Insights Time Dilation and Redshift for a Static Black Hole - Comments

1. Sep 30, 2015

stevebd1

2. Sep 30, 2015

Staff: Mentor

Good post in general and good choice of subject, this is often a source of confusion on PF threads.

I have a few comments, though. First, it's important to be very careful to specify the state of motion of the object that appears to be time dilated, or which is emitting radiation that appears redshifted from far away. The equations you give are for a static object--i.e., an object which is "hovering" at a fixed altitude above the horizon and has no tangential motion at all--and the radiation it emits. However, later in the article, you discuss a spaceship in orbit about the hole at some altitude, and a scout ship free-falling into the hole. The "time dilation" equations for these are different from the one you give.

For someone in a free-fall orbit about the hole, the equation is

$$\tau = t \sqrt{1 - \frac{3GM}{rc^2}} = t \sqrt{1 - \frac{3}{2} \frac{r_s}{r}}$$

There is also a correction to the redshift observed in light emitted by an orbiting observer, but IIRC it is somewhat different (it's similar to the transverse Doppler effect in SR). (Also, if a spaceship is in orbit about the hole, it doesn't "have to have powerful engines to resist the hole's gravity"; a free-fall orbit requires no engine power at all. If you meant that the ship was "hovering", maintaining a constant altitude while having no tangential velocity at all, then "orbit" is not a good word to use.)

For someone free-falling into the hole along a radial trajectory, the "time dilation" factor isn't really well-defined, because this observer is not at rest relative to an observer at infinity, so they don't have a common standard of simultaneity. However, the "redshift factor" for light emitted by the free-falling observer and received by an observer at infinity is well-defined, and can be calculated as the redshift factor for a static observer at altitude $r$, which is what you wrote down, combined with the Doppler redshift for an observer falling inward at velocity $v = \sqrt{2GM/r}$ relative to the static observer.

Also, at the end of the article, there is an incorrect statement:

"At exactly on the event horizon, where gravity is considered to be infinite, if a person was to look outwards from the black hole, they might see some portion of the future of the universe flash before them"

The "future of the universe" is not in the person's past light cone when they cross the horizon (or even when they reach the singularity). In fact, a person free-falling into the hole does not even see light from the outside universe blueshifted; he sees it redshifted (by contrast with a person "hovering" just above the horizon, who does see light from the outside universe highly blueshifted). It's important to be very careful about describing what happens near, at, and inside the horizon, because of the many common confusions people have, based on pop science literature.

3. Sep 30, 2015

SlowThinker

Nice article, with some interesting and hard to find formulas.
Do I understand it correctly that at the horizon, gravity acceleration is infinite, while the tidal forces are manageable/finite?
Second question: I'm in my ship, hovering 10 meters above horizon. In that situation, I turn the engines off. Will the bottom of my ship turn to dust before I hit the horizon?
I'm trying to figure out how gravitational time dilation works on fast moving objects.

4. Sep 30, 2015

Staff: Mentor

Yes. The interpretation of tidal forces as "the difference in acceleration between two neighboring points" only works when you are very far above the horizon; it breaks down close to the horizon (and obviously doesn't work at and below the horizon).

It depends on how large the hole is. For a hole of one solar mass, your ship might well be dust even before you get 10 meters above the horizon. But for a hole of a billion solar masses (such as probably exist in the centers of some quasars), you won't even feel the tidal gravity at the horizon, and your ship will be just fine.

5. Sep 30, 2015

SlowThinker

Sorry for the misunderstaning. I was referring to aging of the ship, due to infinite(?) time dilation.
The mechanical stresses, radiation etc. are a story for another time.

6. Sep 30, 2015

Staff: Mentor

The ship itself doesn't experience any time dilation; to someone on the ship, everything seems perfectly normal, clocks tick at one second per second, etc. This is true even if the ship falls through the hole's horizon. If the ship starts falling 10 meters above the horizon, the time it takes to fall those 10 meters, by the ship's clocks, will be short--the sort of time you would expect an object to take to free-fall 10 meters in a strong gravity field.

Time dilation is something an observer very far away from the hole will observe if they watch what is happening on the ship.

