Apologies in advance for the long question.
My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$ t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.
Why is this so?
Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$. Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$. If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$. If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.
I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$ T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$. Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$ I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align} I_{ba} &= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.
How do others feel about this argument?
I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.
I imagine that in GR a similar argument could be made by invoking local flatness of the metric.
Answer
[This is now a long answer. In summary, generally you need a physical assumption, the clock postulate, which people tend to omit, but is necessary, and can't be argued for a priori. However sometimes special relativity plus a restricted version of the postulate suffices. Mundane experience is sufficient to verify this restricted version.]
Let $\lambda = t$ the time according to inertial $O$, and let $\vec{x}'$ be the spatial position of $O'$ according to $O$ , while $t'$ is the time measured by $O'$. If $O'$ is piecewise inertial, then along each piece, $$c^2(\Delta t'/\Delta t)^2 = c^2 - (\Delta\vec{x}'/\Delta t)^2\qquad[1]$$ and what you are trying to justify is that, even if $O'$ is not piecewise inertial, $$c^2(dt'/dt)^2 = c^2 - (d\vec{x}'/dt)^2\qquad[2]$$ So, the problem is, special relativity strictly speaking only makes claims about inertial observers. And if you don't make any assumptions whatsoever about the experience of accelerated observers, then I think you're just stuck, mathematically I don't think you can go from $[1]$ to $[2]$. (For example, we can't rule out that proper acceleration itself further contributes to time dilation.) I suggest:
The motion of $O'$ is smooth. [A1]
For every $\epsilon>0$, there is a $\delta>0$ such that, if, from the point of view of a certain unaccelerated observer $A$, the magnitude of the velocity of another observer $B$ never exceeds $c\delta$ betwen time $t_0$ and $t_1$, then time lapse $\Delta t_B$ on $B$'s clock satisfies $(t_1-t_0)(1-\epsilon) < \Delta t_B < (t_1-t_0)(1+\epsilon)$. [A2]
Pick $\epsilon>0$, use [A2] to get $\delta$; use [A1] to break the motion of $O'$ into intervals small enough such that, from the frame of reference of an interial observer travelling between the endpoints of a piece, the velocity of $O'$ never exceeds $c\delta$; use [A2] to make $[2]$ true within $\epsilon$. Since this works for all $\epsilon>0$, [2] is simply true.
Now [A1] might look suspect, since we've used a piecewise inertial observer, whose motion is obviously not smooth! So we can't even assume anything about what this piecewise inertial observer experiences at the corners! But that's okay, [A2] only refers to the individual pieces and not the whole. Use a family of (truly) inertial observers that meet at the appropriate points.
As for [A2], it is a bit opaque, but what it says that if you're not moving too fast relative to an inertial observer, your experience of time is almost the same. This doesn't follow logically from anything in particular, it's just a physical assumption. But note that special relativity is so hard for many people to accept precisely because [A2] is a fact of life, for reasonably small $\epsilon$. To make it true for even smaller $\epsilon$ requires more than everyday experience, but it is still "common sense", and presumably testable to quite small values.
Now, to believe it literally for arbitrarily small $\epsilon$ requires quite a leap, but don't take differential equations literally.
(Added:) Aha! I found the clock postulate for accelerated observers, and I do believe [A2] is interchangeable with it. And yes, it is often omitted but cannot be derived from other assumptions. It has been tested.
(Second addendum): Even though they're interderivable, mine is better :-) I've given the accuracy of [2] directly in terms of the accuracy of [A2]. For example, we don't need the full clock postulate for the twin paradox (which you mention as a motivating example in a comment):
- The proper acceleration of $O'$ is continuous and its magnitude is bounded by $a_{max}$. [A1']
(Over any finite interval, [A1] does imply [A1'] for some value of $a_{max}$. And [A1'] is sufficient for the above argument.)
Now, even with mundane accelerations, the twin paradox can produce a sizeable mismatch in ages within a human lifetime. (Besides, if they're not survivable accelerations, the travelling twin's lifetime ends!) So, there's a usable $a_{max}$ for [A1']. And, mundane experience alone proves [A2] up to that $a_{max}$ and down to fairly small $\epsilon$. So [2] holds sufficiently accurately to give the twin paradox. We only need special relativity plus a mundane restricted clock postulate.
(I realise you can bypass the whole acceleration question by altering the paradox so that there are three inertial observers who compare clocks as they pass. But then it's not the twin paradox anymore, duh!)
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