I was thinking a bit earlier:(adsbygoogle = window.adsbygoogle || []).push({});

What if you created a particle at infinity (ignore the normal particle rules for simplicity), and allowed it to fall towards a massive body? Ignoring other effects such as the other forces, would the particle ever gain enough energy to surpass the energy required to create the particle?

Intuitively, we know the answer to be no. But I felt like doing a 5 second back-of-the-envelope calculation anyways. I did the following:

Say the gravitational potential difference from infinity to some distance [itex]r[/itex] is [itex]U=-\frac{GM}{r}[/itex]. Where [itex]M[/itex] is the large mass and [itex]m[/itex] is the mass of the particle. So, by falling from infinity to [itex]r[/itex] the particle would gain energy [itex]ΔE=\frac{GMm}{r}[/itex].

The rest energy of the particle is [itex]E_{rest}=mc^{2}[/itex]. I set the equations equal in order to see what (classical) radius I'd need in order to gain my energy back from creating the particle. The result is:

[itex]r=\frac{GM}{c^{2}}[/itex]

Well, I recognized this as (half) the Schwarzschild Radius, the radius of a black hole! [itex]r_{s}=\frac{2GM}{c^{2}}[/itex]

So now I'm curious! What's this mean? Why would I get half the Schwarzschild Radius? Is it simply telling me it's impossible, or is there a more interesting reason? Does anyone know?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# (What is) Schwarzschild Radius and Potential Energy

Loading...

Similar Threads - Schwarzschild Radius Potential | Date |
---|---|

I Coordinate singularity at Schwarzschild radius | Dec 12, 2017 |

I Schwarzschild radius | Jan 31, 2017 |

B Schwarzschild radius | Nov 25, 2016 |

I Schwarzschild Radius of Known Universe | Aug 2, 2016 |

**Physics Forums - The Fusion of Science and Community**