Kugelblitz at the speed of light misconception

[Andrea Panza]

I have a question regarding Kugelblitz black holes.
I know that they are purely theoretical, and I am perfectly fine with the matter-energy equivalence so I have no problem in assuming that concentrating a sufficient amount of energy in a certain radius might generate an event horizon. However photons do not have mass and travel at the speed of light so I have an issue with the formation of Kugelblizes potentially traveling at the speed of light.

Imagine an observer stationary with a photon source that emits a photon with energy e/2 where 'e' is the energy required to create a kugelblitz
A second source of photons with intensity e/2 is positioned in a way that an object of significant mass bends the trajectory of the photon enough that the trajectory becomes parallel to the trajectory of the first photon (the first trajectory is the geometric tangent to the second trajectory).

If we emit the photons in the way that they meet at the tangent point when their trajectories are parallel they should form a Kugelblitz that moves away from the initial observer at the speed of light, which obviously does not make sense.

I hope you can clarify the issue

 

[Ibix]

Quick answer: the $(t,x)$ components of the four momentum of one of your light pulses are $(E/2,E/2)$. In the case of colliding in opposite directions the other one had $(E/2,-E/2)$, for a total four momentum of $(E,0)$ which implies a total mass (if the four momentum is $(p^t,p^x)$ the mass is $\sqrt{(p^t)^2-(p^x)^2}/c^2$) of $E/c^2$. However, in the case of parallel motion the second pulse has four momentum components $(E/2,E/2)$, for a total four momentum of $(E,E)$ which implies a total mass of zero. So I rather doubt that a black hole would form.

This analysis is somewhat naive, and should really be carried out studying the stress-energy tensor of the EM radiation. But that won't change the result. The underlying point is that two parallel light pulses are a rather different system to two anti-parallel light pulses.

 

[DrStupid]

Imagine an observer stationary with a photon source that emits a photon with energy e/2 where 'e' is the energy required to create a kugelblitz
A second source of photons with intensity e/2 is positioned in a way that an object of significant mass bends the trajectory of the photon enough that the trajectory becomes parallel to the trajectory of the first photon (the first trajectory is the geometric tangent to the second trajectory).

What happens in a frame of reference where each photon has the energy e/4? Can something different happen in another frame of reference?

 

[Andrea Panza]

Quick answer: the $(t,x)$ components of the four momentum of one of your light pulses are $(E/2,E/2)$. In the case of colliding in opposite directions the other one had $(E/2,-E/2)$, for a total four momentum of $(E,0)$ which implies a total mass (if the four momentum is $(p^t,p^x)$ the mass is $\sqrt{(p^t)^2-(p^x)^2}/c^2$) of $E/c^2$. However, in the case of parallel motion the second pulse has four momentum components $(E/2,E/2)$, for a total four momentum of $(E,E)$ which implies a total mass of zero. So I rather doubt that a black hole would form.

This analysis is somewhat naive, and should really be carried out studying the stress-energy tensor of the EM radiation. But that won't change the result. The underlying point is that two parallel light pulses are a rather different system to two anti-parallel light pulses.

Hi,
Thanks for the reply.
If I got correctly what you are saying is that in the case of parallel photon pairs the contributions to the mass cancel out.
Does this mean that in theory, we could have an infinite amount of photon pairs traveling through a vacuum tube carrying an infinite amount of energy without perturbing anything around them (obviously in space they will collide immediately with CMB photons)?

 

[pervect]

A single intense pulse of radiation won't create a black hole / kugelblitz. You'd need a pair of intense pulses, moving in opposite directions, to create the effect.

A simple relativistic boost, a change of reference frame, can change a weak pulse of light into a strong one. But such a relativistic boost, a change of reference, can't change a non-black-hole into a black-hole.

It's really better to talk about the stress-energy tensor than mass, or energy. It's the stress-energy tensor that is the source of gravity in general relativity. Einstein' field equations don't have mass, or energy, on the right hand side - what's on the right hand side is the stress-energy tensor, $T_{\mu\nu}$. The stress energy tensor gives a covariant , frame-independent way, of talking about the needed concepts.

The idea that "mass" causes gravity is basically a Newtonian idea, it gets replaced with the idea of the stress-energy tensor in general relativity.

It's oversimplified to say that if you have a pair of beams of light moving in opposite directions that there is one, and only one, frame of referene that is the "center of momentum" frame, but the idea works well enough to get the right answer in the end. But a really good treatment of the problem wouldn't use such hand-wavy popularied concepts, it'd use the stress-energy tensor and the Einstein field equations.

However, it requires a fair amount of background to do such a good treatment. One can start with Einstein's field equations, $G_{\mu\nu}$ = (some constant) $T_{\mu\nu}$. The first obstacle to using this correct treatment is that is is likely both the left hand side of the equation, representing the Einstein curvature tensor, and the right hand side of the equation, the stress-energy tensor, are likely unfamiliar. The second and bigger problem is actually doing the math, once one has the basic idea of what the stress-energy tensor is.

 

[Ibix]

If I got correctly what you are saying is that in the case of parallel photon pairs the contributions to the mass cancel out.

Not exactly. In relativity mass isn't additive - you can take two massless things and put them together and the result might have mass or not. So the light pulses don't really "contribute" to mass. They do contribute to the four-momentum of the system, and it's the way they contribute to that which leads to the system having zero mass.

