Variant Mass for a Particle which Emits Gravitational Energy for a Particle Orbiting a Large Planet or Sun and for a Binary Star and Variant Frequency for the Light Passing Close a Gravitational Field from a Massive Object (Sun): The Physics and Emission of the Gravitational Energy

Case m1≠m2

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It occurs in binary stars for example. Also, they emit gravitational energy and as a consequence, there is a decrease in their masses and the distance between them due this energy emission.

By considering the center of mass in r=0, it is obtained: m1r1=m2r2 r1=m2r2/m1 r=r1+r2 r= (m2r2/m1)+r2 r=(m1+m2)r2/m1 r2=(m1)r/ (m1+m2) r2=m1r1/m2 r=r1+r2 r=r1+ m1r1/m2 r=(m1+m2)r1/m2 r1=(m2)r/ (m1+m2) $$ K _ {1} = \frac {1}{2} m _ {1} v _ {1} ^ {2} $$

v1=w1 r1 w1= v1/r1

2 1 2 1 1 2 1

$$ G \frac {m _ {1} m _ {2}}{r ^ {2}} = m _ {1} \frac {v _ {1} ^ {2}}{r _ {1}} $$

2 2 2 1 1 2) (

$$ G \frac {m _ {2} ^ {2}}{r \left(m _ {1} + m _ {2}\right)} = v _ {1} ^ {2} $$

$$ G \frac {m _ {2} r _ {1}}{r ^ {2}} = v _ {1} ^ {2} $$

+ =

1 2) 2 1 3 (m m w G r

2 2 1 2 1 2 2 1 1

$$ w _ {1} ^ {2} = \frac {v _ {1} ^ {2}}{r _ {1} ^ {2}} = G \frac {m _ {2}}{r ^ {2} r _ {1}} $$ m r m m r K m v m G G r r = = =

2 2 1 1 2 1 1 1 1 1 2 2 1 1 1 2 2 2

r1=(m2)r/ (m1+m2)

$$K_1 = \frac{1}{2} G \frac{m_1 m_2}{r} \frac{m_2}{(m_1 + m_2)} \text{ kinetic energy for } m_1$$
$$K_2 = \frac{1}{2} m_2 v_2^2 v_2 = w_2 r_2 w_2 = v_2/r_2$$
$$G \frac{m_1 m_2}{r^2} = m_2 \frac{v_2^2}{r^2}$$
$$G \frac{m_1 r_2}{r^2} = v_2^2 G \frac{m_1^2}{r(m_1 + m_2)} = v_2^2$$
$$w_2^2 = \frac{v_2^2}{r^2} = G \frac{m_1}{r^2 r_2} w_2^2 = G \frac{(m_1 + m_2)}{r^3}$$
$$K_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_2 G \frac{m_1 r_2}{r^2} = \frac{1}{2} G \frac{m_1 m_2 r_2}{r^2}$$
$$r_2 = (m_1) r / (m_1 + m_2)$$
$$K_2 = \frac{1}{2} G \frac{m_1 m_2}{r} \frac{m_1}{(m_1 + m_2)} \text{ kinetic energy for } m_2$$

$m_1$ and $m_2$ have different kinetic energies
$$K_T = K_1 + K_2 = \frac{1}{2} G \frac{m_1 m_2}{r} \frac{m_2}{(m_1 + m_2)} + \frac{1}{2} G \frac{m_1 m_2}{r} \frac{m_1}{(m_1 + m_2)}$$
$$K_T = \frac{1}{2} G \frac{m_1 m_2}{r}$$
$$U = -G \frac{m_1 m_2}{r}$$

Therefore, the total mechanical energy of the system of the binary stars which is given as emission of gravitational energy is:
$$E = K_T + U = \frac{1}{2} G \frac{m_1 m_2}{r} - G \frac{m_1 m_2}{r}$$
$$E = -G \frac{m_1 m_2}{2r}$$

It is the same total energy formula of a particle orbiting an almost static massive object. Each mass $m$ has equal angular velocity, equal angular frequency and different kinetic energy with different linear velocities.

