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In case of the Schwarzschild black hole, the $t$ coordinate inside the event horizon becomes spacelike. Hypothetically, if one had too much time inside a grand-super-massive black hole and decided to explore the inner space by traveling back along the $t$ dimension, would his travel be limited by the event of the Big Bang at $t=0$?

I understand that the Schwarzchild solution is eternal and does not account for the existence of the Big Bang in the past of the universe. However, there still seems to be a more general question here regarding the principle bounds of the time coordinate when and if it becomes reversible inside the event horizon.

I would appreciate any information on this matter or alternatively an insight on why this question is not well defined and how it should be phrased instead.

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    $\begingroup$ I don't understand what the question is asking. This is partly for the reasons offered in the second paragraph of the question itself. Also, I don't understand what this means: traveling back along the t dimension. And this: a grand-super-massive black hole. Why does it matter what the black hole's mass is? $\endgroup$ – Ben Crowell Sep 8 '18 at 2:38
  • $\begingroup$ @BenCrowell Hi Ben, thanks for helping out! The BH mass defines the maximum proper time inside as $\pi M$ in geometrized units. This is just one minute inside $\text{Sagittarius A}^{*}$. For a longer travel inside a larger mass is required. So my point on the mass simply means "assume that you have enough time inside for a distant travel". A timelike geodesic of a free fall goes to time of $t=\infty$ before crossing, then comes back from space $t=\infty$. So a free fall inside already is "back" from larger to smaller, but positive $t$. What prevents me from using rocket engines to accelerate? $\endgroup$ – safesphere Sep 8 '18 at 6:31
  • $\begingroup$ @BenCrowell Gravity inside acts to decelerate moving bodies and does not accelerate bodies at rest that only move in time along $r$. I emerge on the inside far away at $t=\infty$ and at a huge speed $\dfrac{dt}{dr}$ come back while slowed down by gravity along $t$. If I can fire my engines against gravity to keep my speed along $t$, I could pass to $t\lt 0$ (when the fall started). There is no causal connection back to the outside time, so this is not a time travel of course. However, $t$ is finite in the backward direction, so the Sch. solution breaks, but with what intuitive consequences? $\endgroup$ – safesphere Sep 8 '18 at 6:46
  • $\begingroup$ @BenCrowell Rephrasing this question, in the white hole solution, ligltlike and timelike geodesics emerge from the singularity, go back in space inside to $t=-\infty$, then cross over to $t$ being time, but still at $t=-\infty$, and come up to us from there. Well, there is no $t=-\infty$ in the universe, so the solution breaks. Could this be the reason for no white holes around? Because the commonly quoted reason is lame, a WH breaks energy conservation no more than BH. The solution is fully symmetric, if one is possible, then the other is too. The only asymmetry is the Big Bang. $\endgroup$ – safesphere Sep 8 '18 at 6:58
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    $\begingroup$ t does not go to minus infinity, when you are at the horizon it goes to plus infinity and then back to a finite and positive t. But that is a mathematical artefact which fails to establish a one way (outside to inside) slice of simultaneity when there is none. t only makes physical sense up to plus infinity, everything beyond that simply never happens in that frame of reference. $\endgroup$ – Yukterez Sep 14 '18 at 1:26
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There is no t direction to travel. The infallen observer has a proper time τ and 3 space dimensions. He can move freely along the transverse θ and φ directions, while his radial coordinate r may only decrease, as his τ must always increase.

t is the coordinate time of a stationary observer far away from the black hole, in terms of this time the journey ends when the path freezes at the horizon, so in this frame of reference nothing behind the event horizon ever happens because it already takes an infinite amount of t for the horizon to even form.

To say the infallen observer with proper time τ shall move in the t direction makes as little sense as to demand the outside observer with proper time t to move back or forth in the τ direction. That is neither his time- nor one of his space coordinates, it's only the time coordinate of an observer he is no longer causally connected with.

It is a mathematical artefact that the time of an outside observer runs backwards again after it took an eternity for the testparticle to even reach the horizon, see MTW, Fig. 32.1

Update after comments:

The time dilation of the test particle from the perspective of the far away observer is

$$ (1) \ \ \ \ \rm \frac{d t}{d \tau} = \frac{1}{\sqrt{1-v^2} \sqrt{1-2/r}} $$

(which would be negative behind the horizon at $r<2$, where the local velocity relative to the singularity $\rm v>c$, since $1/i/i=-1$), and the time dilation of the far away observer from the perspective of the test particle $$ (2) \ \ \ \ \rm \frac{d \tau}{d t} = \frac{\sqrt{1-2/r}}{\sqrt{1-v^2}} $$

(which would be positive even behind the horizon, since $i/i=+1$). Nevertheless, equation $(1)$ is only valid down to $\rm r=2$ and up to $\rm t=\infty$, since it is not physically meaningful to travel to the end of time and back again, as we have seen in the MTW reference (that's what motivated Eddington and Finkelstein to construct their advanced time coordinate which remains valid even behind the horizon).

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  • $\begingroup$ Thank you for the answer. I don't think it's correct, but I appreciate it anyway. A 4D spacetime in any frame is described by 1 temporal and 3 spatial coordinates in a (-1,1,1,1) metric signature. Inside a BH, these coordinates in the Schwarzschild frame are $(dr,dt,rd\theta,r sin{\theta}d\phi)$ where dt is 1 of 3 spatial coordinates. Nothing in this solution prohibits traveling along the t coordinate in space inside a BH in any direction using rocket engines. Even in the frame of a free falling observer, the t coordinate inside is a direction in space (e.g, light goes both ways along it). $\endgroup$ – safesphere Sep 15 '18 at 17:58
  • $\begingroup$ You can also transform into a coordinate system where the proper time τ of a raindrop or some arbitrary observer is used instead of the external observer's coordinate time t, but then still, how is the external observer supposed to move along the τ axis when his own proper time is t? $\endgroup$ – Yukterez Sep 15 '18 at 18:40
  • $\begingroup$ If you look at the light cones on the left diagram you've attached, you'd see that $t$ inside is space and $r$ inside is time, so $\dfrac{dt}{d\tau}$ is no longer time dilation. Speed inside (for radial fall) is $\dfrac{dt}{dr}$ and is never locally superluminnal, as also is evident from the lightcones (or from the light geodesic). Other types of coordinates are math tricks using diffeomorphism, but they have no physical meaning, as they no longer represent physical space and time (the fact people using these coordinates conveniently forget). $\endgroup$ – safesphere Sep 22 '18 at 16:26
  • $\begingroup$ In case you are interested, see the answer of a 110k-reputation member of Math SE that the shape of the Schwarzschild singularity is an infinite spacelike Euclidean line: math.stackexchange.com/questions/2929400/… $\endgroup$ – safesphere Sep 26 '18 at 7:21
  • $\begingroup$ That the singularity is not a point but a line in time just means that it exists longer than an infinitesimal short period. All the trajectories I tried so far have a decreasing t coordinate inside the horizon, so I really see no way how the test particle should move or accelerate to move freely around the t axis as if it were a spatial dimension. The 110k member is thus correct, but it still is an argumentum ad verecundiam to quote his 110k reputation. $\endgroup$ – Yukterez Sep 27 '18 at 0:20

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