# Photon passing through a black hole?

1. Mar 3, 2006

### Graviman

This is something that has bothered me for a while, so i thought to share it with other relativity enthusiasts.

A black hole may accumulate charge, which along with the rest of the matter will work its way to the centre of the black hole. This means that the blach hole will be surrounded by a radially symetrical E-field. Since all EM fields may be seen as a wave travelling out, even if they are effectively 0 Hz, then this means that an EM wave is leaving from the centre of a black hole. Does this mean that a photon can leave from the very centre of the black hole, as long as it is only travelling radially? The charge being form invariant means that to get this to work you could alternately pour in +ve then -ve charges, to make a detectable RF frequency emission.

Put another way if a photon is aimed at the exact centre of a black hole, so that it's velocity always remains a radial vector, does the photon pass straight through? :surprised

Mart

2. Mar 3, 2006

3. Mar 3, 2006

### DaveC426913

"... along with the rest of the matter will work its way to the centre of the black hole..."

I don't know a lot abuut BHs, but my understanding was such that all information about anythnig falling into a black hole is permanently erased from the universe. Thus, no "matter at the centre", no "charge at the centre", etc. thus your premises are in error.

But I can't claim authority on the subject.

4. Mar 3, 2006

### scott1

If white holes existed then that wouldn't permanently erase infromation form it.
http://en.wikipedia.org/wiki/White_hole

5. Mar 3, 2006

### topsquark

There IS such a thing as a charged black hole. Don't ask me how the electric field is supposed to come out of the black hole. Classically I suppose its because the electric field is massless, so the field lines can "escape" the hole. But that doesn't hold up on the Quantum level.

-Dan

6. Mar 3, 2006

### pervect

Staff Emeritus
One way of looking at it is that the virtual particles are not bound by the speed of light limitation. However, they don't cary any actual information.

See for instance

black holes:
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

virtual particles:
http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

Real photons cannot escape a black hole, however. Any electromagnetic waves generated by accelerating particles inside the event horizon would be pulled into the singularity, they would not escape the event horizon.

The above is not intended to address the question of what happens when the black hole evaporates, that's a more complex problem.

Last edited: Mar 3, 2006
7. Mar 3, 2006

### Stingray

For an external observer, objects never really look like they fall into (Schwarzschild) black holes. Instead, they "freeze in" as they approach the horizon. So a realistic "charged black holes" might look like Schwarzschild with a bunch of charges just above the horizon.

So what does the electric field of this kind of system look like? Think of just one charge right outside the horizon. You'd think that the field would look as though it were offset from the center of the black hole, but it actually isn't. Except in a very small "boundary layer" over the horizon, the electric field looks like it is radially centered on the (uncharged) black hole rather than the charge! The horizon acts like an electrical conductor.

8. Mar 4, 2006

### AlphaNumeric

Spinning black holes have a ring singularity, not a point singularity, and if you fall inwards off the equatorial plane (defined by the spin in the black hole) you'll be able to fall through the middle of the ring, which leads to another region of space-time.

9. Mar 4, 2006

### pervect

Staff Emeritus
Some caution is advisable here. The interior structure of rotating black holes is still being debated, though the consesus in the literature as far as I have been able to determine is that the classical Kerr version of the ring singularity is not stable.

See for instance:

http://arxiv.org/abs/gr-qc/9902008

The motivation for studying charged collapse is explained further within this paper:

also

http://arxiv.org/abs/gr-qc/0103012

which makes similar arguments.

The remark that the ring singularity leads to another region of space-time is correct for the classical Kerr geometry, but the actual geometry in the interior of a black hole is probably not given by the Kerr geometry.

If I'm interpreting the remarks in the first paper correctly, it is currently felt that "the stargate is closed".

10. Mar 4, 2006

### Graviman

Hmmm, some good answers. To keep the problem simple i am only considering a static black hole here. Spinning black holes introduce all sorts of interesting possibilities, but also make my original question more complicated than i had intended.

Perhaps throwing charges in was a bad anaolgy, since they will never actually get to the singilarity. Interesting about the charge ending up evenly distributed around the event horizon, Stingray. Yes this does make sence of why the black hole would "appear" to have a charged singularity. I'd like to understand more about how the event horizon acts as a conductor though.

What i was really driving at is that a beam of light is always seen to travel at C by all observers. It may become masively red or blue shifted, but always travels at C. The photon sphere [corrected from "event horizon"] is where a beam of light aimed at missing the black hole gets caught in orbit. What happens if you don't intend to miss the black hole. Instead you aim the beam at its very centre, so that there is never an orbit component to the velocity. Does an outside obsever find that this beam always travels at the speed of light? The beam would become extreme blue shifted, but does it survive then undergo red shifting on its way back out?