The mass of each star is as follows:
$$m_1 = m_{o1} \sqrt{(1 - \frac{v_1^2}{c^2})} G \frac{m_{o2}^2}{r(m_{o1} + m_{o2})} = v_1^2$$

Then, each star has different lost mass given by $\Delta m = m_{o1} c^2 - m_1 c^2$. Also, each star has different gravitational energy emission corresponding to the lost mass of the respective star. It is possible to obtain the total energy at the classic approach:
$$m_1 = m_{o1} \left(1 - \frac{v_1^2}{c^2}\right) E_1 = m_1 c^2 - m_{o1} c^2 = -\frac{v_1^2}{2} m_{o1}$$
$$m_2 = m_{o2} \left(1 - \frac{v_2^2}{c^2}\right) E_2 = m_2 c^2 - m_{o2} c^2 = -\frac{v_2^2}{2} m_{o2}$$
$$E = E_1 + E_2 = -G \frac{m_{o2}^2}{r(m_{o1} + m_{o2})} \frac{m_{o1}}{2} - G \frac{m_{o1}^2}{r(m_{o1} + m_{o2})} \frac{m_{o2}}{2}$$
$$E = -G \frac{m_{o1} m_{o2}}{2r}$$

**Case $m_1 = m_2 = m$:**

$$mr_1 = mr_2 r_1 = r_2$$
Each mass $m$ orbits the same circle of radius $r_1 = r_2$
$$r = r_1 + r_2$$
$$r = 2r_1 r = 2r_2 r_1 = r/2 r_2 = r/2$$
For $m_1$, it is obtained:
$$G \frac{mm}{r^2} = m \frac{v_1^2}{r_1} G \frac{mr_1}{r^2} = v_1^2 G \frac{m}{r^2} = v_1^2 r_1 = r/2$$
$$w_1^2 = \frac{v_1^2}{r_1^2} = G \frac{m}{r^2 r_1} = G \frac{2m}{r^3}$$
$$K_1 = \frac{1}{2} m v_1^2$$
$$K_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} m G \frac{m}{2r} = G \frac{m^2}{4r}$$
For $m_2$, it is obtained:
$$G \frac{mm}{r^2} = m \frac{v_2^2}{r_2} G \frac{mr_2}{r^2} = v_2^2 G \frac{m}{r^2} = v_2^2 r_2 = r/2$$
$$w_2^2 = \frac{v_2^2}{r_2^2} = G \frac{m}{r^2 r_2} = G \frac{2m}{r^3}$$
$$K_2 = \frac{1}{2} m v_2^2$$

Alcocer G. Variant Mass for a Particle which Emits Gravitational Energy for a Particle Orbiting a Large Planet or Sun and for a Binary Star and Variant Frequency for the Light Passing Close a Gravitational Field from a Massive Object (Sun): The Physics and Emission of the Gravitational Energy. Phys Sci & Biophys J 2021, 5(2): 000193.

$$ K _ {2} = \frac {1}{2} m v _ {2} ^ {2} = \frac {1}{2} m G \frac {m}{2 r} = G \frac {m ^ {2}}{4 r} $$ It is possible to obtain the same result, if we replace m1=m2 in the formula:

2 m m m K G r m m $$ = \frac {1}{2} G \frac {m _ {1} m _ {2}}{r} \frac {m _ {2}}{\left(m _ {1} + m _ {2}\right)} $$ $$ K _ {1} = G \frac {m ^ {2}}{4 r} \mathrm {m} _ {1} = \mathrm {m} _ {2} = \mathrm {m} $$

1 2 2 1 1 2

2 m m m K G r m m $$ _ {2} = \frac {1}{2} G \frac {m _ {1} m _ {2}}{r} \frac {m _ {1}}{\left(m _ {1} + m _ {2}\right)} $$ $$ K _ {2} = G \frac {m ^ {2}}{4 r} $$

1 2 1 2 1 2

2 $$ K _ {T} = K _ {1} + K _ {2} = 2 G \frac {m ^ {2}}{4 r} $$

2 $$ U = - G \frac {m ^ {2}}{r} $$

$$ K _ {T} = G \frac {m ^ {2}}{2 r} $$

Each mass m has equal angular velocity, equal angular frequency, equal energy emission, equal linear velocity and equal kinetic energy. Thus, the total mechanical energy of the system of the binary stars is:

