Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can Neutrinos Travel Faster thyan Light?

  1. Oct 27, 2011 #1
    PRELUDE:
    We consider two points A and X . A light ray flashes across an infinitesimally small spatial interval dL at X.

    Local time interval(physical) measured by observer at X:Sqrt[g(X:tt)]dt

    Time interval(physical) measured by observer at A = Sqrt[g(A:tt)]dt Local speed of light at measured at X= dL/[g(X:tt)dt] = c .Speed measured from A ie,

    C[non-local]= dL/[g(A:tt)] = Sqrt[g(X:tt)/ [g(A:tt]] * c . ------------ (1)

    The above value in general is different from c. It might exceed the value of c ,the local speed limit.
    We again consider two points A and X . A particle moves across an infinitesimally small spatial interval dL at X.

    Local time interval(physical) measured by observer at X: Sqrt[g(X:tt)]dt .

    Time interval(physical) measured by observer at A =Sqrt[ g(A:tt)]dt

    Local speed of particle measured at X:v= dL/Sqrt[g(X:tt)dt]

    Speed measured from A ie,
    V[non-local]= dL/Sqrt[g(A:tt)] =Sqrt[ g(X:tt)/ [g(A:tt]] * v -------------------- (2)

    The above value might exceed the local speed limit of c depending on the ratio g(X:tt)/[g(A:tt)] [Chances are greater for fast moving particles like neurtinos]

    But the particle cannot exceed the NON-LOCAL speed barrier since ,locally v<c. We have from relations(1) and (2):

    V[Non-local]< C[non-Local]

    The particle never achieves a superluminal rate in these considerations. It simply crosses[rather reserves the right to cross] the local speed barrier “c” when non-local measurements are being made. Important Points to Observe:

    1. This procedure DOES NOT violate Special Relativity in the local
    context
    2. The Non-local speed barrier is not broken[the local speed barrier is maintained in local experiments]
    3. The procedure/formulation accepts Non-Local speeds/velocities in GR
    [The fact that velocities of satellites constrained in orbits have been accurately measured/estimated demonstrates the existence of non- local velocities in curved spacetime. It is important to take note of
    the fact that the spacetime curvature around the earth is strong enough to maintain satellites[natural and artificial] in their orbits. This is the work of space-time curvature and not something known as the “Force of Gravity”[which is classically known to produce centripetal acceleration]. A person falling out of a roof is seriously in trouble----the acceleration caused is due to the Geometry of space. It is not due to something called the “force” of Gravity! ] . The Geometry of Space causes rivers to flow[provides hydroelectric power], causes the accelerated downward motion of landslides--- the list is quite long.
    [Incidentally the experiment with the OPERA neutrinos was a terrestrial one]

    SAMPLE CALCULATIONS
    Length of path traversed by the OPERA neutrinos: 732km=732000m
    Local time interval for the neutrino as it passes under the ground:
    dT(L)=Sqrt[1- 2GM’/(c^2r)]dt
    M’ is the Mass is the gravitating mass involved at the particular instant concerned, when the neutrino is moving through the earth.
    Local Speed of neutrino=v=dL/ [Sqrt[1- 2GM’/(c^2R)]dt]
    Non-Local time interval considered from the observation stations:
    dT(NL)=Sqrt[1-2GM/(c^2r)]dt
    Non-Local Speed, ie, observed speed=dL/ Sqrt[1-2GM/(c^2r)]dt
    V[non-Local]=Sqrt[1- 2GM’/(c^2R)]/Sqrt Sqrt[1-2GM/(c^2r)]*v
    Time taken[Observed time]=Integral[Sqrt[1- 2GM/(c^2R)]/Sqrt Sqrt[1-2GM’/(c^2r)]*(1/v)dL
    --------------- (3)
    The integral is to be computed between the endpoints of the path
    Now,
    M=4/3 (pi*R^3)*pho
    M’=4/3(pi*r^3)*rho
    M’/M=r^3/R^3
    M’=(r^3/R^3 )*M
    M[Mass of Earth]=5.98*10^24 kg; rho=mean density of the earth.
    R=6400km=6400*1000m
    Length “a“ be the length of the perpendicular from the center of the earth to the chord ,which is the path of the neutrino.
    Substitution:
    Cos(x)=a/r
    X:angle in radians
    L=a tan(x)
    L: distance from the center of the chord[732 km route] to an arbitrary point on the chord.
    dL=a sec^2(x)dx=[a/(Cos(x))^2]dx
    Integral(3)=Time[observed]=Sqrt{(1-GM/(c^2aCos(x)}/{1-GMa^2Cos^(x)/(c^2a^3)}(1/v)dL

