km/s. So this drift is not even a vaguely relativistic velocity.