Back to part 1: Three types of velocities. Rapidity parameter y and rapidity r.
Forward to part 3: Different forms of space-time transformations. Quantum velocities in Special Relativity.
In classical mechanics the length of a rod is determined by the expression r=sqr(x2+y2+z2), which is valid in any system and is invariant value. The values x, y, z are projections of the rod on the axes, and these values are not invariant values, - they can change under transition from one system to another system. In the relativistic physics we loose the r invariant, but find the new invariant, - the interval
s=sqr(x2+y2+z2-c2t2).
The choice of signs (signature) is not yet generally accepted. According to our choice of signature the interval will be some positive number when the r is bigger than ct. In this case the interval is space-like. In another case the interval is some imaginary number; the square of interval is negative; the interval is time-like.
Four-dimensional radius vector of a point:
R = (x, y, z, ict) = (r, ict),
Where r is the three-dimensional radius vector of a point. For clearness the three-dimensional vectors we will write in a solid; four-dimensional - solid underline. The units of components of R are meters. If we divide R by c, we'll receive the 4-vector r, with components measured in seconds,
r = (x/c, y/c, z/c, it) = (r/c, it).
Four-dimensional velocity is usually written in the following form:
U = dR/dt = (gv, icg),
Or in the form, normalized to unit vector:
u = dr/(dt) = (gv/c, ig).
The both formulae look simpler if we change gv by b. Moreover, if there exist two types of three-dimensional velocities, then there must exist two types of four-dimensional velocities. It is logically more correctly to name the values U and u as proper 4-velocity, because they were received by differentiating on proper time. Let's also change the designations U and u by B and b:
B = dR/dt
= (b, icg),
b = dr/dt =
(b/c, ig).
Then it is logically correct to suppose that coordinate 4-velocity can be received at differentiating on coordinate time:
V = dR/dt =
(v, ic),
v = dr/dt = (v/?,
i).
By squaring the 4-velocities we'll receive the following scalars:
B2 = b2 + (icg)2
= - c2,
b2 = (b/c)2 + (ig)2
= - 1,
V2 == v2 + (ic)2 = - (c/g)2,
v2 = (v/c)2 + i2 = - 1/g2.
We can see that the squares of proper 4-velocities are invariants, but the squares of coordinate 4-velocities are not invariants.
If we multiply the vector of proper 4-velocity by invariant mass m, we'll receive the new 4-vector, - the energy-momentum, or 4-momentum:
P = mB = (mb, micg) = (p, iE/c).
The same expression we'll receive by multiplying the coordinate 4-velocity by relativistic mass:
P = Vmr = Vm0g = (m0gv, im0cg) = (p, iE/c),
Where mr - relativistic mass, m0 - rest mass.
By squaring the 4-momentum and using the known formula E2=p2c2+m2c4, we'll receive the new invariant:
(p2 + i2E2/c2) = -E02/c2.
Which under multiplying by -c2 gives the square of rest energy E02, and by dividing by -c2 it gives the square of rest mass m02, or invariant mass m2. The choice between m02 and m2 depends on the used math formalism, does mass depend of velocity, or it is an invariant. Both formalisms are still alive now.
Three-dimensional momentum looks better if we use the proper velocity:
p = mb = mgv.
If we write the momentum through the relativistic mass, the expression looks better, if we use the coordinate velocity:
p = mrb/g = mrv.
Nevertheless, here we can see the deep symmetry between velocities.
If we take the derivative dP/dt, then we'll have the force 4-vector:
F = dP/dt = (f, img3(vdv/dt)/c) = (f, i(vf)/c) = (f, iN/c),
Where, f - three-dimensional force vector; N - is the force power.
If we take the similar derivative but on the proper time dP/dt, then
F = dP/dt = (fg, img4(vdv/dt)/c) = (fg, ig(vf)/c),
Which gives F2 = f2, if v and f are parallel to each other.
Thus, we know that the motion of the material body can be described by two types of velocities v and b, and by the rapidity, measured in m/s, which is also interesting.
But what acceleration is the proper acceleration? Which acceleration must be constant at the uniform accelerated motion, in order the astronaut would not die from overloading, and not become weightless after some time of flight?
