# Thread: Tidal Forces in Supermassive Black Holes

1. Wes
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Mar 2021
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## Tidal Forces in Supermassive Black Holes

Hi Folks:

Another one that's stumping my wife and I.

From Wikipedia: "The point at which tidal forces destroy an object or kill a person will depend on the black hole's size. For a supermassive black hole, such as those found at a galaxy's center, this point lies within the event horizon, so an astronaut may cross the event horizon without noticing any squashing and pulling, although it remains only a matter of time, as once inside an event horizon, falling towards the center is inevitable."

My understanding of the event horizon is that it is the limit at which the forces of gravity are so strong that light cannot escape. How can the gravity be that strong yet not be strong enough to rip apart an object? I just don't get how the Schwarzschild radius can ever be inside the event horizon. Even if the math works, it doesn't make sense inside my head...

Cheers!

2. It's not the gravity that spaghettifies and object, it's the distribution of gravity. In an SMBH, tidal force is spread out over such a vast area that it's locally very weak. In a stellar mass BH, it's concentrated to a deadly degree.

3. To add to that a bit, a stellar mass black hole is a very compact object, only miles (single digits) in diameter. Getting close to the event horizon, the gravity at the lowest point of a small object (like a space capsule or an astronaut) is far higher than the top so it is feeling much more force. Also because the gravity comes from such a small spherical object, the hypothetical astronaut will tend to be compressed in the direction perpendicular to the black hole’s center. The gravity of a super massive black hole doesn’t change so much over a short distance, so wouldn’t be a big issue to an astronaut, but could still rip apart a closely approaching star.

4. And some more...
The force of gravity doesn't bother you if you are falling freely, no matter how strong it is, so long as every part of your body is experiencing the same force, because then every part of your body accelerates at the same rate. That's what "free fall" means.
It's the gravity gradient that causes problems--if the force on your head is substantially different from the force on your feet. Then your feet will try to accelerate at a different rate from your head, and your body will need to generate internal forces to stop that happening--you experience tension forces radially and compression forces transversely.

On a point of terminology, the "Schwarzschild radius" is the radius of the event horizon (for a non-rotating black hole). So by definition it can't be below the event horizon. I wonder if the OP is thinking of the Roche limit, which is the point at which a gravitationally bound object will disintegrate under tidal forces. But that's not actually the right term either, because an astronaut is not primarily held together by her own gravity, and so can get a long way below the Roche limit before internal forces become irresistible.

Grant Hutchison

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Originally Posted by grant hutchison
And some more...
On a point of terminology, the "Schwarzschild radius" is the radius of the event horizon (for a non-rotating black hole). Grant Hutchison
Wouldn't the Schwarzschild radius be the same for rotating black holes? Are non-rotating black holes even possible?

6. Wes
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Mar 2021
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Yep, totally mistook the Schwarzchild radius for the Roche limit - thanks for the correction. Also much clearer now in terms of the gravity gradient - thanks again for all the help!

7. Originally Posted by DaCaptain
Wouldn't the Schwarzschild radius be the same for rotating black holes? Are non-rotating black holes even possible?
The Schwarzschild radius is defined in the Schwarzschild metric, which is the GR description of a non-rotating mass. So it's a property of a non-rotating black hole, whether or not such objects exist in nature.
To describe the gravitational field of a rotating mass you use the Kerr metric, which produces a set of ellipsoidal coordinates . The "radius" of the outer event horizon depends on your coordinate choices, but it isn't spherical. In fact, for rapidly rotating black holes there's no way in Euclidean space to resolve the curvature of the event horizon with the variation in radius with "latitude"--that is, you simply can't produce an accurate model of the shape of the event horizon that you could hold in your hand.

So Kerr black holes don't (by definition) have a Schwarzschild radius, nor do they have a single radius characteristic of their whole event horizon.
Most pop sci discussions of black holes use the Schwarzschild metric and Schwarzschild radius because it's the simplest model, and generally harmless, but all such discussion are of a toy model that (as you point out) isn't entirely applicable to real rotating black holes.

Grant Hutchison

8. Upshot, if falling in you'd be torn to pieces before hitting the Event Horizon of a stellar mass BH.

9. One nice, but apparently little-known, fact about spaghettification applied to humans falling into black holes, is that the person wouldn't have time to be aware of it. Once the tidal forces start to rise to life-threatening levels, they rise so quickly (in the in-faller's reference frame) that a person's body would come apart faster than nerve transmission could inform her brain. True for any size of black hole.

Grant Hutchison

10. Originally Posted by grant hutchison
One nice, but apparently little-known, fact about spaghettification applied to humans falling into black holes, is that the person wouldn't have time to be aware of it. Once the tidal forces start to rise to life-threatening levels, they rise so quickly (in the in-faller's reference frame) that a person's body would come apart faster than nerve transmission could inform her brain. True for any size of black hole.
Good to know. I’ll remember that if I’m ever on death row and allowed to choose the method of execution.

“I’d like to be dropped into a black hole.”

Sent from my iPhone using Tapatalk

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