Rockfall

<continued>  The coffee shop crowd had gotten rowdy in response to my sloppy physics, but everyone hushed when I reached for my holster and drew out Old Reliable.  All had heard of it, some had seen it in action — a maxed-out tablet with customized math apps on speed-dial.

“Let’s take this nice and slow.  Suppose we’ve got an non-charged, non-spinning solar-mass black hole.  Inside its event horizon the radius gets weird but let’s pretend we can treat the object like a simple sphere.  The horizon’s half-diameter, we’ll call it the radius, is rs=2G·M/c²G is Newton’s gravitational constant, M is the object’s mass and c is the speed of light.  Old Reliable says … about 3 kilometers.  Question is, what happens when we throw a rock in there?  To keep things simple, I’m going to model dropping the rock gentle-like, dead-center and with negligible velocity relative to the hole, OK?”

<crickets>

“Say the rock has the mass of the Earth, almost exactly 3×10-6 the Sun’s mass.  The gravitational potential energy released when the rock hits the event horizon from far, far away would be E=G·M·m/rs, which works out to be … 2.6874×1041 joules.  What happens to that energy?”falling rock and black hole

rs depends on mass, Mr Moire, so the object will expand.  Won’t that push on what’s around it?”

“You’re thinking it’d act like a spherical piston, Jeremy, pushing out in all directions?”

“Yeah, sorta.”

“After we throw in a rock with mass m, the radius expands from rs to rp=2G·(M+m)/c².  I set m to Earth’s mass and Old Reliable says the new radius is … 3.000009 kilometers.  Granted the event horizon is only an abstract math construct, but suppose it’s a solid membrane like a balloon’s skin.  When it expands by that 9 millimeters, what’s there to push against?  The accretion disk?  Those rings might look solid but they’re probably like Saturn’s rings — a collection of independent chunks of stuff with an occasional gas molecule in-between.  Their chaotic orbits don’t have a hard-edged boundary and wouldn’t notice the 9-millimeter difference.  Inward of the disk you’ve got vacuum.  A piston pushing on vacuum expends zero energy.  With no pressure-volume work getting done that can’t be where the infall energy goes.”

“How about lift-a-weight work against the hole’s own gravity?”

“That’s a possibility, Vinnie.  Some physicists maintain that a black hole’s mass is concentrated in a shell right at the event horizon.  Old Reliable here can figure how much energy it would take to expand the shell that extra 9 millimeters.  Imagine that simple Newtonian physics applies — no relativistic weirdness.  Newton proved that a uniform spherical shell’s gravitational attraction is the same as what you’d get from having the same mass sitting at the shell’s geometric center.  The gravitational pull the shell exerts on itself originally was E=G·M²/rs.  Lifting the new mass from rs to rp will cost ΔE=G·(M+m)²/r– G·M²/rs.  When I plug in the numbers…  That’s interesting.”

Vinnie’s known me long enough to realize “That’s interesting” meant “Whoa, I certainly didn’t expect THAT!

“So what didja expect and whatcha got?”

“What I expected was that lift-it-up work would also be just a small fraction of the infall energy and the rest would go to heat.  What I got for ΔE here was 2.6874×1041 joules, exactly 100% of the input.  I wonder what happens if I use a bigger planet.  Gimme a second … OK, let’s plot a range …  How ’bout that, it’s linear!”ep-es

“Alright, show us!”

All the infall energy goes to move the shell’s combined mass outward to match the expanded size of the event horizon.  I’m amazed that such a simple classical model produces a reasonable result.”

“Like Miss Plenum says, Mr Moire, sometimes the best science comes from surprises.”

“I wouldn’t show it around, Jeremy, except that it’s consistent with Hawking’s quantum-physics result.”

“How’s that?”

“Remember, he showed that a black hole’s temperature varies as 1/M.  We know that temperature is ΔE/ΔS, where the entropy change ΔS varies as .  We’ve just found that ΔE varies as M.  The ΔE/ΔS ratio varies as M/M²=1/M, just like Hawking said.”

Then Jennie got into the conversation.

~~ Rich Olcott

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