Message boards : LHC@home Science : If Neutrinos have no mass, can they escape a black hole?
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klasm

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Message 9630 - Posted: 24 Aug 2005, 16:52:21 UTC

>May'be the answer is in here. At least i think it is.
>Basically it states that however mass(less) objects cannot "escape" black holes, >black holes still do radiate energy (also in the form of neutrino's).
>It all seems contradictory, but aren't most things in quantum physics. ???

Black holes do indeed radiate energy but that does not contradict the fact that anything inside the event horison is trapped. The Hawking-radiation is caused by the quatum mechanical properties of the vacuum outside the horison. So the particles that radiates out from the black hole does not come from inside the horison but are created in the vacuum outside the event horison. If you want to read more detailed descriptions of this mechanim try gooling for "Hawking raditaion"

One of the intersting properties of hawking radiation is that black holes becomes "hotter" the smaller they are, ie they radiate more energy per time unit. Really small black holes, with mass like an elementary particle, will lose all their energy that way in a very small explosion.
Depending on the some unkown aspects of the underlying physics there is some chance that the LHC wil lactualyl be able to produce tihs kind of miniature black holes and observe their evaporation.
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Message 9651 - Posted: 25 Aug 2005, 13:21:17 UTC

The key here is the quantum uncertainty principle: dE*dt >= h/4*pi

This means that out of nothing you can create a pair of virtual particles (an ordinary particle and its anti particle). As long as they annihilate before anyone has time to notice the energy discrepancy. This happens all the time. Also near the surface of black holes. What can happen is that one of the particles of the pair might enter the black holes event horizon. The other particle is therefor no longer virtual it can no longer disappear and it is outside the black hole and therefore free to fly away. Since the black hole have absorbed the coresponding particle it seems like the particle was created by the black hole.


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klasm

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Message 9652 - Posted: 25 Aug 2005, 13:34:36 UTC

...and by doing so the black hole loses an amount of energy correspoding to the absorbed particle rather than gainging it. So the black hole loses part of its energy/mass in the creation of the the particle which flies away.
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Profile Geoffrey Voeth

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Message 12826 - Posted: 22 Feb 2006, 3:03:51 UTC - in response to Message 8726.  

This is most probably getting off the subject at hand but is it possible to stretch a photons wavelength so that it becomes a DC kind of value and if so how would that be possible ?
That red shift of distant galaxies might not simply be a doppler shift it might also be some sort of decay at work ??
Given enough time and space at what point would the red shift turn into a DC value ?


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Message 12837 - Posted: 22 Feb 2006, 20:40:12 UTC - in response to Message 12826.  
Last modified: 22 Feb 2006, 20:43:44 UTC

This is most probably getting off the subject at hand but is it possible to stretch a photons wavelength so that it becomes a DC kind of value and if so how would that be possible ?

It is well off topic, and restarting a long dead thread.

It is also an interesting question, so I will answer it in a new thread, here
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Message 15830 - Posted: 19 Dec 2006, 1:26:55 UTC - in response to Message 9612.  

>What restricts the massless particles to do not travel faster than light?
>Black holes attributes are mass, charge and angular momentum ... so when a >particle is massless and has no charge, what restricts its movement?

This restriction is not because of the presence of tha black hole. Special relativity tells us that anything with mass must travel slowe than the speed of light, and that massless particles can travel atmost at the speed of light.



I know this is probably a stupid question, but if photons have energy, and energy is mass, and anything with mass must travel slower than speed of light, doesnt that mean that light itself moves slower than the speed of light?

And wouldnt that mean that it then just keeps slowing down, until it doesnt move anymore, and vegetates on the sofa watching tv and drinking beer? :-p


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Message 15954 - Posted: 31 Dec 2006, 18:48:56 UTC

..and what about the tachyon?
Be cool. Join Team Tuna Melt.
http://groups.myspace.com/teamtunamelt
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Message 16015 - Posted: 3 Jan 2007, 22:55:58 UTC - in response to Message 15830.  

I know this is probably a stupid question, but if photons have energy, and energy is mass, and anything with mass must travel slower than speed of light, doesnt that mean that light itself moves slower than the speed of light?


