Acceleration At The Event Horizon

In summary, the object would accelerate at the speed of light when it is at the event horizon of a black hole.
  • #1
mrjeffy321
Science Advisor
877
1
For an object exactly at the event horizon of a black hole, is the acceleration infinite or does the object accelerate at the speed of light?
I have read that the force on the object causing it to accelerate when it is at the event horizon is infitite (so the acceleration would be too), but the reason I am thinking that it would not be is that for a couple reasons,
-wouldnt your acceleration be infinite right when you reach the singularity and not the event horizon?
-you are still a certain distance away from the mass of the black hole, so
r != 0, in the equation for the universal law of gravitation
--but you can't use that here because it is a black hole?
-if a beam of light were pointed staight outward from the black hole, wouldn't it be frozen there, not falling back in and not going out?
 
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  • #2
mrjeffy321 said:
For an object exactly at the event horizon of a black hole, is the acceleration infinite or does the object accelerate at the speed of light?
I have read that the force on the object causing it to accelerate when it is at the event horizon is infitite (so the acceleration would be too), but the reason I am thinking that it would not be is that for a couple reasons,
-wouldnt your acceleration be infinite right when you reach the singularity and not the event horizon?
-you are still a certain distance away from the mass of the black hole, so
r != 0, in the equation for the universal law of gravitation
--but you can't use that here because it is a black hole?
-if a beam of light were pointed staight outward from the black hole, wouldn't it be frozen there, not falling back in and not going out?
For a large BH, the acceleration would not be anywhere near c except for photons already at c. No object with mass would approach c. Acceleration at the EH of a smaller BH would be faster, but still not at c. Infinity is not possible at all, even for photons.
 
  • #3
that was a poor choice of words I used above, "acceleration infinite or does the object accelerate at the speed of light?".
what I was trying to do by finding the acceleration at given points was to find the objects, "weight".
for example, if I was 1 km above the Event Horizon (EH from now on), 1 m above, ..., right on it.
 
  • #4
If you try to hover near the event horizon of a Schwarzschild black hole the required proper acceleration (as measured by the seat of your pants) will approach infinity as you approach the event horizon (when you try to hover, i.e. maintain a constant r coordinate in the Schwarzschild geometry).

Wald gives the proper acceleration at a distance r away from the black hole as being proportional to

[tex]1/\sqrt{r-2M}[/tex]

when one is close to the black hole.

(note this is in geometric units, so that G=c=1, and the Schwarzschild radius is at r=2M)

The length unit here used to measure the acceleration is a meter stick carried by the observer (pointed in the radial direction), and the time unit used to measure the acceleration is a clock carried by the observer. The r coordinate, though, is the Schwarzschild r coordinate.

[add]
The units bothered me, so I worked it out with GRtensor, getting

a = [tex]\frac{m}{r^{\frac{3}{2}}\sqrt{r-2m}}[/tex]

this gives the correct units for acceleration (in geometric units!). This also has the property that a-> m/r^2 when r is very large, which is correct in geometric units (where G=1).

In non-geometric units this is

a = [tex]\frac{Gm}{r^{\frac{3}{2}}\sqrt{r-r_{s}}}[/tex]

where rs= Schwarzschild radius = 2Gm/c^2, and G and c are the gravitatioanl constant and the speed of light, respectively, and m is the mass of the black hole.
 
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  • #5
You used the word singularity. I am having trouble understanding what it really means. I always see that word popping up around these boards. I've seen it as singularity, mathematical singularity, quantum singularity. Theres got to be more...anyways what does/do it/they mean.
 
  • #6
3 entries found for singularity.
sin·gu·lar·i·ty ( P ) Pronunciation Key (snggy-lr-t)
n. pl. sin·gu·lar·i·ties
The quality or condition of being singular.
A trait marking one as distinct from others; a peculiarity.
Something uncommon or unusual.
Astrophysics. A point in space-time at which gravitational forces cause matter to have infinite density and infinitesimal volume, and space and time to become infinitely distorted.
Mathematics. A point at which the derivative does not exist for a given function but every neighborhood of which contains points for which the derivative exists. Also called singular point.
 
  • #7
Thanks for answering, but shortly after I posted it I ended up at dictionary.com finding out what it meant, and then regreted posting WHAT IS SINGULARITY.
 
