[math-fun] faster than light?
Here is a press report http://phys.org/news/2012-10-physicists-special-relativity.html#jCp on paper discussing new way to treat special relativity permitting faster than light particles (tachyons). The momentum of an infinite-velocity particle is in this picture a finite and nonzero consnt depending only on the particle-type. Apparently then you can get conserved real momentum and energy and these kinds of particle could only travel faster than light, never below. I have not yet read the paper itself (behind paywall). It seems to me, though, that such tachyons could interact with exactly-lightspeed=massless particles like photons & gravitons. (Also, e.g, W-bosons have zero mass until they gain effective mass from the Higgs mechanism.) Those in turn could interact with normal particles. Thus effectively tachyons & normals could interact. That would permit transmission of information from normal matter into the past of normal matter, creating paradoxes. If such paradoxes are considered forbidden, we'd presumably still need to conclude that tachyons cannot exist.
"WS" == Warren Smith <warren.wds@gmail.com> writes:
WS> It seems to me, though, that such tachyons could interact with WS> exactly-lightspeed=massless particles like photons & gravitons. Since when do photons have mass (aka energy) == zero? :-/ If they did, their trajectories wouldn't bend around baryonic masses such as planets. And then there's that certain margin note. :) -JimC -- James Cloos <cloos@jhcloos.com> OpenPGP: 1024D/ED7DAEA6
On Sat, Oct 13, 2012 at 1:17 AM, James Cloos <cloos@jhcloos.com> wrote:
"WS" == Warren Smith <warren.wds@gmail.com> writes:
WS> It seems to me, though, that such tachyons could interact with WS> exactly-lightspeed=massless particles like photons & gravitons.
Since when do photons have mass (aka energy) == zero? :-/
E = mc^2 only holds for objects at rest. The more general statement is E^2 = (mc^2)^2 + (pc)^2 which allows for massless particles with momentum travelling at light speed. So the answer to your question is, "Since the Big Bang." -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
"MS" == Mike Stay <metaweta@gmail.com> writes:
MS> E = mc^2 only holds for objects at rest. The more general statement is MS> E^2 = (mc^2)^2 + (pc)^2 But momentum is mass * velocity, so cannot be +ve if mass == 0, so a massless particle is still energy-less. And all photons have energy. (Or is there a concept of non-finite wavelenght?) Put another way, since velocity increases the mass-energy of a particle, a photon would have to have zero mass-energy at rest also to have it at any non-zero speed, including at c. -JimC -- James Cloos <cloos@jhcloos.com> OpenPGP: 1024D/ED7DAEA6
On 10/13/2012 12:05 PM, James Cloos wrote:
"MS" == Mike Stay<metaweta@gmail.com> writes: MS> E = mc^2 only holds for objects at rest. The more general statement is MS> E^2 = (mc^2)^2 + (pc)^2
But momentum is mass * velocity,
Only in the non-relativistic approximation. Brent
so cannot be +ve if mass == 0, so a massless particle is still energy-less. And all photons have energy. (Or is there a concept of non-finite wavelenght?)
Put another way, since velocity increases the mass-energy of a particle, a photon would have to have zero mass-energy at rest also to have it at any non-zero speed, including at c.
-JimC
On Sat, Oct 13, 2012 at 12:05 PM, James Cloos <cloos@jhcloos.com> wrote:
"MS" == Mike Stay <metaweta@gmail.com> writes:
MS> E = mc^2 only holds for objects at rest. The more general statement is MS> E^2 = (mc^2)^2 + (pc)^2
But momentum is mass * velocity, so cannot be +ve if mass == 0, so a
Only for non-relativistic particles.
massless particle is still energy-less. And all photons have energy.
The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency.
(Or is there a concept of non-finite wavelenght?)
No.
Put another way, since velocity increases the mass-energy of a particle, a photon would have to have zero mass-energy at rest also to have it at any non-zero speed, including at c.
There is no rest frame for a photon. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
participants (4)
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James Cloos -
meekerdb -
Mike Stay -
Warren Smith