# Learn Time Dilation and Redshift for a Static Black Hole

The following is an overview of the time dilation and gravitational redshift effects of a static (Schwarzschild) black hole. In accordance with general relativity, a strong gravitational field can slow downtime. The closer you get to the event horizon of a black hole (if you can survive the gravity gradients, g-forces and have some means of propulsion that will keep you stable in the overwhelming gravity), there will appear no different to yourself but time will pass slower for you and quicker for people outside the field (with no apparent difference to them either). On returning from the edge of the black hole, depending on how close you were to the event horizon and how long you were there for, you would have aged less than the people outside the black hole’s strong gravitational field. This can be calculated using the following equation-

[tex]\tau=t\sqrt{1-\frac{2Gm}{rc^2}}[/tex]

which can be rewritten as

[tex]\tau=t\sqrt{1-\frac{r_s}{r}}[/tex]

where [itex]\tau[/itex]- actual (proper) time experienced within black holes gravitational field, [itex]t[/itex]- time observed from infinity, [itex]r_s[/itex]- Schwarzschild radius of black hole (where [itex]r_s=2Gm/c^2[/itex]), [itex]r[/itex]- the coordinate radius time was spent at ([itex]G[/itex]- gravitational constant, [itex]m[/itex]- mass, [itex]c[/itex]- speed of light).

For example, if an observer was to spend an ‘observed’ 3 days at 1 km from the event horizon of a 12,000* solar mass black hole, they would only experience-

[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,445,396}}[/tex]

[tex]\tau=0.01594\ \text{days = 23 minutes}^{**}[/tex]

where t=3 days, [itex]r_s=2Gm/c^2[/itex] which for a 12,000 solar mass black hole equals 35,444,396 m and to be 1 km from the event horizon would mean r=rs+1000 m=35,445,396 m

*A* 12,000 solar mass BH was chosen due to the tolerable tidal forces (dg) near the event horizon, [itex]dg=(2Gm/r^3)dr[/itex] where dr=2 (approx. height of a person), which means dg=71.5 g which is 7.3 earth g from head to toe which is just about manageable.*

***That is, if they instantaneously travel to the mark and travel instantaneously back again, otherwise time dilation would need to be taken into account for the various stages of the trip.*

Likewise, due to the immense gravity, the electromagnetic information being sent out by person would be extremely redshifted. This can be calculated using the following equation-

[tex]z=\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}-1[/tex]

(black hole event horizon = surface of infinite redshift)

and to calculate the actual change in wavelength-

[tex]z=\frac{\lambda_o – \lambda_e}{\lambda_e}[/tex]

which becomes

[tex]\lambda_o=(z \cdot \lambda_e) + \lambda_e[/tex]

where [itex]\lambda_o[/itex]- wavelength observed from infinity and [itex]\lambda_e[/itex]- wavelength emitted from the object.

Therefore, the redshift at 1 km from the event horizon would result in z=187.27. Information at a wavelength of 550 nm (the color green in the visible spectrum) [itex](\lambda_e)[/itex] would be stretched to a wavelength of 103,549 nm (0.10355 mm or 103.55 um) [itex](\lambda_o)[/itex] which would put it in the FIR wavelength (far-infrared light waves) *(in comparison, the sun has a surface gravitational redshift of z=0.0000021 which has near to zero effect on the electromagnetic information. A white dwarf with a 1.14 sol mass and a 4.5×10^6 m radius has a gravitational redshift of z=0.00037. For a neutron star with a 2 sol mass and a 12 km radius, z=0.4034.).* The strong gravitational field would have the opposite effect on light going the other way as it was pulled in towards the event horizon. Viewed from outside the strong gravitational field, the individual would appear to slow down and disappear over time as the light coming from them was redshifted towards radio waves, but from the inside, the observer might see the outside 3 days pass in the 23 minutes they spent at 1 km from the event horizon. *It’s very likely that this light would be blue shifted into the x-ray and gamma-ray spectrum which would no doubt have fairly drastic consequences for the observer.*

Imagine a spaceship is at 1,000,000 km from a 12,000 solar mass black hole (even here, the craft would have to have powerful engines in order for it not to be pulled in by the gravity of the black hole which would still be 1,511,711 m/s^2*** at this point). Gravitational redshift would be minimal (550 nm [itex](\lambda_e)[/itex] would be 560 nm [itex](\lambda_o)[/itex], green would still be green). Here, time dilation would already be beginning to take effect (~59 minutes would pass compared to every hour in regular space outside the black holes gravitational field). A scout ship with the capacity to travel over 1,000,000 km almost instantaneously *(an average velocity of 30,000 km/s, 0.1c, covering the 1 million km in ~35 seconds. Time dilation due to this velocity is virtually undetectable)* is sent out towards the black hole. Due to the velocity of the craft, it would appear orange to the main ship for the majority of the journey but as it reached the 100,000 km mark from the event horizon, it would begin to turn deep red (light wavelengths from the scout ship would be stretched from 600 nm to 650 nm due to gravitational redshifting). At around the 50,000 km mark, it would disappear to the naked eye and the observer ship would have to use infrared equipment in order to monitor the scout ship’s progress. As the scout ship passed the 10,000 km mark it would appear to slow down as time dilation meant that 30 seconds for the scout ship would appear as 1 minute for the main ship (the scout ship would appear to half its speed). By the time the ship reached the 1 km mark from the event horizon, it would be in far-infrared. It would be seen to hover for 3 days, visible only to infrared equipment on the main ship, the crew had plenty of time to monitor the scout ship’s position. After a full 3 days at the 1 km mark, the scout ship would at first appear to slowly make its way back to the ship, picking up speed as it went through far, mid, and near-infrared, back into visible light, 100,000 km from the event horizon, appearing nearly full speed now as the time dilation here is only 0.85 (1 minute for the main ship would be ~52 seconds for the scout ship). Over the remaining 900,000 km, the ship would appear blue due to its velocity. Once the scout ship had returned, it would inform the crew of the main ship that it had been traveling at maximum speed all the way, taking only 35 seconds to arrive at the 1 km mark where it hovered for just 23 minutes before accelerating away at full power to arrive back at the main ship.