  • #8
cosmic singularity is the big bang
 
  • #9
This is actually a fairly simple question. The velocity of an object attracted toward a gravitating body from infinity is the same as the escape velocity at any arbitrary distance from the center of the gravitating body - so the answer is slighly less than the speed of light at the event horizon of a black hole.
 
  • #10
This velocity (of the infalling observer) is measured relative to what?

In Schwarzschild coordinates, the velocity works out to be zero. In the Schwarzschild basis, the velocity at the horizon does turn out to be 'c' for an radially infalling object when the object is at rest at infinity, but not when the object starts with an inital radial velocity .

(Unlesss I've made some error in my calculations, I suppose).

The "schwarzschild basis" can be thought of as the coordinate system of an observer who is "holding station" at some particular Schwarzschild r,theta, phi value.

Because such an observer is necessarily accelerating, one might add that it's actually the local coordinate system of an instantaneously co-moving observer. Another way to describe it is that the local metric is Lorentzian (diagonal with all coefficients +/- 1).

Maybe there's some other way to measure velocity, but I can't think of what it could be. The Schwarzschild basis seems like the most logical candidate for what one means by the "velocity" of an infalling observer.

See for instance the thread

https://www.physicsforums.com/showthread.php?t=73400&page=2&pp=21

where I convince myself that the velocity of an obsrever is not always 'c' in the Schwarzschild basis after originally thinking it should be.
 
  • #11
pervect said:
In the Schwarzschild basis, the velocity at the horizon does turn out to be 'c' for an radially infalling object when the object is at rest at infinity, but not when the object starts with an inital radial velocity .

It must be exactly c regardless how the object fell to arrive at the horizon. If < c at the horizon then the object (or its image or its transmission etc.) can escape, in which case it's not a horizon. If > c at the horizon then > c would be directly measureable by someone hovering just above the horizon, violating special relativity. Both a rocket accelerating in at any rate of acceleration, and a stone hurled from any altitude at any velocity, cross the horizon at a proper velocity of exactly c.

This leads to a problem where the velocity of an object relative to the horizon cannot change while crossing. Nothing special happens when you cross the horizon (e.g. tidal force smoothly increases). So whenever nothing special is happening, you could be crossing a horizon. Then you could cross head-on when you're passing another car on a freeway, in which case for a brief moment your car must be in lockstep with the other car (not passing). So implies general relativity.
 
  • #12
mrjeffy321 said:
For an object exactly at the event horizon of a black hole, is the acceleration infinite or does the object accelerate at the speed of light?

Infinite acceleration is needed by an object hovering at the horizon, as pervect noted. Because of that, an object cannot hover at or rise from the horizon and instead must fall in once it reaches it.

-wouldnt your acceleration be infinite right when you reach the singularity and not the event horizon?

Yes, assuming you mean gravitational acceleration.

-you are still a certain distance away from the mass of the black hole, so
r != 0, in the equation for the universal law of gravitation
--but you can't use that here because it is a black hole?

You can use it at the horizon.

-if a beam of light were pointed staight outward from the black hole, wouldn't it be frozen there, not falling back in and not going out?

Yes, if beamed precisely at the horizon.
 
  • #13
I probably need to think about this some more, but I do have a few more thoughts. This post got a little long, so let me summarize by saying that Zanket has made some excellent points.

I do think it's important to figure out what we mean by "the velocity" of an observer falling into a black hole, though - velocities are relative, we need to define what the velocity is relative to. We can't , for instance, measure the velocity by taking the rate of change of the distance to the singularity - the later isn't well-defined at all.

I think it's reasonable to define the infalling velocity as the velocity measured by a "station-keeping" observer at a constant Schwarzschild coordinates (R, theta, phi). In this problem I think we're mainly interested in the case in which the infalling observer has no angular momentum, and the only component of the velocity is the radial component.

This may be a coordinate dependent definition, but it's the best one I can think of. I'm open to other suggestions as to what the term "infalling velocity" really means.

The station-keeping observer is necessarily accelerating, but the relative velocity should be well-defined when the station-keeping observer is at the same point in space as the infalling observer.

[add-explain]
The relative velocity of an observer and an accelerating observer is a function of time - thus it's not well defined when the observers are not at the same point in space-time, because of issues of the relativity of simultaneity
[end add]

Given that we can agree on this as a defintion of just what the "velocity of an observer falling into a black hole" actually is, I have to agree with Zanket's point, more or less. At any distance R greater than the Schwarzschild radius, the infalling velocity by the defintion above will be less than 'c' because it's an actual "physical" velocity between two objects. The observer exactly at the event horizon isn't "physical", it requires an infinite proper acceleration to hold station there. So we need to take the limit of the velocity as R->R_s to make any sense of the velocity "at the horizon".

