# Beat the speed of light

by henrywang
Tags: beat, light, speed, speed of light
 P: 13 thanks guys!
 P: 13 Awesome, btw i missunderstood the concept of the higgs boson then, since i understood it did not had a mass since is all around the universe dispersed. Probably completely wrong... lol anyhow thanks
Thanks
P: 3,660
 Quote by danshawen I am in agreement with the first answer; Nothing (energy or matter) can travel faster than the speed of light, period. But then you asked "Why?" Well, there's a much better answer to that follow-up question, ever since July 4, 2012.
The Higgs boson has nothing to do with this "why?" question.

 E=mc^2, and that particular derivation by Einstein has only one assumption. It is that the speed of light is invariant and cannot be exceeded by matter or energy.
There are two assumptions, and neither has anything to do with exceeding light speed. You don't have to take my word for it - Einstein's 1905 Special Relativity paper is online and you'll find it pretty quickly if you google for "on the electrodynamics of moving bodies".
Mentor
P: 11,831
 Quote by danshawen But then you asked "Why?" Well, there's a much better answer to that follow-up question, ever since July 4, 2012.
No. The Higgs mechanism is not a reason for special relativity. It is a reason why particles have mass, but even without mass they could not exceed the speed of light, and this has nothing to do with the Higgs mechanism.

 The Higgs mechanism does this without violating conservation of energy by slowing these particles down (from the speed of light).
The particles do not "slow down" as they never travel at the speed of light.
 The Higgs boson also imparts mass to itself, but decays into photons in about 10^-21 seconds in the LHC.
The decay to two photons is possible and interesting for the observation, but it is quite rare (~.3%).

 Heck, the LHC couldn't even have been designed without E=mc^2, and that particular derivation by Einstein has only one assumption. It is that the speed of light is invariant and cannot be exceeded by matter or energy.
I hope the LHC is not based on a wrong (or at least misleading, if quoted like this) equation! See Nugatory's reply for details.

 Quote by repollo BTW: Not about the subject in question but I wonder if two higgs boson can actually collide and wonder what would happen? Super fast telecommunications maybe, anyone?
They could interact and produce other particles, but that process is really unlikely.
I have absolutely no idea how you could think this would have any relation to telecommunications.

 Quote by repollo Awesome, btw i missunderstood the concept of the higgs boson then, since i understood it did not had a mass since is all around the universe dispersed. Probably completely wrong... lol anyhow thanks
What you mean is the Higgs field - this is everywhere. Higgs bosons are rare.
 P: 13 Thanks for the info, I got to the idea of telecommunications with the misleaded(?) idea that higgs bosons can travel faster than light and that the higgs field has different potential depending on the higgs boson quantitative energies imparting on the higgs field. And by now im not sure if Im talking about bosons or whales... Anyhow thanks for the clarification, is difficult to keep up with stuff I cant practice, since right now I dont have the money to make a LHC on my moms basement. ;) And yes i was confusing the higgs field with the bosons.
P: 30
 Quote by phinds No. Nothing travels faster than c. Period.
I have a question about neutrinos in this respect.

Understanding from relativity is that if something has mass, not only can it not travel faster than c, but it must travel at less than c (i.e. cannot even travel at c). Neutrinos qualify as having mass.

However, all recent experiments (since the FTL neutrino imbroglio) seem to report that speed of neutrinos are 'consistent with c', not less than c. They are not able to establish that the speed is less than c.

