# Special relativity paradox and gravitation/acceleration equivalence

One of the features of the black hole complementarity is the following :

According to an external observer, the infinite time dilation at the horizon itself makes it appear as if it takes an infinite amount of time to reach the horizon, while the free falling observer reach the horizon in a finite amount of time.

But, because we cannot differentiate between acceleration and gravitation, it may be equivalent to the following "paradox" in Special Relativity.

Le be the plane $z = 0$ be a kind of "horizon".

Let be an observer $O_1$ which is at $z = 1 \space m$ (1 meter), at $t = 0$, and which is moving left (towards decreasing $z$), with a constant speed $(-1) \space m/s$.

Let be a second observer $O_2$, which has the position $z = 1$ m (1 meter), at $t = 0$, but which is moving right (towards increasing $z$), and which is accelerating, in a precise sense (see below).

Divide the initial distance from the observer $O_1$ to the "horizon" $z=0$, into a serie of distance intervals $L_n$, with $L_n = 1/2^n \space m$(meter): $$1/2, 1/4, 1/8,....,..., 1/2^n,...$$

The initial distance (1 meter) between the observer $O_1$ and the horizon is : $$1 = \sum^{\infty}_{n=1} L_n$$

For each distance interval $L_n$, the corresponding elapsed time (from observer $O_1$ point of view) needed for observer $O_1$ to reach the end of the interval $L_n$, is $\tau_n = 1/2^n s \space (second)$, because its speed is $-1 \space m/s$.

The total time (from observer $O_1$ point of view), necessary for the observer $O_1$ ,to reach the horizon $z = 0$ is then : $$\tau = \sum^{\infty}_{n=1} \tau_n = \sum^{\infty}_{n=1} 1/2^n = 1 s$$

Now, for each interval $L_n$, we may adjust the speed of the accelerating observer $O_2$, such that, due to the time dilatation, the elapsed time necessary for the observer $O_1$ to travel during the interval $L_n$, from the point of view of observer $O_2$, be :

$$T_n = a_n \tau_n$$, where $a_n$ is a coefficient > 1

From the point of view of observer $O_2$, the time necessary for observer $O_1$ to reach the horizon $z = 0$ is then : $$T = \sum^{\infty}_{n=1} T_n = \sum^{\infty}_{n=1} a_n \tau_n = \sum^{\infty}_{n=1} a_n/2^n$$

By choosing, for instance, $a_n = 2^{n+\epsilon}$, where $\epsilon >0$, it is easy to see that, from the observer $O_2$ point of view, the observer $O_1$ needs an infinite amount time to reach the horizon $z=0$, while from the $O_1$ point of view, he reaches the horizon in one second!

Did you agree ?

• I thought that a falling object would follow a de-sitter geodesic to an external observer, meaning that it must continue traveling inward at the horizon. I also thought the infinite time dilatation implied that the object is frozen in time as it passes the horizon, not that it stays there forever. But I'm speaking of the formalization for an external observer, the light you see would show someone frozen above the horizon. – Alan Rominger Oct 9 '12 at 13:52
• Well, if time dilatation is infinite, the object is frozen, but, in my example, the time dilatation, while increasing, is never infinite (at a given external observer time). So the object, from the point of view of the external observer, is never frozen, it moves, but it just don't reach the horizon. – Trimok Oct 9 '12 at 14:09
• I should qualify that I'm not an expert in GR topics, as evidenced by my own questions. I see you describing an asymptotic limit. If, in this formulation, the distance to the horizon is finite, then for this "never crosses" picture hold, then the velocity would have to limit to zero (among other requirements). An inertial object (a "geodesic" if I have terminology right) would always travel inward, and always accelerate to the horizon. Regardless of what happens to the falling observer, the concept of velocity and position still follow their regular rules to the external observer. – Alan Rominger Oct 9 '12 at 14:39
• If one is positioned stationary at 1G on earth, and one measured a light source such as a laser on the moon, one does not measure a constantly changing Doppler effect. If one is in space and is constantly accelerating to maintain a 1G force, and one is approaching the moon, then one DOES detect a constantly changing Doppler effect when observing the laser light. This is what confuses me ! – Sean Nov 20 '17 at 22:03

You ask how things look "from $O_2$'s point of view". It's not clear what this means. Are you asking how things look from $O_2$'s instantaneous inertial frame (and how that changes over time)? Or how things look from some non-inertial coordinate system in which $O_2$'s world-curve is the spatial axis''? If the latter, there is more than one such coordinate system, so the question is not well-specified. I will therefore assume you mean the former.