Here's a link that might help clear up the issue:
http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

Short version is: There's mass, and there's mass. Don't confuse the two :)
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Message 16391 - Posted: 22 Feb 2007, 3:08:02 UTC - in response to Message 6789.  

That´s not right, because a photon has mass, not much but it have (see neutrino has alreadyy a little mass). Only particles with absolutely no mass (i don´t know a particle with no mass) shouldn´t affect by a black hole.



I bet they could. (neutrinios escape a black hole, that is)


http://www.telegraph.co.uk/connected/main.jhtml?xml=/connected/2005/03/30/ecfmice30.xml&sSheet=/connected/2005/03/30/ixconnrite.html

I'm not saying that I can 'prove it' using physics, hand gestures or a combination of physics and handwaving. But, if they have 'no mass' then the black hole gravity shouldn't really affect them.

So, experimentally, a neutrino is a theoretical particle which makes heavy water give of a flash of light. Neutrinos might have an issue of going throught the ground up atomic/subatomic matter inside a black hole, but I suspect that the 'no mass' aspect of neutrinos makes them great at escaping the gravity from a black hole.


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Message 16428 - Posted: 28 Feb 2007, 7:09:36 UTC - in response to Message 16391.  

That´s not right, because a photon has mass, not much but it have (see neutrino has alreadyy a little mass). Only particles with absolutely no mass (i don´t know a particle with no mass) shouldn´t affect by a black hole.



I bet they could. (neutrinios escape a black hole, that is)


http://www.telegraph.co.uk/connected/main.jhtml?xml=/connected/2005/03/30/ecfmice30.xml&sSheet=/connected/2005/03/30/ixconnrite.html

I'm not saying that I can 'prove it' using physics, hand gestures or a combination of physics and handwaving. But, if they have 'no mass' then the black hole gravity shouldn't really affect them.

So, experimentally, a neutrino is a theoretical particle which makes heavy water give of a flash of light. Neutrinos might have an issue of going throught the ground up atomic/subatomic matter inside a black hole, but I suspect that the 'no mass' aspect of neutrinos makes them great at escaping the gravity from a black hole.



You are oh so wrong.

Photons are massless. Theory demands it and it is consistent with experiment. The smallest upper bound on photon mass is something on the order of 10^-17 eV.

Neutrinos are massive - all three of them. WMAP data places the sum of the neutrino masses at about three quarters of an electron volt.

pdg.lbl.gov is a great place to start looking.

Gravity couples to any kind of energy. Mass is not required. Any particle - massive or otherwise - will be affected by a black hole. Unlearn what you learned about Newtonian gravity, it is incorrect.

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Daxa

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Message 16429 - Posted: 28 Feb 2007, 7:50:47 UTC
Last modified: 28 Feb 2007, 8:09:51 UTC

Photons only have "mass" when in relative motion to the observer due to their energy [E=mc^2 --- m=E/c^2]; otherwise they wouldn't "exist" at all. Electrons still have mass even when at rest (except, of course, quantum-level motion) relative to the observer. This was demonstrated by Millikan's Oil Drop Experiment. Electrons at rest can be easily observed as common static electricity.

So here is my question:
Can a neutrino be observed at rest relative to the observer, or is it like a photon in the respect that it must be in motion to be observed. Also, if a neutrino does have be in a state of relative motion to be observer, is that speed c, as with the photon? I think the answers to these questions, which I don't know, may help answer the answer the neutrino / black hole quagmire.


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Daxa

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Message 16432 - Posted: 28 Feb 2007, 12:16:36 UTC - in response to Message 16429.  
Last modified: 28 Feb 2007, 12:52:43 UTC

Sorry about the typos. This is regarding the neutrino question:

Consider photons: They have "mass" when in relative motion to the observer due to their energy [E=mc^2 --- m=E/c^2]; otherwise they wouldn't "exist" at all. In contrast, electrons have mass even when at rest (except, of course, quantum-level motion) relative to the observer. This was demonstrated by Millikan's Oil Drop Experiment. Electrons at rest can be easily observed as common static electricity.