2 2 $$ E = K _ {T} + U = G \frac {m ^ {2}}{2 r} - G \frac {m ^ {2}}{r} $$

2 $$ E = - G \frac {m ^ {2}}{2 r} $$

It is possible to obtain the total energy of the system from the formula:

m m m m m m K K K G G r m m r m m $$ K _ {T} = K _ {1} + K _ {2} = \frac {1}{2} G \frac {m _ {1} m _ {2}}{r} \frac {m _ {2}}{\left(m _ {1} + m _ {2}\right)} + \frac {1}{2} G \frac {m _ {1} m _ {2}}{r} \frac {m _ {1}}{\left(m _ {1} + m _ {2}\right)} $$

1 2 2 1 2 1 1 2 1 2 1 2

$$ U = - G \frac {m _ {1} m _ {2}}{r} $$

For m1=m2=m and r1=r2=r/2 r=r1+r2 $$ E = K _ {T} + U = 2 G \frac {m ^ {2}}{4 r} - G \frac {m ^ {2}}{r} $$

2 $$ E = - G \frac {m ^ {2}}{2 r} $$

The mass of each star is as follows:

$$ m _ {1} = m _ {o 1} \sqrt {\left(1 - \frac {v _ {1} ^ {2}}{c ^ {2}}\right)} $$ v m m c $$ G \frac {m _ {o}}{2 r} = v _ {1} ^ {2} m _ {o 1} = m _ {o 2} = m _ {o} $$ $$ m _ {2} = m _ {o 2} \sqrt {\left(1 - \frac {v _ {2} ^ {2}}{c ^ {2}}\right)} $$ v m m c $$ G \frac {m _ {o}}{2 r} = v _ {2} ^ {2} $$ Then, each star has equal lost mass given by Δm= moc2- mc2. Also, each star has equal gravitational energy emission. It is possible to obtain for the total energy at the classic approach:

$$ = m _ {o 1} \left(1 - \frac {V _ {1} ^ {2}}{2 c ^ {2}}\right) $$

2 1 1 1 2 1 2 o v m m c $$ E _ {1} = m _ {1} c ^ {2} - m _ {o 1} c ^ {2} = - \frac {v _ {1} ^ {2}}{2} m _ {o 1} $$ $$ = m _ {o 2} \left(1 - \frac {V _ {2} ^ {2}}{2 c ^ {2}}\right) $$

2 2 2 2 2 1 2 o v m m c $$ E _ {2} = m _ {2} c ^ {2} - m _ {o 2} c ^ {2} = - \frac {v _ {2} ^ {2}}{2} m _ {o 2} $$ $$ E = E _ {1} + E _ {2} = - G \frac {m _ {o}}{2 r} \frac {m _ {o}}{2} - G \frac {m _ {o}}{2 r} \frac {m _ {o}}{2} \mathrm {m} _ {\mathrm {o} 1} = \mathrm {m} _ {\mathrm {o} 2} = \mathrm {m} _ {\mathrm {o}} $$

1 2 2 2 2 2

2 $$ E = - G \frac {m _ {o} ^ {2}}{2 r} $$

r as the classic result

Thought Experiment 1

Gravitational Red Shift and Effects of the Gravitation on the time measure, frequency, wavelength and on the light velocity: A uniform gravitational field g is directed downward in K, with the photon emitted by an atom A falling a distance d through this field before it is absorbed by the detector D [5]. The system K’ is accelerated uniformly upwards with respect to K with a = g (Figure 4).

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When the photon is emitted in the system K´, it has some velocity u in this system. The velocity of the detector when the photon reaches is u + at, where t is the photon’s flight time: t is approximately d/c and a = g.

The velocity of the detector absorbing it is u + gd /c. If it is introduced the unaccelerated reference system Ko, with respect to which at the instant of light emission, K ‘ has not velocity, then D, at the instant of arrival of the radiation at D has the velocity gd v c = with respect to Ko [6]. Indeed, the detector has an approximation velocity with respect to the emitter of $v = \frac{gd}{c}$, independent of $u$.

