GRAVITATION

Calculating \(L\) and \(\frac{dL}{d\varphi}\)

Replacing \(du\over d\varphi\) by its value (2.k) in the expression \(\tan L=\frac {1}{u}{du \over d\varphi}\), we get:
\(\tan L=\pm\sqrt{\frac{R_s}{b^2u}-1+\frac{1}{b^2u^2}}=\pm\sqrt{\frac{R_sr}{b^2}-1+\frac{r^2}{b^2}}\hspace{2cm}\)(2.u),
with \({dL\over d\varphi}=\frac{d\tan L(\varphi)\over d\varphi}{1+\tan^2 L(\varphi)}=\frac{d\left({1\over u}\frac{du}{d\varphi}\right)\over d\varphi}{1+\tan^2 L(\varphi)}\), then:
\({dL\over d\varphi}=\frac{\frac{1}{u}{d^2u \over d\varphi^2}-\frac {1}{u^2}\left({du\over d\varphi}\right)^2}{\frac{R_s}{b^2u}+\frac{1}{b^2u^2}}\),
so by replacing \(d^2u\over d\varphi^2\) and \(\left({du\over d\varphi}\right)^2\) by their respective values given by (2.l) et (2.k):
\({dL\over d\varphi}=\frac{\frac{1}{u}\left(-u+\frac{R_s}{2b^2}\right)-{1\over u^2}\left(\frac{R_su}{b^2}-u^2+{1\over b^2}\right)}{\frac{R_s}{b^2u}+\frac{1}{b^2u^2}}=-\frac{\left(\frac{R_su}{2}+1\right)}{(R_su+1)}=-\frac{\left(\frac{R_s}{2r}+1\right)}{\left(\frac{R_s}{r}+1\right)}\hspace{2cm}\)(2.v),
applicable to photon trajectories in classical mechanics.

Calculating \(E_{meca}\)

(2.b) results in \(\left({du\over d\varphi}\right)^2=\frac{G^2M^2}{K^4}e^2\sin^2(\varphi-\varphi_0)\),
or with \(e^2\sin^2(\varphi-\varphi_0)=e^2-e^2\cos^2(\varphi-\varphi_0)\)
which according to (2.b) is \(e^2-\left(\frac{K^2}{GM}u-1\right)^2\):
\(\left({du\over d\varphi}\right)^2=\frac{G^2M^2}{K^4}\left(e^2-\left(\frac{K^2}{GM}u-1\right)^2\right)=\frac{G^2M^2}{K^4}(e^2-1)-u^2+\frac{2GM}{K^2}u\hspace{2cm}\).(2.w)
Furthermore, \(v^2(r,\varphi)=\left({dr\over dt}\right)^2+\left(r{d\varphi\over dt}\right)^2\) returns with \(r={1\over u}\):
\(v^2(r,\varphi)=\frac{1}{u^2}\left(\frac{d\varphi}{dt}\right)^2\left(\frac{1}{u^2}\left(\frac{du}{d\varphi}\right)^2+1\right)\) and considering that \(\left(r^2{d\varphi\over dt}\right)^2\) is \(K^2\), we get:
\(v^2(r,\varphi)=K^2\left(\left({du\over d\varphi}\right)^2+\frac{1}{u^2}\right)\),
replacing \(\left({du\over d\varphi}\right)^2\) by its value (2.w) gives:
\(v^2(r,\varphi)=K^2\left(\frac{G^2M^2}{K^4}(e^2-1)+\frac{2GMu}{K^2}\right)=\frac{G^2M^2}{K^2}(e^2-1)+2GMu\hspace{2cm}\)(2.x).
Considering that \(E_{meca}={1\over 2}mv^2(r,\varphi )-GmMu\), (2.x) results in:
\(E_{meca}=\frac{G^2mM^2}{2K^2}(e^2-1)+GmMu-GmMu=\frac{(GmM)^2}{2mK^2}(e^2-1)\hspace{2cm}\)(2.y).

