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   Message 95,205 of 95,770   
   Dawn Flood to Thomas 'PointedEars' Lahn   
   Re: Rest frame of a photon   
   15 Dec 25 09:14:45   
   
   XPost: sci.physics.relativity, alt.atheism, sci.physics   
   XPost: alt.paranormal   
   From: Dawn.Belle.Flood@gmail.com   
      
   On 12/14/2025 7:13 PM, Thomas 'PointedEars' Lahn wrote:   
   > [Supersedes to set F'up2 sci.physics.relativity (again)]   
   >   
   > JTEM *amok*-crossposted across 5 newsgroups and 2 top-level hierarchies   
   > *without* Followup-To:   
   >   
   >> On 12/14/25 9:50 AM, Thomas 'PointedEars' Lahn wrote:   
   >>> JTEM amok-crossposted across 5 newsgroups   
   >>   
   >> You did. You crossposted to 5 newsgroups.   
   >   
   > Because *you* did before.  But I *also* set Followup-To   
   > sci.physics.relativity to contain the crosspost, which you probably   
   > *deliberately* ignored.  That is anti-social behavior on *your* part.   
   >   
   > (If the header of your posting is not forged, you are using Mozilla   
   > Thunderbird the same as I do.  This software observes RFC 5537 "Netnews   
   > Architecture and Protocols", ยง 3.4.3 "Followups", which means that if you   
   > push the "Followup" button, the default for the target newsgroups is *only*   
   > those found in the "Followup-To" header field of the posting you post a   
   > followup to.  The Followup-To header field of my posting was "Followup-To:   
   > sci.physics.relativity" which you can verify by expanding the header pane   
   > or pressing Ctrl+U.)   
   >   
   >> As a typical mental case you're in some narcissistic "Do as I say, not as I   
   do"   
   >> mode...   
   >   
   > No, you simply either have no clue how newsgroups work; or you do, and you   
   > are trolling, and you are the mental, narcissistic case here.  Which one is   
   it?   
   >   
   >>> Which part of "a photon has no inertial rest frame" did you not understand?   
   >>   
   >> Omg! You're HILARIOUS!   
   >>   
   >> The photon is everywhere is can potentially be!   
   >   
   > *If* it *had* an inertial rest frame which it *cannot* have.   
   >   
   >> But to the photon itself that's it -- the one and only frame!   
   >   
   > Such an *inertial* frame of reference does not exist as the speed of a   
   > photon *cannot* be zero.  That would mean that its linear momentum p would   
   > be zero, and by E = p c it would not exist:   
   >   
   > The energy--momentum relation for a free particle in Minkowski space is   
   >   
   >    E^2 = m^2 c^4 + p^2 c^2.   
   >   
   > For a particle to move at the speed c in all inertial reference frames, it   
   > is required that its mass is zero.  Proof: Let us assume that its mass is   
   > not zero, then   
   >   
   >                E^2 = m^2 c^4 + p^2 c^2   
   >                    = m^2 c^4 + gamma^2 m^2 v^2 c^2   
   >                    = gamma^2 m^2 c^4 (1/gamma^2 + v^2/c^2)   
   >                    = gamma^2 m^2 c^4 (1 - v^2/c^2 + v^2/c^2)   
   >                    = gamma^2 m^2 c^4   
   >            gamma^2 = E^2/(m^2 c^4)   
   >    1/(1 - v^2/c^2) = E^2/(m^2 c^4)   
   >       1 - v^2/c^2  = m^2 c^4/E^2   
   >           v^2/c^2  = 1 - m^2 c^4/E^2 ==> (v = c ==> m = 0).   
   >   
   > But then its total energy squared is   
   >   
   >    E^2 = 0^2 c^4 + p^2 c^2 = p^2 c^2,   
   >   
   > so its total energy is (only) given by   
   >   
   >      E = p c.   
   >   
   >   
   > Equivalently, by E = p c = h f, its frequency f would be zero which makes no   
   > sense (or you could say, with frequency zero there is no oscillation of   
   > electric and magnetic fields, so there cannot be photon which is an   
   > excitation state of the electromagnetic field):   
   >   
   > For a photon, P = hbar K ==> p = |P| = hbar k, so   
   >   
   >    E = p c = hbar k c = h/(2pi) 2pi/lambda c = h/lambda c = h f.   
   >   
   >    [Planck--Einstein relation]   
   >   
   >   
   > Another, more robust, way to show that there is no such frame is to show   
   > that there is no Lorentz transformation to such a frame:   
   >   
   > The original Lorentz transformation (as derived by Einstein) for motion of a   
   > "primed" frame in the x-direction of an "unprimed" frame at the velocity v   
   > relative to the latter frame is   
   >   
   >    t' = gamma(v) [t - v/c^2 x]   
   >    x' = gamma(v) [x - v t]   
   >    y' = y   
   >    z' = z.   
   >   
   > But   
   >   
   >    gamma(v) = 1/sqrt(1 - v^2/c^2),   
   >   
   > where v is the speed of the unprimed frame relative to the primed frame (and   
   > vice-versa, and gamma(v) --> inf as v --> c.   
   >   
   > Equivalently, the Lorentz transformation above can be performed conveniently   
   > by multiplication of a four-vector (c t, x, y, z)^T on the left by the matrix   
   >   
   >              [ cosh(w)   -sinh(w)  0  0]   
   >    Lambda := [-sinh(w)    cosh(w)  0  0],   
   >              [       0          0  1  0]   
   >              [       0          0  0  1]   
   >   
   > where w = artanh(v/c) is defined as rapidity.  However, if v = c, then   
   > v/c = 1, and artanh(1) is not well-defined: artanh(x) --> inf as x --> 1.   
   >   
   > So we can calculate the elapsed proper time along a lightlike geodesic; it   
   > is zero.  In Minkowski space (where this is simple), it is (via the   
   > Minkowski metric and the definition of proper time)   
   >   
   >    ds^2 = c^2 (d tau^2) = c^2 dt^2 - dx^2 - dy^2 - dz^2   
   >                         = c^2 dt^2 (1 - v^2/c^2)   
   >               (d tau)^2 = dt^2 sqrt(1 - v^2/c^2)   
   >    ==> (v = c ==> d tau = 0 ==> Delta tau = int_W d tau = 0).   
   >   
   > But that does not mean that we can make any scientifically solid statements   
   > about what "a photon experiences".  In fact, not only does the existence of   
   > such an inertial frame contradict special relativity and quantum theories;   
   > but also, if special relativity and quantum theories are correct theories   
   > (and there is strong indication that they are), we will never be able to   
   > *falsify* any statements about this because *according to the theory*   
   > material objects *cannot* move at c through space (as their mass is not   
   > zero).  But hypotheses that cannot be falsified are not scientific.   
   >   
   >    [The situation is very different if that frame is non-inertial in   
   >     the Newtonian sense.  Is there such a frame?  Absolutely: The relative   
   >     speed of a photon propagating radially outwards from the event horizon   
   >     of a Schwarzschild black hole is zero.  But notice that I said   
   >     "propagating": it is still moving, but space is falling in as fast, so   
   >     its position does not change (this river model is one way to understand   
   >     it).  This is just its *coordinate* speed, NOT its local speed.  And   
   >     the geometry of our universe is NOT the Schwarzschild geometry.]   
   >   
      
   Wow, amazing!!   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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