Therefore the received frequency $f'$ at the detector is greater than the emitted one $f$. It is giving by the formula of the Doppler effect:

$$\frac{f'}{f} = \sqrt{\frac{c + v}{c - v}} = \sqrt{\frac{c + \frac{gd}{c}}{c - \frac{gd}{c}}} \frac{f'}{f} = \sqrt{\frac{1 + \frac{gd}{c^2}}{1 - \frac{gd}{c^2}}}$$

$$\left(1 + \frac{gd}{c^2}\right)^{1/2} \cong 1 + \frac{gd}{2c^2} \quad gd << c^2$$

$$\left(1 - \frac{gd}{c^2}\right)^{-1/2} \cong 1 + \frac{gd}{2c^2} \quad gd << c^2$$

$$\left(1 + \frac{gd}{c^2}\right)^{1/2} \left(1 - \frac{gd}{c^2}\right)^{-1/2} = \left(1 + \frac{gd}{2c^2}\right) \left(1 + \frac{gd}{2c^2}\right) \left(\frac{gd}{2c^2}\right)^2 \approx 0 \text{ for } gd << c^2$$

$$\frac{f'}{f} \cong 1 + \frac{gd}{c^2} \quad f' \cong f \left(1 + \frac{gd}{c^2}\right) \text{ for system } K'$$

$$f' \cong f \left(1 + \frac{v}{c}\right) \text{ where } v = \frac{gd}{c}$$

It is possible to obtain the same formula with the equivalence principle for the system $K$ as follows: Let’s consider that $A$ and $D$ are at rest and there is no Doppler effect which explains the increase in frequency. But, there is a gravitational field and this field acts on the photon [5].

Then, it is possible to obtain the same formula with the equivalence principle for the system $K$ where there is a gravitational field ($A$ and $D$ are at rest). It is given to the photon a gravitational mass equal to its inertial mass $E/c^2$.

When falling a distance $d$, the photon gains an energy $(E/c^2)gd$. In quantum theory, the relationship for $E$ is $E = hf$. It can be shown that $E$ is proportional to $f$ and that from the relativistic transformation of energy and momentum. Thus, it is obtained: $\Delta m = E/c^2 = hf/c^2$.

The energy of the photon being absorbed at $D$ is its initial emission energy plus the energy gained by falling from $A$. This absorption energy in $D$ is:

$$E' = E + (E/c^2)gd \quad E' = E \left(1 + \frac{gd}{c^2}\right) \quad E' = hf$$

Therefore, it is necessary to add to the energy $E$ the potential energy due to the gravity. It probes also that the increase/decrease of the inertial mass in the accelerated system $K'$ corresponds to the increase/decrease of the gravitational mass in the gravitational system $K$.

Thus, when light falls in a gravitational field at the same direction of the gravity acceleration, the light acquires greater energy and frequency, its wavelength decreases and it is said to deviate towards the blue. If the emitter and detector had been reversed and light rises in opposite direction to the gravity acceleration, it is concluded that the energy and frequency of the light decrease, its wavelength increases and it is said to drift towards the red. Thus, if the photon instead of descending from $A$ to $D$ in the accelerated system, it ascends from $D$ to $A$, the formula is:

$$f' \cong f \left(1 - \frac{gd}{c^2}\right) \text{ light rises in opposite direction to the gravitational field}$$

The value of the gravitational potential is replaced and it is obtained:

$$E' = E \left(1 + \frac{\Phi}{c^2}\right)$$

$$U = m_0 gd \quad U = -\frac{GMm_0}{R}$$

$$\frac{GMm_0}{R} = m_0 \dot{g} \quad \frac{GM}{R} = gd$$

$$f' \cong f \left(1 - \frac{GM}{Rc^2}\right) \text{ gravitational redshift for the light}$$

where $M$ is the mass of the planet or Sun and $R$ is the respective radius. The minus sign is because the photon losses energy to pass near the gravitational field of the Sun or the Planet. This is known as gravitational red shift, because light in the visible part of the spectrum will be shifted toward the red end [5].

By using the variant Mass due the Gravitational Potential Energy, it is obtained:

The gravitational potential energy is equal to the negative work done by the gravitational force because it is necessary to give energy or to do work on the system to separate the particle m.

$$c^2 dm = dU$$

$$c^2 dm = G \frac{Mm}{R^2} dR$$

$$\frac{dm}{m} = \frac{GM}{c^2} \frac{dR}{R^2}$$

$$\ln \frac{m}{m_o} = -\frac{GM}{c^2} \left( \frac{1}{R} - \frac{1}{R_o} \right)$$

The reference point is arbitrary for the gravitational potential energy. It is chosen the point where the force is 0 at $R_o = \infty$.