My question is aimed more at plugging gaps in my own knowledge than finding any flaws...

Mart

Last edited: Mar 4, 2006
11. Mar 4, 2006

### George Jones

Staff Emeritus
This is true locally, i.e., if light goes whizzing by any observer, then the observer will measure (with respect to his local orthonormal frame) its speed to be $c$.

Light doesn't go into orbit at the event horizon. For a mass $M$ (uncharged) black hole, the (Schwarzscild) coordinate of the event horizon is $r = 2M$ ($G = c = 1$), while light orbits happen at $r = 3M$, which is called the photon sphere.

The coordinate speed of the light is

$$v = 1 - \frac{2M}{r},$$

so, at the event horizon, the coordinate speed of the light is zero.

Just as for material objects, for a distant observer, the light takes an infinite amount of time to cross the event horizon.

However, things are different for other observers that the light passes on the way.

Consider a series observers, each using rockets to hover at a constant $r$ above the event horizon. A hovering observer is possible, in principle, at any $r$ larger than $2M$. As the light passes, each of these observers measures the light's speed. Each result is $c$, even when the observer is hovering infinitesimally close to the event horizon.

Conisder another series of observers, each of whom falls freely along the same radial path taken by the light. Each observer stars falling before the light starts - just how much before depends on the observer. As the light proceeds, it will pass each observer in succession. Again, each observer measures the speed of the light, and, again, each result is $c$. This is true for observers outside, at, and inside the event horizon.

Note that in this scenerio, if the observer starts out early enough, the light will pass the observer after both have crossed the event horizon. This, in spite of the fact that a distant observer "sees" neither the light, nor any of the freely falling observers, cross the event hroizon.

Regards,
George

Last edited: Mar 4, 2006
12. Mar 4, 2006

### Graviman

Thanks George, i suspected it was an error in my understanding. Event horizon has been corrected to photon sphere, my memory is acting up again.

So an observer trying to see if the light came out would have an infinately long wait....

Mart

13. Mar 4, 2006

### pervect

Staff Emeritus
One way of understanding what happens inside a Schwarzschild black hole is this:

The Schwarzschild 'r' coordinate, inside the event horizon changes its sign in the metric.

This means that the 'r' coordinate essentially becomes a time coordinate inside the event horizon. So for a particle inside the event horizon, the "future" of the particle always points towards the central singularity.

Imagine 4 one-way streets, all meeting together at a point, with the direction of traffic being towards that point. This is the situation particles face inside a black hole. In the analogy above, the cars cannot stop moving down the street, because the street represents the flow of time, and all the particles must go from their past into their future.

What happens when the particles reach the singularity? Where do they go? Classical GR does not have a definte answer to this question, especially in the case of realistic physical collapse. (The pure mathematical solution can be "extended" to allow the particles to enter another universe at the singularity, but this mathematical solution is not felt to be particularly likely solution for a realistic physical collapse.) The safest answer is to say that the theory itself breaks down at the singularity, that the infinites in the theory are a sign that a better theory like quantum gravity is needed to answer the question.

As far as the observable universe, goes, it does not particularly matter what happens "afterwards", because it will not affect the observable universe, i.e. what we can see outside the black hole.

Note that the analogy above describes the interior geometry of a static Schwarzschild black hole. There is still considerable debate as to what the geometry of a 'real' imploding black hole will be, especially if it has non-zero angular momentum, as per the previous comments.

14. Mar 5, 2006

### Graviman

In order for my simple mind to comprehend what is happening, i have always visualised gen rel as a theory describing the rate of travel through cT of a point in space against a reference far from any matter. Time dilated near an object means it's rate goes up. The spacial contraction being just a natural side effect of time slowing down in a region.

From what has been said in this thread, perhaps this is a bad analogy inside a black hole itself since time has already stopped at the horizon (or rate of flow of time is infinite compared to far away reference). This probably is the source of my confusion about what happens inside the event horizon. From what i now understand most matter ends up at the event horizon for the duration of our universe. The singularity is more of a mathematical concept.

What maths would you recommend i study to gain a fuller understanding? Tensors clearly, but anything else? I would not describe myself as a natural mathematician, although engineering mathematics was easy since i could visualise what was happening. Is there any way i will ever be able to just "see" what is happening?

