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In have always believed that the centripetal force that one experiences on a roundabout is due to ones movement relative to the general gravitational field of the whole universe.What would cause centripetal forces if the whole universe was rotating?.
This Mach Principle is quite hard to get your head around. If you are out on your own in deep space and you fire two rockets, tangentially, on two opposite sides of the ship it is hard to realise that the resulting rotational effects are only experienced because of the rest of the Universe.
Rotation generates forces that can be measured. The idea that Mach's principle applies here is not correct I think. If you are on a ship and fire rockets to rotate the ship you will feel a gravitational effect thought to be (according to general relativity) equal to the required cetripetal force exerted by the outer walls of your ship. You can tell you are moving without reference to the exterior world or you could posulate a gravitational field acting outwards. The force you feel varies with your distance from the centre of rotation. If you say that in an empty universe you would not experience these forces, which may or may not be the case as it is unclear what is meant by an empty universe (less clear than in Mach's day) and then say that the universe as a whole may be rotating but also not experience such forces, then surely the concept of rotation has no meaning in this case.
I would be interested to know if when we experience centripetal forces due to rotation it is due to our rotation relative to the local galaxy or due to the universe as a whole (the fixed stars as Mach put it), I can conceive of no way this could be measured.
which, like in the EM case, goes as 1/R instead of 1/R2
Quotewhich, like in the EM case, goes as 1/R instead of 1/R2Could you enlarge on that? The Coulomb force and radiation law drop off as 1/r2, don't they? Have I missed something? I agree that potential in both cases follows 1/r.
Physical background.The electric field created by an electric charge is: E = (-q/4πε0)[er'/r'2 + (r'/c)(d/dt)(er'/r'2) + (1/c2)(d2/dt2)(er')].where:q is the accelerating chargeE is the electric field vector.c is the speed of light in vacuum.ε0 is the permittivity of free space.r' is the distance from the observation point (the place where E is evaluated) to the point where the charge was r'\c seconds before the time when the measure is done.er' is the unit vector directed from the observation point (the place where E is evaluated) to the point where the charge was r'\c seconds before the time when the measure is done.The "prime" in this formula appears because the electromagnetic signal travels at the speed of light. Signals are observed as coming from the point where they were emitted and not from the point where the emitter is at the time of observation. The stars that we see in the sky are no longer where we see them. We will see their current position years in the future; some of the stars that we see today no longer exist.The first term in the formula is just the electrostatic field with retarded time.The second term is as though nature were trying to allow for the fact that the effect is retarded (Feynman).The third term is the only term that accounts for the far field of antennas.The two first terms are proportional to 1\r^2. Only the third is proportional to 1\r.
But a 1/r law, or at least a non-inverse square law, would be reasonable if the spherical symmetry of the situation no longer held and it wouldn't be a 'point source' situation because it would involve a very distributed cause.
If you have a rotating universe in any meaningful sense of the word rotation, you have an axis hence anisotropy. In an "empty" universe you can postulate that you would not feel any forces when rotating because you are not rotating with respect to anything and angular momentum would not have any magnitude. So how would this apply to a rotating universe? For argument's sake, assuming a closed universe (which may not be the case), would it not be also difficult to say with what the universe is rotating with respect to?
Quote from: graham.d on 13/08/2008 13:59:39If you have a rotating universe in any meaningful sense of the word rotation, you have an axis hence anisotropy. In an "empty" universe you can postulate that you would not feel any forces when rotating because you are not rotating with respect to anything and angular momentum would not have any magnitude. So how would this apply to a rotating universe? For argument's sake, assuming a closed universe (which may not be the case), would it not be also difficult to say with what the universe is rotating with respect to? I agree, it's exactly what I claimed: inertial forces exist within a universe, that is when we are in a ref frame which rotates with respect to the other masses of the universe; but the universe as a whole doesn't rotate with respect to anything, so for the same reason no inertial forces, no anysotropy should appear, in my opinion.
I think if there were a rotation it would define a special point in the universe about which rotation occurred. None has ben found and is not in any models as far as I know. Also the universe would have an "up" and a "down" along any rotational axis so would no longer be isotropic. This would not fit in with observation but it is possible it is sufficiently small that it is unnoticable. The concept of an overall net rotation would mean that there was a very large angular momentum just after the Big Bang and I suppose an infinite one at some point. I don't think any models have this and it maybe not even a practical supposition, I'm not sure. Certainly the models that already exist are complicated enough without one that introduces artifacts to cope with a possible feature that has never been observed.