$$m = m_o e^{-\frac{GM}{c^2} \left( \frac{1}{R} \right)}$$

(Variant Mass due the Gravitational Potential Energy)

$$\frac{GMm}{R^2} = m \frac{v^2}{R} \frac{GM}{R} = v^2$$

$$GM \left( \frac{1}{R} \right) = c^2 v^2 << c^2 v << c$$

$$e^{-\frac{GM}{c^2} \left( \frac{1}{R} \right)} = 1 - \frac{GM}{c^2} \left( \frac{1}{R} \right) + \frac{1}{2} \left[ \frac{GM}{c^2} \left( \frac{1}{R} \right) \right]^2 + \dots$$

$$m = m_o \left[ 1 - \frac{GM}{c^2} \left( \frac{1}{R} \right) \right]$$

$$mc^2 = m_o c^2 \left[ 1 - \frac{GM}{c^2} \left( \frac{1}{R} \right) \right]$$

For the light, it is necessary to replace $mc^2 = hf$:

$$f' \cong f \left( 1 - \frac{GM}{Rc^2} \right) \text{ gravitational redshift for the light}$$

It is possible to have the next question due the gravitational redshift: if there is constant transmission of light from A to D, how can any other number of periods per second other than that emitted at A reach D? The answer is because we cannot consider f or f' simply as frequencies (as the number of periods per second) since we have not yet determined the time in the system K. What f denotes is the number of periods per second with reference to the unit of time of the clock U in A, while f' denotes the number of periods per second with reference to the identical clock in D [6].

Nothing forces us to suppose that U-clocks at different gravitational potentials must be considered to be running at the same rate. On the contrary, we must certainly define the time in the gravitational system K in such a way that the number of wave crests and valleys between A and D is independent of the absolute value of time. Thus, the two clocks at A and D measures the time at the same rate. It means that If we measure the time in D with the clock U, then we must measure the time in A with a clock running $\left( 1 - \frac{GM}{Rc^2} \right)$ times slower than the clock U:

$$t_A = t_D \left( 1 - \frac{GM}{Rc^2} \right)$$

when compared to U in D and in the same place [1]. Then, the relation of the both times is as follows:

$$t_D = t_A / \left( 1 - \frac{GM}{Rc^2} \right) t' = t / \left( 1 - \frac{GM}{Rc^2} \right)$$

Then, when we measure by this clock, the frequency of the light ray in its emission at A is equal to:

$$f \left( 1 - \frac{GM}{Rc^2} \right) \text{ (less count of number of wave crests and valleys due the time dilation measures in A), and therefore it is equal to the frequency f' of the same light ray upon arrival at the detector D.}$$

Thus, it is necessary to use U clocks of different constitution to measure time in places of different gravitational potential. If we want to measure time in a place that, with respect to the origin of coordinates, has the gravitational potential $\Phi$, it is necessary to use a clock that (when brought to the origin of coordinates) runs $\left( 1 - \frac{GM}{Rc^2} \right)$ times slower than the clock used to measure time at the origin of coordinates.

Also, the formula for the decreased distance $d$ of a particle orbiting a Planet due the gravitational energy emission is as follows:

$$d' = d \left( 1 - \frac{GM}{Rc^2} \right)$$

The wavelength is given by the next formula:

$$\tilde{e} = \frac{-\frac{GM}{Rc^2}}{\lambda_c} : \text{increased wavelength due to gravitational potential.}$$

The light velocity $v' = \lambda' f'$ is obtained by replacing the respective formula $\lambda'$ and $f'$ or $v' = \frac{d'}{t'}$:

$$v' = v_o \left( 1 - \frac{GM}{Rc^2} \right)^2$$

$$v' \cong v_o \left( 1 - \frac{GM}{Rc^2} \right) \text{ light velocity in a gravitational field } \Phi$$

It is also known as the refraction index for the light [1].

The variation velocity is because the different gravitational potential between different places. It is consequence of the time dilation of the place with gravitational field. The light has lower velocity in a place with gravitational velocity than in a place without the gravitational field [6].

This seems to contradict the principle of constancy of the speed of light. But there is no contradiction, because this light velocity variation is due to the gravitational field. The light velocity remains constant at the place where there is the same gravitational potential but with different value in relation with other location where there is other gravitational potential.