****Gravity for a static black hole is* [itex]g=Gm/(r^2\sqrt(1-r_s/r))[/itex].

Time dilation and gravitational redshift become more extreme the closer you get to the black hole, at 1 m from the event horizon (r=rs+1 m), you would only experience-

[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,444,397}}[/tex]

[tex]\tau=0.0005\ \text{days = 43 seconds}[/tex]

The redshift would be z=5,957.83 (550 nm (green) [itex](\lambda_e)[/itex] would be at 3,277,358 nm or 3.277 mm [itex](\lambda_o)[/itex], which would put it in EHF, Extremely high frequency radio waves, detectable only by radio receivers)

and at 5 mm from the event horizon (r=rs+0.005 m)-

[tex]\tau=3\sqrt{1-\frac{35,444,396}{35,444,396.005}}[/tex]

[tex]\tau=0.000036\ \text{days = 3 seconds}[/tex]

The redshift would be z=104,962.52 (550 nm (green) would be at 57,729,938 nm or 57.730 mm, which would put it in SHF, Super high frequency radio waves).

Within the last 5 mm as you reached the event horizon, you would shift through medium to a very low frequency to infinite wavelength, and time dilation would go from 3 days passing in 3 seconds to centuries passing in mere fractions of a second. A lifetime would pass for anyone observing you from the outside with sensitive radio equipment.

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 before they free fall at close to the speed of light towards the singularity, the time of this fall (for a static Schwarzschild black hole) can be calculated by the following *(which seems derivative of the antipode distance of a hypersphere based on the gravitational radius [itex](M=Gm/c^2)[/itex])-*

[itex]M \pi/c = Gm \pi/c^3 = 1.548×10^{-5}\text{ x solar mass = time in seconds to singularity}[/itex]

For a 12,000 solar mass black hole, this would be a full 0.1857 seconds to admire the view before reaching the singularity.

*Note: As stated, the spaceship would experience a slight time dilation of its own. If the spaceship was hovering at r=10^6 km, then 3 days at infinity would be 2 days, 22 hours, 45 minutes on the spaceship. If it was in orbit at r=10^6 km, then the time dilation would be 2 days, 22 hours, 7 minutes where-*

*[tex]\tau_o=t\sqrt{1-\frac{r_s}{r}}\cdot\sqrt{1-v_s^2}[/tex]*

*where [itex]\tau_o[/itex] represents the time dilation for an object in stable orbit and [itex]v_s[/itex] is velocity required for stable orbit where [itex]v_s=\sqrt(M/(r(1-r_s/r)))[/itex].*

*The above equation is equivalent to*

*[tex]\tau_o=t\sqrt{1-\frac{3}{2}\cdot\frac{r_s}{r}}[/tex]*

*where [itex]\frac{3}{2}r_s[/itex] is considered the absolute last stable orbit, it also defines the photon sphere.*

*Another interesting equation is the proper distance to the event horizon for a hovering observer-*

*[tex]\Delta r’=\frac{\Delta r}{\sqrt{1-\frac{r_s}{r_s+\Delta r}}}[/tex]*

*where [itex]\Delta r'[/itex] is the proper distance as observed by the hovering observer and [itex]\Delta r[/itex] is the coordinate distance as observed from infinity.*

*At 1 km from the EH (as observed from infinity), the hovering observer will see himself to be 188 km from the EH, at 1 m, the hovering observer will see himself to be 5.954 km from the EH and at 5 mm, 421 m from the EH.*

*Also note, the equations shown for time dilation are for an object hovering at a specific r, for an object that has fallen from rest at infinity, the time dilation would be-*

*[tex]\tau=t\left(1-\frac{r_s}{r}\right)[/tex]*

*where [itex]\tau=t\sqrt(1-r_s/r)\cdot\sqrt(1-v^2)[/itex] where for an object that has fallen from rest at infinity, [itex]v=\sqrt(r_s/r)[/itex]*

Early life spent working and studying in York UK, 3 year architecture degree at Oxford polytechnic, 2 year architecture diploma at Oxford polytechnic, part-time in US. Worked in both York and London within architectural profession.

[QUOTE=”PeterDonis, post: 5257136, member: 197831″]

As Nugatory said, events on and inside the horizon can never be in your past light cone if you are outside the horizon, even if you live forever, and that statement is invariant, independent of coordinates..[/QUOTE]

Ok, that sums up what I have been trying to say all along. There are no event horizons in my past light cone. That means there are no event horizons in the [I]visible [/I]universe, [I]as far as I can see. [/I]So any supermassive objects we [I]can [/I]see are presumably eternally collapsing objects. And as I plan to live forever, that will be the casew forever (provided I can avoid falling into one of these objects)..

Thanks for your responses. I will leave it at that.

