Riemann surface
matlab

Consider the following thought:
“Special relativity holds for frames moving at constant relative velocity, but of course acceleration requires general relativity because the frames are noninertial.”

Ineffable twaddle. Special relativity certainly doesn’t cower before simple kinematical acceleration. On the other hand, acceleration, even just uniform acceleration in one dimension, is not without its connections with general relativity. We shall explore some of them here. For ease of notation, we set $c=1$. In part (d) we’ll put $c$ back.
1a) Let us first ask what we mean by “uniform acceleration.” After all, a rocket approaching the speed of light $c$ can’t change its velocity at a uniform rate forever without exceeding $c$ at some point. Go into the frame moving instantaneously at velocity $v$ with the rocket relative to the “lab.” In this frame, by definition the instantaneous rocket velocity $v$ vanishes. Wait a time $d t^{\prime}$ later, as measured in this frame. The rocket now has velocity dv’ in this same frame. What we mean by constant acceleration is $d v^{\prime} / d t^{\prime} \equiv a^{\prime}$ is constant. The acceleration measured in the fixed lab is certainly not constant! The question is, how is the lab acceleration $a=d v / d t$ related to the truly constant $a^{\prime}$ ? To answer this, let $V=v / \sqrt{1-v^2}$, the spatial part of the 4-vector $V^\alpha$ associated with the ordinary velocity $v$. The same relation holds for $V^{\prime}$ and $v^{\prime}$. Assume for the moment that the primed and unprimed frames differ by some arbitrary velocity $w$. The 4-velocity differentials are then given by:
$$d V^{\prime}=\left(d V-w d V^0\right) / \sqrt{1-w^2}$$
Explain.
1b) Now, set $w=v$. We thereby go into the frame in which $v^{\prime}=0$; the rocket is instantaneously at rest. Prove that $d v=d v \prime\left(1-v^2\right)$. (Remember, $v$ and $v^{\prime}$ are ordinary velocities.) From here, prove that
$$\frac{d v}{d t}=a^{\prime}\left(1-v^2\right)^{3 / 2} .$$

Solution:
This is the standard material of Special Relativity. For a particle moving in an interial frame S along the trajectory $x^\mu=x^\mu(\tau)$ parametrised by the proper time $\tau$, one defines the 4-velocity $U^\mu=d x^\mu / d \tau$ and 4 -acceleration $A^\mu=d U^\mu / d \tau$. Since $d s^2=-c^2 d t^2+d \mathbf{x}^2=-c^2 d \tau^2$, we have $d \tau=d t \sqrt{1-\mathbf{v}^2 / c^2}=d t / \gamma$, so $U^\mu=(\gamma c, \gamma \mathbf{v})$ and
$$A^\mu=\left(\gamma^4 \frac{\mathbf{v} \cdot \mathbf{a}}{c}, \gamma^4 \frac{\mathbf{v} \cdot \mathbf{a}}{c^2} \mathbf{v}+\gamma^2 \mathbf{a}\right)$$

FIG. 1: A particle moving in an inertial frame $\mathrm{S}$ along the trajectory $x^\mu=x^\mu(\tau)$ parametrised by the proper time $\tau$. The particle’s 4-velocity $U^\mu$ is orthogonal (in Minkowski metric) to its 4 -acceleration $A^\mu: U_\mu A^\mu=0$.
In particle’s own frame, $A_0^\mu=\left(0, \mathbf{a}0\right)$. Computing the invariant $A\mu A^\mu=\left(A_0\right)_\mu A_0^\mu$, we find
$$\mathbf{a}_0^2=\gamma^6 \frac{(\mathbf{v} \cdot \mathbf{a})^2}{c^2}+\gamma^4 \mathbf{a}^2$$
If $\mathbf{v}$ and a have the same direction, eq. (2) simplifies to $a_0^2=a^2 \gamma^6$. One can show that for a particle moving in $\mathrm{S}$ under the action of a constant force, $a_0=$ const. This is the situation considered in the present problem (we shall use the notation $a_0$ rather than $a^{\prime}$ in the solutions).
1c) Show that, starting from rest at $t=t^{\prime}=0$,
$$v=\frac{a^{\prime} t}{\sqrt{1+a^{\prime 2} t^2}}, \quad a^{\prime} t=\sinh \left(a^{\prime} t\right)$$
and hence show that (for $x=0$ at $t=t^{\prime}=0$ ):
$$v=\tanh a^{\prime} t^{\prime}, \quad x=\frac{1}{a^{\prime}}\left[\cosh a^{\prime} t^{\prime}-1\right] .$$
The integrals are not difficult; do them yourselves.

Problem 2
Recognising tensors. One way to prove that something is a vector or tensor is to show explicitly that it satisfies the coordinate transformation laws. This can be a long and arduous procedure if the tensor is complicated, like $R_{\mu \nu \kappa}^\lambda$. There is another way, usually much better! Show that if $V_\nu$ is an arbitrary covariant vector and the combination $T^{\mu \nu} V_\nu$ is known to be a contravariant vector (note the free index $\mu$ ), then
$$\left(T^{\prime \mu \nu}-T^{\lambda \sigma} \frac{\partial x^{\prime \mu}}{\partial x^\lambda} \frac{\partial x^{\prime \nu}}{\partial x^\sigma}\right) V_\nu^{\prime}=0$$
Why does this prove that $T_{\mu \nu}$ is a tensor? Does your proof actually depend on the rank of the tensors involved?

Solution:
If $T^{\mu \nu} V_\nu$ and $V_\nu$ are known to be tensors (vectors), we can write
$$T^{\prime \mu \nu} V_\nu^{\prime}=\frac{\partial x^{\prime \mu}}{\partial x^\lambda} T^{\lambda \kappa} V_\kappa=\frac{\partial x^{\prime \mu}}{\partial x^\lambda} \frac{\partial x^{\prime \nu}}{\partial x^\kappa} T^{\lambda \kappa} V_\nu^{\prime}$$
which implies
$$\left(T^{\prime \mu \nu}-T^{\lambda \sigma} \frac{\partial x^{\prime \mu}}{\partial x^\lambda} \frac{\partial x^{\prime \nu}}{\partial x^\sigma}\right) V_\nu^{\prime}=0$$
and
$$T^{\prime \mu \nu}=T^{\lambda \sigma} \frac{\partial x^{\prime \mu}}{\partial x^\lambda} \frac{\partial x^{\prime \nu}}{\partial x^\sigma}$$
since $V_\nu$ is arbitrary. The same reasoning applies to tensors of arbitrary rank.

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