Also, the velocity should increase as R decreases.

Given this, it seems almost certain that the limit as R-> schwarzschild radius of the "infalling velocity" should be equal to "c" if it exists.

So I need to go back over my calculations when I have more time to figure out what's going wrong.
 
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  • #14
pervect said:
The observer exactly at the event horizon isn't "physical", it requires an infinite proper acceleration to hold station there. So we need to take the limit of the velocity as R->R_s to make any sense of the velocity "at the horizon".

Yes. The same is noted in the book Exploring Black Holes. On pg. 3-15 it says:

The prediction of [an equation that returns the velocity of an in-falling particle as measured by an observer hovering at a given altitude] at the horizon must be taken as a limiting case, an extrapolation.
 
  • #15
I think you should use the derivative with respect to 'c', not infinity.
 
  • #16
Ok I have a curious question. What if half my body was above the event horizon and half my body was beneath it. Would I be able to lift my arm?
 
  • #17
You could have crossed an event horizon while you wrote your post, so yes. You can't hover draped across the horizon, according to the theory. You must move across it at the speed of light.
 
  • #18
Zanket said:
You could have crossed an event horizon while you wrote your post, so yes. You can't hover draped across the horizon, according to the theory. You must move across it at the speed of light.
Agreed that you can't hover, but the rest (moving at c) is just not correct. Massless photons move at c whether near a BH or not, but any particle or object cannot be accellerated to c no matter how strong the gravity. A BH is just a big source of a strong gravitational field, and one of the basics of GR is that nothing with mass can be accellerated to c as that would require an infinite amount of energy.

Infinite energy is not available from any source. Even neutrinos, with their very small mass, do not travel at c. It is close, but not quite c. A lot of outdated websites are still out there stating that neutrinos are massles and travel at c, but ignore them.
 
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  • #19
In relativity theory, the speed of light limit applies only to direct measurements of velocity.

Objects (assume material; i.e. not photons) at the horizon have been accelerated to c in the limit. All objects must move inward at exactly c at the horizon, precluding any velocity <> 0 relative to another object to be directly-measurable. But an infalling object can directly measure the velocity of a object hovering arbitrarily close to and above the horizon to be arbitrarily close to c.

To say that one cannot move at c at the horizon puts the cart before the horse. The reason that c is not directly measurable there is because all objects must move inward at c there, because gravity has there accelerated to c all objects.
 
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  • #20
Zanket said:
To say that one cannot move at c at the horizon puts the cart before the horse. The reason that c is not directly measurable there is because all objects must move inward at c there, because gravity has there accelerated to c all objects.
Sorry, but this is simply not the case at all. Maybe some other "expert" will chime in with an opinion.
 
  • #21
You must be right, by your mere declaration of it.

The book Exploring Black Holes on pg. 3-23 says:

The shell [hovering] observer just outside the horizon measures the in-falling particle to move at less than the speed of light. AT THE HORIZON no shell is possible, because the “local acceleration of gravity” increases without limit. So no shell observer can be stationed at the horizon to verify that the in-falling particle moves at light speed.

Looks like Taylor and Wheeler are wrong too.
 
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  • #22
At: https://www.physicsforums.com/showthread.php?t=71009&highlight=falling+black+hole the subject was already covered (falling into a black hole), but I didn't read all 22 posts.

I would agree that "c is not directly measurable there" (at the EH) but not that infalling material actually reaches c. I think that it is semantics or the definition of the reference frame causing us to disagree on what is actually correct by both of us. At the EH, light coming back to reach us at a distance is "infinitely redshifted" causing the time dilation effect so the light that does reach us (stationary in the Schwarzschild metric) does tell us that infalling material is at v=c.

But, it is the information (light) we receive that says c is attained, and this does not mean that the infalling material itself is at c unless you want to directly correlate the light reaching us with a measurement of "true velocity". I would say that the material "appears" to be at c but actually is not. This is because of SR's requirement of infinite energy to accellerate any mass to v=c. IOW, it is what the gravity of the BH does to the light, not what it does to the infalling material.

