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Ask A Nerd : Time Dilation and Blueshift At Event Horizons

This is a question that comes up often, and seems to be the source of neverending confusion among people : is time dilation – and hence red/blue shift – really infinite at the event horizon of a black hole ?

First of all, let us assume that we are in a Schwarzschild spacetime, and are hence dealing with a Schwarzschild black hole. The relationship between far-away coordinate time, and local proper time, is then given via the metric as

(1)   \begin{equation*} \displaystyle{d\tau =\sqrt{1-\frac{r_s}{r}}dt} \end{equation*}

wherein r_{s} is the Schwarzschild radius ( i.e. the event horizon radius ). As is easily seen, \int dt \rightarrow \infty for r \rightarrow r_{s}, so you would be forgiven to think that “time dilation is infinite at the event horizon”, and that observers there are fried to a crisp by infinitely blueshifted light from distant stars.

But this is wrong, for a very simple reason : the relation (1) – and, in fact, the entire metric written in Schwarzschild coordinates – is valid only as a relationship between stationary observers, specifically between a stationary far-away observer, and a stationary shell observer in the vicinity of the black hole. The problem here is of course that there are no stationary observers at the event horizon, or even in a small region just above it; in fact, the Schwarzschild coordinate system does not even extend to the horizon, it is defined only outside of it. With the exception of photons ( which are not valid reference frames here, since for them ds=0 ), anything that touches the horizon must fall in, and cannot remain stationary there. The only type of observer we can physically find at r \rightarrow r_{s} is one moving radially inwards – and that motion with respect to the distant light source changes things radically, so that (1) is no longer valid. The original argument is nonsensical, and the question itself is not meaningful.

Let us work through this in detail to see what actually happens; for reasons of simplicity let us assume that the motion is a straight radial free fall, for which case all maths become straightforward.

The total time dilation of an observer in free fall with respect to a distant reference clock is made up of gravitational time dilation and kinematic time dilation ( due to relative motion ), and turns out to be

(2)   \begin{equation*} \displaystyle{\frac{d\tau }{dt}=1-\frac{r_s}{r}} \end{equation*}

The absense of the square root is not a typo – this time dilation is greater than the one due to gravity alone, because of the extra contribution by relative motion. As a relation between coordinate and proper time, this relation is also precisely equal to the coordinate (!) velocity of light, as seen by a far away observer

Next, we get a coordinate (!) velocity for the freely falling observer himself of

(3)   \begin{equation*} \displaystyle{\frac{dr}{dt}=\left ( 1-\frac{r_s}{r} \right )\sqrt{\frac{r_s}{r}}} \end{equation*}

Put (2) and (3) into the formula for the Doppler shift, and we get

(4)   \begin{equation*} \displaystyle{\frac{f}{f_0}=1-\frac{v_{obs}}{v_{light}}=1-\sqrt{\frac{r_s}{r}}} \end{equation*}

This is the frequency shift as calculated by a far-away observer. For the observer in free fall, the shift is then

(5)   \begin{equation*} \displaystyle{\frac{f}{f_0}=\frac{1-\sqrt{\frac{r_s}{r}}}{1-\frac{r_s}{r}}=\frac{1}{1+\sqrt{\frac{r_s}{r}}}} \end{equation*}

At the event horizon, this goes as

(6)   \begin{equation*} \displaystyle{\frac{f}{f_0}\overset{r\rightarrow r_s}{\rightarrow}\frac{1}{2}} \end{equation*}

Not only is the final result finite and well defined, but it is less than unity, meaning that light from distant sources is actually redshifted for an observer in free fall, despite the influence of gravity !

In summary : there are no stationary observers at ( or even near ) the event horizon, so no physical observer will ever experience infinite time dilation and blueshift at the horizon; that is a naive view that stems from a misunderstanding of the physical meaning of Schwarzschild coordinates. Furthermore, a proper treatment that accounts for the kinematic effects of motion reveals that the net effect is really a redshift, and not a blueshift at all.


References

http://iopscience.iop.org/article/10.1088/1742-6596/104/1/012008/pdf

http://physics.stackexchange.com/questions/213441/do-free-falling-observers-see-gravitational-blueshift


 

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