Mart

Last edited: Mar 5, 2006
15. Mar 5, 2006

### ZapperZ

Staff Emeritus
Actually, this is incorrect.

A time-dilated frame S' will measure a shorter time to traverse the same distance as viewed by another inertial frame S. In S, it will take a time T to move through a distance L, whereas according to S, S' will only take T'<T to move the same distance.

On the other hand, according to S', the length L is actually L' in this frame, a contracted length (remember, L was measured in S). So while S' it sees its time as the proper time in S' frame, it only sees a shorter distance to move, i.e. L'. Again, this means that it takes less time to move that "distance".

One needs to make sure one understands the basics of SR before jumping into GR.

Zz.

16. Mar 5, 2006

### George Jones

Staff Emeritus
This is true from the perspective of a distant observer, but it is not true from the perspect of somebody/something that falls into the black hole!

Very counterintuitive from an everyday perspective!

I recommend putting off the study of tensors for quite some time. Much about black holes can be studied and calculated using nothing more than first-year calculus.

"Relativistic" intuition is needed to "see" what is happening, and intuition is based on experience. As ZapperZ has said, developing this intuition requires getting a good grounding in special relativity. I think a good way to get this grounding is by working your way through https://www.amazon.com/gp/product/0...02-7908719-9192156?s=books&v=glance&n=283155" by Thomas Moore. This book is a short, readable, interesting, and pedagically sound introduction to special relativity.

The book by Taylor and Wheeler requires only first-year calculus, and, as indicated by the title, is largely restricted to black holes. Many interesting calculations and examples are provided, and the exercises are useful and doable. In a forthcoming post in this thread, I hope to give an example of what can be achieved form an understanding of the material in Taylor and Wheeler.

Hartle's book is pitched at a slightly higher level, is more general, and requires slightly more mathematics. Late in the day, Hartle introduces tensors. Before doing this, Hartle concentrates on deveoping a feel for relativistic geometry, and on physical applications of relativity, including black holes.

I would say that:

Moore is at the level of first-year/second year students;

Taylor and Wheeler is at the level of second-year/third year students;

Hartle is at the level of third-year/fourth-year students.

Each of these books is a pedagogical masterpiece, and none of them have mathematical prerequisites beyond the second-year level.

Finally, http://www.arxiv.org/PS_cache/gr-qc/pdf/0506/0506075.pdf" each have articles available on-line expressing their views about the teaching of relativity.

Regards,
George

Last edited by a moderator: May 2, 2017
17. Mar 5, 2006

### Graviman

I could have been clearer in what i meant here - see corrections. Please let me know if this is still wrong! :yuck:

I'll have a look at the books you recommend. It's been a while since i studied SR, as my daft mistakes show. I'd like to eventually gain a good understanding, but only really wish to study the maths required without drowning the original intention....

Mart

Last edited: Mar 5, 2006
18. Mar 5, 2006

### George Jones

Staff Emeritus
My experience is that relying on time dilation and Lorentz contraction too early to solve problems is unsound pedagogically and easily leads to misconceptions This is precisely what Moore does not do. Much better to use invariance of the interval, as this leads to better intuition that more easily transfers to general relativity.

For more on my views on this matter, see my post in https://www.physicsforums.com/showthread.php?t=106450&highlight=contraction".

Regards,
George

Last edited by a moderator: Apr 22, 2017
19. Mar 5, 2006

### George Jones

Staff Emeritus
Consider an observer that hovers at coodinate $r=R$ from a black hole, and a series of mirrors that hover between the observer and the black hole. The observer sends light towards the black hole, the light is reflected by one of the mirrors, and the light returns to sender. The amount of time taken, as recorded by the observer's wristwatch, for this roundtrip depends on the location of the mirror off which the light reflected.

The geometry of spacetime is given by the metric, and, in spherical coodinates, the metric for a spherically symmetric black hole is given by

$$ds^{2}=\left( 1-\frac{2M}{r}\right) dt^{2}-\left[ \left( 1-\frac{2M} {r}\right) ^{-1}dr^{2}+r^{2}\left( d\theta^{2}+\sin^{2}\theta d\phi ^{2}\right) \right] .$$

Compare the stuff in the square brackets with the metric (in spherical coordinates)

$$ds^{2}=dr^{2}+r^{2}\left( d\theta^{2}+\sin^{2}\theta d\phi^{2}\right)$$

for Euclidean 3-dimensional space. The notation $dr^{2}$, e.g., is short form for $\left( dr\right) ^{2}$, the square of a(n infinitesimally) small change in the $r$ coordinate.

