<p>When we talk of kinetic energy you need to specify a frame of reference.  In its own reference frame a body is at rest and its kinetic energy is zero.</p>

<p>So there is no special effect as a body crosses the event horizon. From the point of view of an infalling object, there is nothing at the event horizon, and the body will just continue to fall (though not for long, because there is a singularity in every possible future). If the black hole was big and quiet enough, you might not notice tidal forces as you crossed the horizon</p>

<p>Hawking radiation is not an observed anomaly that needs a theoretical explanation, it is a theoretical prediction, that is need observational support. The maths of Hawking radiation are well understood. It doesn't need an explanation; it is an explanation</p>

CC BY-SA 4.0
2020-04-30T17:22:58.873Z
2020-04-30T17:22:58.873Z
James K
<p>Everything is relative. The infalling object has a very high KE as measured by a distant observer stationary relative to the Black Hole. But as measured by another observer falling alongside it, it has zero KE. If it collided with a particle that was somehow stationary just above the event horizon a huge amount of energy would be released, and if you were very close indeed to the event horizon you might see the kind of particle physics effects you describe. But if it doesn't hit anything, nothing special happens. Just because a distant observer thinks it has a very high KE doesn't mean that it spontaneously breaks down in some way. </p>

CC BY-SA 4.0
2020-04-30T17:19:16.780Z
2020-04-30T17:19:16.780Z
Steve Linton
<p>An object reaching the event horizon has almost infinite kinetic energy. When the total energy of the object surpasses the breaking energy of the electroweak field , the Higgs mechanism would stop working for this object and the subatomic particles of the object will get 0 mass.</p>

<p>This means that electrons would no longer form orbitals and what is left from the object will be massless electrons moving randomly and a quark-gluon soup.</p>

<p>I guess because the object will have lost its mass(though energy) , it will radiate some energy away. Could this be an explanation for the Hawking radiation or what's really going on in this case?</p>

CC BY-SA 4.0
2020-05-03T20:59:42.623Z
Why do you think there is a "breaking energy of the electroweak field"? Have you tried calculating the kinetic energy of a falling object?
Anders Sandberg
CC BY-SA 4.0
2020-05-03T22:01:36.677Z
The kinetic energy of a falling object right before the event horizon approaches infinite because inertial mass and velocity are both really high.When the universe was hot(things moving fast) electromagnetism was merged with the weak nuclear interaction.
Jelly Strawberry
CC BY-SA 4.0
2020-05-25T09:29:14.187Z
No, actually do the calculation. You will get a finite value: high, but not crazy.
Anders Sandberg
<black-hole><event-horizon>
Objects falling in a black hole radiate their mass?
Jelly Strawberry
# Objects falling in a black hole radiate their mass?

(Asked by: username_0 on 2020-04-30)

An object reaching the event horizon has almost infinite kinetic energy. When the total energy of the object surpasses the breaking energy of the electroweak field , the Higgs mechanism would stop working for this object and the subatomic particles of the object will get 0 mass.

This means that electrons would no longer form orbitals and what is left from the object will be massless electrons moving randomly and a quark-gluon soup.

I guess because the object will have lost its mass(though energy) , it will radiate some energy away. Could this be an explanation for the Hawking radiation or what's really going on in this case?

**Tags:** black-hole, event-horizon
> **Comments:**
> **username_1**: Why do you think there is a "breaking energy of the electroweak field"? Have you tried calculating the kinetic energy of a falling object?  
> **username_0**: The kinetic energy of a falling object right before the event horizon approaches infinite because inertial mass and velocity are both really high.When the universe was hot(things moving fast) electromagnetism was merged with the weak nuclear interaction.  
> **username_1**: No, actually do the calculation. You will get a finite value: high, but not crazy.  

---

## 2 answers

### Answer by username_2 on 2020-04-30. Score: 1

When we talk of kinetic energy you need to specify a frame of reference. In its own reference frame a body is at rest and its kinetic energy is zero.

So there is no special effect as a body crosses the event horizon. From the point of view of an infalling object, there is nothing at the event horizon, and the body will just continue to fall (though not for long, because there is a singularity in every possible future). If the black hole was big and quiet enough, you might not notice tidal forces as you crossed the horizon

Hawking radiation is not an observed anomaly that needs a theoretical explanation, it is a theoretical prediction, that is need observational support. The maths of Hawking radiation are well understood. It doesn't need an explanation; it is an explanation


---

### Answer by username_3 on 2020-04-30. Score: 0

Everything is relative. The infalling object has a very high KE as measured by a distant observer stationary relative to the Black Hole. But as measured by another observer falling alongside it, it has zero KE. If it collided with a particle that was somehow stationary just above the event horizon a huge amount of energy would be released, and if you were very close indeed to the event horizon you might see the kind of particle physics effects you describe. But if it doesn't hit anything, nothing special happens. Just because a distant observer thinks it has a very high KE doesn't mean that it spontaneously breaks down in some way.


---

