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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.

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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 we are allowed the assumption $c^2(dt'/dt)^2 = c^2 - (d\vec{x'}/dt)^2$ then isn't the result immediate? And if we are not allowed the assumption how could we calculate the example of an accelerating worldline in Minkowski space? –  Retarded Potential Feb 7 '13 at 20:26
    
Yes I definitely agree that the result is immediate given that assumption. In fact, most of my post is an attempt to justify that assumption by approximating an arbitrary worldline by a piecewise non-accelerating (inertial) worldline. The problem I have with the assumption is that it certainly holds for inertial observers since Minkowski arc-length is Poincare invariant; but it's not clear to me why it should hold for accelerating observers without some sort of argument like the one I'm attempting to make. –  joshphysics Feb 7 '13 at 20:39
    
You ask for "an example of an accelerating worldline ... calculated explicitly", but how can we calculate this without the assumption? I.e. just what assumptions are we allowed to make about an accelerating observer, if none, then it seems there is no argument that can be made. –  Retarded Potential Feb 7 '13 at 20:46
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Good point; I'll remove that part since what I'm actually asking is for arguments making that assumption plausible. Let me add that I'm completely comfortable simply taking this fact as an axiom of relativity that is ultimately justified by experiment. What bothers me is that people rarely state the assumption in such terms, and it seems to me that they often even suggest that it's some obvious/trivial consequence of how events transform between inertial frames. For example, the resolution of the twin paradox ultimately rests on this assumption (as far as I can tell). –  joshphysics Feb 7 '13 at 20:52

3 Answers 3

up vote 2 down vote accepted

[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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Are you sure you mean to say "deviation of velocity from inertial"? Recall that even inertial observers can have speeds arbitrarily close to $c$; are you sure you didn't mean to phrase the assumption in terms of acceleration? If so, I think my initial argument is essentially in the same spirit as what you're attempting here, in particular, the assumption I call $(\star)$. –  joshphysics Feb 7 '13 at 22:08
    
Mine is different. For one, it doesn't assume there is any transforms applicable to the accelerated observer, nor any bound on acceleration. From the point of view of the inertial A, their own velocity is 0, and B's velocity is bounded to within $c\delta$ in any direction (in A's frame). Then A2 asserts A and B experience the same time lapse, to within a factor $(1 \pm \epsilon)$. You only need to believe that there is some small $\delta$ for which this is true. Then the only Lorentz transforms you need are for inertial observers, which I took to be given. –  Retarded Potential Feb 7 '13 at 22:21
    
Ok I'll read it more closely and try to get a feel for your assumption [A2]; I still feel that the phrase "deviation of velocity from inertial" is a bit misleading. By the way; thanks for all of your attention and comments; I really do appreciate it: +1. –  joshphysics Feb 7 '13 at 22:36
    
I rephrased. I think you need something like [A2] to connect the experiences of inertial and accelerated observers, or else the latter's experiences can be quite arbitrary. –  Retarded Potential Feb 7 '13 at 22:47
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I don't know if you saw this as well, but it was posted in a related Baez page cartan.e-moka.net/content/download/252/1495/file/… –  joshphysics Feb 8 '13 at 0:23

I think it's obvious from the pic:

enter image description here

(My secondary argument would be that the integral is the parametrization-independed geometrical measure, minimizing the energy functional and thus being the natural generalization of the flat case time translation. You often encounter these exponential maps in dynamics, here given by the Geodesic flow. Locally you can flatten out the metric to obtain $g(x)=\eta+O(x^2)$ and try to find a limit of smaller and smaller patches. But this really is equivalent solving the geodesic equation, which should be viewed as giving the spacetime direction of the moving object as it's pushed through spacetime. It is also equivalent to minimizing the curve between the points, as motivated above.)

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Hahhahha. I'm tempted to +1 for the picture alone. As for the secondary argument; it's not clear to me how this answers the question about time measured by an observer moving along a curve given that I'm looking for plausibility of (arc length) = (time measured by observed moving on a curve) even in the flat space case. –  joshphysics Feb 7 '13 at 21:45
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@joshphysics: The geometric length is parameter independed, so as soon as you've chosen the curve in the manifold (requirement of extremal length) the quantity depends only on the two points $x_a,x_b$ in spacetime. If your Newtonian wolrd is $\mathbb{R}\times\mathbb{R}^3$, then $t_{ab}=f(I_{ab})$ is the obvious choice. –  NikolajK Feb 7 '13 at 21:51
    
That's an interesting argument; I like it for plausibility quite a lot actually. –  joshphysics Feb 7 '13 at 22:02
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+1: At the very least for your picture answer. –  joshphysics Feb 7 '13 at 22:38
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@RetardedPotential: It's all explained in the movie/documentary. –  NikolajK Feb 8 '13 at 8:11

I definitely thing this question is much simpler.