On that interpretation, there is of course never any event that $O_2$ says liesinfinitely far in the future, because, in any inertial frame, every point is labeled with finite coordinates.

But it is possible for $O_2$ to permanently believe that $O_1$ has not yet reached the horizon. It is even possible for $O_2$ to continuously revise upward his calculation of how long it will take for $O_1$ to get there.

More explicitly:

Let $O_1$ hit the horizon at the event $E=(t=1,x=0)$ (in his own frame). (I am setting $v=1$, so his journey started at some location $x<1$.)

Suppose $O_2$ follows the trajectory $x=\sqrt{1+t^2}$ (as expressed in $O_1$'s frame). Let $F=(t_0,x_0=\sqrt{1+t_0^2})$ be any point on this trajectory, so that $O_2$'s instantaneous velocity (relative to $O_1$) is $v=t_0/\sqrt{1+t_0^2}$. (Note that this is less than $1$, so our postulated trajectory is feasible.)

In $O_1$'s frame, the vector $E-F$ has coordinates $(1-t_0,-x_0)$. Lorentz-transforming this to $O_2$'s instantaneous frame at $F$, the same vector has time coordinate $$t_0'={1-t_0+x_0v\over\sqrt{1-v^2}}=\sqrt{1+t_0^2}>0$$

Therefore, at every point in his journey, $O_2$ says that event $E$ lies in the future. In other words, he always says that $O_1$ hasn't reached the horizon yet.

Moreover, because the expression $\sqrt{1+t_0^2}$ is increasing in $t_0$, $O_2$ is always upward-revising his calculation of how long it's going to take $O_1$ to get where he's going.

Well, an heuristic solution would be that it is actually impossible for anybody to cross the event horizon. As you noticed, Schwarzschild black hole is dual (at least locally, that is, if you don't make full turn) to Rindler's observer (the uniformly accelerated one).

The stationnary observer in the Schwarzschild case is the uniformly accelerated one in Rindler's case. The free falling observer for Schwarzschild black hole is therefore the inertial one in Rindler's case, that is, the one that is not moving at 0.

Now, from the point of view of the uniformly accelerated observer, there exists an event of horizon that cut half of minkowski space. Clearly, the accelerated observer will never see anybody crosses it. Obviously, from the point of view of the inertial observer, there is no such horizon (the horizon is "fictive"), and he would have no problem to cross this fictive line in a finite amount of time.

The point is that for the event horizon to keep existing (at the same place and time) for the accelerated observer, the latter has to keep accelerating up to infinity at a constant rate! This is impossible, this would mean that it is possible to bring an infinite amount of energy with you. There exists therefore a time $t_c$ in the accelerated frame such that, after this time, it becomes impossible to accelerate at a uniform rate anylonger. Clearly, accelerating less will "move" the event horizon, and inertiality will simply remove it fully. The event horizon either moves further and further from the inertial observer or vanishes, in such a way that nobody ever crossed it.

Now, what plays the role of this in General Relativity? Well, I would expect (and want) that Hawking's radiation proves that this is exactly what happens: any black hole is either evaporating "too fast" for anybody to ever cross the horizon, or either vanishes totally. This solves all paradoxes.

Any object or mass emitting light that enters the horizon won’t be seen passing it just like you can’t see the sun pass the horizon. Therefore the observer will only see the object reach the horizon and never fall into the center of the black hole. The traveler however would be able to see the observer past the horizon. The object approaching or passing the horizon will experience time dilation inversely proportional to the quantity of energy gained relative to their rest mass (conservation of energy and momentum). A further deduction can be made that because an object of mass accelerates towards the speed of light c in or around a black hole, an objects own displacement would appear as if it happened instantaneously, assuming one arrived on the opposite side of the black hole or wherever its destination is, unharmed and whole. It should be noted that black holes are observed to intake mass and expel energy. It’s highly probable mass consumed by a black whole would be transformed into energy at a rate of c² and perhaps vice versa.

If we are allowed to solve the problem of accelerating bodies in Special Relativity - which postulates relativity of motion - than we can always revert the situation, and assume it is $O_1$ and $z=0$ accelerating relative to $O_2$.

In such case, and as there is always constant $v'$ (I used primed variable to denote motion) between $O_1$ and $z=0$, then regardless of how big the acceleration $a$ is, the $O_1$ will cross $z=0$ after time $t'=1s$. This will take place at a distance $x'$ between $O_1$ and $O_2$ which we can calculate based on acceleration $a$. Ultimately, we will get the dilated, albeit finite (since $t'$ is finite), $t$ for this moment from the perspective of $O_2$ .

