<p>When some object is moving, it appears smaller and more massive. What if it's moving so fast it should become a black hole? Some calculations:</p>

<p>Let observed radius $R_0\gamma$ ($R_0$ - radius at rest) be equal Schwarzschild radius for object with observed mass $M_0\gamma$ ($M_0$ - mass at rest):</p>

<p>$$R_0 / \gamma = \frac{2GM_0\gamma}{c^2}$$</p>

<p>$$\gamma^2=\frac{R_0c^2}{2GM_0}$$</p>

<p>$$\frac{v}{c}=\sqrt{1-\frac{2GM_0}{R_0c^2}}$$</p>

<p>For Earth it gives $v=0.999999999304c$ and for Sun $v=0.999997877903c$. Is my reasoning correct? How would we see such objects?</p>

CC BY-SA 3.0
2016-11-21T14:47:06.457Z
I think this question has been asked before on SE. The Schwarzschildt solution only applies to a mass at rest (in the references frame in which its calculated). If an object at high velocity could collapse into a BH, then for reasons of symmetry you would be able to make any object collapse simply by moving past it at high velocity.
pela
CC BY-SA 3.0
2016-11-21T19:18:31.360Z
@pela - - but why would that be strange?  we live in a universe where *causality itself* (for hell's sake!) is totally relative
Fattie
CC BY-SA 3.0
2016-11-21T20:38:46.280Z
@JoeBlow: Hmm… _simultaneity_ is relative, but causality isn't. If event A is the cause of event B for some observer, then A precedes B for all inertial observers. The laws of physics should be the same for all observers. If observer A is at rest with respect to some star and concludes that pressure and gravity is in equilibrium, then to observer B, traveling nearly at the speed of light, _must_ conclude the same. For instance, in the reference frame of the [Oh-My-God particle](https://en.wikipedia.org/wiki/Oh-My-God_particle), everything should collapse into black holes, which hasn't happened
pela
CC BY-SA 3.0
2016-11-21T20:59:32.420Z
(simultaneity is relative, but causality isn't - sorry :O )   But surely it would just be the case that, the speedy observor sees a black hole; others do not see a black hole.  What's the problem with that man?
Fattie
<black-hole><special-relativity>
Relative black hole?
Michał Wójcik
# Relative black hole?

(Asked by: username_0 on 2016-11-21)

When some object is moving, it appears smaller and more massive. What if it's moving so fast it should become a black hole? Some calculations:

Let observed radius $R\_0\gamma$ ($R\_0$ - radius at rest) be equal Schwarzschild radius for object with observed mass $M\_0\gamma$ ($M\_0$ - mass at rest):

$$R\_0 / \gamma = \frac{2GM\_0\gamma}{c^2}$$

$$\gamma^2=\frac{R\_0c^2}{2GM\_0}$$

$$\frac{v}{c}=\sqrt{1-\frac{2GM\_0}{R\_0c^2}}$$

For Earth it gives $v=0.999999999304c$ and for Sun $v=0.999997877903c$. Is my reasoning correct? How would we see such objects?

**Tags:** black-hole, special-relativity
> **Comments:**
> **username_1**: I think this question has been asked before on SE. The Schwarzschildt solution only applies to a mass at rest (in the references frame in which its calculated). If an object at high velocity could collapse into a BH, then for reasons of symmetry you would be able to make any object collapse simply by moving past it at high velocity.  
> **username_2**: @pela - - but why would that be strange?  we live in a universe where *causality itself* (for hell's sake!) is totally relative  
> **username_1**: @JoeBlow: Hmm… _simultaneity_ is relative, but causality isn't. If event A is the cause of event B for some observer, then A precedes B for all inertial observers. The laws of physics should be the same for all observers. If observer A is at rest with respect to some star and concludes that pressure and gravity is in equilibrium, then to observer B, traveling nearly at the speed of light, _must_ conclude the same. For instance, in the reference frame of the [Oh-My-God particle](https://en.wikipedia.org/wiki/Oh-My-God_particle), everything should collapse into black holes, which hasn't happened  
> **username_2**: (simultaneity is relative, but causality isn't - sorry :O )   But surely it would just be the case that, the speedy observor sees a black hole; others do not see a black hole.  What's the problem with that man?  

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## 0 answers