7. Oct 1, 2015

stevebd1

The figures provided at a distance of 1km, 1m and 5mm from the EH of the black hole are for an object (i.e. the scout ship) hovering at these particular radii. The statement about seeing 'some portion of the future of the universe' is based on the notion of the scout ship trying to hover closer to the EH within the last 5mm but as it reaches the event horizon, were there is no stable r, being pulled into the BH. Within those last 5mm, just before being pulled into the BH, the occupants would see the outside universe speed up significantly, albeit via immensely blueshifted and highly radiative light. As they were pulled across the event horizon, I imagine that this image of the outside universe would begin to redshift as they fell towards the singularity. I've also added some footnotes and supplementary equations.

8. Oct 1, 2015

Staff: Mentor

Ah, ok. That wasn't really clear to me from the description; you might want to emphasize it more at the start of the scout ship portion of the article.

Yes, but the term "future of the universe" is still misleading. It's not the "future" that is being seen from the viewpoint of the occupants; it's still the past. Being close to a black hole's horizon doesn't allow you to see things outside your past light cone.

Yes.

9. Oct 1, 2015

jartsa

Consider a fleet of spaceships landing on a planet that has a frictionless surface, very deep gravity well, and a few moons.

The spaceships will slide around into different directions pulled by the gravity of the moons, slowly according to far away observers, at great speed according to surface observer.

10. Oct 1, 2015

SlowThinker

I was, uh, thinking some more.
1. The tidal forces are a difference between gravity at my head, minus gravity at my toes, right? (Lets define gravity as the acceleration necessary to stay in place relative to a distant observer)
2. If my ship goes lower so that my head is where my toes were, the difference is nearly the same as before.
3. But when I reach the horizon, the acceleration needed is infinite.

Does that mean that, as long as my engines can keep the ship in place, I can go down and down and never reach the horizon? I think this must be the case, and something funny must be happening with the height of my ship.

Sadly this is not related to the topic any more, is there a chance for a follow-up article discussing the horizon thoroughly? Or is there such an article already?

11. Oct 1, 2015

Boing3000

I have also got a question regarding two free falling ship case:
1) vertically "falling"
2) orbiting "falling"

In both those frame there is no local gravity/acceleration, but I think the gravitational potential (hence time ratio) is somewhat different along the vertical axis in both ship (and rapidly changing in ship 1
When both ship are at the same altitude, do they experience (locally) the same time dilatation ? Or does the "horizontal" movement of the orbiting ship 2 change it slightly compare to ship 1 ?

From the article, it is not evident to guess the difference in time dilatation for a 12,000 solar mass black, between the foot and the head (let's say 2 meter) of someone free falling, and whose foot are at 1 meter of the event horizon. If the tidal force (the radial angle difference) is manageable, I suppose the time dilation must be quite weak too.

I think SlowThinker has inverted the gravitational time dilatation effect, because his toes are not supposed to turn to dust, but his head is

12. Oct 1, 2015

SlowThinker

I'm pretty sure that for an object in free fall, the time dilation is the same throughout the volume, or at least nearly the same - it follows from the equivalence principle. Further, a free falling object is gaining speed, which, due to Lorentzian time dilation, is exactly equal to the gravitational dilation at any lower height (or at least I am sure it is), so no paradox there.
For ship 2, you just need to add "standard" Lorentzian time dilation due to orbital speed.

Yes you are right, my head would age fast while my legs are OK.

13. Oct 1, 2015

Staff: Mentor

Only in the Newtonian approximation, where you can view "gravity" as a "force". This approximation is certainly not valid near the horizon of a black hole.

The more general definition of tidal gravity involves geodesic deviation: how fast a pair of freely falling objects that start out at rest relative to each other separate or converge with time (where "time" here means proper time along either geodesic). It is quite possible for tidal gravity, by this definition, to be negligible at the horizon of a black hole, if the hole is massive enough. But the proper acceleration required to "hover" goes to infinity at the horizon regardless of the hole's mass. This makes it obvious that tidal gravity and proper acceleration are not related in the general case.

No. See above.

14. Oct 1, 2015

Staff: Mentor

Strictly speaking, "gravitational potential" is a property of the spacetime, not the object, and it is only well-defined to begin with in a special class of spacetimes, called "stationary spacetimes". In such spacetimes, any two objects at the same spatial location have the same gravitational potential.

However, this does not mean they necessarily have the same time dilation factor. See below.