We can then go back to our stationary $O_2$ by calculating $x$ (based on $x'$) between $O_1$ and $O_2$. This will allow us to calculate the time $t_1$ necessary for the light to travel this distance $x$. Obviously, it will be finite, and so the total time of $T=t + t_1$ will also be finite. Therefore $O_2$ will see $O_1$ reach $z=0$ in finite time $T$.

If someone says we cannot do that, than we simply cannot apply SR to this situation, which means we do not have SR paradox here.

EDIT: In the first sentence of my answer I wrote: "If we are allowed to solve the problem of accelerating bodies in SR ...". There are numerous claims (not only on this forum) that acceleration can be easily and rightfully handled by SR, so I decided to show what happens, when you do that.

Now, I personally prefer to stick to Einstein's postulate which means: "Thou shall not consider non-inertial frames in Special Relativity":

"If, relative to K, K' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to K' according to exactly the same general laws as with respect to K." And then: "In order to attain the greatest possible clearness, let us return to our example of the railway carriage supposed to be travelling uniformly. We call its motion a uniform translation ('uniform' because it is of constant velocity and direction."

However, in the discussion below John Rennie maintains that acceleration can be rightfully considered in SR, and also that by doing so we invalidate the postulate that no frame of reference is preferred. So we get rid of two basic postulates of SR (inertial frames and no preferred frames), and we still keep calling it SR? To me it's like putting up a cow with a plate "Cow", and then replacing the cow with a goat, but still keeping the same plate. Excuse my trivial example, but that's how I see it.

John Rennie even cited John Baez in his own answer to the question (now removed). However, if one follows this link and clicks on the "accelerating clocks", they will find this as explanation: "... the accelerated clock's rate is identical to the clock rate in a 'momentarily comoving inertial frame' (MCIF), which we can imagine is holding an inertial clock that for a brief moment slows to a stop alongside the accelerated clock, so that their relative velocity is momentarily zero. At that moment they are ticking at the same rate. A moment later, the accelerated clock has a new MCIF, again one that is moving momentarily to match its speed, and there is a new inertial clock that briefly slows to a stop alongside the accelerated clock." Which means, in plain English, that the clock is stopped, and yet at the same time it is ticking. Now, that's not SR, that's SF to me ... (I have seen two other explanations for acceleration in SR on this forum, and they both used either exactly the same or a very similar trick).

Einstein, when deriving his field equations for GR said (page 98) here: "For infinitely small four-dimensional regions the theory of relativity in the restricted sense is appropriate, if the coordinates are suitably chosen." "Relativity in the restricted sense" is simply Special Relativity. So he believed he needed to go down to "infinitely small regions" in order to get rid of acceleration (i.e. gravity - which he - through his equivalence principle - postulates is the same to prove his GR theory), and be allowed to use SR. And he also said in this same book (page 90): "By the word special it is signified that the principle [of relativity] is limited to the case, where K' has uniform translatory motion with reference to K". Here we go! Special, because there are no accelerations. If we introduce accelerations, we are no longer on the grounds of "relativity in the restricted sense" called "special".

This is not to say that what Einstein had said 100 years ago cannot ever be questioned. He is no god whatsoever, and science moves on. I have my own doubts about various of his claims. But then, if one wants to use his theory, and yet get rid of his basic postulates, then he needs to show it is a valid move. And, obviously, I'm not saying accelerations cannot be considered by physics. Sure they can. But in order to claim it can be done on the grounds of SR, one must simply prove it. I must see it to believe it.

What I actually proved in my answer is that if we introduce acceleration to SR, and yet do what the theory allows us to do - i.e switch the frames of reference - then we will obtain two different results. The interpretation of this fact seems just all to obvious.

• Acceleration is absolute in SR, and it is always unambiguous who is accelerating and who is not - the person accelerating is the one who feels a force. That means the assumption in your first paragraph isn't justified. – John Rennie May 6 '14 at 16:45
• There are on absolute frames of reference in SR. If there is one, than it is not SR case. – bright magus May 6 '14 at 16:49
• Velocity is not absolute in SR, but acceleration is absolute in SR. – John Rennie May 6 '14 at 16:50
• In SR inertial frames are defined as having a constant velocity relative to each other i.e. their acceleration is zero. That's why there are no preferred (or to use your term absolute) inertial frames in SR. If you consider accelerating frames then there is a preferred frame i.e. the frame in which the acceleration is zero. Acceleration can be handled perfectly well in SR, but you need to be careful about the calculation. – John Rennie May 6 '14 at 16:54
• If you want to post this as a question I can demonstrate (using the Minkowski metric) why acceleration doesn't affect the clock rate. – John Rennie May 6 '14 at 17:08