Mike

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]So there are actually coordinate sets where a black hole and event horizon appear within someone’s past light cone?[/QUOTE]

That’s not what I said. I said there are charts in which events on the horizon are assigned a finite time. In other words, there are charts in which events on the horizon are simultaneous with events on a distant observer’s worldline. “Simultaneous” is not the same as “in the past light cone”.

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]Eddington-Finkelstein are identical to Schwarzschild outside the event horizon.[/QUOTE]

No, they aren’t. This is obvious from looking at the line elements.

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]Kruskal-Sekeres also show that you have to pass over all future times to reach the event horizon.[/QUOTE]

No, they don’t. The horizon is at a finite “time” coordinate on any worldline in the K-S chart.

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]Gullstrand-Painleve are the coordinates for the person/object falling into the black hole, and not the distant observer[/QUOTE]

Wrong. G-P coordinates cover all of both the exterior region (outside the horizon) and interior region (inside the horizon), so they can describe the worldlines of the distant observer just as well as they can describe an observer falling into the hole. Furthermore, it is the distant observer who is at rest in G-P coordinates, [I]not[/I] the observer falling into the hole!

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]Having the event horizon/black hole form in a finite time leads inevitably to the problem of loss of information[/QUOTE]

This issue is an issue about invariants, and is present regardless of what coordinates we adopt. Adopting Schwarzschild coordinates, which can’t cover the horizon or the region inside it, doesn’t magically make issues involved at or inside the horizon go away. It just means your coordinates can’t describe them.

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]The Schwarzschild solution describes a universe with no singularities,. no information loss, no infinite densities, no places where physics breaks down (except maybe the Big Bang), and one coordinate system which describes ALL of the universe[/QUOTE]

This is all wrong. First of all, you should say “the Schwarzschild coordinate chart” (even more precisely, the Schwarzschild [I]exterior[/I] coordinate chart, since there is also a Schwarzschild chart covering the interior of the black hole!), not “the Schwarzschild solution”; the latter is a geometric object whose properties are independent of the coordinates you choose, and which certainly includes the event horizon and the region inside it, whether or not your chosen coordinates cover that region.

Second, the Schwarzschild exterior chart is geodesically incomplete: in other words, it has a boundary (as ##t rightarrow infty##) into which physical objects, as far as the coordinates are concerned, simply “disappear”. The fact that the ##t## coordinate goes to infinity before this happens is irrelevant; coordinates are not physics. The physics is contained in the invariants, and it is simple to compute invariants (such as the proper time elapsed for an observer free-falling to the horizon) which are finite, and clearly show that physical objects simply “disappear” from this chart. (Note that these invariants can be computed using Schwarzschild coordinates, even though those coordinates don’t cover the horizon; you just take limits as ##r rightarrow 2M##, and these limits are perfectly well-defined). But physical objects can’t just disappear; if your coordinates make it appear that they are, then your coordinates are incomplete–they can’t possibly describe ALL of the universe, and they certainly can’t tell you that the universe contains no singularities, etc.

[QUOTE=”Mike Holland, post: 5256977, member: 429699″]Yes, you can always choose the coordinates of an infalling body. I think that is the only way to define a set of coordinates for an event horizon that will never end up in your past light cone (assuming you will live forever).[/QUOTE]

As Nugatory said, events on and inside the horizon can never be in your past light cone if you are outside the horizon, even if you live forever, and that statement is invariant, independent of coordinates.

You appear to have some serious misconceptions about the various coordinate charts on Schwarzschild spacetime, and also about how the physics of the Schwarzschild solution actually works. These misconceptions are not new; they have been discussed many, many, many times in previous PF threads. I realize that these misconceptions can be intuitively plausible, but they’re still misconceptions, and they’re still wrong, and this thread is not the place to rehash them. If you insist on asking further questions about them, please start a new thread, and please do not simply restate your incorrect statements. Ask questions.

[QUOTE=”Nugatory, post: 5256885, member: 382138″]But that has little do with my freedom to choose coordinates in which events on the horizon have a finite time coordinate.[/QUOTE]

Yes, you can always choose the coordinates of an infalling body. I think that is the only way to define a set of coordinates for an event horizon that will never end up in your past light cone (assuming you will live forever). Every other point with a finite timestamp must, unless it is moving away at high speed. Can you suggest any alternatives?

[QUOTE=”Mike Holland, post: 5256856, member: 429699″]So there are actually coordinate sets where a black hole and event horizon appear within someone’s past light cone? Which are these? I haven’t heard of them.[/quote]

What is in or not in the past light cone of an event is the same no matter what coordinates you adopt. Some events on the horizon will be in the past light cone of an observer inside the horizon and no event on the horizon will be in the past light cone of an observer outside the horizon; and these relationships will hold no matter what coordinates we use. But that has little do with my freedom to choose coordinates in which events on the horizon have a finite time coordinate.

[quote]Kruskal-Sezeres also show that you have to pass over all future times to reach the event horizon.[/quote]They show that I have to pass through all curves of constant Schwarzschild ##t## to reach the event horizon, but that just tells us something about curves of constant Schwarzschild ##t##.

[QUOTE=”PeterDonis, post: 5256798, member: 197831″]But if you adopt a different set of coordinates, then the horizon forms at a finite time, and you will experience events after that time, at which point the horizon will have formed, according to those coordinates..[/QUOTE]

So there are actually coordinate sets where a black hole and event horizon appear within someone’s past light cone? Which are these? I haven’t heard of them.

Eddington-Finkelstein are identical to Schwarzschild outside the event horizon.

Kruskal-Sekeres also show that you have to pass over all future times to reach the event horizon.