Photon emitted from a light source not located at infinity

The invariance of mechanical energy makes it possible to determine the theoretical speed of the photon \(p\) at infinity by considering that at emission its distance \(r_{em}\geq R_s\) and its speed is \(c\):
\({1\over 2}v_{\infty}^2={1\over 2}c^2-\frac{GM}{r_{em}}\) or \(v_{\infty}=c\sqrt{1-\frac {Rs}{r_{em}}}\) which gives with (2.c) and (2.d) respectively:
\(u(\varphi)=\frac{R_s}{2b^2\left(1-\frac{R_s}{r_{em}}\right)}(1+e\cos(\varphi-\varphi_0))\) with pericentre \(r_{per}=\frac{2b^2\left(1-\frac{R_s}{r_{em}}\right)}{R_s(1+e)}\) and
\(e=\sqrt{1+\frac{4b^2}{R_s^2}\left(1-\frac{R_s}{r_{em}}\right)^2}\).
The axis of symmetry \(\varphi_0\) is given by the initial condition \(u(\varphi_{em})=\frac {R_s}{2b^2(1-\frac{R_s}{r_{em}})}(1+e\cos(\varphi_{em}-\varphi_0))=\frac{1}{r_{em}}\) or:
\(\varphi_0=\varphi_{em}+\arccos\left(\frac{\frac{2b^2\left(1-\frac{R_s}{r_{em}}\right)}{r_{em}R_s}-1}{\sqrt{1+\frac{4b^2}{R_s^2}\left(1-\frac{R_s}{r_{em}}\right)^2}}\right)\).
The very specific case of a circular trajectory of radius \(R_c\) around the massive object and at the speed of light in vacuum is found by applying \(K=cR_c\) (emission of the photon tangentially to the circle of radius \(R_c\)) and \(e=0\) in (2.b) and writing \(u=const={1\over R_c}\), that is:
\({1\over R_c}=\frac{GM}{c^2R_c^2}\) hence \(R_c=\frac{GM}{c^2}={R_s\over 2}\).
In classical mechanics, this is the only case where the photon trajectories are performed with an invariant speed equal to \(c\).

Particle with zero speed at infinity and non-zero \(K\)

For \(v_{\infty}=0\), the impact parameter \(b=\frac{K}{v_{\infty}}\) is not defined.
Mechanical energy \(E_{meca}\) is zero, which leads according to (2.y) to an eccentricity \(e=1\) that is, a parabolic trajectory.
(2.b) gives: \(u(\varphi)=\frac{c^2R_s}{2K^2}(1+\cos(\varphi-\varphi_{0}))\) with pericentre \(r_{per}=\frac{2K^2}{c^2R_s}\).
The axis of symmetry with the initial condition \(u(0)=0\) is given by \(\cos\varphi_0=-1\) that is: \(\varphi_0=\pi\), which gives \(u(\varphi)=\frac{c^2R_s}{2K^2}(1-\cos\varphi)\) and a total deflection \(2\varphi_0-\pi=\pi\).
If the massive object is a sphere of radius \(R\), the condition for the particle not to impact the massive object is \(R>r_{per}\) or \(K>K_{lim}=c\sqrt{RR_s}\).

Cartesian equation

\(r=\frac{2K^2}{c^2R_s}+r\cos \varphi\) or \(\sqrt{x^2+y^2}=\frac{2K^2}{c^2R_s}+x\).
Squared, this equality turns into: \(x^2+y^2=\frac{4K^4}{c^4R_s^2}+\frac {4K^2}{c^2R_s}x+x^2\), or:
\(x=\frac{c^2R_s}{4K^2}y^2-\frac{K^2}{c^2R_s}=\frac{y^2}{4r_{per}}-r_{per}\).

Particle emitted from \(R_s\ (r_{em} = R_s)\) at speed \(c\)

\(E_{meca}\) is \({1\over 2}mc^2-\frac{GmM}{R_s}\) which is zero since by definition \(R_s=\frac{2GM}{c^2}\) which leads according to (2.w) to an eccentricity \(e=1\) and a parabolic trajectory.
The speed of the photon \(p\) at infinity \(v_{\infty}\) will be zero since \(E_{meca}=0\).
If \(\alpha_{em}\) is the emission angle with respect to the \(O_x\) axis, the coordinates of the initial speed vector \(c\) of \(p\) in the reference frame \(O_{xy}\) are \(c_x=c\ \cos\alpha_{em}\) and \(c_y=c\ \sin\alpha_{em}\) and
\(K=||\overrightarrow{OP}\land\vec{v}||=x_{em}c_y-y_{em}c_x\) is
\(cR_s(\sin\alpha_{em}\cos\varphi_{em}-\cos\alpha_{em}\sin\varphi_{em})\) or \(cR_s\sin(\alpha_{em}-\varphi_{em})\).
With the emission angle \(L_{em}=\frac{\pi }{2}-(\alpha_{em}-\varphi_{em})\) (angle with the plane tangent to the sphere of radius \(R_s\) at the emission point), and \(L_{em}\neq\frac{\pi}{2}\), (2.b) then gives
\(u(\varphi)=\frac{1+\cos(\varphi-\varphi_0)}{2R_s\cos^2L_{em}}\) with \(r_{per}=R_s\cos^2L_{em}\).
The axis of symmetry \(\varphi_0\) is determined by the initial condition \(u(\varphi_{em})=\frac{1+\cos(\varphi_{em}-\varphi_0)}{2R_s\cos^2L_{em}}={1\over R_s}\) or
\(\varphi_0=\varphi_{em}+\arccos(2\cos^2L_{em}-1)\),
and \(K\) is \(cR_s\cos L_{em}\).
For \(L_{em}=0\) (emission in the tangent plane to the sphere of radius \(R_s\)), the trajectory of the photon is a half-parabola with \(K=cR_s=\frac{2GM}{c}\), tangent to the sphere since \(r_{per}=R_s\), and can be travelled in both directions (particle emitted at speed \(c\) from \(R_s\) or entering from infinity at zero speed).
The total deflection is \(\pi\) for the full parabola.
Note: the case \(L_{em}=\frac{\pi}{2}\) means a radial trajectory (\(\varphi=const=\varphi_{em}\)).

