# I Survival time in a black hole

Tags:
1. Nov 26, 2017

In a thread a decade ago https://www.physicsforums.com/threads/how-to-survive-in-a-black-hole-myth-debunked.170829/, there was a discussion about the paper https://arxiv.org/abs/0705.1029v1, in which the authors discuss the way to maximize one's survival (proper) time after passing the event horizon of a (non-spinning) black hole; they treat radial motion only. This is a closed thread, and I am not trying to prolong that discussion by re-opening it, but rather to pose a question that was not posed in that thread, even though I am using the same article as reference.
What no one in that earlier thread asked was: would this survival time increase if one's rocket was heading towards the black hole not radially, but at an angle, say (at first) tangentially? (If so, would this also change the basic strategy of when one should or should not accelerate?) Intuitively, I would think that it would, but as the article and that thread shows, intuition is a horrible guide in this situation.

2. Nov 26, 2017

### Ibix

3. Nov 26, 2017

Thanks, Ibix, I hadn't seen that one. If I understand some of the comments correctly , the short answer to my first question is "yes", although I am still unclear as to whether there is consensus on the answer to my second question. I shall post further questions there.

4. Nov 28, 2017

### pervect

Staff Emeritus
I would guess no, and that's what I calculate, though I haven't really spent enough time with it to make sure I'm not making a calculation error. (When I make errors, I tend to make them in favor of what I guess, perhaps unsurprisingly).

But here is the argument. To get the proper time , one needs to integrate the differential equations of motion. From https://www.fourmilab.ch/gravitation/orbits/ we can get the equations of motion. (Or alternatively, go to their source, which happens to be MTW's book "Gravitation").

$$\left( \frac{dr}{d\tau} \right) ^2 + \left( 1- \frac{2m}{r} \right) \left( 1+ \frac{L^2}{r^2} \right) = E^2$$

We can re-write this to solve for $\frac{d\tau}{dr}$, and we know that r varies from 0 to 2m. So we can divide the total fall into a bunch of small intervals as r goes from 2m to zero, and find the proper time that it takes for each interval, and add them up togeter

$$\tau = \int_{r=2m..0} \frac{d\tau}{dr} dr$$

Because $\tau$ looks so much like r, this is a bit hard to read, at least for me. It's not too confusing later, but I'll rewrite it with T instead of $\tau$ to make it easier to read to get the basic setup accross.

$$T = \int_{r=2m..0} \frac{dT}{dr} \, dr$$

At this point we'll substitute in m=1/2, to make the calculation slightly easier. We need to solve the first equation for $d\tau / dr$. We get

$$\frac{d\tau}{dr} = \sqrt{ \frac{1}{E^2 - \left(1-\frac{1}{r} \right) \left( 1 + \frac{L^2}{r^2} \right) } }$$

Because r < 2m and m=1/2, r<1, and it's helpful to re-write this as:

$$\frac{d\tau}{dr} = \sqrt{ \frac{1}{E^2 + \left(\frac{1}{r} - 1 \right) \left( 1 + \frac{L^2}{r^2} \right) } }$$

We know that (1/r)-1 is never negative, it can be zero or positive. So, how do we maximize this? Well, we can see that setting E^2>0 only makes this quantity smaller, so we set E=0, the same as we did before.

[add]. Let me clarify this a bit. 1/X is a maximum when X is a minimum. To make X a minimum, where X is a positive number, we don't add a positive number to it, and we also don't multiply it by a number greater than 1. Making E>0 adds a positive number to the denominator which I've represtend as X, and making L>0 multiplies X by a number greater than 1.

We see this by noting that setting E>0 makes each term in the integral smaller, so the infinite sum of the integral is also smaller. The same argument applies for L.

The limiting case with E=L=0 is just

$$\int_{0..1} \frac{1}{\sqrt{ \frac{1}{r} - 1 }}$$

(I've done a bit of re-arranging of the signs, which is an opportunity for error, but we know we want the result to be positive).

This turns out to be finite and equal to $\frac{\pi}{2}$ according to my symbolic integration package, but I don't have any closed-form integrations for when L>0 or E>0. We can still argue, as I said previously, that by inspection the integral is maximized when E=L=0, though, as each term of the infinite sum is maximized, which makes the infinite sum (the integral) itself a maximum.

This is not striclty related, but there is a simper argument that leads to a possibly more intuitive though less rigorous explanation. Light orbits a black hole at the photon sphere, at r=3M. Inside the photon sphere, one needs to apply thrust to hold station, there is no possible orbit. Furthermore, if one attempts to orbit the black hole in the photon sphere, one needs MORE thrust to hold station than if one stays still with no angular motion. This counter-intiutive fact about orbiting inside the photon sphere is mentioned in a few papers (that I don't have the references handy for), and in Kip Thorne's book, "Black holes and Time Warps".

Last edited: Nov 28, 2017
5. Nov 29, 2017

Thanks for these calculations, pervect. I look forward to working through them. My initial guess of the affirmative answer was also shot down in the other thread to which I went after posting the above. (That is, the thread cited by Ibix, above.) There they present different calculations to also come up with a negative answer.

Since I happen to have a copy of Thorne's book, I tried to look up your reference to photon spheres, but the index was no help, so perhaps you have a chapter (or even better, a page for the Norton 1995 softcover edition)? [In the meantime, I went to Wikipedia https://en.wikipedia.org/wiki/Photon_sphere].

6. Nov 29, 2017

### pervect

Staff Emeritus
Thorne has a section where he talks about a trip to a several black holes , of differeing sizes. In part of the trip, the ship attempts to orbit within the photon sphere, and starts to drop until counter-thrust is applied. IIRC.

I also found an "alternate view" science fact column by Cramer, https://www.npl.washington.edu/av/altvw55.html. It's a popularization, but he gives some references too.

7. Nov 29, 2017

Thanks, pervect. In Thorne's book, you are referring to the prologue. In it, there are several places in which he has to abandon the descent towards a black hole for various reasons, such as the growing tidal forces, or the difficulty in getting out past a certain point; his brief and implicit mention of the photon sphere (he never either uses the term or explains it in terms of the place where a photon will orbit) is in the section "Sagittario" (Page 38 in my edition) when he finds that a blast of the rocket engine in the direction of the orbital motion "instead of driving you into a slightly tighter circular orbit, sends you into a suicidal plunge toward the horizon." (Trivia: His computer for some reason likes to say in Russian "тихий, тихий" , wanting to say "calm down" (lit.: "quiet, quiet"), but gets the form wrong, using the adjectival rather than the adverbial form; correct would be "тихо, тихо") But very picturesque description of some of the properties of the space around black holes.
I also looked briefly at the references in the Alternative View column; they look worth investigating, especially as they are freely available on the Internet . Thanks again.

8. Nov 29, 2017

### timmdeeg

It seems you have raised this question in #15 talking about "spiral in" in the recent thread @Ibix mentioned in #2.

9. Dec 7, 2017