# Time for an object's afterimage to disappear at the event horizon

My understanding is that as someone/something falls into a black hole, they would get dimmer and dimmer until disappearing entirely as they cross the event horizon. Most non-radiant objects would probably become invisible to outside observers well before hitting the event horizon. But if we had something very bright (like a hot young star) hitting a black hole's event horizon, about how long would we see its "afterimage floating near the horizon" before it disappeared? (Assume the star's trajectory into the black hole was directly along our line-of-sight distance to the black hole itself.)

• "until disappearing entirely as they cross the event horizon" They would disappear "way before" that. Note that an observer at infinity will never see anything crossing the event horizon as that would take an infinite amount of time from the observer's point of view. So what I'm saying is: They would see the object disappear way before $t = \infty$, i.e. at a finite time. This is due to the fact that, at some point, the star's light will get redshifted so much that it won't be detectable anymore.
– balu
Oct 6, 2020 at 21:27
• @balu Yes, I'm well aware that they would see the object disappear way before $t = \infty$. As you should know from reading the question, I was asking about the time for the image to disappear, not about "whether or not" the image would disappear. Oct 6, 2020 at 21:32
• Right, maybe I was confused by the fact that in that sentence you are saying the object would disappear "entirely as [it crosses] the event horizon" when the whole point of your question is that the disappearance happens (much) earlier, i.e. after a finite amount of time. I apologize if my comment came across the wrong way. Anyway, it sounds like you understand all the details. What is stopping you from calculating the time for the image to disappear?
– balu
Oct 6, 2020 at 22:29

If you look at my answer to What is the general formula for time-dilation due to velocity and gravity together? this explains the equation for the time dilation of an object falling radially inwards towards a Schwarzschild black hole. We end up with the result:

$$\frac{d\tau}{dt} = \sqrt{1 - \left(1 - \frac{r_s}{r}\right)^2\frac{r_s}{r} - \frac{r_s}{r} } \tag{1}$$

You need to choose some value of the time dilation that you consider to be disappeared. For example if we choose a factor of a hundred that redshifts all wavelengths by a factor of a hundred and reduces the intensity by a factor of a hundred. That would certainly be undetectable to the naked eye though sufficiently sensitive instruments could still detect the falling object. Using equation (1) we find this happens at a distance:

$$\frac{r}{r_s} \approx 1.0001$$

i.e. when the distance to the horizon is $$0.01\%$$ of the horizon radius. Then we just need to find the time taken to reach this distance.

The velocity of an object falling freely from infinity is discussed in my answer to Will an object always fall at an infinite speed in a black hole? and we get the result:

$$v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{2}$$

and this looks like: The velocity initially rises as the object falls inwards, but then very near the horizon the time dilation dominates and the object slows again and after infinite time comes to a halt at the horizon. We get the time taken to reach a distance $$r$$ by integrating this expression, but sadly it doesn't have a closed form integral and we'd have to do it numerically.

At this point I have run out of enthusiasm for doing the calculation, but I did calculate the time taken to fall to $$1\%$$ of the horizon distance in my answer to Can you have a giraffe shaped black hole? and the result was under a millisecond.

The time will be proportional to the light crossing time of the black hole, which is less than a millisecond for a stellar-mass black hole, or around a day for the M87 black hole.

I don't have a source for the following derivation and it's possible I made a mistake.

An object radially infalling into a Schwarzschild black hole satisfies $$d^2r/ds^2 = -GM/r$$. Integrating this once gives $$dr/ds = -\sqrt{r_s/r-r_s/r_0}$$. For simplicity I'll take $$r_0=\infty$$ (the object falls in from infinity). Arbitrarily choosing $$r=r_s$$ at $$s=0$$, the solution is $$r = r_s (1 - \frac32 s/r_s)^{2/3}$$.

I'll use Eddington-Finkelstein outgoing coordinates $$ds^2 = (1-r_s/r)\,dv^2 + 2\,dv\,dr$$ because the redshift as seen by someone at rest at infinity is just $$1{+}z = dv/ds$$. Plugging in the formula for $$dr/ds$$ gives us $$1 = \left(1-\frac{r_s}{r}\right)\left(\frac{dv}{ds}\right)^2 - 2\,\frac{dv}{ds}\,\sqrt{\frac{r_s}{r}}$$ which has the solution $$dv/ds = \left(1-\sqrt{r_s/r}\right)^{-1}$$. Plugging in the formula for $$r$$ gives us $$\frac{dv}{ds} = 1{+}z = \left(1-\left(1-\frac{3\,s}{2\,r_s}\right)^{-1/3}\right)^{-1}$$

I'd rather have $$z(v)$$ than $$z(s)$$ so I multiply by $$ds/dz$$ to get $$\frac{dv}{dz} = 2\,r_s\,(1+z)^3/z^4$$ $$v(z)=2\,r_s\left(\ln z - 3/z - 3/2z^2 - 1/3z^3\right) + \text{const}$$

I doubt that's invertible. At any rate, you can see that at late times ($$z\gtrsim 10$$) the redshift increases exponentially with a time constant of $$2\,r_s$$ (33 hours for the M87 BH). It takes about $$4\,r_s$$ to get from $$z=e$$ to $$z=e^2$$, about $$9\,r_s$$ to get from $$z=1$$ to $$z=e$$, and about $$44\,r_s$$ to get from $$z=1/e$$ to $$z=1$$.