The $\left(t,r\right)$ coordinate labels of the key events are: $\left( 0,R\right)$ for emission by the observer; $\left( t_{m},r_{m}\right)$ for reflection by the mirror; and $\left( t_{r},R\right)$ for reception by the observer. Since all motion is radial, $d\theta=d\phi=0$, and the Schwarzschild metric is

$$ds^{2}=\left( 1-\frac{2M}{r}\right) dt^{2}-\left( 1-\frac{2M}{r}\right) ^{-1}dr^{2}.$$

For a lightlike worldline, $ds=0$, which, when used in the metric, gives

$$\frac{dr}{dt}=\pm\left( 1-\frac{2M}{r}\right) .$$

The $-$ sign gives radially inward travel, and the $+$ sign gives radially outward travel. So, during the inward trip by the light,

\begin{align*} dt & =-\frac{dr}{1-\frac{2M}{r}}\\ \int_{0}^{t_{m}}dt & =-\int_{R}^{r_{m}}\frac{dr}{1-\frac{2M}{r}}\\ t_{m} & =2M\ln\left( \frac{R-2M}{r_{m}-2M}\right) +R-r_{m} \end{align*}

During the outward trip by the light,

\begin{align*} dt & =\frac{dr}{1-\frac{2M}{r}}\\ \int_{t_{m}}^{t_{r}}dt & =\int_{r_{m}}^{R}\frac{dr}{1-\frac{2M}{r}}\\ t_{r}-t_{m} & =2M\ln\left( \frac{R-2M}{r_{m}-2M}\right) +R-r_{m} \end{align*}

Combining these results gives

$$t_{r}=4M\ln\left( \frac{R-2M}{r_{m}-2M}\right) +2\left( R-r_{m}\right)$$

Now, $t_{r}$ is a coordinate, which is just a label, and which is not the time measured by the observer's wristwatch. All events on the observer's worldline occur at $r=R$, so for the observer, $dr=d\theta=d\phi=0$, and the metric becomes

$$d\tau=\sqrt{1-\frac{2M}{R}}dt,$$

where $s=\tau$ is wristwatch time and $\sqrt{1-2M/R}.$ Hence, the time measured by the observer's wristwatch is

$$\tau=\sqrt{1-\frac{2M}{R}}t_{r}.$$

This gives radar coordinates for the the reflection events.

Clearly, $\tau\rightarrow\infty$ as $r_{m}\rightarrow2M$, i.e., the time elapsed on the observer's wristwatch between emission and reception of the light goes to infinity as the reflection event approaches the event horizon.

Regards,
George

20. Mar 5, 2006

### pervect

Staff Emeritus
While it won't answer your questions about what happens inside a black hole, there is a good website that talks about how to compute orbits around black holes.

http://www.fourmilab.ch/gravitation/orbits/

Thorne's book "Black holes and time warps: Einstein's outrageous legacy" is an excellent popular book.

https://www.amazon.com/gp/product/0393312763/103-7179970-5511867?v=glance&n=283155

"Exploring black holes"
https://www.amazon.com/gp/product/020138423X/103-7179970-5511867?v=glance&n=283155

and

"General relativity from A to B"
https://www.amazon.com/gp/product/0...103-7179970-5511867?s=books&v=glance&n=283155

though I haven't read either of them myself.

What happens at the event horizon is different from what happens at the singularity itself.

As George mentioned earlier, time slows down for someone hovering outside the event horizon, and the rate of slow-down approaches infinity.
This is true from both perspectives - the hovering observer will see the universe age very quickly, and the outside observer will see the hovering observer age very slowly.

This is not, however, true for an obserer who falls into a black hole. The outide obsever will see much the same thing, but the experience of the falling observer will be quite different. He will not see the universe aging more quickly, and he will not notice anything in particular "odd" when he falls through the event horizon - with the exception of the tidal forces one usually expects from a massive body. For a small black hole these tidal forces may be a problem for anything other than a small proble, for a large black hole they will be small enough that a human being could sruvive them.

Most of the weirdess at the event horizon of the black hole is a so-called "coordinate singularity". This is an artifact of the Schwarzschild coordinate system, a lot like the fact that it's only possible to go south at the north pole using lattitude/longitude coordinates on the Earth's surface.

The same is not true of the inner singularity at the center of the black hole. This is a true singularity, not just a coordinate singularity. The observer cannot escape being destroyed by infinite tidal forces. It's a one-way trip.

Last edited by a moderator: May 2, 2017