So here is my question:
Can a neutrino be observed at rest relative to the observer, or is it like a photon in that it must be in motion to be observed. Also, if a neutrino does have be in a state of relative motion to be observed, is that speed c, as with the photon? I think the answers to these questions, which I don't know, may help answer the neutrino / black hole quagmire.

Feel free to flame me. I'm only posing as a physicist.


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Message 16441 - Posted: 1 Mar 2007, 7:28:23 UTC

Because the neutrino has a finite mass there exists a system where it is in rest to the observer, which has a speed smaller than c. (in comparison to the photon or any other massless particle, which always has the speed c to the observer)
Consider how neutrinos are produced, e.g. the beta decay - its an threebody reaction n -> p e- nybar so there is the possibility that the neutrino doesn't pick up any impulse in this reaction, so it will be in rest. This is actually the way KATRIN (Karlsruhe Tritium Neutrino Experimen) wants to measure the neutrino mass with the tritium beta decay.
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Message 16447 - Posted: 2 Mar 2007, 1:24:49 UTC - in response to Message 16429.  

Photons only have "mass" when in relative motion to the observer due to their energy [E=mc^2 --- m=E/c^2]; otherwise they wouldn't "exist" at all. Electrons still have mass even when at rest (except, of course, quantum-level motion) relative to the observer. This was demonstrated by Millikan's Oil Drop Experiment. Electrons at rest can be easily observed as common static electricity.


Many o' misconceptions on this fine day.

1) m = 0 for photons. Period.

The full energy-momentum relation in special relativity is E^2 = [pc]^2 + [mc^2]^2, not E = mc^2. So please don't argue that photons have mass through E=mc^2.

Read this page. This details the theory and consequences of a massive photon.

http://www.phys.lsu.edu/students/kristina/PhMass/PhMass.html

2) Millikan's oil drop experiment showed that charge is quantized. It had nothing to do with the mass of the electron.
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Daxa

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Message 16448 - Posted: 2 Mar 2007, 3:49:18 UTC - in response to Message 16447.  
Last modified: 2 Mar 2007, 4:01:47 UTC

Millikan's oil drop experiment showed that charge is quantized. It had nothing to do with the mass of the electron.
Touché. You have proven that my memory of physics curriculum is rusty.

However, I do sense that you you believe that mass and energy are well separated concepts. That, I am still confident, is shaky ground. Consider wave-particle duality; it is clear that these concepts are NOT well separated. the Schrödinger equation, the de Broglie hypothesis, and the Heisenberg uncertainty principle all point to this idea.

But how does wave-particle duality relate to mass-energy duality? Notice that in modern physics, the terms "mass, velocity" and even "energy" are often replaced by the concept of momentum. Momentum allows us to side-step the problems of thinking about particle 'x' having mass, velocity, energy, or some combination of these. It's a handy theoretical (and for now, mathematical) shortcut. Your full energy-momentum relation in special relativity was certainly a shot in the arm for me, but is there a mass-momentum relation in special relativity? Enlighten me if you have the time; I'm eager to learn.
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Message 16655 - Posted: 1 Apr 2007, 17:40:24 UTC - in response to Message 16447.  

> Read this page. This details the theory and consequences of a massive photon.
>
> http://www.phys.lsu.edu/students/kristina/PhMass/PhMass.html
>
Very interesting page (except for that the equations where nearly unreadable). However this gives an idea for a speculation. If the photon has a rest mass, then perhaps we could also have three types of photons, just like there are three types of neutrinos. As the muon and tauon photons are likely to be
very rare, i think the measurments described above is not possible to do
for them. But the rest-mass for the muon and taoun photons may be much
larger than for the electron-photon, so perhaps it is possible to
detect in some other experiment.

Any ideas on this??

/Thord.


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David C.

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Message 17459 - Posted: 20 Jul 2007, 19:22:44 UTC - in response to Message 16448.  

Millikan's oil drop experiment showed that charge is quantized. It had nothing to do with the mass of the electron.
Touché. You have proven that my memory of physics curriculum is rusty.

However, I do sense that you you believe that mass and energy are well separated concepts. That, I am still confident, is shaky ground. Consider wave-particle duality; it is clear that these concepts are NOT well separated. the Schrödinger equation, the de Broglie hypothesis, and the Heisenberg uncertainty principle all point to this idea.