What is the current interpretation of this result? There may be an underlying assumption that perhaps neutrinos travel very, very close to c but not at c (as GR would require). However, given the level of accuracy of current experiments, and given that so many have been conducted independently, why is the statement still being made 'consistent with c within the margin of error' rather than 'very close to but just short of c within the margin of error'?
 Mentor P: 11,831 The expected difference between neutrino speeds and the speed of light is extremely tiny - several orders of magnitude below the experimental uncertainties of all current and even all planned experiments. "Consistent with c" and "consistent with the predicted speed slightly below c" are the same.
 P: 72 On the other hand, galaxies move away from each other faster than c due to expansion of the universe. While the Dark Energy increases this expansion rate, billions of years from now, if not now there would even be matter moving away from us not only above c but with an acceleration rate over c.. But this is quite a different situation. As in, an object can not move next to another object with a speed faster than c. The inflation is taking affect only when the two objects is away from each other. Around 4,200 megaparsecs away. And that is quite away from each other. Enough that they can't share any information in any form anyway.
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P: 6,283
 Quote by ExecNight On the other hand, galaxies move away from each other faster than c due to expansion of the Around 4,200 megaparsecs away. And that is quite away from each other. Enough that they can't share any information in any form anyway.
That is wrong in two regards. First, expansion takes place on a much smaller scale. 4,200 parsecs is more than enough, to say nothing of 4,200 MEGAparsecs. Second, even objects which are 4,200 megaparsecs away from us are sharing light (information) with us right now. In fact, objects at the edge of our observable universe are something like 14,000 megaparsecs away and we can still see them.
 P: 72 It's true we do receive light from such galaxies. Yet, the information is still of its past self that haven't crossed the c threshold for us yet. The updating of the light will cease and the galaxies we now see will freeze in space time and then red shift into darkness. In the end it will be a lonely place in this universe as the expansion rate increases, as every object will redshift to darkness except close objects. Anyway, i don't know what OP was trying to get at, but there are objects in this universe when you take reference point as earth they accelerate away from that point faster than c. How that information is useful is beyond me though.
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P: 6,283
 Quote by ExecNight The updating of the light will cease and the galaxies we now see will freeze in space time and then red shift into darkness.
I agree that they will red shift into darkness but I have no idea why you think they will "freeze in space time" since they won't.

 How that information is useful is beyond me though.
Agreed, but the point I was addressing had nothing to do with whether or not the information was useful just whether or not it gets here.
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P: 11,831
 Quote by phinds That is wrong in two regards. First, expansion takes place on a much smaller scale. 4,200 parsecs is more than enough, to say nothing of 4,200 MEGAparsecs. Second, even objects which are 4,200 megaparsecs away from us are sharing light (information) with us right now. In fact, pbjects at the edge of our observable universe are something like 14,000 megaparsecs away and we can still see them.
4200 parsecs away is still within our galaxy, where expansion does not happen.
We can see objects 4,200 Mpc away, but only in a state how they looked like several billion years ago. The border where we will never be able to see their current state is somewhere at this distance. They don't freeze in spacetime, but our view on them will freeze.
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P: 6,283
 Quote by mfb 4200 parsecs away is still within our galaxy, where expansion does not happen.
OOPS. My bad. Thanks for that correction. I can add, but I can't multiply

 We can see objects 4,200 Mpc away, but only in a state how they looked like several billion years ago. The border where we will never be able to see their current state is somewhere at this distance. They don't freeze in spacetime, but our view on them will freeze.
Hm ... I don't follow. How does our view of them freeze? Wouldn't they just fade into darkness with greater and greater redshift?
 P: 72 I would expect them to freeze and redshift into darkness because; Well for conservation of information i would expect them to act like an object falling into the black holes event horizon. Otherwise, that would raise many questions. Like Stephen Hawking did back then
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P: 11,831
 Quote by phinds Hm ... I don't follow. How does our view of them freeze? Wouldn't they just fade into darkness with greater and greater redshift?
Into darkness, but also into slower evolution (as seen by us) due to the redshift. The effect is very similar to objects falling into black holes (as seen by outside observers), just on a completely different timescale.
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P: 6,283
 Quote by mfb Into darkness, but also into slower evolution (as seen by us) due to the redshift. The effect is very similar to objects falling into black holes (as seen by outside observers), just on a completely different timescale.
OK, that I understand. I think the fading to darkness would occur before the "freezing" got too severe, but I guess you could say that depends on the sensitivity of the instruments "seeing" the objects.
P: 193
 Quote by henrywang If a a object is falling in a gravity field with infinitly long radius. can it eventually travel faster than the speed of light?
No, the coordinate speed of a test probe falling from infinity is:

$$v=c(1-\frac{r_s}{r}) \sqrt{\frac{r_s}{r}}$$ for $r>r_s$

where $r_s$ is the Schwarzschild radius of the "attracting" gravitational mass and $r$ is the radial Schwarzschild coordinate. So, $v<c$ for all $r>r_s$.

If the test probe is dropped from $r_0$ the formula becomes:

$$v=c(1-\frac{r_s}{r}) \sqrt{\frac{r_s}{r}-\frac{r_s}{r_0}}$$ for $r_0>r>r_s$

For light, the coordinate speed is:

$$v=c(1-\frac{r_s}{r})$$

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