If a mass moves close to the speed of light, does it turn into a black hole?

Question

I'm a big fan of the podcast Astronomy Cast and a while back I was listening to a Q&A episode they did. A listener sent in a question that I found fascinating and have been wondering about ever since.

From the show transcript:

Arunus Gidgowdusk from Lithuania asks: "If you took a one kilogram mass and accelerated it close to the speed of light would it form into a black hole? Would it stay a black hole if you then decreased the speed?"

Dr. Gay, an astrophysicist and one of the hosts, explained that she'd asked a number of her colleagues and that none of them could provide a satisfactory answer. I asked her more recently on Facebook if anyone had come forward with one and she said they had not. So I thought maybe this would be a good place to ask.

Answer 1

The answer is no.

The simplest proof is just the principle of relativity: the laws of physics are the same in all reference frames. So you can look at that 1-kg mass in a reference frame that's moving along with it. In that frame, it's just the same 1-kg mass it always was; it's not a black hole.

Answer 2

No, a 1kg mass would not turn into a black hole, even if it were zipping past you at very close to the speed of light.

The principle of relativity is a fundamental idea in physics, and one consequence of it is that we can understand the physics of something that's moving by imagining we're moving alongside it.

For example, you are watching people play pool on a train as it rushes past you. You want to know whether a certain shot that's just been made will sink the 8-ball. You figure it out by imagining you're inside the train and calculating everything you'd expect to happen from that simpler viewpoint where the pool table is stationary. If the 8-ball goes into a certain pocket from that point of view, you can rest assured it will go into the same pocket if you analyze the situation again from your original vantage point on terra firma.

Applying the same principle to the 1kg mass, we see that moving along side it, it just looks like a normal mass, not a black hole. Hence, from another point of view in which it moves close to the speed of light, it still looks like a normal mass, not a black hole.

Answer 3

While good, I think the other answers are currently missing one ingredient, so I'll post this answer.

For particles traveling at constant velocity there is no event horizon, and so they act nothing like a black hole. Light from other regions of space will eventually reach it, unlike a black hole. Further, the forces between atoms in what ever matter constitutes the mass are co-moving and so there is no increased gravitational interaction between them. While the distances between them appear to change to an outside observer (as the mass is accelerated) once it reaches constant velocity they are fixed.

What has not been mentioned in other answers is the effect of acceleration. When a particle is continuously accelerated there is an apparent event horizon. See the relevant Wikipedia page here. So this has some features that we associate with a black hole, however there are still major differences. An object undergoing constant acceleration does indeed behave like it is static in a constant gravitational field. However, in the case of such an object the direction of the equivalent field is constant (and in a constant direction) throughout the object. This is not true for the gravitational field of a black hole, which is spherically symmetric.

Of course once the particle stops accelerating the apparent horizon disappears.

Answer 4

I am presuming the idea is the 1kg mass will length contract to below the Planck length. It is either that or the relativistic energy (mass) $E~=~\gamma mc^2$ would be so large it would gravitationally implode. The question though can be thought of according to what would happen to an observer on the mass. The question could be turned around: Would the universe implode? If a mass $M$ passes by a smaller mass $m~<<~M$ then one might think that $M$ could become a black hole and the small mass $m$ if close enough would become trapped in the black hole. However, from the frame of the big mass $M$ the small mass is not a black hole. This is a contradiction.

A ultra-relativistic mass will behave similar to a gravity wave as it passes another reference point. This Aichelburg-Sexl ultraboost has a plane wave pulse of spacetime. The relativistic mass will result in a gravity wave pulse as detected by a stationary observer. So there is a gravitational implication to such extreme relativistic boosts.

Answer 5

None of the existing answers mention one of the most salient points, which is that Newtonian gravity is only an approximation valid at low speeds. When relativistic considerations enter (as they certainly do for a mass moving close to light speeds) then the relativistic theory of gravity, i.e. general relativity, must be used.

The source of gravity in GR is not just mass but the stress energy tensor, which includes terms for energy, momentum, and pressure. In a sense the momentum terms for a moving object cancel out the extra kinetic energy from motion.

Answer 6

It can be noted that particles (cosmic rays) pass the Earth at close to the speed of light every day. From their point of view, the Earth IS moving at very close to the speed of light, yet clearly, the Earth does not turn into a black hole. That is the short answer, but how do we prove that is what the equations of relativity predict when the object appears to be moving close to the speed of light?

In Newtonian physics, inertial mass and gravitational mass are equal, and Newton considered this a mystery as he could find no reason why this should be the case. Things are slightly different in relativity. The gravitational formula is:

$$F = \frac{GMm}{R^2}$$
where $M$ is the active gravitational mass (the attracting mass), and $m$ is the passive inertial mass (which is the proportionality factor between how a particle accelerates relative to the force applied to it). The relativistic equation for force transverse to the motion is:

$$F'_{\perp} = \frac{G(M\gamma^{-2})(m\gamma)}{R^2}$$

where $\gamma$ is the usual relativistic gamma factor equal to $1/\sqrt{1-v^2/c^2}$ where $-v$ is the velocity of the observer orthogonal to the gravitational force vector. The end result is that the observer sees the transverse gravitational force to be weaker by a factor of gamma such that $F_{\perp}' = F/\gamma$. This is in agreement with the Lorentz transformation of transverse force and the old-fashioned concept of transverse inertial mass.

Consider the following scenario: A scientist measures the time for a mass to fall from a height L above the surface as $t_0$. He calculates the acceleration as $a_0 = L_0 t_0^{-2} = g \ \text{m/s}^2$. To an observer (Anne) moving East to West at a velocity such that the gamma factor is $\gamma = 2 $, the time to fall a vertical distance of $L$ is $2 \ t_0$ seconds due to time dilation. Using the equation of motion, $ L =1/2 \, a t^2 \iff a = 2L/t^2$, she calculates the acceleration to be $2L_0 t_0^{-2} \gamma ^{-2} = g/4 \ \text{m/s}^2 $. In other words, $a =a_0/\gamma$ and the force of gravity on the surface of the Earth appears to be weaker to the passing observer. Objects appear to fall in slow motion.

To Anne, the passive inertial mass of the falling test object has increased by a factor of gamma. We can now calculate the force acting on the test object with a rest mass of $m_0$ and acceleration $a_0$ in the rest frame as $F = m a = (m_0 \gamma )(a_0 \gamma^{-2}) = F_0 \gamma^{-1}$ in agreement with the result obtained earlier. For a large object like the Earth to become a black hole, it has to be compressed by a force. I have shown that the transverse compression force is actually weaker from the point of view of an observer who sees the system as moving with relativistic velocity. The Lorentz transformation also tells us the compression force parallel to the motion is the same as the force measured in the rest frame of the system, so, overall, there is no increased compression force. The proper volume of the Earth remains the same in its rest frame, so there is no increase in the mass density, so there is no reason to think the Earth should become a black hole as a result of being accelerated to relativistic velocities or equivalently, as a result of an observer passing by at close to the speed of light. It is not predicted by the equations of relativity.