XPost: sci.skeptic, alt.atheism, alt.conspiracy   
   XPost: sci.physics   
   From: PointedEars@web.de   
      
   [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", § 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.]   
      
   --   
   PointedEars   
      
   Twitter: @PointedEars2   
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