Gullstrand-Painleve are the coordinates for the person/object falling into the black hole, and not the distant observer, and we know they describe a swift fall to the singularity.

Having the event horizon/black hole form in a finite time leads inevitably to the problem of loss of information, which all he astrophysicists are still battling over. The Schwarzschild solution describes a universe with no singularities,. no information loss, no infinite densities, no places where physics breaks down (except maybe the Big Bang), and one coordinate system which describes ALL of the universe – at least until the end of eternity! But it doesn’t describe what happens to the collapsing star/falling object after eternity (or in the local time frame).

Mike

[QUOTE=Mike Holland]they will see the contraction slow down and effectively come to a stop.

[QUOTE=”PeterDonis, post: 5256798, member: 197831″]Yes, this is what remote observers will [I]see[/I], in the sense of the light signals they receive.[/QUOTE]

[/QUOTE]

I don’t agree. A remote observer will [I]never[/I] see anything “effectively coming to a stop”. Only the event horizon is at time infinity. Everything outside is at finite time and is not stopped. Beside the object will effectively be out of “sighting” range way before due to the extreme red shifting of the signals

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]The time dilation calculated from GR is real for clocks in orbit which run faster than ours, affecting GPS systems.[/QUOTE]

Yes; more precisely, the fact that GPS clocks have to be adjusted in order to provide correct time references is an invariant, independent of any coordinates we adopt.

However, there is no reason why we have to adopt the view that the adjustment that needs to be made to GPS clocks must be decomposed into a “gravitational” part and a “Doppler shift due to motion” part. That is how it is usually analyzed, but that is a matter of convenience, not physics.

Similar remarks apply to all your other examples. Nobody is saying that invariant effects that are usually attributed to “time dilation” are just illusions, or are not real, or don’t really happen. But the fact that something really happens does not mean it has to be analyzed in a particular way.

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]When radioactive subatomic particles are created in the upper atmosphere by cosmic rays, some of them make it to Earth’s surface although their half-lives are to short for them to survive the trip. They only make it because of their high speed, and time dilation extends their half-lives. But in their local frame they are not dilated, and they make it because the distance they have to travel is reduced by Lorentz contraction. So is the time dilation story an illusion? I hold that both views are equally true, complementary descriptions of what happens.[/QUOTE]

I agree.

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]In the same way I see no conflict between gravitational contraction being time dilated from our point of view, and proceeding extremely swiftly to a singularity from the local point of view.[/QUOTE]

You’re right, there is no conflict.

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]So I agree, nothing is “frozen” when viewed locally, but who is viewing a collapsing black hole locally?[/QUOTE]

Someone who is falling in with the collapsing matter.

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]All the observers I know of are remote[/QUOTE]

Which says nothing about whether all possible observers must be remote.

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]they will see the contraction slow down and effectively come to a stop.[/QUOTE]

Yes, this is what remote observers will [I]see[/I], in the sense of the light signals they receive.

[QUOTE=”Mike Holland, post: 5256773, member: 429699″]So for all the observers I know of, event horizons don’t exist yet.[/QUOTE]

No. This is where you go wrong. This statement, unlike the other ones I agreed with above, is [I]not[/I] an invariant; it depends on the coordinates you adopt. If you, as a remote observer, adopt a particular set of coordinates, yes, a collapsing object has not “yet” formed a horizon, because in those coordinates, the “time” assigned to any event on the horizon is ##+ infty##. But if you adopt a different set of coordinates, then the horizon forms at a finite time, and you will experience events after that time, at which point the horizon will have formed, according to those coordinates.

In other words, if we restrict ourselves to only talking about invariants (and all of the actual physics is contained in invariants), then [I]there is no answer[/I] to the question “when does the horizon form?” [I]except[/I] for observers who actually fall in and experience the horizon locally. To them, they cross the horizon at some definite time by their clock. But a remote observer, who never actually crosses the horizon, can only assign a “time” to any event on the horizon by adopting a simultaneity convention, and that is a matter of convention, not physics.

By the way, this limitation is not particular to black holes and event horizons; it arises whenever we try to assign a “time” to events that are spatially separated from us. There is no invariant way to make such assignments; it always requires adopting a simultaneity convention.

[QUOTE=”PeterDonis, post: 5256072, member: 197831″]

Which is true, regardless of whether you split the observed redshift into gravitational and Doppler parts or not. Nothing is “frozen” when viewed locally; that is an invariant, independent of how we describe what the distant observer sees.[/QUOTE]

The time dilation calculated from GR is real for clocks in orbit which run faster than ours, affecting GPS systems. The time dilation measured on the sun and on white dwarf stars is real. The gravitational time dilation me assured in the laboratory is real. None of these have anything to do with successive photons emitted from a falling body, and none of them are illusions.. They really happen. Admittedly the effects will appear different to observers in other reference frames, but that does not make them illusions.

When radioactive subatomic particles are created in the upper atmosphere by cosmic rays, some of them make it to Earth’s surface although their half-lives are to short for them to survive the trip. They only make it because of their high speed, and time dilation extends their half-lives. But in their local frame they are not dilated, and they make it because the distance they have to travel is reduced by Lorentz contraction. So is the time dilation story an illusion? I hold that both views are equally true, complementary descriptions of what happens. In the same way I see no conflict between gravitational contraction being time dilated from our point of view, and proceeding extremely swiftly to a singularity from the local point of view.

So I agree, nothing is “frozen” when viewed locally, but who is viewing a collapsing black hole locally? All the observers I know of are remote, and they will see the contraction slow down and effectively come to a stop. So for all the observers I know of, event horizons don’t exist yet.

