My question is similar to many others, Infalling light signals seen by a free falling observer for instance, but I have not seen it addressed exactly as I want to ask it.

It is clear for me (I am familiar with the principle of the Kruskal-Szekeres coordinates, but I never got down to precise calculations using them) that the last light signal that the free-falling observer can receive before hitting the singularity has only been sent a finite time after the beginning of the fall.

But what is its exact frequency as seen by the observer ? Assume the observer falls from a spaceship that hovers at a fixed distance from the black hole and sends a laser at a fixed frequency towards the free-falling observer. Of course, if there was another spaceship hovering at a fixed but smaller distance, light from the farther one would be blue-shifted when received by the closer one, and light from the closer one red-shifted when received by the farther one by exactly the same ratio.

But for the observer in free fall, there is no such symmetry : after he crosses the horizon, light he would try to send out won't come out at all. Contrariwise he keeps receiving light from his departure point. Before crossing the horizon, but when crossing the position of the closer spaceship, he has already a large velocity away from his original position, so the light he receives from there is obviously less blue-shifted than as seen from the stationary spaceship the position of which he is just crossing.

So my question is: with respect with the frequency when emitted at the starting point, is the light as seen by the free-falling observer still somewhat blue-shifted, or already red-shifted, or at exactly at the original frequency ?

Please note that the comparison with the frequency of the blue-shifted light received by the closer spaceship only makes sense up to the horizon, but my question keeps making sense all the way to the singularity, regardless.

What is the frequency seen by the observer in free fall from a given distance with respect to the frequency of the light emitted at the starting point ? Is it blue- or red-shifted ? Does it change monotonically during the fall ? In particular, I wonder if by monitoring this frequency the observer can determine when the horizon is crossed ?


Thanks to safesphere, I checked another question and found timm’s answer there.

It does answer my question for the red-shift (yes, it is a red-shift) for an observer free-falling from infinity starting from there with zero speed. (It would take some time to reach the horizon… ;) ) , with respect to light incoming from infinity.

This, however, is not the full answer to my question, which was about an observer free-falling from a finite radius $r_1$ starting with no speed, with respect to light coming from the departure point.

The last part, starting point of light, is easy to correct. That timm's observer would have already a non-zero speed at $r_1$ if it were falling from infinity, that too I have to correct for, and I am not sure I know how to do it.

As I have said, I am not an expert in GR, but I have solid experience with special relativity and a good deal of physical intuition.

I apologize for sometimes using the term "red-shift" in an improper way. Unless I call it $z$, at the end, in which case I do mean the larger than one ratio of wavelengths of the observed light with respect to that of the emitted one, I mostly use it as the smaller than one ratio of frequencies, while I always use blue-shift as the larger than one ratio of frequencies.

I knew from the beginning that the frequency shift I want to find is the product of the gravitational blue-shift for a stationary observer at distance $r$ from the black hole, with the Doppler red-shift for my free-falling observer due to his velocity relative to the stationary observer there.

I already knew that, at distance $r$ the gravitational blue-shift is $1/ \sqrt {1-r_S/r}$ where $r_S$ is the Schwarzschild radius

Note that this blue-shift increases indefinitely as $r$ gets closer and closer to $r_S$. It becomes a finite but imaginary number if $r \lt r_S$ but this is not a paradox, since inside the horizon there can be no stationary observer. Curiously, this unphysical quantity is still a useful information inside the horizon.

All this was already clear to me.

I also knew that the Doppler red-shift is $\sqrt{\frac {c-v}{c+v}}$ where $v$ is the infalling speed of the observer with respect to a stationary observer, the source of light being stationary. Again, I knew that $v$ increases all the way to $c$ when the position $r$ of the observer gets closer and closer to $r_S$, so the Doppler red-shift goes to zero, but I did not know the exact value of $v$ in terms of $r$ and $r_S$.

Inside the horizon, since there can be no stationary observer $v$ is physically meaningless. So there is no problem for the formal quantity $v$ to become larger than $c$ leading to an imaginary formal Doppler red-shift.

The only quantity that remains meaningful is the product of these two quantities, product which is always real and finite. And continuous across the horizon.

From timm’s comment in answer to safesphere’s I learned only one thing that I did not know before, but an essential one: the dependence of $v$ in terms of $r$ and $r_S$, namely $v=c\sqrt{r_S/r}$.

The change of sign with respect to timm’s comes from the fact that in his notation this speed is negative because it describes a diminishing $r$. I just look at its absolute value. This change of sign is compensated by the change of sign in my expression of the Doppler shift with respect to timm’s. As expected, $v$ is formally larger than $c$ inside the horizon, but $v$ has no physical meaning there in the absence of stationary observer.

The math is easy to perform and leads to a net red-shift. The frequency ratio $$f’/f={\frac 1 {1+\sqrt{r_S/r}}}$$ is less than one, so this is interpreted as a redshift


As expected, $z$ remains finite inside the horizon and only becomes infinite at the singularity at $r=0$. Exactly at the horizon $z=2$.

But this is just for an observer in free fall starting with zero speed at infinity.

The full answer to my question is for an observer in free fall starting with zero speed from radius $r_1$

Correcting the blue shift is easy : just divide by the finite, real blueshift at the starting point : $${\frac {\sqrt {1-r_S/r_1} }{\sqrt {1-r_S/r}}}$$.

The expression of the Doppler shift is the same, but I have to substitute the speed $V$ of my observer, starting with zero speed from radius $r_1$, to the speed $v=c\sqrt{r_S/r}$ of the timm’s observer.