In resume, in a gravitational field, the light has lower velocity, lower rates of the clock-time (less time measured) than the light in a place without gravitational field. Then, the light passing near the Sun for example experiments the gravitational redshift when the light is observed the Earth (less frequency measured, deviation to the red), more wavelength, because of the time clock measure and the real lost energy.

**Thought Experiment 2**

Inertial Mass=Gravitational Mass: It is considered the following sequences from the previous thought experiment [6]:

1. The energy $E$ measured in $A$ is emitted in the form of radiation from $A$ towards $D$, where by the result just obtained, the energy absorbed and measured in $D$ is:

$$E' = E \left( 1 + \frac{gd}{c^2} \right)$$

2. An object of mass $M$ is lowered from $A$ to $D$, doing work $W$ in the process.

3. The energy $E$ is transferred from $D$ to the object of mass $M$ while it is in $D$. The mass $M$ is now $M'$ due to this addition of the energy.

4. The object is raised again to $A$, doing work in this process.

5. Let the energy $E$ transferred back from the object to $A$.

6. The effect of this cycle is simply that $D$ has experienced the increase in energy $E \frac{gd}{c^2}$ of the pulse emitted in $A$ and that the amount of energy $M'gd-Mgd$ has been transmitted to the system in the form of mechanical work. By the principle of energy, we must have:

$$E \frac{gd}{c^2} = M'gd - Mgd$$

Thus, the increase in gravitational mass is equal to $E/c^2$ and equal to the increase in inert mass which is established by the theory of relativity. This result is in accordance with the equivalence principle.

**Gravitational Redshift**

The predicted fractional change in frequency from the thought experiment 4.1 is:

$$f' \cong f \left( 1 + \frac{gd}{c^2} \right) \frac{f'-f}{f} = \frac{\Delta f}{f} = \frac{gd}{c^2}$$

If we consider the distance $d$ from sea level to the highest mountain on Earth, its value is approximately only $10^{-12}$. In 1960, Pound and Rebka were able to confirm the prediction using the Jefferson Tower at Harvard with a height of 22,5 m [5]:

$$\frac{f'-f}{f} = \frac{\Delta f}{f} = \frac{gd}{c^2} = \frac{(9,8)(22,5)}{(3 \times 10^8)^2} = 2,45 \times 10^{-15}$$

Using the Mossbauer Effect (which allows a very sensitive measurement of frequency variations) with a gamma ray source and taking great care to control the variables, Pound

and Rebka observed this gravitational effect on photons and confirmed the quantitative prediction [5]. Then, Pound and Snider in a subsequent experiment found the same result with more accuracy [5].

The result can be generalized to photons emitted from the surface of stars and observed on Earth. It is assumed that the gravitational field does not need to be uniform and that the result depends only on the potential difference between the source and the observer. The formula for the gravitational redshif was calculated at the thought experiment 4.1:

$$ \prime \cong F \left(1 - \frac {G M}{R c ^ {2}}\right) $$

2 ´ 1 GM f f Rc

It is possible to explain with the next example: f represents the number of vibrations of an elemental light generator measured by a delicate relay in the same place. This is done on the surface of the Sun where A is located as the thought experiment 4.1. A portion of the light that is emitted from the Sun reaches the Earth (detector D) where we measure the frequency of the light that arrives with a clock U [5]. The measured frequency in D is given by the demonstrated formula:

$$ \prime \cong f \left(1 - \frac {G M}{R c ^ {2}}\right) $$

2 ´ 1 GM f f Rc

$$ \Phi = \frac {G M}{R} = \mathrm {g d} $$

In this case, M is the mass of the Sun and R is the radius of the Sun and Φ is the gravitational potential of the Sun. Thus, the spectral lines of sunlight must be somewhat redshifted which is the gravitational redshift, compared to the corresponding spectral lines of terrestrial light sources by the approximate amount:

$$ \frac {f - f ^ {\prime}}{f} = \frac {G M}{R c ^ {2}} \approx 2, 1 1 x 1 0 ^ {- 6} $$

6 2 ´ 2,11 10 f f GM x f Rc

This shift of the spectral lines of sunlight can be measured, but due to other influences (pressure, temperature) that affect the position of the centers of the spectral lines, it is difficult to measure this spectral shift [1].

