# Exact definition of momentarily comoving reference frame (MCRF)

Consider a particle $P$ in the framework of special relativity with position $r(t)=(ct,x(t),y(t),z(t))$ respect to an inertial reference frame $\Sigma=(ct,x,y,z;O)$. I need to know if the following is a correct definition of a MCRF for $P$:

A momentarily comoving reference frame (MCRF) for P is a frame reference $\Sigma'=(ct',x',y',z';O')$ such that if $r'$ is the position of $P$ respect to $\Sigma'$, the following two conditions are satisfied:

1) $r'(t'_0)=(ct'_0,0,0,0)$ for every $t'_0$ fixed.

2) $\frac{dr'}{dt'}(t'_0)=(c,0,0,0)$ for every $t'_0$ fixed.

Moreover the motion of $P$ is uniformly accelerated if for every choice of a MCRF $\frac{d^2r'}{{dt'}^2}(t'_0)$ is constant $\forall t'_0$.

If the above definition is wrong I would appreciate a formal correct definition of a MCRF.

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There is no such thing as a "momentarily comoving frame of reference", if by "momentarily" you mean $\Delta t=0$ (do you?), because in such a case there is no movement at all. There is no movement without the passage of time, by definition. – bright magus Dec 31 '14 at 13:39
Is this a block quote? If yes, from which reference? – Qmechanic Feb 27 at 22:37

Consider a particle $P$ in the framework of special relativity with [...] respect to an inertial reference frame $\Sigma$

(It is of course an important requirement that this question should be considered within the scope of SR; i.e. the setup prescription that an inertial frame can be found in the region containing particle $P$, and thus many different inertial frames may be found such that members of one moved uniformly wrt. members of another. It might be quite another question to ask about an MCRF $\Sigma'$ of particle $P$ if in the region containing particle $P$ there couldn't be found any inertial frame $\Sigma$ to begin with.)

A momentarily comoving reference frame (MCRF) for $P$ is a frame reference $\Sigma'$ [...]

Is the/any "MCRF for $P$" whose further specification is sought therefore necessarily an inertial frame? (After all, reference frames which are not inertial frames may be considered within the framework of special relativity, too.)

Let's first consider the case that the "MCRF for $P$" is required to be an inertial frame (and discuss the other case only briefly later).

[...] (MCRF) for $P$ is a frame reference $\Sigma' = (ct',x',y',z';O')$ such that if $r'$ is the position of $P$ respect to $\Sigma'$, the following two conditions are satisfied:

1) $r'(t'_0)=(ct'0,0,0,0)$ for every $t'_0$ fixed.

The essential requirement "part 1)" to be expressed seems that
- "$O'$" refers to a "principal identifiable point" (in the sense of MTW Box 13.1), such as "particle $P$" itself; - that this particular principal identifiable point had been a member of an inertial system, namely $\Sigma'$, throughout the experimental trial under consideration; and - that $O'$ and $P$ had been coincident at least momentarily (i.e. "in passing").

An assignment of coordinates "$x',y',z'$" to members of inertial system $\Sigma \, '$ (and "$ct'$" to their simultaneous indications) is only subsequent and incidental.

2) $\frac{dr'}{dt'}(t'_0)=(c,0,0,0)$ for every $t'_0$ fixed.

The essential requirement "part 2)", in coordinate-free expression, seems to be that
${\mathbf \beta}_{\Sigma \, '}(P) = 0$ as evaluated at the coincidence event of $O'$ and $P$, which incidentally had been given the coordinates "$r'(t'_0)=(ct'0,0,0,0)$".

Or, based on the setup presciption, equivalently that
${\mathbf \beta}_{\Sigma}(P) = {\mathbf \beta}_{\Sigma}(O')$, also evaluated at the coincidence event of $O'$ and $P$; i.e.
${\mathbf \beta}_{\Sigma}(P) = {\mathbf \beta}_{\Sigma}(O') \mid_{\mathcal{E}_{O \, ' P}}$.

Now to consider the case that a momentarily comoving reference frame (MCRF) for $P$ is not necessarily an inertial frame, one could consider some other principal identifiable point, say $Q$, which was not member of any inertial frame throughout the experimental trial under consideration, but whose geometric relations with others are instead summarized as the (non-inertial) reference system $\Theta$. Then it may still be required, and found satisfied, that
${\mathbf \beta}_{\Sigma}(P) = {\mathbf \beta}_{\Sigma}(Q) \mid_{\mathcal{E}_{P \, Q}}$.

(Where of course $\mathcal{E}_{PQ} \equiv \mathcal{E}_{O \, ' PQ}$ and
${\mathbf \beta}_{\Sigma}(P) = {\mathbf \beta}_{\Sigma}(O \, ') = {\mathbf \beta}_{\Sigma}(Q) \mid_{\mathcal{E}_{O \, ' PQ}}$.)

In this sense, reference frame $\Theta$ could be called/considered a momentarily comoving reference frame (MCRF) for $P$ as well. (Similarly one could even consider reference frames of which $O'$ is a member, but all other members, or at least some, were not members of any inertial frame.)

Moreover the motion of P is uniformly accelerated if for every choice of a MCRF [...]

In order to characterize motion at all, it seems useful to refer first of all to one particular reference frame; such as the surely well suited inertial reference frame $\Sigma$, with its already specified member $O$.

In terms of the quantity ${\mathbf \beta}_{\Sigma}(P)$ used above already, and in terms of the (even more basic) distance between members of $\Sigma$, such as $OS$ as the distance value between (the fixed) $O$ and (a variable) $S$,
the relevant motion of $P$ is characterized by
$(\sqrt{ 1/(1 - ({\mathbf \beta}_{\Sigma}(P) \! \! \mid_{\mathcal{E}_{PS}})^2) } - 1) / OS$ := constant.

And this particular kind of motion, namely hyperbolic motion,

would be relevant because thereby

$(\sqrt{ 1/(1 - ({\mathbf \beta}_{\Sigma \, '}(P) \! \! \mid_{\mathcal{E}_{PS'}})^2) } - 1) / O'S'$ := constant, too
for any (variable) member $S'$ of inertial reference frame $\Sigma \, '$ who had been coincident with $P$ at least momentarily (i.e. "in passing").

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For an accelerated particle, a comoving (inertial) reference frame is (generally) only available at a fixed time $t_0$, so we could call it $\Sigma(t_0)$. The four-velocity, at time $t_0$, relatively to $\Sigma(t_0)$, is (in units $c=1$):

$u(t_0)= (1,0,0,0)$

If the instantaneous acceleration is on a $x$ axis, the quadrivector acceleration of the particle is, relatively to $\Sigma(t_0)$ :

$a(t_0) = (0,a,0,0)$

You may check that the (quadrivectors) speed and acceleration are orthogonal, because $u^\mu u_\mu = 1$, so, you have $a^\mu u_\mu=0$

You may transform the quadrivectors speed and acceleration, in any other inertial frame $\Sigma$, by simply applying Lorentz transformations ($\Sigma(t_0)$ and $\Sigma$ are two inertial frames with constant relative speed, which is the instantaneous (galilean) speed $v(t_0)$ of the particle relatively to $\Sigma$).

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