It is an a priori assumption upon which GR is built, that this quantity shall be independent of the observer:

$$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

that is, two different observers shall measure the same spacetime interval between two infinitesimally close events.

On the other hand, the $x^0$ coordinate of any observer is by definition the lecture in his rest clock, (times $c$ or not, times $+1$ or $-1$, all depending on every modern author's bad taste).

Since the clock of the moving observer is at rest with respect to his proper reference frame, he measures no variation in the spatial coordinates, i.e. $dx^1=dx^2=dx^3=0$. Therefore, as measured from the moving observer, $$ds^2=g_{00}dx^{0}dx^{0}$$

That is his proper time (squared) because $g_{00}=\pm 1$ (the metric of a free falling observer is locally flat). Since $ds^2$ shall be the same for both observers, then the lecture on the free-falling clock equals the spacetime interval measured by the distant observer. That's all.

Now, if you prefer so, you may rewrite the $dx^{\mu}$ as functions of a variation in the parameter $\lambda$ describing the curve, and integrate along that curve, in order to arrive trivially at the expressions in your question.

I don't mean that the other answers are wrong, but merely that this is a straightforward result from the original theory, as exposed very early by Einstein himself in the Princeton lectures of 1921, that doesn't need much mathematical sophistication or additional postulates. Einstein wanted to extend the invariance of $ds^2$ from Special Relativity, to the realm of non-inertial observers. In his effort to find how $g_{\mu\nu}$ had to be for that invariance to hold too in the presence of acceleration/gravity and curved coordinates, he then used parallel transport to arrive at the geodesics equation. That is why the geodesics equation was a postulate in the early formulation of GR (the next step was to relate $g_{\mu\nu}$ to matter/energy, and that is why the Field Equations with the Einstein tensor is the other postulate of the theory, but that is another question)

I think I remember having read somewhere, that the geodesics is no longer a postulate, because it can be derived from the Field equations. I would be thankful to any user that provides the corresponding reference. In 1921, however, it was a postulate.

This Einstein's early exposition of GR of the Princeton lectures is of a raging beauty, full of fresh insights and brilliant heuristic thinking. I don't know why it is almost systematically ignored in every standard GR bibliography.

Now, if I am wrong and this question is not so naive as it seems, I hope that someone points out where I am wrong and what I am missing, so that I learn something new.

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I agree that it's part of the very construction of GR that the metric is a coordinate-independent quantity in the sense that you can perform arbitrary diffeomorphisms on the spacetime manifold without changing the metric. As far as I can tell, however, it is not a priori clear that such transformations allow one to say something about what arbitrary timelike observers measure. I'd have to think more about your argument to be more clear on where (if at all) you're making the "clock postulate" assumption. –  joshphysics Feb 8 '13 at 17:11
    
Again, there must be something here I don't understand. For me, it is crystal clear that $dx^0$ is the measure on the clock of any observer. That is a definition that goes back to Minkowski. Einstein did a great effort to maintain that definition and postulated that the distance between two infinitesimally close spacetime events $ds^2$ is the same for all observers, even in the presence of gravity and curved coordinates. In his effort to find how $g_{\mu\nu}$ had to be for that assumptions to be true not only in the absence of gravity, he then used parallel transport to arrive at... –  Eduardo Guerras Valera Feb 8 '13 at 22:46
    
(...he then used parallel transport to arrive at) the geodesics equation and eventually postulated the Einstein tensor in the field equations. That is how GR is exposed in the Princeton lectures in 1921. The invariance of $ds^2$ is postulated, and the meaning of $dx^0$ as the clock measure (multiplied with c) is simply an a priori definition. –  Eduardo Guerras Valera Feb 8 '13 at 22:49
    
Yea your point is well taken, and I completely understand what you're saying; let me think about it for a while. Thanks for the detailed comments! –  joshphysics Feb 8 '13 at 22:50
    
@joshphysics, something tells me that you have completely forgotten this question... –  Eduardo Guerras Valera Feb 11 '13 at 13:19

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