Does this mean that in theory, we could have an infinite amount of photon pairs traveling through a vacuum tube carrying an infinite amount of energy without perturbing anything around them

No. Light does, in principle, have gravitational effects (you can google pp-wave spacetimes, but you probably won't find much non-technical and it's all theoretical because the energy densities needed for an experiment are silly). It just doesn't necessarily lead to a black hole just because the energy density is high. The source term for gravity in general relativity is the stress-energy tensor, which is a more complex structure than just "mass" or "energy", and can have much more complicated effects than intuition trained with Newtonian gravity might suggest.

 

[pervect]

The p-p wave space-time is an excellent way of looking at what happens.

The wiki overview at https://en.wikipedia.org/w/index.php?title=Pp-wave_spacetime&oldid=950243734 will tell the reader that the space-time associated a beam of light or other radiation is a p-p wave spacetime, though it won't get into all the technical details.

The pp-waves solutions model radiation moving at the speed of light.

I don't think that the wiki explicitly mentions that this p-p wave spacetime is not the space-time of a black hole, , so you'd have to take our word for that part without an explicit reference.

Colliding p-p waves, however, CAN form singularities (black holes). There's a paper by Tipler on this that can serve as a reference, another by Yurtsever.

Google finds a couple of (technical) papers on the topic, one by Tipler, another by Yurtsever.
https://inspirehep.net/literature/160486
https://inspirehep.net/literature/268146

I believe there is an entire book devoted to colliding p-p waves by Griffiths.

Note that Tipler's paper, that says that colliding plane p-p waves must form singularities applies to plane waves that are infinite in extent. Yurtsever's paper talks about a more realistic, non-infinite waves.

 

[PeterDonis]

I have an issue with the formation of Kugelblizes potentially traveling at the speed of light.

They don't. A Kugelblitz black hole is not a black hole that moves at the speed of light. It is a black hole that gets formed by radiation coming together in a compact enough region of spacetime to form a black hole, instead of matter doing so. Once it is formed, the black hole does not move at the speed of light; it has nonzero invariant mass and behaves like any other black hole.

 

[Andrea Panza]

They don't. A Kugelblitz black hole is not a black hole that moves at the speed of light. It is a black hole that gets formed by radiation coming together in a compact enough region of spacetime to form a black hole, instead of matter doing so. Once it is formed, the black hole does not move at the speed of light; it has nonzero invariant mass and behaves like any other black hole.

Hi Peter, thank you for your reply.
I did not explain myself properly in the original post: my issue was that matter coming together in a compact enough region is necessarily traveling below c, while radiation coming together in a given region is traveling at c.

My initial thought was that radiation must be converging to a given point so that opposite contributions to the final speed of the object will partially cancel out, but then I realized that it was possible for two beams of radiation to converge while remaining parallel (given gravitational curvature due to the presence of a massive object). I know that massive objects cannot move at c and I posted my question here to understand where I was wrong

Ibix in his reply (#2) explained to me that for parallel beams of radiation the momentum components imply that the total mass is zero

 

[PeroK]

Hi Peter, thank you for your reply.
I did not explain myself properly in the original post: my issue was that matter coming together in a compact enough region is necessarily traveling below c, while radiation coming together in a given region is traveling at c.

My initial thought was that radiation must be converging to a given point so that opposite contributions to the final speed of the object will partially cancel out, but then I realized that it was possible for two beams of radiation to converge while remaining parallel (given gravitational curvature due to the presence of a massive object). I know that massive objects cannot move at c and I posted my question here to understand where I was wrong

Ibix in his reply (#2) explained to me that for parallel beams of radiation the momentum components imply that the total mass is zero

A useful concept here is the invariant mass of a system (of particles). This is given (in $c = 1$ units) by:
$$M = \sqrt{E^2 - |\vec P|^2}$$
Where $E$ is the total energy and $\vec P$ the total momentum of the system. This quantity, as implied by the name, is the same in all reference frames.

Now, a photon or photons all moving in the same direction has zero invariant mass. But, two photons moving in opposite directions has a non-zero invariant mass. If they have the same energy, $E$, and equal and opposite momenta, then:
$$M = \sqrt 2 E$$
Note that this two-particle system does not move at the speed of light.

Likewise, if you confine radiation in a box, then the system has invariant mass through the energy of the confined radiation; and this contributes to the inertial and gravitational mass of the box.

 

[Andrea Panza]

A useful concept here is the invariant mass of a system (of particles). This is given (in $c = 1$ units) by:
$$M = \sqrt{E^2 - |\vec P|^2}$$
Where $E$ is the total energy and $\vec P$ the total momentum of the system. This quantity, as implied by the name, is the same in all reference frames.

Now, a photon or photons all moving in the same direction has zero invariant mass. But, two photons moving in opposite directions has a non-zero invariant mass. If they have the same energy, $E$, and equal and opposite momenta, then:
$$M = \sqrt 2 E$$
Note that this two-particle system does not move at the speed of light.

Likewise, if you confine radiation in a box, then the system has invariant mass through the energy of the confined radiation; and this contributes to the inertial and gravitational mass of the box.

Hi PeroK,
Thank you very much for the explanation!