Timeline for Why is the momentum of a particle $\gamma mv$?
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| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
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| Jul 2, 2015 at 15:29 | comment | added | user12262 | @0celo7: "This is a deterministic classical system, not a probabilistic quantum one." -- As far as I understand the OP question it is concerned with attributing and measuring momentum values, e.g. of objects (such as $A$) wrt. to suitable systems (such as $\mathcal S$). There doesn't seem to be any requirement or exclusion being made concerning some categorizazion as "classical or quantum"; whatever you might mean by that. "doesn't $\hbar =0$ [...]?" -- Perhaps you mean cases or limits in which $$\frac{m_A~c^2~\Delta \tau_A}{\hbar} \gg 1 $$ ? ... | |
| Jul 2, 2015 at 10:17 | comment | added | Ryan Unger | Some quick thoughts before I head out to work...why is there a wave function in a classical system? This is a deterministic classical system, not a probabilistic quantum one. Also, doesn't $\hbar\equiv0$ because this is classical mechanics? Have I mislearned quantum mechanics? | |
| Jul 1, 2015 at 21:46 | comment | added | user12262 | @0celo7: $\Delta\tau_A$ -- the duration of object $A$ throughout the trial (as spelled out in the answer). $\Delta\tau_{\mathcal S}$ -- the duration of system $\mathcal S$ throughout the trial (I may add that to my answer ...); $\Delta\mathbf r_{\mathcal S}$ -- the (spatial) separation between the member of system $\mathcal S$ who met/passed object $A$ at the beginning of the trial, and the member of system $\mathcal S$ who met/passed object $A$ at the end of the trial (I may add that, too.) "If this actually pans out, fantastic work!" -- Hmm ... What'ya mean by "panning out"?? | |
| Jul 1, 2015 at 21:46 | comment | added | user12262 | @0celo7: "where the phases and the $\hbar$ go" -- The "phases" can go into the description of states (see my above comment); and if so, the $\hbar$ goes both into the denominator of the "angle", and into the numerator of the applicable operator; so the $\hbar$ cancels (see my first comment) upon application of the operator to the state description. (So why bother inserting the $\hbar$ symbol at all? ...) "you saying special relativity follows from quantum mechanics?" -- I'd consider QM the general framework for "measurement", and GR/SR specifying particular geometric/kinematic operators. | |
| Jul 1, 2015 at 15:35 | comment | added | Ryan Unger | I'm very confused...are you saying special relativity follows from quantum mechanics? I don't see where the phases and the $\hbar$ go... Some more questions: to what do we apply the energy and momentum operators and what are $\Delta \tau_A$, $\Delta\tau_\mathcal{S}$ and $\Delta r_\mathcal{S}$? If this actually pans out, fantastic work! | |
| Jun 25, 2015 at 16:45 | comment | added | user12262 | @Kyle Kanos: "You've given a definition of the phase, but shown nowhere where it ought to be." -- Hmm ... (I hadn't persued the mentioned afterthought up to this point) ... I suppose the "phase" $\phi_A$, as specified above, might/should be part of the state description/representation (colloquially a.k.a. "wave function") of object $A$; formally: $$|\psi_A\rangle := |\phi_A~\varphi_A\rangle = \phi_A~|\varphi_A\rangle,$$ where $|\varphi_A\rangle$ is (the part of the description of $A$ which is) explicitly independent of $\Delta \tau_A$ (in the suitable, short trial under consideration). | |
| Jun 25, 2015 at 16:44 | comment | added | user12262 | @Kyle Kanos: "But you don't seem to invoke your phase [...]" -- Correct, IIUYC. "Invoking the phase" is explicitly done only in the equation of my above comment. In my answer itself, "invocations" are only of $$\frac{d}{d \mathbf r_{\mathcal S} }[~\Delta \tau_A~],$$ or incl. conventional coeff.s: $$-m_A~c^2~\frac{d}{d \mathbf r_{\mathcal S} }[~\Delta \tau_A~].$$ "Phase" (and "energy") are considered merely as afterthoughts; though adding e.g. the equation mentioned above may not hurt. Btw.: I trust you're aware of the distinction between "operator" (with a "hat") and "(measured) value". | |
| Jun 25, 2015 at 13:25 | comment | added | Kyle Kanos | But you don't seem to invoke your phase in $E_S[A]$, or really anywhere, for that matter. You've given a definition of the phase, but shown nowhere where it ought to be. If it's designed to cancel the quantum mechanical aspect of special relativity, shouldn't you show more explicitly where it should be, to clarify matters? | |
| Jun 25, 2015 at 6:21 | comment | added | user12262 | @Kyle Kanos: "How does your $i\hbar$ magically disappear?" -- You mean the $i\hbar$ in operator $$\hat E_{\mathcal S}:=i~\hbar~\frac{d}{d\mathbf\tau_{\mathcal S} }[~]$$? This (magically! ;) **cancels** against the coefficient $\frac{-i}{\hbar}$ which had been (cleverly! ;) first inserted into the "phase": $$\phi_A :=\text{Exp}[~-i~\frac{m_A~c^2}{\hbar}~\Delta\tau_A~],$$ such that $$\hat E_{\mathcal S}[~\phi_A~]=m_A~c^2~\frac{d}{d\mathbf\tau_{\mathcal S} }[~\sqrt{(\Delta\tau_{\mathcal S}[~A~])^2-(\Delta \mathbf r_{\mathcal S}[~A~] / c)^2}~]~\phi_A=m_A~c^2~\gamma_{\mathcal S}[~A~]~\phi_A.$$ | |
| Jun 24, 2015 at 22:53 | comment | added | Kyle Kanos | How does your $i\hbar$ magically disappear? | |
| Jun 24, 2015 at 20:40 | history | answered | user12262 | CC BY-SA 3.0 |