And this is what I do not know. I have never got my hands “dirty” in GR calculations.

My intuition, based on special relativity, is that the value of $V$ satisfies the equation

$$1-V^2/c^2={\frac {1-r_S/r}{1-r_S/r_1}}$$

leading to $V/c=\sqrt{ {\frac {r_S} r} {\frac {r_1-r}{r_1-r_S}}}$

Note that this value of $V$ coincides with $c$ on the horizon and becomes larger beyond, giving an imaginary formal Doppler shift leading again to a real and finite red-shift inside the horizon.

The remainder of the math I can do myself, do not bother to do it for me.

My question now reduces to this.

Can anyone with a real understanding of GR confirm my guess for the speed $V$ of my observer, or contrariwise give me the correct value ?

  • 1
    $\begingroup$ It is redshifted exactly twice at the horizon in the radial direction: physics.stackexchange.com/q/420721 $\endgroup$
    – safesphere
    Commented Dec 29, 2021 at 5:19
  • $\begingroup$ @safesphere Thanks ! By looking at timm's answer there I was almost able to answer my own question. There is just one bit where I am guessing and I need confirmation by someone who has a better grasp than me on GR $\endgroup$
    – Alfred
    Commented Dec 29, 2021 at 10:32
  • $\begingroup$ @safesphere Stationary in space $t$, yes. In time $r$, no. What I want is to compute the redshift for a free-falling observer. When out of hte horizon, it can be computed thanks to the Doppler effect with respect to an observer fixed in $r$. After that, it is impossible. But the expression of the product of a formal (imaginary) Doppler effect and a formal imaginary gravitational blue-shift used by timm apparently works also inside the horizon. $\endgroup$
    – Alfred
    Commented Dec 29, 2021 at 17:23
  • $\begingroup$ @safesphere Besides, for the use I am making of $V$ its expression is not $dr/dt$ nor $(dt/dr)^{-1}$. It is not at all in the variation of $r$ that I am interested, but rather in the relative velocity with respect to a stationary observer. For light, for instance this relative velocity is always $c$ for any observer whatsoever, at least outside the horizon. Which is famously not the case for $dr/dt$. $\endgroup$
    – Alfred
    Commented Dec 29, 2021 at 17:26
  • $\begingroup$ @safesphere BTW, for light, since nothing really serious happens when one crosses the horizon, the relative velocity is also always $c$ for every observer inside the horizon, free-falling or not until one hits the singularity... But of course this fact does not help to compute the red-shift for a free falling observer that falls from a finite distance rather than from infinity. $\endgroup$
    – Alfred
    Commented Dec 29, 2021 at 17:29

2 Answers 2


Alfred asked: "What is the frequency of the light seen by an observer free-falling into a black hole?"

The equations are already given in the answer above, but since an image says more than a thousand words here's the $360° \times 180°$ full panorama:

Schwarzschild Background

Schwarzschild Redshift Blueshift

That is the frequency shift of the background sky in the frame of an observer crossing the horizon with the negative escape velocity, i.e the speed of light.

The gravitational and kinematic time dilation cancel in the frame of the free faller (therefore $g^{\rm t t}=1$ in Raindrop coordinates) and the pure Doppler remains.

In the back the redshift of light emitted by the stars far away from the black hole is $\rm f_{obs}/f_0=1/2$, and in the front the blueshift would be infinite if it wasn't obscured by the shadow of the black hole.

If you look $90°$ to the side, $\rm f_{obs}/f_0=1$ since the transverse Doppler with a cancelled time dilation vanishes.

With rotating and/or charged black holes the picture is a little bit more complicated and when you cross the inner Cauchy horizon you can indeed receive an infinite blueshift which is not obscured by the shadow, but the $\rm f_{obs}/f_0=1$ perpendicular to the direction of motion is always the case for a free faller from rest at infinity.


In response to my other question Javier confirmed that my intuition that the intuition that led me to $$V=c\sqrt{ {\frac {r_S} r} {\frac {r_1-r}{r_1-r_S}}}$$ was correct, where $V$ is defined as in my question.

For an observer in free-fall starting with zero velocity at a distance $r_1$ from a black hole of Schwarzschild radius $r_S$, the shift in frequency (written formally as a blue-shit, even if it is a redshift in which case it ls less than one) of the light originating from his departure point at $r_1$ is

$${\frac {\sqrt {1-r_S/r_1} } {\sqrt {1-r_S/r}} } {\frac {\sqrt {c-V}}{\sqrt {c+V}}}$$ with $V$ as above.

When $r$ crosses $r_S$, both $(c-V)$ and $(1-r_S/r)$ become negative. Taking their square root separately would lead to imaginary numbers, but in fact their ratio remains positive and finite, and the expression

$${\frac {\sqrt {1-r_S/r_1} }{\sqrt {c+V}}}\sqrt{{\frac { {c-V}} {{1-r_S/r}}}}$$

makes sense both for $r\gt r_S$ and $r\lt r_S$ and is continuous at $r=r_S$. I have computed the value at $r=r_S$ and found it is 1/2 whatever $r_1 \gt r_C$.

So it is indeed the same redshift $$1+z=2$$ whether $r_1$ is at infinity or at a finite distance.

So the conclusion is : whatever distance from the black hole you start your jump in free-fall, you can tell that you crossed the horizon precisely when the light emitted from your starting point reaches you with exactly half the frequency it is emitted.

A very useful bit of information... !


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.