Hi yor_on,

What follows is an excerpt from:

http://www.thenakedscientists.com/forum/index.php?topic=46034.msg400132#msg400132 [nofollow]It is about how "the light and gravity communicate".

If you do not like it, I will remove it. I would appreciate your comments, anyway.

Regards,

zordim

So, here is an easy and accurate derivation of the true relativity equations, which could (should) had been done as soon as Einstein predicted the gravitational red shift of a photon.

Actually, the first scientists who were in position to derive the true relativity equations, in the year 1900, were both Max Planck and Henry Poencare. Because, for that derivation, one needs only the following:

- Newton’s gravitation law [tex]F = G \cdot \frac{m \cdot m_p}{r^2}[/tex]

- Planck’s equation for elementary EM-radiation energy [tex]\Delta E = h \cdot \nu[/tex],

- Poencare’s derivation of [tex]E = m \cdot c^2[/tex] (on the basis of Maxwell’s and Poynting’s works (

http://arxiv.org/ftp/physics/papers/0608/0608289.pdf [nofollow], page 2))

Moving of the mass [tex]m_p[/tex] radially away from a gravitation source for some infinitesimal distance [tex]dr[/tex], would require an infinitesimal work, which – according to the Newton’s gravitation law – is

[tex]dA = G \cdot \frac{m}{r^2} \cdot m_p \cdot dr[/tex].

In the case of a photon, [tex]m_p[/tex] would be the photon’s non-inertial mass. Or, we can put instead [tex]\Delta E / c^2[/tex]. From the energy conservation principle, follows that when a photon moves radially away from a source of gravitation, energy of the photon [tex]\Delta E[/tex] has to reduce for the amount [tex]dA[/tex]. Hence:

[tex]d \Delta E(r) = - dA = - G \cdot \frac{m}{r^2} \cdot \frac{\Delta E(r)}{c^2} \cdot dr \Longrightarrow \frac{d \Delta E(r)}{\Delta E(r)} = - \frac{G}{c^2} \cdot \frac{m}{r^2} \cdot dr[/tex]

And this equation gives the theoretical explanation of the Pound-Rebka experiment:

[tex]\frac{d \Delta E(r)}{\Delta E(r)} = \frac{d E_{ph.}}{E_{ph.}} = - \frac{a(r)}{c^2} \cdot dr \Longrightarrow \frac{\Delta E_{ph.}}{E_{ph.}} = - \frac{g}{c^2} \cdot \Delta r \approx -2.456 \cdot 10^{-15}[/tex], where [tex]g = 9.81 \frac{m}{s^2}[/tex], [tex]\Delta r = 22.5 m[/tex]

Follows the energy equation of a photon:

[tex]\displaystyle \int_r^{\infty} \frac{d \Delta E(r)}{\Delta E(r)} = - \frac{G \cdot m}{c^2} \int_r^{\infty} \frac{dr}{r^2} \Longrightarrow \Delta E(r) = \Delta E \cdot e^{\frac{G \cdot m}{c^2 \cdot r}}[/tex] ,

where [tex]\Delta E[/tex] is the photon’s energy very, very far away from the gravitation-source.

From [tex]\Delta E = h \cdot \nu[/tex] follows [tex]\Delta E \cdot \Delta t = h[/tex]. Namely, since the EM-oscillation frequency is [tex]\nu[/tex] then the period of that oscillation is [tex]\Delta t = \frac{1}{\nu}[/tex]

From [tex] \Delta E(r) = \Delta E \cdot e^{\frac{G \cdot m}{c^2 \cdot r}}[/tex], and from [tex]\Delta E = h \cdot \nu[/tex] follows [tex] \Delta t(r) = \Delta t \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}}[/tex].

At some fixed radius, an infinitesimal part of the time period [tex]d \Delta t(r)[/tex] will be [tex]d \Delta t(r) = d \Delta t \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}}[/tex].

Now, the velocity of a photon can be derived:

[tex]dv(r) = - a(r) \cdot dt(r) = - \frac{G \cdot m}{r^2} \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}} \cdot d \Delta t = - \frac{G \cdot m}{r^2} \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}} \cdot \frac{d \Delta r}{c} \Longrightarrow \frac{dv(r)}{dr} = - \frac{G \cdot m}{c \cdot r^2} \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}}[/tex]

[tex]\frac{d}{dr} (e^{ - \frac{G \cdot m}{c^2 \cdot r}}) = \frac{G \cdot m}{c^2 \cdot r^2} \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}} = \frac{1}{c} \cdot \frac{G \cdot m}{c \cdot r^2} \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}} = \frac{1}{c} \cdot ( - \frac{dv(r)}{dr})[/tex]

[tex]\Rightarrow \frac{dv(r)}{dr} = c \cdot \frac{d}{dr} (e^{ - \frac{G \cdot m}{c^2 \cdot r}}) \Rightarrow dv(r) = c \cdot d(e^{ - \frac{G \cdot m}{c^2 \cdot r}}) \Longrightarrow v(r) = - c \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}}[/tex]

So, obviously, it cannot be like this. The only explanation is that we made wrong starting assumption. Namely, according to the equation for time period, the time period of a photon increases as the photon moves radially away. We know that acceleration is negative for a body which moves away, that is, the change of its velocity [tex]dv[/tex] is negative. But, that is the case for inertial-mass bodies. The velocity change of a photon which moves radially away is positive. The key point is to realize that [tex]\epsilon[/tex] and [tex]\mu[/tex] are the properties of space, and that matter affects the space in the way that it causes the [tex]\epsilon[/tex] and [tex]\mu[/tex] to increase within the space occupied by matter. Outside matter, and further away, [tex]\epsilon[/tex] and [tex]\mu[/tex] decrease. Therefore, according to the Maxwell’s equation [tex]v(r) = \frac{1}{\sqrt{\epsilon (r) \cdot \mu (r)}}[/tex], the allowed velocity, that is, the velocity of our photon, would increase.

Hence, in the previous velocity-equation derivation, the starting equation should be

[tex]dv(r) = a(r) \cdot dt(r)[/tex],

and we would then get that the velocity of the photon which moves radially away from the source of gravitations is

[tex]v(r) = c \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}}[/tex]

And, the length of a photon would change as:

[tex]ds(r) = v(r) \cdot dt(r) = c \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}} \cdot e^{ - \frac{G \cdot m}{c^2 \cdot r}} \cdot dt = e^{ - 2 \frac{G \cdot m}{c^2 \cdot r}} \cdot dr[/tex]

Shapiro time delay [nofollow]