# Is the potential energy of falling into a black hole already included into $E=mc^2$?

Each object in the Universe has a huge potential energy, because you can drop it into a black hole, accelerating it to nearly the speed of light.

Is this potential energy already included into Einstein's $$E=mc^2$$ equation, or does it come on top of that?

• "drop it into a black hole, accelerating it to nearly the speed of light" - This depends on the point of view. In the view of any outside observer, the speed of light at the horizon is zero. So, while falling things approach the speed of light near the horizon, they actually slow down instead of speeding up. See the first red curve here: mysearch.org.uk/website1/html/270.Free-Fall.html – safesphere Apr 22 at 7:51
• $E=mc^2$ isn't a single, specific fact. It has multiple interpretations in different contexts. – Ben Crowell Jun 20 at 15:18

$$E=mc^2$$ for a stationary particle in special relativity, where there are no black holes (and no gravity). So black holes (and gravity more generally) have nothing to do with this $$E$$.

• Consider a weightless mirror box filled with light. It is well known that the mass-energy relation holds for it, so $E=mc^2$. The mass of the normal baryonic matter consists in 99% of the energy of (virtual) gluons, which like photons are massless. So the normal baryonic matter is not that different from the mirror box after all. Now, drop this box to a black hole. The photons get redshifted down to zero energy at the horizon. Thus the mass of the box there in the view of a remote observer is zero. Push the box away and it is $E=mc^2$ again. So clearly $E$ has something to do with gravity. – safesphere Apr 22 at 7:33
• binding energy contributes to E, instead of atoms think of gravitational systems – Wolphram jonny Apr 22 at 11:03
• physics.stackexchange.com/questions/66359/… – Wolphram jonny Apr 22 at 11:05

You could say that the energy of a small test-body in the vicinity of a spherically symmetric black hole in general relativity in the case when the test-body has no velocity compared to the center of the gravitational field can be written as:

$$E=mc^2{\sqrt{1-\frac{2GM}{rc^2}}}.$$

So if you drop an object from rest at infinity so that if falls down and crashes into standstill on the surface of a planet the energy lost as heat/radiation at impact can be written as:

$$\Delta E=mc^2-mc^2\sqrt{1-\frac{2GM}{rc^2}}$$

Answer: You could say that the potential energy is not baked in the expression $$E=mc^2$$ but it is baked into $$E=mc^2\sqrt{1-\frac{2GM}{rc^2}}$$.

• This doesn't really answer the OP's conceptual question, and the extremely awkward and ugly way of expressing the energy is a distraction. – Ben Crowell Jun 20 at 15:15