But maybe the Buddhists were right all along, and [I]all [/I]is illusion!

Mike

[QUOTE=”Mike Holland, post: 5255741, member: 429699″]After his fable about ants pushing balls out of a hole, he concludes that what we see is the Doppler effect for successive photons from a falling body taking longer and longer to escape[/QUOTE]

IIRC, that’s because he is not using the concept of gravitational time dilation in that passage; he is viewing the entire redshift as “Doppler” (though you still have to be careful with that term because it’s not the ordinary SR Doppler shift in an inertial frame–there is no inertial frame that covers both the free-faller near the horizon and the distant observer). The splitting that I describe, into gravitational and Doppler parts, is not the only possible way of describing what happens; it just happens to be a useful one. Thorne’s way, which is a different way, can be useful too.

[QUOTE=”Mike Holland, post: 5255741, member: 429699″]he concludes “that it appears to freeze as seen from far away is an illusion.”[/QUOTE]

Which is true, regardless of whether you split the observed redshift into gravitational and Doppler parts or not. Nothing is “frozen” when viewed locally; that is an invariant, independent of how we describe what the distant observer sees.

[QUOTE=”PeterDonis, post: 5244402, member: 197831″]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 [I]r[/I] r, which is what you wrote down, combined with the Doppler redshift for an observer falling inward at velocity [I]v[/I]=2[I]GM[/I]/[I]r[/I] − − − − − − √ v = sqrt{2GM/r} relative to the static observer.[/QUOTE]

Peter, I think Kip Thorne gets this wrong in his book “Black Holes and Time Warps”. After his fable about ants pushing balls out of a hole, he concludes that what we see is the Doppler effect for successive photons from a falling body taking longer and longer to escape, and he concludes “that it appears to freeze as seen from far away is an illusion.” (p249). As you point out above, what we would see is the combination of the two effects.

Mike

[QUOTE=”Bernard McBryan, post: 5251570, member: 567315″]Would there be a difference? Who would be “younger”? And what would the external observer record?[/QUOTE]

The only thing I might add to PeterDonis’s excellent answer is that this problem responds well to the method described in the “Doppler Shift analysis” in the [URL=’http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html’]twin paradox FAQ[/URL]. Imagine that both spaceships carry a strobe light that flashes once a second while they’re separated. Clearly the total number of times each ship’s light flashes is the amount of time that elapsed on that ship while they wrere separated. Furthermore, both ships can see and count the number of flashes from the other ship’s strobe. Thanks to time dilation, gravitational effects, and light travel time it may not be clear when the flashes will get to the other ship, but it is clear that they will get there and be counted eventually (and before the reunion).

Thus, if I see my light flash eight times while we separated, and I count ten flashes from your ship…. I know that you aged ten seconds while I only aged eight. All observers every must agree about the number of times each strobe flashed; we can put a counter on the strobe just to be sure.

[QUOTE=”Bernard McBryan, post: 5251570, member: 567315″]The first continues to hover in place, but the second case momentarily turning their engine off, and then free falls long enough to create some additional time dilation from one (or both perspectives), but not long enough to cross the event horizon or reach excessive speeds. Then, the engines are re-engaged, first to stop the downward fall, and then slowly to rise to the original altitude.[/QUOTE]

Ok, this is a good scenario. We’ll call the two ships A (the hovering ship) and B (the traveling ship, that falls down and then climbs back up).

[QUOTE=”Bernard McBryan, post: 5251570, member: 567315″]To simplify the scenario, a larger 1 billion or 1 trillion solar mass black hole could be used to reduce the tidal gravitational forces across the top/bottom of each ship, and even across the change in altitudes of the two ships.[/QUOTE]

Yes, this is fine; nothing will depend on there being measurable tidal gravity involved, so we can assume it’s negligible.

[QUOTE=”Bernard McBryan, post: 5251570, member: 567315″]Would there be a difference? Who would be “younger”?[/QUOTE]

Yes, there would be a difference. The person on ship B would be younger when the two ships meet up again. This is easily seen from the fact that ship B does two things relative to ship A, both of which create increased time dilation relative to A: ship B goes to a lower altitude (so ship B has more gravitational time dilation), and ship B moves while ship A remains stationary (so ship B has additional time dilation due to motion, which ship A does not have).

[QUOTE=”Bernard McBryan, post: 5251570, member: 567315″] And what would the external observer record?[/QUOTE]

The external observer records just what I described above. The difference in aging between ships A and B when they meet up again is an invariant; all observers must agree on it.

Mentor said: “First somebody needs to state a “twin paradox” scenario–that is, a scenario where two observers start out together, separate for a while, then come back together and compare the elapsed times on their clocks. Nobody has yet done that in this thread. Once a specific scenario is stated, explaining how it works from the three perspectives will be straightforward.”

The scenario is a simple extension of the above case (or original cases) where two twin ships start at the same altitudes in a gravitational field, both initially hovering. The first continues to hover in place, but the second case momentarily turning their engine off, and then free falls long enough to create some additional time dilation from one (or both perspectives), but not long enough to cross the event horizon or reach excessive speeds. Then, the engines are re-engaged, first to stop the downward fall, and then slowly to rise to the original altitude. To simplify the scenario, a larger 1 billion or 1 trillion solar mass black hole could be used to reduce the tidal gravitational forces across the top/bottom of each ship, and even across the change in altitudes of the two ships. The second twin ship, as she returns may only be slightly different due to the small distance and time traveled (rather than years in special relativity based twin paradox). Would there be a difference? Who would be “younger”? And what would the external observer record?

