Acceleration At The Event Horizon

  • Thread starter mrjeffy321
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  • #26
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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
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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
pervect
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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
pervect
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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
George Jones
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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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