    Integral=integration[Sqrt{(1-GM/(c^2aCos(x)}/{1-GMa^2Cos^(x)/(c^2a^3)}(1/v)[a/(Cos(x))^2]dx ------------------ (4)
    The limits of integration are the angles[in radians] between the end points of the path

    The relation:
    r=a/Cos(x) is important
    r is a coordinate value so a should also be a coordinate value
    The physical value corresponding to “a” = Sqrt[(6400*1000)^2 – (732/2 * 1000)^2]=6389526.117m
    I am using the physical value in place of the coordinate value as an approximation
    The limits of integration are from - ArcCos(a/R) to +ArcCos(a/R) : -0.0572187 to +0.0572187 radians
    The integral works out[by numerical methods] to: 730403.299/v
    Let us take v=0.997845c
    =0.997845*3*10^8 m/s
    Observed time: 730403.299/ [0.997845*3*10^8] seconds
    Expected time for a light ray:732000/[3*10^8] seconds
    Difference=
    732000/[3*10^8]-730403.299/[.997845*3*10^8]
    =6.427*10^(-8) Seconds.
    =64 nano- seconds earlier
    If the neutrinos travel with the speed “c” they would be reaching even earlier.
    All this should hold with the assumption that the earth is a homogeneous body
    We have also taken the physical values of a and R
    COORDINATE VALUES
    Solving the equation:
    R=Sqrt[r^2 – 2mr] + 2m ln[Sqrt(r)-Sqrt(r-2m)]+C
    with R=6400000m

    Link for the equation: https://www.physicsforums.com/showpost.php?p=3458428&postcount=126
    We have R=6399924.39 m
    Coordinate value of “a” may be determined by using the following relation[from similar triangles]
    a/732000 = 6399924.39/6400000
    The value of “a” works out to:6389455m
    The angles (+/-)ArcCos(a/R) remain the same as before as expected[from similar triangles]
    One may apply these values
    Conclusion remains the same: The neutrinos can reach earlier than a signal travelling at the speed “c”
    But these neutrinos are not super luminal. They always lag behind the Non local speed barrier for non-local observations. For local observations they are behind the local barrier.
     
    Last edited: Oct 27, 2011
  2. jcsd
  3. Oct 27, 2011 #2
    Anamitra, why after one and a half year on this forum do you still not bother use Latex?
     
  4. Oct 28, 2011 #3
    Re: Can Neutrinos Travel Faster than Light?

    Many of my postings are in a disrupted state:

    Examples:
    https://www.physicsforums.com/showpost.php?p=2923989&postcount=14
    https://www.physicsforums.com/showpost.php?p=2951580&postcount=22

    I dont know for how long.

    Possibly plain text is more resistant to software disruption.[I am saying this with a view to the fact that some new software might be loaded in future-a possibility can always be there]

    I would do a reposting of #1 in latex, keeping the first one as a stand by. I would do this only for the important postings
     
  5. Oct 28, 2011 #4
    Re: Can Neutrinos Travel Faster than Light?

    As we can see from post #1 the gravitational effects are not as ignorable as me might be incloned to think of.
    The OPERA experiment results are possibly not the consequences of any experimental error. They simply speak the truth.
    Repeation of similar experiments is necessary.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can Neutrinos Travel Faster thyan Light?
Loading...