Light Deflection

Einstein calculated that the light from the stars should be deflected by approximately 1.75 arc seconds when passing close to the sun. Stars that appear near the Sun can only be visible in a total solar eclipse. In 1919, the Royal Society of London organized an expedition to Africa and another to Brazil to observe an eclipse of the sun. The observations showed with great accuracy that the apparent deviation of the position of the stars near the Sun corresponded to that predicted by Einstein [7] (Figures 5 & 6).

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Light Deflection by using the velocity formula for the light The light velocity is given by the next formula (from 4.1):

$$ ^ {\prime} \cong v _ {o} \left(1 - 2 \frac {G M}{R c ^ {2}}\right) $$

2 ´ 1 2 o GM v v Rc

$$ d v ^ {\prime} = 2 \frac {v _ {o} G M}{c ^ {2} r ^ {2}} d r \frac {d v ^ {\prime}}{d t} = 2 \frac {v _ {o} G M}{c ^ {2} r ^ {2}} \frac {d r}{d t} \frac {d r}{d t} = v ^ {\prime} $$ $$ d v ^ {\prime} = 2 \frac {v _ {o} G M}{c ^ {2} r ^ {2}} v ^ {\prime} d t \frac {d v ^ {\prime}}{v ^ {\prime}} = 2 \frac {v _ {o} G M}{c ^ {2} r ^ {2}} $$ o v GM dv dt v c r = dz dt c ≅ (photon time) $$ \frac {d v ^ {\prime}}{v ^ {\prime}} = 2 \frac {v _ {o} G M}{c ^ {2} r ^ {2}} \frac {d z}{c} \frac {d v ^ {\prime}}{v ^ {\prime}} = 2 \frac {G M}{c ^ {2} r ^ {2}} d z v _ {o} = c $$

2 2 ´ 2 ´ $$ d \theta = 2 \frac {d v ^ {\prime} \cos \varphi}{v ^ {\prime}} $$

$$ d \theta = 2 \frac {G M}{c ^ {2} r ^ {2}} d z \cos \varphi \cos \varphi = \frac {R}{r} $$

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$$d\theta = 2 \frac{GM}{c^2} r^2 dz \frac{R}{r} d\theta = 2 \frac{GMR}{c^2} \frac{dz}{\left(R^2 + z^2\right)^{3/2}} r = \sqrt{\left(R^2 + z^2\right)}$$

$$\theta = 2 \frac{GMR}{c^2} \int_{-R}^{+R} \frac{dz}{\left(R^2 + z^2\right)^{3/2}}$$

$$\theta = 2 \frac{GMR}{c^2} \frac{1}{R^2} \int_{-R}^{+R} \frac{Rsec^2 \theta}{sec^2 \theta} d\theta$$

$$\theta \approx \frac{4GM}{Rc^2}$$

If we replace the values of the mass and radius of the Sun and the light velocity, we obtain:

$$\dot{e} = 1.75 \text{ arc seconds of light deflection}$$

Light Deflection by using Relativity Theory

The four dimensional space time $ds$ in flat space according to spatial relativity is given by the next equation (1), [8]:

$$ds^2 = c^2 dt^2 - dl^2$$

In this case, light follows straight paths which are expressed by the relation:

$$ds^2 = 0$$

Therefore, we obtain:

$$c^2 dt^2 - dl^2 = 0 \quad c = \frac{dl}{dt} \text{ (constant velocity)}$$

But our world is no flat, then it is necessary to do some corrections due to the curvature of the space time. Therefore, we consider a spherical distribution of mass $M$ and the four dimensional space time is given by the next expression (1), [8]:

$$ds^2 = \left(1 - \frac{2GM}{rc^2}\right) c^2 dt^2 - \left(1 + \frac{2GM}{rc^2}\right) dl^2$$

