Neutrinos Interaction with EH of Black Holes


by PVastro
Tags: black, holes, interaction, neutrinos
PVastro
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#1
Feb16-12, 05:39 PM
P: 24
Hi all,
While the articles I have been reading are about a year old (which can be considered relatively outdated in terms of particle physics) I have been looking at the possibility of Neutrinos being able to travel at FTL (Experiments conducted by CERN and OPERA). I understand in the articles how it states the difficulty in proving such a claim (due to measurement limitations, and possible systematic errors) but the concept of something obtaining FTL has made me think of an interesting concept. It is my understanding that when anything enters the EH of a Black hole nothing can escape (including light). With this in mind I suppose I need clarify parts of what im about to ask.

1) IF something can travel FTL does that mean it could "escape" the EH of a Black Hole or would it still not be enough (if thats even possible to mathematically determine)?

2) IF FTL is enough to "escape" the gravitational tide of the EH then would that suggest neutrinos have the ability to "pass" through a BH or just warp around it (Or if neither how would they interact)?

3) Lastly if FTL is possible then what would that entail for a neutrinos "timescale" in the EH (This last question may be bunk due to my lack of understanding of the function of time regarding the EH of a BH)

I understand that this topic is under quite a bit of dispute in the world of particle physics (since it would practically crush Special Relativity), but even if you do NOT support the claim of neutrinos traveling FTL please at least consider the question posed as more of a thought problem, or at least enlighten me as to why it would not be possible.

Thanks!
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Chronos
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#2
Feb16-12, 10:38 PM
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So far as we know, a BH does not know about the speed of light barrier. More mass = higher escape velocity. The problem is with neutrino emissions - just what would be emitting neutrinos from inside the EH?
PVastro
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#3
Feb16-12, 10:55 PM
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Quote Quote by Chronos View Post
So far as we know, a BH does not know about the speed of light barrier. More mass = higher escape velocity. The problem is with neutrino emissions - just what would be emitting neutrinos from inside the EH?
Again this falls under me having only a basic amateur understanding of this subject but i was under the impression that neutrinos were present relatively "everywhere" where there is matter (again i could be completely wrong) but as far as the neutrinos being present in the EH:
1) Would they have to be emitted FROM inside the EH or could they be traveling INTO it the EH from another direction?
2) If the above statement isn't possible then would Hawking radiation explain where they could be emitted from?

twofish-quant
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#4
Feb17-12, 03:22 AM
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Neutrinos Interaction with EH of Black Holes


Quote Quote by PVastro View Post
HI understand in the articles how it states the difficulty in proving such a claim (due to measurement limitations, and possible systematic errors) but the concept of something obtaining FTL has made me think of an interesting concept. It is my understanding that when anything enters the EH of a Black hole nothing can escape (including light). With this in mind I suppose I need clarify parts of what im about to ask.
The answer to all of these questions is "don't know yet."

If it turns out that neutrinos are traveling FTL, then at that point we'd have to rewrite all of the old models, and how to rewrite them depends on how exactly the neutrinos are traveling.

Even if you do NOT support the claim of neutrinos traveling FTL please at least consider the question posed as more of a thought problem, or at least enlighten me as to why it would not be possible.
Once we have figured out that are models are wrong, then the next questions is "how wrong are they" and it's only after we come up with new models can we figure out the implications on black holes. It's possible that if we have FTL neutrinos that black holes are not possible.

Right now theorists are waiting for more information on what exactly is going on. If it turns out that neutrinos are traveling FTL, we need information on exactly how they are traveling FTL.