But how does wave-particle duality relate to mass-energy duality? Notice that in modern physics, the terms "mass, velocity" and even "energy" are often replaced by the concept of momentum. Momentum allows us to side-step the problems of thinking about particle 'x' having mass, velocity, energy, or some combination of these. It's a handy theoretical (and for now, mathematical) shortcut. Your full energy-momentum relation in special relativity was certainly a shot in the arm for me, but is there a mass-momentum relation in special relativity? Enlighten me if you have the time; I'm eager to learn.


Your sense about him believing mass and energy are well separated is incorrect. He's simply stating one of the basic equations of special relativity, which is, again

E^2 - p^2c^2 = m^2c^4

where m denotes the invariant or rest mass. A photon has no rest mass, so the equation reduces to

E = pc

where p is the momentum of the photon. Even though the photon has no rest mass, that doesn't make it an exception to general relativity. In fact, because the photon has no invariant mass, it will travel along a perfect geodesic since it does not distort spacetime.

To answer your other question, yes, there is a mass-momentum relationship in special relativity, but it only applies to particles that actually have mass (read: invariant or rest mass).

It is:

p = mv/Sqrt[1-v^2/c^2]

I am unsure if you are familiar with the concept of a Taylor series, but it is a common technique used in physics to approximate an equation about some value by an infinite series. The idea is to find a polynomial whose derivatives all match a given function at a point of interest. The upshot is that the more terms you add, the better the approximation is, and for small perturbations, the approximation is extremely accurate. In this case, we can take the above expression and expand it into this series:

p = mv + mv^3/(2c^2) + 3mv^5/(8c^4) + O(v^6)

where the O term denotes whatever higher order terms I have left out in this expansion (the higher the order, typically the less important the term). The interesting point is that we see the first term (typically the most important) is mv, which is the classical expression for momentum, so we know then that the special relativity relation for momentum-rest mass reduces to what we classically expect in the small velocity regime v << c.
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Message 17461 - Posted: 20 Jul 2007, 19:58:02 UTC - in response to Message 17459.  
Last modified: 20 Jul 2007, 20:10:44 UTC

...stating one of the basic equations of special relativity, which is, again

E^2 - p^2c^2 = m^2c^4

where m denotes the invariant or rest mass. A photon has no rest mass, so the equation reduces to

E = pc

where p is the momentum of the photon. Even though the photon has no rest mass, that doesn't make it an exception to general relativity. In fact, because the photon has no invariant mass, it will travel along a perfect geodesic since it does not distort spacetime.

To answer your other question, yes, there is a mass-momentum relationship in special relativity, but it only applies to particles that actually have mass (read: invariant or rest mass).

It is:

p = mv/Sqrt[1-v^2/c^2]

I am unsure if you are familiar with the concept of a Taylor series, but it is a common technique used in physics to approximate an equation about some value by an infinite series. The idea is to find a polynomial whose derivatives all match a given function at a point of interest. The upshot is that the more terms you add, the better the approximation is, and for small perturbations, the approximation is extremely accurate. In this case, we can take the above expression and expand it into this series:

p = mv + mv^3/(2c^2) + 3mv^5/(8c^4) + O(v^6)

where the O term denotes whatever higher order terms I have left out in this expansion (the higher the order, typically the less important the term). The interesting point is that we see the first term (typically the most important) is mv, which is the classical expression for momentum, so we know then that the special relativity relation for momentum-rest mass reduces to what we classically expect in the small velocity regime v << c.

Mercy. The forums (actually, "fora" being plural of "forum", if I'm correct) are getting deep again. Time for me to get an Extension Course catalog and go back to school. - - Ah bugger; I used to feel so smart here.



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Message 20729 - Posted: 31 Oct 2008, 0:51:01 UTC

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Pokud se klient pokusí alespoň zpracovat Vámi poslanou jednotku , měl by být za tuto snahu nějak odměněn !!!
S pozdravem Váš člen !!
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Message boards : LHC@home Science : If Neutrinos have no mass, can they escape a black hole?


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