A Cornell U. site states that infalling material for a 30 solar mass black hole:
Relative to an observer stationary in the Schwarzschild metric, our velocity has now reached the speed of light. Relative to an observer freely falling radially from rest at infinity, our velocity is (8/9)1/2 c = 0.94 c.
and:
But the tides wouldn't be so bad for a very massive black hole. The tide at 1 Schwarzschild radius would be less than 1g if the black hole exceeded 30,000 solar masses.
http://origins.colorado.edu/~ajsh/singularity.html
So, I don't think that a 1g accelleration has any chance of causing any matter to reach v=c. Also there (somewhere) is:
when something falls into a black hole it is as if it is falling into the neck of an infinitely long funnel, down which it will spiral, getting more and more energy, and heavier and heavier, and faster and faster. But the speed increases more and more slowly (and the weight increases more and more quickly to keep the energy increasing) as it nears the speed of light.
I probably didn't explain my point very well because I have now even confused myself... :confused: , but I guess my main point is that no gravity or other source of energy can cause a violation of SR (I said GR earlier.. :yuck:) Not because SR is the Gospel, it has just been verified too many times to leave much doubt. On this subject, it is all dependant on reference frames.
 
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  • #23
I am not sure what happens inside the event horizon, but outside the event horizon the radial speed of a massive particle falling from rest at infinity towards the black hole will approach c when reaching the event horizon.

I may be wrong but I don't think one can make a reasonable distinction between "true" speed and "apparent" speed, as speed is always "apparent", in the sense that it is always relative.

The statement above about the speed of infalling particles is valid in Schwarzschild coordinates and therefore for stationary remote observers. The speed is given by:

[tex]\frac {dr}{d\tau} = \sqrt{\frac{2M}{r}}[/tex]

This follows from the equation [tex]\inline{ p \cdot p = - m^2}[/tex] in a Schwarzschild spacetime and E = 1 (particle at rest at infinity), see for example Schutz section 11.2.
 
  • #24
Note that dr/dtau can be greater than '1' when the particle starts out with E>1. If L=0

dr/dtau = sqrt(2M/r + (E^2 - 1))

dr/dtau is not really a standard velocity. I believe it can be described as a "rapidity", though.
 
  • #25
Thank you for your correction pervect. How can be the radial speed computed?
 
  • #26
Labguy said:
I probably didn't explain my point very well because I have now even confused myself... :confused: , but I guess my main point is that no gravity or other source of energy can cause a violation of SR (I said GR earlier.. :yuck:) Not because SR is the Gospel, it has just been verified too many times to leave much doubt.

And this is compatible with what Taylor and Wheeler say above. The easy way to think about this is the intuitive way: if you fall across the horizon at < c then the definition of a horizon is not met, for info from you could then escape the horizon. If you fall across the horizon at > c then someone hovering above the horizon could directly measure your velocity to be >= c, violating SR. Then you must cross the horizon at exactly c (and that is supported by GR's math, and it applies regardless how you got to the horizon--you could have plunged into it or been hurled to it at relativistic velocity or dropped from any height or accelerating into it with rocket engines). Then if you fall across the horizon next to something else, your velocity relative to it is zero (because it crosses at your same exact speed); that's the only velocity you can directly measure at the horizon.
 
  • #27
hellfire said:
How can be the radial speed computed?
Sorry, I noticed now that the answer is already in the thread linked by pervect.
 
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  • #28
hellfire said:
Sorry, I noticed now that the answer is already in the thread linked by pervect.

On my "things to do, someday" list is to review some of my calculations in that linked thread, as I have become a bit suspicious of them (if it's the same linked thread that I think you mean).

I think that dr/dtau is actually one of the most convenient measures of speed for actual calculation, BTW, even though it's not strictly speaking a velocity. It's convenient because there is a simple formula for it.

I also think that idea of "the velocity" of an object falling into a black hole requires some careful definition before it can be answered.

As I said earlier in the thread, I think that one precise formulation of the question is the velocity at which an infalling object would have relative to a "station-keeping" observer at constant R, theta, and phi in Schwarzschild coordinates. Schwarzschild coordinates are used here because the metric does not change with respect to time (so the station-keeping observer thus sees a static metric).

This velocity is to be measured when the infalling object is at the same point in space as the station-keeping observer to avoid simultaneity problems.

I'm still open to alternative defintions of how "the" velocity is to be computed, as long as they are precise enough to allow actual calculation.

It's already been pointed out in some other thread that the coordinate velocity as seen by a distant observer is dr/dt = 0. I don't think that that's the answer people who are asking the question are looking for, though.
 
