The upgrade from conventional 3d velocity, which is (dx/dt,dy/dt,dz/dt), to a 4-vector is not as trivial as it may seem. The first temptation would be to keep these components and add the 0 component as dt/dt=1. But then, its interval would be 1-v², which is not a Lorentz invariant. But this attempt gives us the clue for how to build a 4-velocity: just divide our previous guess by √.
The book uses the letter u for the four-velocity, but we are using it for relative velocity of isor. But letters are limited, so we are using u too. We define
uⁱ = dxⁱ/ds
where recall that ds=c·dτ, where dτ is the proper time. In other words, four-velocity divides by proper time dτ instead of using coordinate time dt. Recall also that dτ = dt·√, so ds = cdt√. Here, √ is √{v}. Then, we get
uⁱ = (1/√ , |v/c>/√).
A surprising thing here is that the four-velocity is a dimensionless quantity! I think this is quite confusing, and that we should have taken uⁱ/c as the dimensionless version of it.
The magnitude of this vector is uⁱuᵢ = (1/√)²·(1-v²) = 1.
The four velocity is, then, a *unit* four-vector with a direction always tangent to the world line of the moving particle.
Can we get a 2nd derivative here, as we do in 3d to get the acceleration? Of course, and again we differentiate over proper time, so
wⁱ = d²xⁱ/ds² = duⁱ/ds,
where wⁱ is called the four-acceleration.
We know from classical mechanics that the unit vector who is tangent to a path is always perpendicular to its derivative. Let's review this. Recall that a vector is written here as |v>, that its magnitude can be written as v. But we need also a notation for unit vector |v>/v. We can write |^v^> for an explicit unit vector.
Then, every vector |v> = v|^v^>. If we now perform a total derivative
d|v> = dv|^v^> + vd|^v^>.
However, if our vector is unitary from the beginning (|v>=|^v^>), dv=0, since v=constant. Then,
d|v> = vd|^v^> = vd|v> = d|v>. (trivial)
If we now multiply |v>·d|v> we see how we can write
(cosϕ,sinϕ)·(-sinϕ,cosϕ)dϕ = 0.
In general,
|^v^>·d|^v^>=0.
In our relativistic case, the equivalent is
uⁱwᵢ = 0.
We can see how d(uⁱuᵢ) = wⁱuᵢ = uⁱwᵢ = 0.
So the four-velocity and the four-acceleration of a particle are "perpendicular" in this geometry.
Determine the relativistic uniformly accelerated motion, i.e. the rectilinear motion for which the acceleration a in the proper reference frame (at each time) remains constant.
For the proper frame, v=0, and if a is constant along x, then wⁱ=(0,a/c²,0,0). Why a/c²? Because of dimensions. Remember that w is the derivative d/ds of the dimensionless 4-velocity, so w has dimensions 1/L, and then, a/c² has dimensions L·T²/(T²L²) = 1/L.
The magnitude of the 4-acceleration is now -a²/c⁴. This must be Lorentz invariant.
In the fixed frame, the acceleration a must be the same, but now it is defined as
a = d/dt(v/√).
As you can see, while in classical mechanics we have a=dv/dt, in relativity we have a=d(v/√)/dt.
Integrating, we get
v/√ = a·t + C
where we can set C=0 if v=0 at t=0. Then, we can isolate v from this expression to get
v = a·t / (1 + a²t²/c²)¹⸍².
In order to get x we must integrate over time once again, since dx/dt = v. We see how the numerator contains the derivative of what is inside the square root of the denominator, so we can directly write
x = c²/a·(1 + a²t²/c²)¹⸍² + C.
If we want x=0 for t=0, then C=-c²/a, and
x = c²/a·[ (1 + a²t²/c²)¹⸍² - 1 ].
If you miss the second degree polynomial of classical mechanics, we need to consider low speeds wrt c. However, no speed appear here in an explicit way. We can consider that a·t gives a good estimator of its order of magnitude, so we take the limit a·t << c. Then we get v = a·t and x = at²/2, the usual formulae.
For a·t → ∞ we see how v → c, as it should be.