No. First let's take the simplest case, where the vertically free-falling ship has just started falling from rest at that altitude (it has just turned off its rockets and stopped "hovering", or it was free-falling upward and has just reached its maximum altitude and is starting to fall back down again). In this case, the (instantaneous) time dilation factor of the vertically free-falling ship will be the same as that of a "hovering" ship at the same altitude, i.e., $\sqrt{1 - 2M / r}$ (in units where $G = c = 1$). But the orbiting ship will have a time dilation factor of $\sqrt{1 - 3M / r}$, i.e., a larger one.

The extra time dilation is due to the orbiting ship's motion; in general, in a stationary spacetime, we can split up any object's time dilation factor into two pieces, a gravitational piece (which depends only on its position, as above), and a piece due to velocity relative to a static observer (i.e., an observer whose position is not changing) at the same position. The general formula for the time dilation factor is then $\sqrt{1 + 2 \phi - v^2}$, where $\phi$ is the gravitational potential. In the case we're discussing, where the gravitating mass is spherically symmetric and static, we have $\phi = - M / r$, and an object in a free-fall circular orbit has $v = \sqrt{M / r}$; plugging these in will give you the specific formulas I quoted above.

No. The two are not related, for much the same reasons that the tidal gravity and proper acceleration are not related. See my previous post.

No. Tidal gravity, when it gets strong enough, will affect both his head and his feet; it will "try" to make them separate rapidly, and at some point the structural strength of his body will be overcome and his whole body will turn to dust. Assuming his body is small compared to the size of the black hole, one end won't be affected significantly before the other end is.

Yes; more precisely, it follows from the EP plus the assumption that the object's size is very small compared to the size of the black hole (roughly speaking, to the radius $r = 2M$ of its horizon).

I'm not sure what you mean by this. Gravitational time dilation and "Lorentzian" time dilation (by which I assume you mean time dilation due to velocity, which I explained above) are two different things.

15. Oct 1, 2015

SlowThinker

Are you 100% sure that the height contraction does not play a crucial role? I still don't see a flaw in my reasoning:
1. The acceleration needed to keep my head in place is $g_{head}$. Similarly define $g_{feet}$.
2. The difference $g_{head}-g_{feet}$ seems to fit your definition of tidal effects, and so is finite everywhere except singularity.
3. If the ship is descending slowly, $g_{feet}$ eventually reaches infinite value.
4. Therefore, there has to be an infinite number of heights before horizon.
Which of these steps are wrong?

I was imagining this scenario: Let's have 2 ships, one hovering low, another high. The high ship turns engines off. By the time it meets the lower ship, what are their relative time speeds, and tallness? I thought they are the same, but now I'm not so sure...

16. Oct 1, 2015

Staff: Mentor

What is "height contraction?"

Yes, these can be defined. But they aren't the same as tidal gravity. Read my description of tidal gravity again, carefully. Note that I specifically talk about freely falling objects, and how they separate or converge. If you are accelerating to stay in the same place, you are not in free fall, and the difference in acceleration across your body is not the same as tidal gravity.

No, it doesn't. See above.

No; the acceleration needed to "stay in the same place" goes to infinity at the horizon. Once more: the change in the acceleration needed to stay in the same place, and tidal gravity, are two separate things. They don't go together.

Since they are at the same altitude, the contribution of gravitational potential to their "time speeds" is the same in both cases. So the only difference is their relative velocity. Each will measure the other's clock to be running slow, just as two observers in relative motion in SR do. An observer at infinity, watching them both, will see the falling ship's clock running slower than the stationary ship's clock.

I'm not sure what you mean by "tallness".

17. Oct 2, 2015

jartsa

So clearly on the surface the situation is such that huge tidal forces are mangling the fleet. If the spaceships are chained together the chains will break.

18. Oct 2, 2015

Boing3000

OK. I am still struggling with that. You introducing many other trajectory for other ship 'free fall' being in the same place helped me very much to understand "stationary space time". So I suppose the static observer (you mentioned below) would be in the same "instantaneous" frame. And all situations would be equivalents.
But I get lost again when you say a hovering ship would experience the same time dilatation (I understand it is minuscule on the size of a ship). Ok this ship is a the same place, but both top and bottom clock will experience a very strong and uniform/parallel gravitational field. Does it not somewhat influence those accelerating clocks by creating for them a new gravitational potential to be added to the one created by the massive body ?