Thanks for your help thus far. It has been very helpful.

Bernie.

[QUOTE=”Bernard McBryan, post: 5250835, member: 567315″]Regarding the hovering case 1 (above) and free fall case 2 gravitational time dilation being equal seems to possibly violate the equivalence principle.[/QUOTE]

No, it doesn’t. The EP doesn’t say that free fall and proper acceleration can’t yield the same local experimental results in some particular cases. It only says that free fall compared to free fall, or proper acceleration compared to proper acceleration of the same magnitude, can never yield different experimental local results.

[QUOTE=”Bernard McBryan, post: 5250835, member: 567315″]Case 2 is in free fall, and assuming small enough to ignore the tidal forces, does not feel the gravity, nor her acceleration downward.[/QUOTE]

Correct.

[QUOTE=”Bernard McBryan, post: 5250835, member: 567315″]I suppose she sees the hovering case 1 time dilate due to the gravity, or perceived acceleration away from her.[/QUOTE]

No. Remember we are talking about the instant where the free faller is momentarily at rest, right next to the hoverer. They are motionless at that instant with respect to each other, and they are both at the same altitude, so neither one sees the other as time dilated at that instant.

Of course, after that instant, the free faller and the hoverer will start moving relative to each other, and consequently they will start seeing the effects of time dilation (due to both relative motion and being at different altitudes). But I specifically restricted attention, in my example, to the instant where they are momentarily at rest relative to each other, to eliminate those effects.

[QUOTE=”Bernard McBryan, post: 5251174, member: 567315″]Does a clock in free fall, and a hovering clock at the same gravitational altitudes experience the same gravitational slowdowns from the three perspectives[/QUOTE]

At the instant they are at rest relative to each other, yes. Otherwise no. See above.

[QUOTE=”Bernard McBryan, post: 5251174, member: 567315″]how do they explain the twin paradox from the three perspectives (hovering, free fall, far away outside observer)[/QUOTE]

First somebody needs to state a “twin paradox” scenario–that is, a scenario where two observers start out together, separate for a while, then come back together and compare the elapsed times on their clocks. Nobody has yet done that in this thread. Once a specific scenario is stated, explaining how it works from the three perspectives will be straightforward.

I tried to make some simplifying assumptions to simplify this scenario to focus just on the time dilation aspects:

1) assumed ship (or very small clock) small enough to ignore the tidal forces: thus the ship is small enough in length (and width) so that the tidal forces can be ignored. ( e.g. head to feet height less than .25 inches). Alternately, one can assume that the black hole is larger (e.g. 1 Billion or 1 Trillion solar masses, such that the g force and tidal forces are weaker and can be neglected).

2) To further simplify and eliminate velocity based terms, the falling ship can reverse course after a very short time, before it reached relativistic speeds. I eliminated the orbiting case 3, so velocities can be kept at non-relativistic speeds.

3) I also intentionally did all this outside the event horizon to avoid its entanglements.

I guess I did not even need a black hole, but the question would be true outside any sun or planet (but with time slowdowns at the microsecond level).

The basic question: Does a clock in free fall, and a hovering clock at the same gravitational altitudes experience the same gravitational slowdowns from the three perspectives and how do they explain the twin paradox from the three perspectives (hovering, free fall, far away outside observer).

[QUOTE=”Bernard McBryan, post: 5250835, member: 567315″]Regarding the hovering case 1 (above) and free fall case 2[/QUOTE]I’ve realized there are too many things going on to keep track of, without a very careful analysis: Lorentzian time dilation, length contraction and simultaneity shift, plus these same effects caused by gravity.

Ideally, we’d name 5 points (top and bottom of a hovering ship, top and bottom of a falling ship, and a distant observer), and describe how each sees the other 4.

I gave up :sorry:

Regarding the hovering case 1 (above) and free fall case 2 gravitational time dilation being equal seems to possibly violate the equivalence principle.

Case 2 is in free fall, and assuming small enough to ignore the tidal forces, does not feel the gravity, nor her acceleration downward.

I suppose she sees the hovering case 1 time dilate due to the gravity, or perceived acceleration away from her.

Case 1 feels the gravity but thinks he is stationary, and may perceive time dilation of Case 2 due to her acceleration downward.

But it is unclear how an external observer would see the identical time dilation for each of these cases.

Additionally, if case 2 stops her descent downward towards the event horizon, and reversed course back upward to the exact height, would the clocks be the same when they meet in the same reference frame? And how does the external observer perceive the clocks?

[QUOTE=”Mike Holland, post: 5250356, member: 429699″]SlowThinker has commented a couple of times that the gravitational acceleration at the horizon is infinite! This is surely incorrect. The escape velocity at that point is c

[/quote]

It’s the other way around. The escape velocity at the horizon is undefined, as there are no timelike worldlines that get you from a point on the horizon to anywhere outside the horizon. Conversely, the proper acceleration required to hold you stationary at the horizon would indeed be infinite (and hence unrealizable). A more informal way of seeing that the escape velocity at the horizon is not ##c## is to consider that a flash of light moving radially outwards from the horizon will not escape even though it moving at ##c## – so whatever ##c## is, it’s not the escape velocity from the horizon.

[quote]Approaching the singularity the gravitational force would tend to infinity, but it becomes meaningless at the singularity.[/QUOTE]

The singularity is not at the horizon, it’s at the center of the black hole. There’s no meaningful definition of gravitational force anywhere inside the horizon, as the spacetime is not stationary there so there’s no potential to take the gradient of to find a force.