One possibility is that it's a local effect. We think based on neutrinos from 1987A, that distant neutrinos aren't travelling FTL, so it's possible that it only happens under rare circumstances that don't apply to black holes. Or not.
Chalnoth
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#5
Feb17-12, 03:58 AM
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Quote Quote by PVastro View Post
1) IF something can travel FTL does that mean it could "escape" the EH of a Black Hole or would it still not be enough (if thats even possible to mathematically determine)?
Depends upon how much faster. In the extraordinarily unlikely event that neutrinos actually do travel faster than light, the most likely situation is that they have some other maximum speed that is somewhat greater than c, call it c'. This new speed limit would cause there to be a second event horizon for black holes just inside the one for light. Objects that fall through the first event horizon and then emit neutrinos could have those neutrinos escape, until the object falls through the second event horizon. Then even neutrinos couldn't escape (due to the fact that they can't travel faster than c').
phinds
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#6
Feb17-12, 05:05 AM
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I think one of the things that some of the posters in this thread are missing is the following:

The event horizon of a black hole is the boundary point that separates where thing that travel at light speed can/cannot escape the gravitational pull of the BH. Just outside the EH, photons can escape, just inside the BH they cannot.

What has not been pointed out (although it is a direct result of what Chronos correctly stated in post #3) is that WAY inside the EH, the escape velocity is WAY more than the speed of light (just as way OUTSIDE the EH, it is way LESS than the speed of light).

Escape velocity is a continuous function of the depth of the gravity well that an object is in. The EH of a BH is just a point on that curve, not a discontinuity.
Drakkith
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#7
Feb17-12, 05:25 AM
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Quote Quote by phinds View Post
Escape velocity is a continuous function of the depth of the gravity well that an object is in. The EH of a BH is just a point on that curve, not a discontinuity.
That's what I was thinking earlier myself, I just wasn't sure.
lpetrich
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#8
Feb19-12, 09:47 PM
P: 514
It's easy to find the tachyonic versions of Schwarzschild geodesics - Wikipedia, the free encyclopedia

It's easiest with affine parameter [itex]\lambda[/itex]:

[itex] \tau = m\lambda ,\ \frac{dt}{d\lambda} = E \left( 1 - \frac{2M}{r} \right)^{-1} ,\ \frac{d\phi}{d\lambda} = \frac{L}{r^2} ,\ \left(\frac{dr}{d\lambda}\right)^2 = E^2 - \left( 1 - \frac{2M}{r} \right) \left( m^2 + \frac{L^2}{r^2} \right) [/itex]

The proper time, the coordinate time, the angle in the orbit plane, the radius.
All in the usual Schwarzschild coordinates.
Test-particle mass: m, Energy: E, Angular momentum: L
Source mass: M

Going to the tachyonic case, replace the mass m by i*m, and the proper time by i * proper distance. The affine parameter stays real.

[itex] s = m\lambda ,\ \frac{dt}{d\lambda} = E \left( 1 - \frac{2M}{r} \right)^{-1} ,\ \frac{d\phi}{d\lambda} = \frac{L}{r^2} ,\ \left(\frac{dr}{d\lambda}\right)^2 = E^2 + \left( 1 - \frac{2M}{r} \right) \left( m^2 - \frac{L^2}{r^2} \right) [/itex]

When observed from outside, any particle will take an infinite amount of time to reach the event horizon. Inside the event horizon, a tachyon can bounce and go outward again if its angular momentum is low enough.
PVastro
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#9
Feb19-12, 10:15 PM
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Quote Quote by lpetrich View Post
It's easy to find the tachyonic versions of Schwarzschild geodesics - Wikipedia, the free encyclopedia

It's easiest with affine parameter [itex]\lambda[/itex]:

[itex] \tau = m\lambda ,\ \frac{dt}{d\lambda} = E \left( 1 - \frac{2M}{r} \right)^{-1} ,\ \frac{d\phi}{d\lambda} = \frac{L}{r^2} ,\ \left(\frac{dr}{d\lambda}\right)^2 = E^2 - \left( 1 - \frac{2M}{r} \right) \left( m^2 + \frac{L^2}{r^2} \right) [/itex]

The proper time, the coordinate time, the angle in the orbit plane, the radius.
All in the usual Schwarzschild coordinates.
Test-particle mass: m, Energy: E, Angular momentum: L
Source mass: M

Going to the tachyonic case, replace the mass m by i*m, and the proper time by i * proper distance. The affine parameter stays real.