As a side note, my intend was to ask if to clocks, one on the facing side, and one on the opposite side of the moon, would experience a different time rate. Moon wise they are in the same potential (altitude). But earth wise, they are at different potential...

I must also says that I understood one of SlowThinker's point. I lost track of the EP when thinking off free falling object (which would imply no different time between the top and bottom lock). But I think you clear-up the situation by reminding us that this EP is "local" up to "a point", not between two spaced-up clocks or head/toes. Did I get that right ?

That's a very clear explanation, thank you !

I still think that SlowThinker 'dust' is related to time ratio difference and not the unrelated tidal 'spagettification'. Maybe I could add my layman picture to precise that gravitational vertical line are not parallel, but convergent, and that tidal force are borned from this differential. And that for small BH those lines are very convergent, but for large BH those line are more like parallel (at the EH).

I think I will try myself to compute the time ratio difference part for both a stellar and a massive BH. You have provided enough formulas, it is time to get dirty

Thank you again for all those explanations.

19. Oct 2, 2015

SlowThinker

I've read in some derivations of time dilation, that objects in gravitational field are contracted vertically, in addition to time dilation. The Note added to the original article also suggests that something is happening to the hovering ship's height.
In this scenario, there is the height of the ships above horizon, and the height of the ships themselves. I was trying to distinguish between the two. I'm not native English speaker and could not find a better word.

This was the base of my argument. They seem to be very closely related, which is confusing. I have to admit I still don't quite see the difference, but I'll try to search around for a few days for an explanation.

Somehow I believed that for a distant observer, the time speeds would be the same. But there's a lot going on in that situation, it's easy to get lost.

20. Oct 2, 2015

Staff: Mentor

No. The difference in rate of time flow between the top and bottom clocks is due to the potential created by the massive body. The "gravitational field" due to the acceleration of the clocks is not something different from the "gravitational field" created by the massive body; it's the same thing, just viewed from a different perspective.

Since the Moon's distance from Earth is 60 times the Earth's radius, and the Moon's radius is only 1/4 the Earth's radius, the variation in the Earth's potential across the Moon's radius is pretty small, so the two clocks on opposite sides of the Moon will have the same time rate to a pretty good approximation. But if you made accurate enough measurements, yes, you would see a slight difference due to the difference in the Earth's potential from one to the other. (Strictly speaking, even that is an approximation, because potentials don't add linearly, but for weak fields like those of the Earth and the Moon the nonlinearities are too small to detect.)

The EP is only valid within a small enough patch of spacetime that tidal gravity is negligible. So for the EP to apply to two clocks or your head and toes, the distance between them would have to be small enough for tidal gravity effects to be undetectable.

It isn't.

That's one aspect of tidal gravity around a large massive body, yes. Another aspect is the fact that two objects that start falling along the same vertical line, but at different heights, will separate.

Yes, this is true. But again, it is something different from the variation in "rate of time flow". There is no way to even define "rate of time flow due to gravitational potential" at or below the horizon of a black hole, because spacetime is not stationary there. But tidal gravity can be defined anywhere.

21. Oct 2, 2015

Staff: Mentor

This misconception does appear in some pop science sources, but it is indeed a misconception.

22. Oct 3, 2015

Boing3000

One of the coolest explanation I ever get, thank you !

And again I learn something new. Is it also the case for solution for the (hypothetical) non rotating BH ?

23. Oct 3, 2015

SlowThinker

I believe we were talking about a non-rotating BH this whole time.
Since the field is stronger close to the BH, your feet (try to) fall faster than your head. I believe this is the case with a rotating BH as well.

24. Oct 3, 2015

Staff: Mentor

Yes.

Yes.

Yes, but how much faster depends on how large the hole is and how tall you are. For a large enough hole, you wouldn't notice the difference even if you were falling through the horizon. (This is not just the case for a black hole; if you jump off a platform on Earth and free-fall downwards, your feet try to fall faster than your head as well. But the difference is too small for you to notice.)

The same tidal gravity effects exists outside a rotating BH, yes, but there are also additional effects due to the rotation. Basically, as an object falls, it not only gets stretched or squeezed by tidal gravity, it also gets twisted.

25. Oct 5, 2015

Staff: Mentor

A interesting but unrelated digresion has been moved to another thread, and this thread is now open again

Last edited: Oct 5, 2015