Impact of a photon entering from infinity at speed \(c\)

In classical mechanics, when the impact parameter \(b\) of the photon trajectories is less than \(b_{lim}\), the photon \(p\) impacts the massive object of radius \(R\), and (2.j) with \(u={1\over R}\) gives:
\(\frac{RR_s}{2b^2}(1+e\cos(\varphi_{impact}-\varphi_0))=1\hspace{2cm}\)(2.z),
which leads, after development by replacing \(e\) and \(\varphi_0\) by their respective value given by (2.g) et (2.h), to:
\(\varphi_{impact1}=\arccos\left(\frac{1}{1+\frac{4b^2}{R_s^2}}\left(1-\frac{2b^2}{RR_s}-\frac{4b^2}{RR_s^2}\sqrt{R^2+RR_s-b^2}\right)\right)\),
and \(\varphi_{impact2}=\arccos\left(\frac{1}{1+\frac{4b^2}{R_s^2}}\left(1-\frac{2b^2}{RR_s}+\frac{4b^2}{RR_s^2}\sqrt{R^2+RR_s-b^2}\right)\right)\hspace{2cm}\)(2.aa),
with \(\varphi_{impact}=\min(\varphi_{impact1},\varphi_{impact2}\)), the other angle corresponding to the « exit » of the massive object if \(p\) could pass through it.
On the other hand, the expressions of \(\varphi_{impact1}\) and \(\varphi_{impact2}\) in \(\arcsin\) can be written:
\(\varphi_{impact1}=\arcsin\left(\frac{2b}{R_s}\frac{1}{1+\frac{4b^2}{R_s^2}}\left(\frac{2b^2}{RR_s}-1+\frac{4b^2}{RR_s^2}\sqrt{R^2+RR_s-b^2}\right)\right)\),
and \(\varphi_{impact2}=\arcsin\left(\frac{2b}{R_s}\frac{1}{1+\frac{4b^2}{R_s^2}}\left(\frac{2b^2}{RR_s}-1-\frac{4b^2}{RR_s^2}\sqrt{R^2+RR_s-b^2}\right)\right)\hspace{2cm}\)(2.ab).
\(L\) defined above represents for \(\varphi=\varphi_{impact}\) the impact angle of \(p\) (angle with the plane tangent to the point of impact with the massive object).
For \(r=R\), (2.u) gives \(\tan L_{\varphi_{impact}}=\pm\frac{\sqrt{R^2+RR_s-b^2}}{b}=\pm\sqrt{\frac{b_{lim}^2}{b^2}-1}\hspace{2cm}\)(2.ac).
When \(b=0\), \(\varphi_{impact}=0\) and \(L_0=\frac{\pi}{2}\).
When \(b=b_{lim}\), \(\varphi_{impact}\) has a double root found by replacing \(b_{lim}\) by its value (2.n):
\(\varphi_{impact}=\arccos\left(\frac{-R_s}{R_s+2R}\right)\hspace{2cm}\)(2.ad), which is none other than \(\varphi _0\),
and \(L_{\varphi _{impact}}=0\), which means that \(p\) tangents the surface of the massive object at the point of impact.

Contraction of angular directions

An observer located on the surface of the massive object at the point of impact notes a direction \(\varphi_{apparent}=\frac{\pi}{2}-L_{\varphi_{impact}}\) for an actual direction of emission \(\simeq\varphi_{impact}\) (considering that \(R\ll\) the distance of emission of the photon).
When the radius \(R\) of the massive object is \(R_s\), exact numerical values can be found for specific conditions.

