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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.

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.

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