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  • #29
I think I'm getting a sensible answer to this question this time around.

We start with two original equations in geometric units which describe the motion of a particle in a Schwarzschild geometry

[tex]
\frac{dr}{d\tau} = \sqrt{E^2 - 1 + \frac{2M}{r}}
[/tex]
[tex]
\frac{dt}{d\tau} = \frac{E}{1-\frac{2M}{r}}
[/tex]

These can be found many places, one online reference is
http://www.fourmilab.ch/gravitation/orbits/

This also defines the symbols used, such as the fact that E is the energy/ unit mass, and is equal to 1 if the particle is at rest at infinity.

The problem is that the tangent vectors in the Schwarzschild coordinate system are not normalized.

There are several ways to correct this, one straightforwards but slightly tedious way is to introduce new scaled coordinates

rr = a*r, tt = b*t

(we don't have to worry about theta and phi in this example) so that the tangent vectors are normalized.

We will achieve the normalization we seek when the metric can be written as

ds^2 = d rr^2 - dtt ^ 2

The Schwarzschild metric is

ds^2 = dr^2 / (1 - 2M/r) - (1-2M/r) dt^2

we substitute drr = a*dr, dtt = b*dt, and solve for a and b.

We then take drr/dtt as the velocity, which we compute by the following

[tex]
\frac{drr}{dtt} = (\frac{drr}{dr}\frac{dr}{d\tau}) \, / (\frac{dtt}{dt} \frac{dt}{d\tau}) = (a \frac{dr}{d\tau}) \, / (b \frac{dt}{d\tau})
[/tex]

Much computer-aided symbolic manipulation later, we come up with

[tex]
v = \frac{drr}{dtt} = \frac{\sqrt{E^2 - (1 - \frac{2M}{r})}}{E}
[/tex]

Which has the property that it's equal to 1 (the speed of light in geometric units) when r=2M, and less than 1 when r>2M, so it looks sensible at first glance. It also gives a reasonable-looking expression for the velocity when r=infinity, for instance E=1 implies that the velocity at infinity is zero.

The form of the expression suggests that there should be a simpler way to derive it than the path I took.
 
  • #30
Let me try again.

Here's another look at velocities similar to the one pervect gave. Aside from the motivation for [itex]\frac{dt}{d\tau} = E(1-\frac{2M}{r})^{-1}[/itex], which can be skipped, my presentation is similar to stuff in Exploring Black Holes by Taylor and Wheeler.

Motivation
A timelike geodesc between causally related events [itex]A[/itex] and [itex]B[/itex] is found by extremizing proper time for timelike worldline that connect [itex]A[/itex] and [itex]B[/itex]. Since [itex]d\tau^2 = g_{\mu \nu} dx^\mu dx^\nu[/itex],

[tex]
\tau_{AB} = \int_{A}^{B} d\tau = \int_{A}^{B}d\tau \left( g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \right)^{\frac{1}{2}}.
[/tex]

Using a variational principle to extremize this line integral gives Lagrange's equations

[tex]
\frac{d}{d\tau} \frac{\partial L}{\partial \dot x^\mu} = \frac{\partial L}{\partial x^\mu}
[/tex]

with [itex]L = (g_{\mu \nu} dx^\mu dx^\nu)^{\frac{1}{2}}[/itex] the Lagrangian, and where dot denotes differentiaition with respect to proper time.

Using the Schwarzschild metric, which has no explicit [itex]t[/itex] dependence, and after simplifying by using [itex]L = 1[/itex], the [itex]t[/itex] Lagrange equation becomes

[tex]
\frac{d}{d\tau} \left[ \left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau} \right] = 0,
[/tex]

so,

[tex]
\left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau}
[/tex]

is a conserved quantity.

As [itex]r \rightarrow \infty[/itex], the Schwarzschild metric goes to flat Minkowski spacetime and the conserved quantity goes to [itex]\frac{dt}{d\tau}[/itex], which in Minkowski spacetime is just [itex]E/m[/itex]. Consequently, call the conserved quantity

[tex]
\left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau} = \frac{E}{m}
[/tex]

the energy per unit mass of a mass [itex]m[/itex]test particle in Minkowski spacetime. From now on, take [itex]m = 1[/itex], or, equivalently, redefine [itex]E[/itex] to be energy per unit mass.
End Motivation

