Sine the gravity near a black hole is so high, the time moves slower than the observer(in our case we from the earth).In that case, imagine if are viewing a black hole's accretion disk(the innermost rim) a million light years away and if we are observing it now then since it took the light a million years to reach us, we are looking at the past state of the black hole(how it was a million years before). But light is traveling in time that is faster than it used to the moment light escapes the event horizon.

So if we imagine one photon at a time, the first photon reaches us and then as we observe the second photon, we have moved so much into the future as our time travels faster compared to the time the photon originated.

Are millions of years old black holes actually even more older than they appear to be?

  • $\begingroup$ "Sine the gravity near a black hole is so high, the time moves slower" - You have the cause and result reversed. Gravity near a black hole is so high, because time there moves slower. Not the other way around. Gravity is a consequence of a spacetime curvature, more specifically, of time curvature, which is known as time dilation (time moving slower). $\endgroup$
    – safesphere
    Commented Apr 23, 2019 at 3:50
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    $\begingroup$ @safesphere Not quite; it's a case where more than one approach is mathematically coherent and complete. You can start from spacetime and geometry, or you can start from the concept of a field with a Lagrangian. You end up with the same predictions for observable quantities. $\endgroup$ Commented Apr 23, 2019 at 13:03
  • $\begingroup$ @AndrewSteane In General Relativity you first find the spacetime metric from the gravitational equations. Then the metric defines the Lagrangian. So the concept of a field you are referring to is a direct consequence of the spacetime geometry. The only case where your comment would make sense, professor, is the Newtonian gravity, but (1) it does not explain the nature of gravity and (2) doesn't correctly define black holes, which are a subject here. $\endgroup$
    – safesphere
    Commented Apr 23, 2019 at 14:51
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    $\begingroup$ @safesphere If you want to find out more about this, look up "Einstein Hilbert action" in the context of Lagrangian mechanics. General relativity can itself be derived from a suitable Lagrangian. $\endgroup$ Commented Apr 27, 2019 at 22:22
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    $\begingroup$ @ Andrew Steane Shouldn't we just say that gravitation and time dilation go together, rather than saying that one causes the other? (I think I can infer that from your 1st comment above.) $\endgroup$
    – D. Halsey
    Commented Jun 24, 2019 at 19:00

2 Answers 2


Consider the following. You ask a young volunteer to come to live near a black hole, where time passes much slower. You observe them to age much less than you. Are they "stuck in the past"? It might seem so, since you were their biological age years ago. But this is not the correct interpretation, because at one point they can get on a rocket and come greet you. Surely they have not "arrived from the past", since they could have come to you at any point, and you could have come to them as well. So it really makes sense to talk about things happening synchronously, at the same time at your location as near the black hole. So really, the most accurate thing to say is that time passes slower near a black hole, but still we can synchronize!

Another, somewhat more everyday example. The Earth has formed some 4 billion years ago. The difference in the gravitational field at one position on the surface of the Earth and another one about one meter higher causes time to flow very very slightly differently. To be precise, the difference is about one second less flows one meter lower in $10^{16}$ seconds. But guess what, 4 billion years is $\sim 10^{17}$ seconds, so some $\sim 10$ seconds less have elapsed per every meter lower you go since the birth of Earth.

But now take a look at your feet. Are you looking ten seconds in the past? Are your feet traveling in and out of the past as you swing on a swing? Of course not. You are looking at things that happen (almost) synchronously in every physically meaningful way. So once again, slower flow of time means exactly that, clocks tick slower, but the notion of synchronicity can stay!

The additional question you are asking can be formulated in the following way: How longer will it take for a photon fired off from near the black hole to arrive then when a photon would be fired off from the "same position" without the black hole nearby? Notice that I put the "same position" in quotation marks, because the translation dictionary between a position near a black hole and a position in the space-time without the black hole will not be unique. We are after all in curved space-time, and the geometry is simply different from Euclidean space!

For instance, you can define the position near the black hole by infinity-signaling distance - the distance that is defined by how much time it takes for a light signal to reach infinity. Naturally, points at the same infinity-signaling distance in the black hole space-time as in flat Minkowski will take the same time to reach infinity. So in this operational definition of position the presence of the black hole does not push the signal "further into the past".