[QUOTE=”Bernard McBryan, post: 5249666, member: 567315″]1) hovering (engines accelerating fighting 1,511,711 m/s^2 ): experience listed gravitational time dilation.[/QUOTE]

Yes.

[QUOTE=”Bernard McBryan, post: 5249666, member: 567315″]2) engines just turned off, same height but now in free fall but still at slow velocity: experience same listed gravitational time dilation.[/QUOTE]

Yes, because the velocity relative to a static observer is negligible.

[QUOTE=”Bernard McBryan, post: 5249666, member: 567315″]3) in circular orbit (free fall at same height but with high tangent orbit velocity): experience slightly higher gravitational time dilation (perhaps due to the same listed gravitational time dilation but with extra velocity time dilation due to the orbital tangent velocity.[/QUOTE]

No: experience the same listed gravitational time dilation (because that depends only on altitude, and the orbiting observer is at the same altitude as the static observer), but also experience additional time dilation due to velocity (relative to the static observer at the same location). The time dilation due to velocity is not considered “gravitational”, because it depends only on velocity relative to a static observer, not on location.

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

[QUOTE=”Edriven, post: 5249564, member: 572993″]I’m sorry. I don’t want to measure time without a clock. I just want to use a different clock. One free of gravity and speed, if there is such a thing. We both know enironment would determine this.[/QUOTE]

[QUOTE=”SlowThinker, post: 5249065, member: 572662″]The contrary. In the space you age fastest. On the Earth you age at about 99.99999993% of that speed. Close to the horizon of a black hole, you can age at 1% speed or even slower.[/QUOTE]

Thank you. Can u pls tell me why the clocks had two different times? [URL]https://en.m.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment[/URL]

[QUOTE=”Nugatory, post: 5248475, member: 382138″]The “clocks” of relativity are any time-dependent process: the graying of my hair, sand falling through an hourglass, the decay of radioactive atoms, the decay of unrefrigerated meat, the motion of the earth around the sun, the height of the tree in my front yard… Google for “Time is what a clock measures” to see the importance of this insight from Einstein.

I think Einstein’s brilliance is that he has discovered how the change of environments effect out time pieces. I think he was telling us that our measurements will change will going fast or in different gravities. We are comparing the standard clock with the rest of the entire universe. The earth’s surface is such a small percentage of the universe that this comparison doesn’t make sense.

How do you explain the twin “paradox” ([URL]http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html[/URL]) then? Basically two identical twins by definition the same age shake hands; then take different journeys; then reunite to shake hands again. At the reunion, one of the twins has experienced less time and is physically less aged than the other. It’s really hard to see how the difference in aging can be attributed to anything except having lived through different amounts of time between handshakes.

(Note: It’s called the twin “paradox” because it’s often presented as a paradox in elementary classes as a teaching tool, but in fact it’s not a paradox – the correct and non-paradoxical resolution has been understood since the birth of relativity).[/QUOTE]

The twin paradox is fiction. NASA has had plants in space since 1960. It can be theorized, by this fiction, that the plants should have a slower growth rate and metabolism. The opposite is occurring.

[QUOTE=”Boing3000, post: 5247122, member: 559056″]Is it also the case for solution for the (hypothetical) non rotating BH ?[/QUOTE]

Yes.

[QUOTE=”SlowThinker, post: 5247167, member: 572662″]I believe we were talking about a non-rotating BH this whole time.[/QUOTE]

Yes.

[QUOTE=”SlowThinker, post: 5247167, member: 572662″]Since the field is stronger close to the BH, your feet (try to) fall faster than your head.[/QUOTE]

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.)

[QUOTE=”SlowThinker, post: 5247167, member: 572662″]I believe this is the case with a rotating BH as well.[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5247122, member: 559056″]

[QUOTE=”PeterDonis, post: 5246808″]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.[/QUOTE]

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

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.

[QUOTE=”PeterDonis, post: 5246808, member: 197831″] it’s the same thing, just viewed from a different perspective.[/QUOTE]

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

[QUOTE=”PeterDonis, post: 5246808, member: 197831″]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.[/QUOTE]

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

[QUOTE=”SlowThinker, post: 5246073, member: 572662″]I’ve read in some derivations of time dilation, that objects in gravitational field are contracted vertically, in addition to time dilation.[/QUOTE]

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

[QUOTE=”Boing3000, post: 5246052, member: 559056″]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 ?[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5246052, member: 559056″]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…[/QUOTE]

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.)

[QUOTE=”Boing3000, post: 5246052, member: 559056″]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.[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5246052, member: 559056″]I still think that SlowThinker ‘dust’ is related to time ratio difference[/QUOTE]

It isn’t.

[QUOTE=”Boing3000, post: 5246052, member: 559056″]gravitational vertical line are not parallel, but convergent, and that tidal force are borned from this differential[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5246052, member: 559056″]for small BH those lines are very convergent, but for large BH those line are more like parallel (at the EH).[/QUOTE]

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.

[QUOTE=”PeterDonis, post: 5245967, member: 197831″]What is “height contraction?”[/QUOTE]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.

[QUOTE=”PeterDonis, post: 5245967, member: 197831”]I’m not sure what you mean by “tallness”.[/QUOTE]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.

[QUOTE=”PeterDonis, post: 5245967, member: 197831″]Once more: the change in the acceleration needed to stay in the same place, and tidal gravity, are [I]two separate things[/I]. They don’t go together.[/QUOTE]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.