Consider an observer hovering with constant [itex]r[/itex], [itex]\phi[/itex], and [itex]\theta[/itex] coordinates, which means [itex]0 = dr = d\phi = d\theta[/itex]. Plugging these into the Schwarzschild metric results in

[tex]
d\tau_S = \left(1 - \frac{2M}{r} \right)^{\frac{1}{2}} dt
[/tex]

for the relationship between proper time [itex]\tau_s[/itex] for the hovering observer and coordinate time [itex]t[/itex]. Now consider that occurs a small spatial distance away from the hovering observer. The hovering observer measures spatial (r) distance by considering simultaneously where the ends of his metre stick lie, i.e., now [itex]dr \neq 0[/itex] and [itex]0 = d\tau_S = dt = d\phi = d\theta[/itex]. The Schwarzschild metric gives

[tex]
dr_S = \left(1 - \frac{2M}{r} \right)^{-\frac{1}{2}} dr
[/tex]

as the hovering observer's proper distance between himself and the event.

Now consider an observer freely falling radially, so [itex]0 = d\phi = d\theta[/itex], and the metric gives

[tex]
d\tau^2 = \left( 1 - \frac{2M}{r} \right) dt^2 - \left( 1 - \frac{2M}{r} \right)^{-1} dr^2
[/tex]

for his proper time. Using the expression for the conserved [itex]E[/itex] gives

[tex]
d\tau^2 = \left( 1 - \frac{2M}{r} \right) E^2 \left( 1 - \frac{2M}{r} \right)^{-2} d\tau^2 - \left( 1 - \frac{2M}{r} \right)^{-1} dr^2,
[/tex]

and, after mutliplying by [itex]\left( 1 - \frac{2M}{r} \right)[/itex] and rearranging,

[tex]
dr^2 = \left[ E^2 - \left( 1 - \frac{2M}{r} \right) \right] d\tau^2
[/tex]

is obtained. Therefore,

[tex]
\frac{dr}{d\tau} = -\sqrt{E^2 - \left( 1 - \frac{2M}{r} \right)},
[/tex]

where the negative sign reflects that, for the freely falling observer, [itex]r[/itex] decreases as proper time increases.

Finally, consider 2 events - one where the two observers are coincident and the other after they have separated a short distance. The hovering observer records [itex]d\tau_S[/itex] and [itex]dr_S[/itex] as proper time and space separations for these events. Using the above gives

[tex]
\begin{equation*}
\begin{split}
\frac{dr_S}{d\tau_S} &= \left( 1 - \frac{2M}{r} \right)^{-1} \frac{dr}{dt}\\
&= \frac{1}{E} \frac{dt}{d\tau} \frac{dr}{dt}\\
&= \frac{1}{E} \frac{dr}{d\tau}\\
&= -\frac{1}{E} \sqrt{E^2 - \left( 1 - \frac{2M}{r} \right)}
\end{split}
\end{equation*}
[/tex]

for the velocity with which the freely falling observer sails past the the hovering observer.

This is the same expression derived by pervect. As he noted, [itex]E = 1[/itex] for an observer that freely falls from rest at infinity.

I hope to make anothe post that treats this stuff more at the level of Misner, Thorne, and Wheeler, i.e., that uses 1-forms and tangent vectors like [itex]e_t = \frac{\partial}{\partial t}[/itex]. Then, orthonormal frames can be set up for both observers, and an interesting relations between the frames can be established.

Regards,
George
 
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What is acceleration at the event horizon?

Acceleration at the event horizon refers to the rate at which an object gains speed as it approaches the event horizon of a black hole. This acceleration is caused by the intense gravitational pull of the black hole.

How is acceleration at the event horizon calculated?

The acceleration at the event horizon can be calculated using the formula a = GM/r^2, where G is the gravitational constant, M is the mass of the black hole, and r is the distance from the center of the black hole to the object.

Can acceleration at the event horizon be observed?

No, acceleration at the event horizon cannot be directly observed because it occurs within the black hole's event horizon, which is the point of no return for anything that enters it.

What happens to an object at the event horizon?

An object at the event horizon will experience extreme gravitational forces, causing it to accelerate towards the black hole at an increasing rate. Eventually, the object will reach the singularity at the center of the black hole, where the gravitational forces become infinite.

How does acceleration at the event horizon affect time?

According to Einstein's theory of relativity, time slows down as an object approaches the event horizon due to the intense gravitational forces. This phenomenon is known as time dilation and is a result of the distortion of space-time near the black hole.

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