[itex] s = m\lambda ,\ \frac{dt}{d\lambda} = E \left( 1 - \frac{2M}{r} \right)^{-1} ,\ \frac{d\phi}{d\lambda} = \frac{L}{r^2} ,\ \left(\frac{dr}{d\lambda}\right)^2 = E^2 + \left( 1 - \frac{2M}{r} \right) \left( m^2 - \frac{L^2}{r^2} \right) [/itex]

When observed from outside, any particle will take an infinite amount of time to reach the event horizon. Inside the event horizon, a tachyon can bounce and go outward again if its angular momentum is low enough.
Thanks the wiki article helped sum it up fairly well as well as your last statement... unfortunately due to my incredible lack of understanding of math (especially particle physics) your posted formulas completely escape me sorry . But when you mean a tachyon can bounce back out due to its angular momentum how does that give it the ability to escape the EH? (Again your equations probably explained it but that may as well be a different language to me).
lpetrich
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#10
Feb19-12, 11:19 PM
P: 514
I'll do the radial-infall case, since it's the easiest. The event horizon is at r = 2M. Outside the event horizon, (dr/dl) > E, while inside it, (dr/dl) < E, and for some value of r, it is zero. That's the bounce point. When the particle gets there, it turns around and retraces its inward path, and it passes the event horizon as it goes outward.
PVastro
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#11
Feb20-12, 12:40 AM
P: 24
I see well I think I understand it better but when you say it bounces and retraces its path geometrically (at least as far as I can "picture it") it suggests that it partially goes in it and then goes back out the exact same path or more more "mirroring" its own path as it travels through the EH. Again this falls under my lack of understanding of the math again but in terms of "shape" of the EH is it more "spherical" or a disk? I only picture is being a disk due to the fact that it seems to hit a "bounce" point which makes me picture it "hitting" off something (again most likely a complete lack of understanding of the math and I could just be reading to much into the term bounce).
Chalnoth
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#12
Feb20-12, 03:29 AM
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Quote Quote by lpetrich View Post
It's easy to find the tachyonic versions of Schwarzschild geodesics - Wikipedia, the free encyclopedia

It's easiest with affine parameter [itex]\lambda[/itex]:

[itex] \tau = m\lambda ,\ \frac{dt}{d\lambda} = E \left( 1 - \frac{2M}{r} \right)^{-1} ,\ \frac{d\phi}{d\lambda} = \frac{L}{r^2} ,\ \left(\frac{dr}{d\lambda}\right)^2 = E^2 - \left( 1 - \frac{2M}{r} \right) \left( m^2 + \frac{L^2}{r^2} \right) [/itex]

The proper time, the coordinate time, the angle in the orbit plane, the radius.
All in the usual Schwarzschild coordinates.
Test-particle mass: m, Energy: E, Angular momentum: L
Source mass: M

Going to the tachyonic case, replace the mass m by i*m, and the proper time by i * proper distance. The affine parameter stays real.

[itex] s = m\lambda ,\ \frac{dt}{d\lambda} = E \left( 1 - \frac{2M}{r} \right)^{-1} ,\ \frac{d\phi}{d\lambda} = \frac{L}{r^2} ,\ \left(\frac{dr}{d\lambda}\right)^2 = E^2 + \left( 1 - \frac{2M}{r} \right) \left( m^2 - \frac{L^2}{r^2} \right) [/itex]

When observed from outside, any particle will take an infinite amount of time to reach the event horizon. Inside the event horizon, a tachyon can bounce and go outward again if its angular momentum is low enough.
I'll just say that it's not possible for neutrinos to be tachyons. If they were tachyons, then neutrino interactions would lead to a continuous growth of energy, basically causing the universe to explode.

The only remote possibility for FTL neutrinos is that they have a different speed limit than c.


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