We can calculate the proper time measured by the travelling clock. We know it is ∫dτ=∫(1-v²/c²)¹⸍²dt, and now we have an expression for v(t), so we can plug it here. We arrive to the integral
Δτ = ∫ 1 / (1 + a²t²/c²)¹⸍² dt .
At this point, one choice is to remember the Moses/Leibniz integration tables, which are God Given, and recall that
dsinhθ/dθ = 1 / (1+θ²)¹⸍².
This immediately leads to the result
Δτ = (c/a)·arsinh(a·t/c).
Mind the ar in arsinh. We don't use "arc", since the function of a hyperbolic function is not related to arcs.
If you don't like following tables, you can think of a right-angled triangle with hypotenuse 1, as
·
/|
/_|
1 / α| (1-x²)¹⸍²
/ |
/\θ__|
x
Then, it is clear that 1/sinθ = 1/(1-x²)¹⸍². The integral ∫1/(1-x²)¹⸍²dx needs dx. But x=cosθ, so dx=-sinθ·dθ. Then, the integral becomes very easy, just -∫dθ= -θ + C. We need to determine C. For very small x we see that the integral is ∫dx = x (1st order), and for this limit, θ tends to π/2. This means x ≃ -π/2 + C ≃ 0, so C = π/2. Once we know this, we can write π/2 - θ = α, but sinα = x, so π/2 - θ = arcsinx. Conclusion, ∫dx/(1-x²)¹⸍² = arcsinx. Without tables! Unless you consider that d(cosθ) = -sinθ·dθ also requires a table!
Once we know this, it is easy to go for the case of ∫dx/(1+x²)¹⸍². If I=arcsinx, then sinI=x. Now we want to transform into sinh, and we know that i·sinI = sinh(iI), so x = -i·sinh(iI) and arsinh(ix)=iI= ∫d[ix]/(1+[ix]²)¹⸍² = ∫dy/(1+y²)¹⸍² = arsinh(y) where y = ix. Done!
Always keep in your memory this:
sin(ix) = i·sinhx sinh(ix) = i·sinx,
where the i going OUT just flips the character of the function. For cosines,
cos(ix) = coshx cosh(ix) = cosx,
where i does not go out but changes the character of the function anyway.
If we do the same for -1 instead of i, and always begin by having our test subject INSIDE the trig function, then sin(-x) = -sinx, we see how the test subject, -1, goes OUT of the function. In cos(-x) = cosx it never goes out. If the test subject is now i, it's the same but with the function changing character. So, in sin(ix), the i must go out and give isinhx. And in sinh(ix), we must get isinx. For cos(ix), we must get coshx, and for cosh(ix), we must get cosx. And for sin(-ix)? No problem, there are two test subjects together, and they are independent, so we get -isinhx, and so on.
Going back to the physics, consider how interesting is this. We have a clock travelling with constant acceleration a, and now we have an expression to compute the total accumulated proper time as a function of t.
We may be interested in how the expression Δτ=(c/a)·arsinh(a·t/c) behaves for low t and for big t. There is no need of magic formulas again.
For small t, as we already know that arsinhx = ∫dx/(1+x²)¹⸍², we can expand 1/(1+x²)¹⸍² with the binomial theorem, obtaining ≃ 1 - x²/2. Then, we just need to integrate this wrt x to get x - x³/6. For 1st order only, we get arsinhx ≃ x, which implies Δτ ≃ (c/a)·(a·t)/c = t. So for small t's, coordinate time and proper time grow similarly.
And for big t? Again, arsinhx = ∫dx/(1+x²)¹⸍² is the key. But now we want x² >> 1, so we can rewrite get rid of the 1 in the denominator and have just ∫dx/x = ln(x)+ln(C) = ln(C·x). The problem, of course, is to know C here.
We know that sinhx = (eˣ - e⁻ˣ)/2, so x = arsinh[(eˣ-e⁻ˣ)/2]. For big x, e⁻ˣ tends to disappear, so x ≃ arsinh[eˣ/2]. Defining y=eˣ/2 we get ln(2y) ≃ arsinh[y]. From here we learn that C=2.
Then we can write Δτ ≃ (c/a)·ln(a·t/(2c)). So as t increases, the proper time increases more slowly than coordinate time.