86 reputation
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location El Cerrito, CA
age 35
visits member for 2 years
seen Apr 13 at 20:06

I am a Ph.D. candidate in Experimental High Energy Particle Physics at Princeton University. My dissertation is titled "A Search for Heavy Resonances Decaying to Long-Lived Exotica in the Displaced Lepton Channel".


Jul
24
answered Why does the water-pressure of my shower fall if I hold the shower head high?
Jul
10
comment Can wireless power transfer harm organic life by anyway?
You are correct. However, microwave ovens and lasers operate at much higher energy density levels than the system in the article. If the question was about wireless power transfer in general, yes, intensity is something to consider.
Jul
9
answered Can wireless power transfer harm organic life by anyway?
Jul
9
awarded  Supporter
Jul
6
answered From where this number of 14 Tev has arised for proton-proton head on collision in LHC?
Jul
6
awarded  Teacher
Jul
6
comment How do photons travel at a speed that should be impossible to attain?
Sorry, I can't tell if that was a serious question or if you are trolling. I've been giving you the benefit of the doubt so far.
Jul
6
comment How do photons travel at a speed that should be impossible to attain?
$E^2=(mc^2)^2+(pc)^2$. Since photons are massless, the equation reduces to $E=pc$. In this case, $p\neq\gamma mv$. The momentum of a photon is $p=h/\lambda$ and $\lambda>0$. Therefore, there is no divide-by-zero issue. Also, since $\lambda>0$, the energy of a photon is always finite. I should have just written that as my answer rather than trying to correct your misconception. Oh well.
Jul
6
comment How do photons travel at a speed that should be impossible to attain?
When we measure or quote the mass of a particle, we are only talking about the invariant mass. No one who actually does this for a living would ever think about a particle in terms of relativistic mass. I wrote two equations I thought would improve the clarity of your answer: $E^2=(mc^2)^2 + (pc)^2$, which would have avoided the divide-by-zero problem, and $E=\gamma mc^2$. Now that you mention it, if I were to write the momentum in terms of velocity, I would write $p=\gamma mv$. I did not state $p=mv$, nor did I write $E=mc^2$.
Jul
6
awarded  Editor
Jul
6
revised How do photons travel at a speed that should be impossible to attain?
added 25 characters in body
Jul
6
comment How do photons travel at a speed that should be impossible to attain?
The problem I have with your answer is simple. Relativistic mass is not a physical quantity. In reality, the mass of a particle doesn't not increase with Energy. E^2 = (mc^2)^2 + (pc)^2 applies to both massive and massless objects and is boost invariant. I suggest using it in the future. E=mc^2 creates confusion and misconceptions and does not apply to massless particles. I wish I could have just commented on your answer, but for some reason I don't have that ability. Since I could only create a new answer and you are not supposed to reference another answer in your answer, that's what I did
Jul
6
comment How do photons travel at a speed that should be impossible to attain?
In both the equations I posted, I did not use relativistic mass. I used rest mass.
Jul
6
comment How do photons travel at a speed that should be impossible to attain?
There are two problems here. 1) Photons don't have mass; rest, relativistic or otherwise. 2) The mass of all particles is invariant. It doesn't increase with speed. The correct equation is E^2 = (mc^2)^2 + (pc)^2 or E = \gamma mc^2, where \gamma is the Lorentz factor. If you learned that the mass of a particle increases with speed, you were misinformed. All the particle physics experiments have shown otherwise. Ask me how I know (hint: I work at the LHC).
Jul
6
answered How do photons travel at a speed that should be impossible to attain?