2

  1. Is it me who have a poor understanding, or does all matter have to become 'pure energy' in order to achieve speed-of-light speed?

  2. If so, does that mean that no material can achieve the speed of light and remain in its original state of matter?

Qmechanic
  • 201,751
Yuran Pereira
  • 175

2 Answers2

4

Your understanding is spot on, as is PhotonicBoom's Answer. Something that might give you a bit more insight along the lines that you are thinking is if I answer your question backwards: the property we call "mass" (or "rest mass") is acquired by a particle with a rest mass of nought when that particle is confined in some way. If you look at my thought experiment here you can understand that if you put light into a perfectly reflecting box, the box's inertia increases by $E/c^2$, where $E$ is the energy content. This is the mechanism for most of your body's mass: massless gluons are confined by being coupled to, and by coupling, quarks in the nucleusses of your body's atoms and are accelerating backwards and forwards all the time, so they have inertia just as the confined light in a box did. Indeed this is how most rest mass in the World arises; the Higgs mechanism itself (as I understand it - this is outside my field) is an interaction between otherwise massless particles with the Higgs field. "Coupling", aside from being cross coupling terms in Schrödinger and other quantum state evolution equations, is physically a kind of tethering of particles: they are no longer free to run off at the speed of light but are held back by each other. Even the fundamental particle the electron can be thought of in this way, if you look at my other answer.

Community
  • 1
Selene Routley
  • 88,112
  • Upvoted, exceptional insight on how rest mass is acquired! – PhotonBoom Apr 05 '14 at 22:42
  • WetSavannaAnimal: "[...] light into a perfectly reflecting box, the box's inertia increases by $E/c^2$, where $E$ is the energy content" -- Fine. ""rest mass" is acquired by a particle with a rest mass of nought when that particle is confined" -- ?!? By Planck's analysis, the frequency of this photon is $\nu = E/h$. Now, what do you suggest is its wavelength: $\lambda = \frac{c}{\nu} \times \sqrt{1 - \left(\frac{m , c^2}{E}\right)^2 }$ for some non-zero "acquired rest mass $m$"? Even for $m = E/c^2$?? p.s. virtual +1 for excellent use of "nucleusses" in a sentence about nuclei. – user12262 Apr 05 '14 at 23:39
  • @user12262 That of course depends on the reference frame. In the photon picture, if the box is moving past you at speed $v$ left-to-right the photon is then in a quantum superposition of a rightwards propagating, higher frequency eigenstate and a leftwards, lower frequency state. You can no longer describe it as a pure energy eigenstate. The classical analysis, although much harder, is more enlightening: what you get out of Maxwell's equations for the field within the moving box is a solution which can be construed as the quantum state for a lone photon within the box ... – Selene Routley Apr 06 '14 at 02:28
  • @user12262 ... I analyse in full a similar, but not identical situation in my answer here; sometime I hope to put up the full calculation of the moving box with light in it. – Selene Routley Apr 06 '14 at 02:29
  • WetSavannaAnimal: "[...] depends on the reference frame" -- 1. My comment/question referred to $\nu$ and $\lambda$ "measured by the box walls; incl. equipment at rest wrt. both box walls". (I presumed the box walls remaining at rest to each other.) That seems pretty much implied by your identifying "$E/c^2$" as "the box's inertia increase". So my question stands; "in the box picture". 2. Whatever is frame dependent surely doesn't deserve to be called "some sort of invariant mass". Does your questionable "acquired "rest mass" of the photon" not mean "acquired invariant mass"?? – user12262 Apr 06 '14 at 06:25
  • @user12262 I categorically did not say the acquired inertia was frame dependent. Read what I have said again, this time carefully, and then state what you do not understand. I would recommend that you do the calculations along the lines in this question here (although that's not the same thing, it shows you how you would do the calculation of what impulse it would take to change the box's state of motion. – Selene Routley Apr 06 '14 at 10:29
  • WetSavannaAnimal: "I categorically did not say the acquired inertia was frame dependent." -- Then the start of your first comment "That of course depends on the reference frame." was a poor choice of wording. To repeat: I object to the phrase ""rest mass" is acquired by a particle with a rest mass of nought when [...] confined" of your answer above; and I ask you again which value of wavelength $\lambda$ "in the box picture" you suppose for a confined photon of energy $E = h , \nu$. – user12262 Apr 06 '14 at 10:57
  • @user12262 You asked about wavelengths and frequencies of the light, which are frame dependent. In frames other than the box's, there isn't a single frequency: the photon, if you imagine its state in the box's frame to be a pure energy eigenstate - you do not need to do this BTW, is not in an energy eigenstate in other frames. Therein, it is a (pure) quantum superposition of a redshifted "right-to-left" running state (if the box is moving rightwards) and a blueshifted "left-to-right) running state. The difference between the two momentums is the nett momentum of the system, which is .... – Selene Routley Apr 07 '14 at 00:36
  • @user12262 ...how you calculate the impulse needed to get the system to change inertial frames and thus the effective inertia. The rest mass is then the value of this inertia for changes from the box's rest frame. These ideas are pretty standard and date from Einstein's initial thoughts about the subject (between STR and GTR) - of course he thought in the classical terms. Another take on this is Method 2 of my answer here and Einstein's original is here – Selene Routley Apr 07 '14 at 00:41
  • WetSavannaAnimal: "You asked about wavelengths and frequencies of the light, which are frame dependent." -- True. (So is $E$.) But starting from my first comment I only intended to consider "the box at rest". (Due to being limited to 600 characters.) I recognized later that you're used to considering box and fields in other setups. (Me too, btw.) "[...] The rest mass is then the value of this inertia for changes from the box's rest frame." -- To me, non-zero rest mass is $E \times \sqrt{1 - \left(\frac{v}{c}\right)^2}$. And [...to be contd] – user12262 Apr 07 '14 at 05:24
  • @WetSavannaAnimal: (Oops!, btw. I set $c = 1$ above) ... And I'd be happy to agree on saying that "the rest mass (and even the inertia) of the box-with-photon-system is larger than that of the box alone". "In frames other than the box's, there isn't a single frequency [...] superposition of a redshifted "right-to-left" running state (if the box is moving rightwards) and a blueshifted "left-to-right"" -- This certainly doesn't quite fit my "standing wave" intuition. But to learn and resolve that I really should read the two links you indicated carefully again, and possibly comment there. – user12262 Apr 07 '14 at 05:34
2

Any body as you say with rest mass cannot fully reach the speed of light, as you would need to supply an infinite amount of energy to accelerate it to that exact speed. We do know thought that all massless particles do travel at the speed of light. Are they pure energy? They are, but then, everything is as we know from Einstein's relation $E = mc^2$. I assume by matter you mean solids, liquids and gases, and in that sense you are right, as I said above nothing with rest mass can achieve the speed of light. They can go infinitely close to it, but never fully at $c$.

PhotonBoom
  • 5,320