[QUOTE=”PeterDonis, post: 5245967, member: 197831″]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.[/QUOTE]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.

[QUOTE=”PeterDonis, post: 5245804, member: 197831″]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.[/QUOTE]

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 [U]hovering[/U] 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 ?

[QUOTE=”PeterDonis, post: 5245804, member: 197831″]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.[/QUOTE]

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:smile:

Thank you again for all those explanations.

[QUOTE=”jartsa, post: 5245478, member: 386340″]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.[/QUOTE]

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.

[QUOTE=”SlowThinker, post: 5245842, member: 572662″]Are you 100% sure that the height contraction does not play a crucial role?[/QUOTE]

What is “height contraction?”

[QUOTE=”SlowThinker, post: 5245842, member: 572662″]The acceleration needed to keep my head in place is ##g_{head}##. Similarly define ##g_{feet}##.[/QUOTE]

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 [I]freely falling[/I] 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.

[QUOTE=”SlowThinker, post: 5245842, member: 572662″]The difference ##g_{head}-g_{feet}## seems to fit your definition of tidal effects[/QUOTE]

No, it doesn’t. See above.

[QUOTE=”SlowThinker, post: 5245842, member: 572662″]and so is finite everywhere except singularity.[/QUOTE]

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 [I]two separate things[/I]. They don’t go together.

[QUOTE=”SlowThinker, post: 5245842, member: 572662″]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[/QUOTE]

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.

[QUOTE=”SlowThinker, post: 5245842, member: 572662″]and [I]tallness[/I][/QUOTE]

I’m not sure what you mean by “tallness”.

[QUOTE=”PeterDonis, post: 5245797, member: 197831″]The more general definition of tidal gravity involves geodesic deviation[/QUOTE]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?

[QUOTE=”PeterDonis, post: 5245804, member: 197831″]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.[/QUOTE]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 [i]tallness[/i]? I thought they are the same, but now I’m not so sure…

[QUOTE=”Boing3000, post: 5245706, member: 559056″]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[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5245706, member: 559056″]When both ship are at the same altitude, do they experience (locally) the same time dilatation ?[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5245706, member: 559056″]If the tidal force (the radial angle difference) is manageable, I suppose the time dilation must be quite weak too.[/QUOTE]

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.

[QUOTE=”Boing3000, post: 5245706, member: 559056″]I think SlowThinker has inverted the gravitational time dilatation effect, because his toes are not supposed to turn to dust, but his head is[/QUOTE]

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.

[QUOTE=”SlowThinker, post: 5245748, member: 572662″]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.[/QUOTE]

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).

[QUOTE=”SlowThinker, post: 5245748, member: 572662″]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.[/QUOTE]

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.

[QUOTE=”SlowThinker, post: 5245630, member: 572662″]The tidal forces are a difference between gravity at my head, minus gravity at my toes, right?[/QUOTE]

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.

[QUOTE=”SlowThinker, post: 5245630, member: 572662″]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?[/QUOTE]

No. See above.

[QUOTE=”Boing3000, post: 5245706, member: 559056″]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[/QUOTE]

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.

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:nb)

[QUOTE=”SlowThinker, post: 5244835, member: 572662″]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?

[/QUOTE]

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.

[QUOTE=”stevebd1, post: 5245211, member: 68677″]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.[/QUOTE]

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.

[QUOTE=”stevebd1, post: 5245211, member: 68677″]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.[/QUOTE]

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.

[QUOTE=”stevebd1, post: 5245211, member: 68677″]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.[/QUOTE]

Yes.

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.

[QUOTE=”SlowThinker, post: 5244862, member: 572662″]I was referring to aging of the ship, due to infinite(?) time dilation.[/QUOTE]

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.

[QUOTE=”PeterDonis, post: 5244850, member: 197831″]It depends on how large the hole is.[/QUOTE]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.

[QUOTE=”SlowThinker, post: 5244835, member: 572662″]Do I understand it correctly that at the horizon, gravity acceleration is infinite, while the tidal forces are manageable/finite?[/QUOTE]

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).

[QUOTE=”SlowThinker, post: 5244835, member: 572662″]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?[/QUOTE]

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.

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.

SlowThinker has commented a couple of times that the gravitational acceleration at the horizon is infinite! This is surely incorrect. The escape velocity at that point is c. Approaching the singularity the gravitational force would tend to infinity, but it becomes meaningless at the singularity..

So, outside the event horizon in the three cases:1) hovering (engines accelerating fighting 1,511,711 m/s^2 ): experience listed gravitational time dilation.2) engines just turned off, same height but now in free fall but still at slow velocity: experience same listed gravitational time dilation.3) in circular orbit (free fall at same height but with high tangent orbit velocity): experience slightly higher gravitational time dilation (perhaps due to the same listed gravitational time dilation but with extra velocity time dilation due to the orbital tangent velocity.

In addition. I think of it like this. Light traveling from a distant star can be used as a historical record of time. Much like archeologists date objects based on depth of dirt. But no one is saying we can bend the dirt-time continuum

Loved the post, but is respectfully disagree with use of time dilation. Doesn't gravity affect light and matter? Is it not having an affect on our clock? Clock was designed to work on Earths surface. A change in function of this machine is only a loss of calibration to a different environment. Thus this time piece can work correctly in a stable environment, if it is recalibrated to that environment. So it would be closer to the truth if you said time dilation is only a relationship of the rest of our universe using Earths time. But time is a constant and can't be bent or altered.

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?

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.