B Special relativity - frame of reference

[Grinkle]

length-contraction argument is "only" a heuristic

To me calling it a heuristic and jumping straight to a calculation for it means that the more complete way of thinking about it requires an understanding of relativity of simultaneity.

There is a bit of judgement involved imo when one calls something a heuristic.

 

[pervect]

To be clear, I'm aware that charge density is the temporal component of the four-current-density.

What I was thinking of (but not expressing well) is the relation $V = V_0 / \gamma$ between a volume $V$ and the corresponding proper (rest-frame) volume $V_0$ (because of length contraction along the axis of observers' relative motion). This is relevant to the relation between a charge density and the corresponding proper charge density.

The volume element $V_0$ associated with the rest frame of an object is a different entity than the volume element V in a different frame, a frame in which the object is moving, due to the relativity of simultaneity. Writing that $V = V_0 / \gamma$ treats them as if they were the same object, but they're really not the same object in a geometrical sense. Hence the cautions.

The most elementary way I'm aware of showing this is to draw a space-time diagram. One colors in the set of points represented by the volume V in some moving frame, and contrasts it to the set of points representing the volume $V_0$ in some stationary frame. (Moving and stationary are of course regarded as a matter of convention). Then one shows that the two sets of points are different sets.

The fundamental issue here is the relativity of simultaneity, and the space-time diagram is the most elementary tool that I'm aware of that illustrates the issues associated with the relativity of simultaneity.

Of course it's difficult to draw 4-d diagrams, so I advocate reduce the number of dimensions to get the insight. I suggest starting with a 2d diagram, and then drawing (or just imagining) a 3-d diagram. The volume element in the 1+1 diagram is a line segment, in the 2+1 diagram it's a segment of a plane.

 

[vanhees71]

No, it's not only a heuristic, but in a way it specializes to very special "foliations" of spacetime, namely hypersurfaces of constant coordinate times in inertial reference frames.

The most general case is however much easier formulated in terms of covariant objects, i.e., the four-vector current and, for the corresponding integrated quantities via the covariant integration over space-like 3D hypersurfaces, which lead to scalar quantities, no matter whether you integrate a conserved or some non-conserved current.

The trouble with the integrated quantities is that they are not local quantities like the fields. That lead to very confusing problems before the full theory of relativity has been discovered and even for some years after 1905. The breakthrough was Minkowski's mathematical analysis of SRT spacetime, leading to Minkowski vectors, tensors, and all that.

The most (in)famous classic problem within SRT is the definition of (integrated) electromagnetic field energy at the presents of charge-current distributions. When integrating the energy-momentum-stress tensor naively over an arbitrary constant-coordinate-time hypersurface of an arbitrary intertial frame you get a contradiction with the total-energy-momentum relation ("on-shell condition"), but that's of course because you integrate over a non-conserved current. Only the total energy, i.e., the electromagnetic an mechanical energy and momentum are of course conserved, and you can get these quantities by naive integrating over these special hypersurfaces, and then energy and momentum build together a proper four-vector as it must be.

See Jackson, Classical Electrodynamics for a very careful treatment of these problems.

 

[Alan McIntire]

I get the following from a book by Paul Davies.

Assume Alpha Centari is exactly 4 light years away, and one twin is
traveling there at 4/5 speed of light. (Using a 3,4,5 triangle I avoid
complicated messy fraction in my computations.

Traveling at 4/5 the speed of light, from the point of view of the
stay at home twin, the trip will take 10 years, 5 years there, 5 years
back.

Time for the traveller T' = T( sqrt( 1- (v^2/c^2))) = 3/5 T
Likewise, the distance for the traveler, D' = 3/5 D

The traveler on the spaceship sees himself traveling a distance of
4*3/5 = 2 2/5 light years in a time of 3 years, and likewise the 2 2/5
light years back
in a time of 3 years, so the traveler will see the trip as lasting 6
years.

Suppose the twins have super telescopes and can see each other throughout
the trip.
As long as they are traveling apart, the twins will see each other as
aging at 1/3 speed. As long as they are traveling towards each other,
the twins will see each other as aging at triple speed.

The difference is, the traveling twin will see the stay at home twin
as aging at 1/3 speed for the 3 years to Alpha Centauri,
and at triple speed for the 3 year trip back to Earth for a total of
3* 1/3 + 3*3= 1 + 9= 10 years.
The stay at home twin will see the travel age at 1/3 speed for 9
years, the 5 years it takes the traveler to get to Alpha Centauri,
plus the 4 years it takes the light to get back to earth. Since the
total trip will take 10 years, the stay at home twin will see the
traveler age at triple speed during the one year he observes the
traveler coming back to earth. The Earth observer sees the traveler
age at 1/3 speed for 9 years, and at triple speed for 1 year, for a total of
1/3*9 + 3*1 =3+3=6 years.
Both observers see each other aging at the same slow rate while moving
apart, they see each other aging at the same fast rate while moving
together. The difference lies in one observer deliberately changes
the relative motion of his rocket from moving away from Earth to
moving towards earth, and the other observer remaining passive, and
not seeing the change until the light from the
traveler reaches earth. If the Earth could be accelerated like a
rocket ship, and the earthbound observer decided to change his frame
so the rocket appeared to be moving towards him at 4/5 lightspeed
rather that away at 4/5 lightspeed, while the rocket remained in
motion past Alpha Centauri, then it would have been the Earth twin who
appeared to age less.
Of course you could have some intermediate situation where BOTH
observers decide to change their relative motion before they see the
other observer change his motion.

Say A and B are flying apart at 12/13 the speed of light. As long as
they are receeding from each other, they will see each other as moving
at ((1 - 12/13)/(1+ 12/13))^(1/2) = 1/5 normal speed. When they
approach each other, they will
see each other as moving at 5 times normal speed.((1 + 12/13)/(1 -
12/13)^(1/2).
If they are separated by 10 light years when A changes direction, A
will see B approaching immediately, and see B aging at 5 times speed
immediately. B, who hasn't done anything to change relative motion,
will see A continue to recede at 1/5 speed, and won't see A approach
at 5 times normal speed until 10 years have passed, the 10 years it
takes light from A to reach B. Whoevher changes direction, A or B,
will IMMEDIATELY see the other change direction, and start aging more
rapidly than normal. The other party will not see the change until
the light reaches him or her. It doesn't matter who fires their
rocket or feels acceleration, but if they're ever going to meet again,
one of them must do so.
With special relativity, two observers traveling apart will see each
other appear to age at a rate slower than normal, each will see the
other age at the same slower rate. Two observers approaching will see
each other appear to age at a rate faster than normal, the same fast
rate. The DIFFERENCE is, if they do nothing, they will continue to
travel on different paths, never meeting again, so no one could tell
who is "REALLY" aging faster. One of the two will have to fire his
rocket and change direction. This is the observer who will age less.