\(b\) close to \(0\)

(2.aa) leads to: \(\varphi_{impact1}\simeq\frac{2b}{R_s}\left(\frac{b_{lim}}{R}-1\right)\)
and (2.ac) leads to: \(\cot L_{\varphi_{impact1}}\simeq\frac{b_{lim}}{b}\), or \(\varphi_{apparent}=\frac{\pi}{2}-L_{\varphi_{impact1}}\simeq\frac{b}{b_{lim}}\).
Hence \(\frac{\varphi_{apparent}}{\varphi_{impact1}}\simeq\frac{Rs}{2b_{lim}}\left(\frac{1}{\frac{b_{lim}}{R}-1}\right)\),
and for \(R=R_s\) considering that \(b_{lim}=\sqrt{2}Rs\):
\(\lim_{b \to 0}\frac{\varphi_{impact1}}{\varphi_{apparent}}=2\sqrt{2}(\sqrt{2}-1)\simeq 1.172\) which is the contraction factor for angular directions in the vicinity of \(0\), independent of the mass of the black hole.

\(\varphi_{impact}=90^\circ\)

For \(R=R_s\), (2.j) gives:
\(\frac{du}{d\varphi}=-\frac{Rs}{2b^2}e\cos\varphi_0=\frac{R_s}{2b^2}\) or \({1\over u}\frac {du}{d\varphi}=\frac{R_s^2}{2b^2}\) since \(u={1\over R_s}\) on impact.
Furthermore, (2.u) gives: \(\tan L_{\varphi_{impact}}=\sqrt{\frac{2R_s^2}{b^2}-1}\) which also is according to paragraph 2.2 of the study \({1\over u}{du\over d\varphi}\) or \(\frac{R_s^2}{2b^2}\) as seen above, hence the equation \(\frac{R_s^4}{2b^4}-\frac{2R_s^2}{b^2}+1=0\).
This equation in \(\frac{R_s^2}{b^2}\) has two roots: \(4\pm 2\sqrt{3}\), hence the two possible values of \(L_{90^\circ}=\arctan(2\mp \sqrt{3})=15^\circ\) or \(75^\circ\), the 1st value corresponding as seen above to the impact and the 2nd to the « exit » if \(p\) could pass through the black hole.
Hence: \(\frac{\varphi_{impact}}{\varphi_{apparent}}=\frac{90^\circ}{90^\circ -15^\circ}=1.2\), which is the contraction factor of angular directions for \(\varphi_{impact}=90^\circ\), independent of the mass of the black hole.

\(b=b_{lim}\)

\(\varphi_{impact1}=\varphi_0\) and \(L_{\varphi_{impact1}}=0\) as seen above, hence:
\(\frac{\varphi_{impact1}}{\varphi_{apparent}}=\frac{2\varphi_0}{\pi}\)
and for \(R=R_s\), (2.h) gives:
\(\varphi_0=\arccos\left(-{1\over 3}\right)\)
hence \(\frac{\varphi_{impact1}}{\varphi_{apparent}}=\frac{2\arccos\left(-{1\over 3}\right)}{\pi}\simeq 1.216\)
which is the contraction factor for angular directions in the vicinity of \(\varphi_0\), independent of the mass of the black hole.

Contraction of apparent star diameters

The contraction of the apparent diameters of the stars corresponds to the ratio \(\frac{Actual\ \Delta L}{Apparent\ \Delta L}\) for a specified elevation \(L\).
For \(b=b_{lim}\) and \(r=R_s\), this ratio can be calculated with (2.v):
\({dL\over d\varphi}=-\frac{\left(\frac{R_s}{2R_s}+1\right)}{\left(\frac{R_s}{R_s}+1\right)}=-\frac{3}{4}\) hence \(\frac {Actual\ \Delta L}{Apparent\ \Delta L}={-d\varphi \over dL}={4\over 3}\).
In summary, the analytical calculation applied to the photon trajectories according to classical mechanics gives the following results on the event horizon:

\(Actual\ L\)	\(Apparent\ L\)	\(Contraction factor\)	\(\frac{Actual\ \Delta L}{Apparent\ \Delta L}\)
\(90^\circ\)	\(90^\circ\)	\(2\sqrt{2}(\sqrt{2}-1)\simeq 1.172\)	\(2\sqrt{2}(\sqrt{2}-1)\simeq 1.172\)
\(45^\circ\)	\(51.8^\circ\)	\(\simeq 1.178\)	\(\simeq 1.189\)
\(0^\circ\)	\(15^\circ\)	\(\arctan(2-\sqrt {3})=1.2\)	\(\simeq 1.261\)
\(90^\circ-\arccos\left(-{1\over 3}\right)\simeq 19.5^\circ\)	\(0^\circ\)	\(\frac{2\arccos(-{1\over 3})}{\pi}\simeq 1.216\)	\(4\over 3\)