Let's write comprehensible expressions for different types of acceleration.
dv/dt=d2r/dt2.
Let's note in advance, that dv/dt = 1*dv/dt =
g0dv/dt.
dv/dt=d(dr/dt)/dt
= gd2r/dt2.
dv/dt =
g1dv/dt.
db/dt = d(dr/dt)/dt
= g3v(vdv/dt)/c2
+ gdv/dt.
If v || dv/dt, then db/dt =
g3dv/dt.
If v is perpendicular to dv/dt, then db/dt
= gdv/dt.
db/dt = d(dr/dt)/dt
= g4v(vdv/dt)/c2
+ g2dv/dt.
If v || dv/dt, then db/dt
= g4dv/dt.
If v is perpendicular to dv/dt, then db/dt
= g2dv/dt.
dr/dt = (c*arth(v/c))' =
g2dv/dt.
dr/dt =
g3dv/dt,
Comparing the upper expressions, we can conclude that dr/dt = db/dt, at least if v || dv/dt.
It is well known from the relativistic mechanics that the second Newton low, written in the form f=ma, is nor correct. It's correct form is
f = m (g3v(va)/c2 + ga),
Which is the basis for engineering computations for relativistic accelerators. If we compare this low with the acceleration db/dt:
db/dt = g3v(vdv/dt)/c2 + gdv/dt,
We'll see, that they differ only by the factor m. Consequently
f = m*db/dt.
By the way, the mass again has the status of the measure of inertia. The Newton law is the relativistic law again, but it must be written through acceleration db/dt.
It' simple to receive the following expressions:
dV/dt = (dv/dt, 0),
dV/dt = (gdv/dt,
0),
dB/dt = (db/dt, i(bdb/dt)
/ gc),
dB/dt = (gdb/dt,
i(bdb/dt) / c),
dv/dt = (dv/(cdt), 0)
dv/dt = (gdv/(cdt),
0),
db/dt = (db/(cdt), i(bdb/dt)
/ gc2),
db/dt = (gdb/(cdt),
i(bdb/dt) / c2).
If we multiply B = (b, icg), or b=(b/c,ig) by any of dB/dt, dB/dt, db/dt, db/dt, we'll have zero, i.e., vectors B and b are orthogonal to vectors dB/dt, dB/dt, db/dt, db/dt, and are parallel to each other in their own group.
(b, db/dt) = 0, (b, db/dt) = 0, and so on for any combinations of b and B.
But 4-vectors of v-velocities and v-accelerations are not orthogonal.
The relativistic rocket is the rocket in which the acceleration filling by the astronaut does not change with time. It is easy to show (look my Russian-language page) that proper uniformly accelerated rectilinear motion is described by the constant acceleration
db/dt = dr/dt = const.
The rocket must be made of two tanks, one is of matter, and another is of antimatter. In the combustion chamber the constantly decreasing portion of fuel are burning in every equal dt. The decreasing of portion of fuel is the result of rocket mass decreasing. Thus we can write
dp(M/M0) = const (relatively time t).
Such process of fuel burning reminds us the law of radioactive decay. Computation shows that, if the start mass of a rocket is 1000 kg, if the proper acceleration of a rocket is 10 m/s2, then all fuel will be used in the proper time of about 69 years.
From the other hand, if the Universe is closed and if it's curvature radius is equal to 2*1025m, then the proper time needed for the Universe-round trip is 44 years, and coordinate time for this trip will be near to the "age" of the Universe, i.e. 13,3 billion years. If our Universe is the space-time crystal, and if the rocket will be launched, for example, in 2010, then we can see it today, approaching, from the opposite point of the sky. But what if we change our mind to launch the rocket? That will be defect of a crystal, and our rocket will not get to the future equivalent point, but we'll meet the rocket launched 13,3 billion years ago, with astronauts, who are about 40-years older, than their equivalent copies, remaining on the Earth.
If in the real Universe the proper acceleration is not db/dt=dr/dt, but dr/dt, for example, then the Universe-round trip will take the shorter time interval, but here we can not manage without the additional hypotheses.
Back to part 1: Three types of velocities. Rapidity parameter y and rapidity r.
Forward to part 3: Different forms of space-time transformations. Quantum velocities in Special Relativity.
To index of Space Genetics.