There are other definitions that can be used. For instance, you define distances by how much string of vanishing mass you have to use to reach a certain point. Then you would get that signals from the "same points" in the black hole geometry do arrive later than in a Minkowski one. The precise numbers will depend on your translation dictionary, but for reasonable choices and for light sources orbiting at astrophysically realistic distances from the black hole, the time delays will be of the order of $10 \mu {\rm s} (M/M_\odot)$, where $M$ is the mass of the black hole, and $M_\odot$ is the solar mass. If you plug in the mass of M87 observed by the Event horizon telescope, you get a delay of the order of days.

  • $\begingroup$ Thanks for the response, it does help but I have a double, but what if I use a very powerful telescope to view that person (it's hypothetical as the telescope size should be so massive but if I somehow manage to do so), I look at the person's past. I want to know the difference between that person placed near a black hole and not near a black hole, somewhere else where gravity is not as strong. How does the black hole's massive gravity affect what we see? $\endgroup$ Commented Apr 21, 2019 at 19:29
  • $\begingroup$ @dhayanandbaskar You don't see their past. You just see them moving slow. Good answer +1. $\endgroup$
    – safesphere
    Commented Apr 22, 2019 at 8:11
  • $\begingroup$ @safesphere. By past I meant - when the light reaches us after millions of years, we are looking at millions of year old state, not what is currently happening. Can you please elaborate on how we see them move slow? $\endgroup$ Commented Apr 22, 2019 at 10:52
  • $\begingroup$ @dhayanandbaskar I was responding to your comment above, not answering your question. In the comment you have two observers, one near and one away from a black hole. No "millions of years" were mentioned. As you look at these observers, you'd see the one closer to the black hole moving slower, like in a slow motion film. $\endgroup$
    – safesphere
    Commented Apr 23, 2019 at 3:45

If the true shape and texture of the accretion disk (viewed from 45°) around a rotating black hole at the time when the photon whose origin is closest to the observer is emitted is this:

accretion disk, true shape

and the rings of which the disk is composed of are rotating with the prograde orbital velocity, taking the light travel time and the counterclockwise rotation of the disk components into account the observed distortion would be

Kerr Newman accretion disk, distorted shape, orbital velocity

As you can see the inner parts of the disk are observed farther in the past than the outer parts, therefore the clockwise distortion. If the texture of the disk was static in a way that the radial white stripes would represent constant φ, the observed distortion would look like

Kerr Newman accretion disk, distorted shape, static texture

That's also how it would look if all the photons would need the same time to reach the observer, but still took the curved paths. The images were generated with the relativistic raytracer from here. The redshift is not shown above, since some colors would shift outside the visible spectrum:

Kerr Newman accretion disk, red- and blueshift

The white line in the green area marks the parts which are neither redshifted nor blueshifted. If there is enough space between the atoms that light can shine through, you would see some even older images of the disk getting bent around the photon sphere:

Kerr Newman black hole accretion disc, transparent texture

For the correspondig texture renderings with a transparent disk see here.

  • $\begingroup$ Thanks a lot for taking time to answer. I really appreciate the amount of effort you have spent on this. This exactly answers the question. $\endgroup$ Commented Jun 26, 2019 at 11:36
  • $\begingroup$ What is the scale on these pictures? An accretion disc does not extend below about $3 r_s$. $\endgroup$
    – ProfRob
    Commented Jul 1, 2019 at 6:58
  • $\begingroup$ @Rob Jeffries - Who told you that, the ISCO for a maximally rotating Kerr black hole (a=1) goes all the way down to r=1; set r(6r-r²-9Q²+3a²)+4Q²(Q²-a²)-8a(r-Q²)^(3/2)=0 and solve for r. In this scenario we have a Kerr Newman black hole with a=0.95, Q=0.3 where the prograde ISCO is at r=1.313818. If the scale you are asking for is the field of view, that is 30°×24° with the observer at a distance of r=50 and an outer disk radius of r=10, in units of G=M=c=kB=1. For other parameters and angles see here: notizblock.yukterez.net/viewtopic.php?p=904#p904 $\endgroup$
    – Yukterez
    Commented Jul 1, 2019 at 8:02

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