We can solve the equation by using power series [8]:

$$0 = \left(1 - \frac{2GM}{rc^2}\right) c^2 dt^2 - \left(1 + \frac{2GM}{rc^2}\right) dl^2$$

$$\frac{dl}{cdt} = \frac{\sqrt{\left(1 - \frac{2GM}{rc^2}\right)}}{\sqrt{\left(1 + \frac{2GM}{rc^2}\right)}} \left(1 - \frac{2GM}{rc^2}\right)^{1/2} \approx 1 - \frac{GM}{rc^2} \text{ for } \frac{GM}{r} \quad c^2$$

$$\left(1 + \frac{2GM}{rc^2}\right)^{-1/2} \approx 1 - \frac{GM}{rc^2} \text{ for } \frac{GM}{r} \quad c^2$$

$$\frac{dl}{cdt} \approx 1 - \frac{2GM}{rc^2} \text{ for } \left(\frac{GM}{rc^2}\right)^2 \text{ is neglected for } \frac{GM}{r} \quad c^2$$

$$dl = cdt \left(1 - \frac{2GM}{rc^2}\right)$$

If we divide by $dt$, we obtain:

$$v' = v_o \left(1 - \frac{2GM}{rc^2}\right) \text{ where } v_o = c$$

By applying the same procedure of before, it is possible to obtain:

$$\theta \approx \frac{4GM}{Rc^2} \dot{e} = 1.75 \text{ arc seconds of light deflection}$$

**Perihelion Precession of Mercury**

The perihelion or the ellipse of Mercury has a small rotation and changes a small angle $\theta$ after every turn around the Sun [1]. A prediction of the Einstein Theory is that the Perihelion Precession of Mercury must differ of the classic prediction for about 43 arc seconds by century [6].

It is possible to obtain by using the formula demonstrated from 4.1:

$$d' = d \left(1 - \frac{GM}{ac^2}\right) r = a \text{ for an ellipse}$$

where $M$ is the mass of the Sun, $d$ is the perihelion and $c$ is the light velocity: $M = 1.989 \times 10^{30} \text{ kg}$, $d = 45926524600 \text{ m}$, $c = 3 \times 10^8 \text{ m/s}$

a: semi major axis of the ellipse $= 57894348600 \text{ m}$

The formula for the period of every Planet is given by Keple Law [2]:

$$T^2 = \frac{4\pi^2}{GM_s} a^3 \quad T = 7598971,805 \text{ s for circular motion a=r radius of the orbit}$$

$d = r$ perihelion distance $d'$: the decrease perihelion distance $d' = 45926523431 \text{ m} \quad \Delta d = d - d' = 1169,352684 \text{ m}$

$a'$: the decrease semi mayor axis of the ellipse $= a - \Delta d = 57894347431 \text{ m}$ $$T' = \frac{4\pi^2}{GM_s} a'^3$$

$$T' = 7598971,575 \text{ s}$$

Because of the different time measured in a gravitational field, it is used the formula:

$$t = t' \left( 1 - \frac{GM}{Rc^2} \right) t' = T' t = T'' : \text{time in the gravitational field}$$

$$T'' = 7598971,381 \text{ s}$$

The perihelion of Mercury has moved an angle $\theta'$:

$$\theta' = \frac{2\pi T''}{T} \left( 1 - \frac{GM}{ac^2} \right) \theta' = 6,28319949 \text{ Rad}$$

The factor $(1 - \frac{GM}{ac^2})$ is because the value of $\frac{\text{Perimeter of Circumference}}{\text{radius}}$ in a euclidean geometry is $2\pi$, but it has an increase value due the decrease of the radius because of the gravitational field.

$$\Delta\theta = 2\pi - \theta'$$

$$\Delta\theta = 5,1032 \times 10^{-7} \text{ Rad/rev}$$

$$1 \text{ arcsecond} = 4,844 \times 10^{-6} \text{ Rad}$$

$$T_{\text{mer}} = 0,241 T_{\text{Earth}} \text{ (relation between periods of Mercury and Earth)}$$

$$1 \text{ century} = 100 \text{ years}$$

By applying the factors of arcsecond, $T_{\text{mer}}$ and century, it is obtained:

$$\Delta\theta = 43,75 \text{ arcseconds/century}. \text{This value is in agreement with prediction of the value of General Relativity Theory of 43 arcseconds.}$$

The value by using General Relativity Theory is obtained by using the formula [1]:

$$\Delta\theta = \frac{24\pi^3 a^2}{T^2 c^2 \left( 1 - e^2 \right)} \text{ (Rad/Rev)} \text{ where } e = 0,2056, \text{ r is the perihelion of Mercury and } M_s \text{ is the mass of the Sun and by applying the factors of before, it is obtained:}$$

$$\Delta\theta = 43 \text{ arcseconds/century}$$