Many times, people often claim nothing "weird" occurs at the event

horizon of a black hole. They point to the case of an astronaut in

freefall falling through and experiencing no ill effect at the crossing,

which occurs in finite proper time. But what of an astronaut NOT in

freefall? In a related problem, people often ask something to the

effect of, "Why can't you lower a string through the event horizon and

then pull it back out again." Often the answer given is muddled,

incorrect or a brush off. In fact, the static force or weight of a

static or suspended object diverges as one approaches the event horizon.

Two references are:

Gravitation, by Misner, Thorne and Wheeler; aka MTW

Essential Relativity, Wolfgang Rindler; aka Rindler

Fact #1. The radial proper distance to the event horizon (r=2M) from

any Schwarzchild "r" is finite. See MTW, eq 31.9, pg 824.

Fact #2. The LOCAL static acceleration or "weight" of a stationary

observer diverges or goes to infinity as "r" approaches 2M.

The per unit weight of an object (ie, local proper acceleration)

in conventional units is:

(GM/r^2) (1 - 2GM/(c^2 r))^(-1/2)

See Rindler, eq 8.71, pg 148.

[note: An easy way to remember the above is Newton's formula devided by

the redshift factor.]

By now the cogs in the brain of the astute physicist or mathematician

should be turning and he or she should begin to realize that the WEIGHT

DIVERGES OVER A FINITE DISTANCE! And this diverging is occurring

OUTSIDE the event horizon. So the STATIC tidal forces, defined to be:

d(weight)/d(proper distance), or the weight gradient, also diverges as

one approaches the event horizon.

So now the budding general relativist is thinking to himself, "I'm so

confused. How can this be? I thought there are finite tidal forces at

the event horizon?"

Well let's look at this situation from a free-falling observer. Since

the event horizon represents a stationary light-like surface, by

definition any infaller falls thru the horizon locally at the speed of

light. So he is locally Lorentz contracted to zero thickness with

respect to a stationary observer at the horizon (for the purists, use

limits and pretend you are a small distance epsilon outside the horizon,

and then let epsilon -> 0.) Also at the event horizon the angle of

escape for light (and thereby anything else as well) is zero degrees

with respect to the vertical (in 3 space coordinates), all freefalling

trajectories are vertical at the event horizon. Since the infaller is

of zero thickness in the radial direction (ie, FLAT) his d(proper

distance) in the radial direction is ZERO. To get the tidal force on

the faller we multiply the static tidal force times the faller's radial

extent. We get:

infinity * zero

which is an indeterminate form. If you work this out in detail you

will see that the above product remains finite for a free falling

object. So the tidal forces on a freefaller remain finite, although

they vary depending on his 4-velocity. Since the faller is in

freefall, his static weight is of course zero.

Now an astronaut who is able to stick his arm thru the event

horizon, but*remain* himself outside the horizon, *cannot* be in

freefall. Therefore he will experience*inifinite* tidal forces across

the length of his arm, and it will of course be torn off. The same

thing with a string.

horizon of a black hole. They point to the case of an astronaut in

freefall falling through and experiencing no ill effect at the crossing,

which occurs in finite proper time. But what of an astronaut NOT in

freefall? In a related problem, people often ask something to the

effect of, "Why can't you lower a string through the event horizon and

then pull it back out again." Often the answer given is muddled,

incorrect or a brush off. In fact, the static force or weight of a

static or suspended object diverges as one approaches the event horizon.

Two references are:

Gravitation, by Misner, Thorne and Wheeler; aka MTW

Essential Relativity, Wolfgang Rindler; aka Rindler

Fact #1. The radial proper distance to the event horizon (r=2M) from

any Schwarzchild "r" is finite. See MTW, eq 31.9, pg 824.

Fact #2. The LOCAL static acceleration or "weight" of a stationary

observer diverges or goes to infinity as "r" approaches 2M.

The per unit weight of an object (ie, local proper acceleration)

in conventional units is:

(GM/r^2) (1 - 2GM/(c^2 r))^(-1/2)

See Rindler, eq 8.71, pg 148.

[note: An easy way to remember the above is Newton's formula devided by

the redshift factor.]

By now the cogs in the brain of the astute physicist or mathematician

should be turning and he or she should begin to realize that the WEIGHT

DIVERGES OVER A FINITE DISTANCE! And this diverging is occurring

OUTSIDE the event horizon. So the STATIC tidal forces, defined to be:

d(weight)/d(proper distance), or the weight gradient, also diverges as

one approaches the event horizon.

So now the budding general relativist is thinking to himself, "I'm so

confused. How can this be? I thought there are finite tidal forces at

the event horizon?"

Well let's look at this situation from a free-falling observer. Since

the event horizon represents a stationary light-like surface, by

definition any infaller falls thru the horizon locally at the speed of

light. So he is locally Lorentz contracted to zero thickness with

respect to a stationary observer at the horizon (for the purists, use

limits and pretend you are a small distance epsilon outside the horizon,

and then let epsilon -> 0.) Also at the event horizon the angle of

escape for light (and thereby anything else as well) is zero degrees

with respect to the vertical (in 3 space coordinates), all freefalling

trajectories are vertical at the event horizon. Since the infaller is

of zero thickness in the radial direction (ie, FLAT) his d(proper

distance) in the radial direction is ZERO. To get the tidal force on

the faller we multiply the static tidal force times the faller's radial

extent. We get:

infinity * zero

which is an indeterminate form. If you work this out in detail you

will see that the above product remains finite for a free falling

object. So the tidal forces on a freefaller remain finite, although

they vary depending on his 4-velocity. Since the faller is in

freefall, his static weight is of course zero.

Now an astronaut who is able to stick his arm thru the event

horizon, but

freefall. Therefore he will experience

the length of his arm, and it will of course be torn off. The same

thing with a string.