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Raparicio
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Dear Friends,
I want to indicate a question on photons. Imagine that a photon goes in a direction and hits in a mirror to 45º. This photon is later observed like it has 90º respect its original direction.
Well, now imagine that this mirror is closed in a black box, and you don't know that's going to happen into this closed black box, but you see that the photon acts exactly like in first case. But someone says to you that, into the black box, there is a little BlackHole that has changed the direction on the photon.
It seems the same, but it's not, becouse in first case, is a reflection, but in second, is a change of the direction of the photon.
c is constant, and it can't have acceleration (neither normal or tangencial), but what has really happened is that photon has adquired a new direction by the action of the black hole (gravity atraction).
Well, now if you take one second, and you draw the Russell picture, u have this: first you draw the distance that you have seen that one object has
<br /> \begin{picture}(200,200)(0,0) <br /> \put(50,50){\textcolor{red}{\line(1,1){15}}} <br /> \put(50,50){\textcolor{black}{\line(0,1){50}}} <br /> \put(50,50){\textcolor{black}{\line(1,0){50}}} <br /> \put(50,51){\textcolor{green}{\line(1,0){15}}} <br /> \put(65,65){\textcolor{blue}{\line(0,-1){15}}} <br /> \put(50,50){\circle{100}}<br /> \end{picture}<br />
If the center is namened O, the green line is OM, the blue line is MP and the red line is OP.
Bertrand Russel says (the ABC of the relativity) that if you want measure a body that has coming to us, in a second it traces the distance OM. If we make a circle arround O of radius of the distance of light in one second, we can trace MP, perpendicular to MO, crossing the circle on P. Thus OP is the distance that light runs in one second, and the relationship within OP and OM is the relationship of the velocity of light with every one of they. The relationship on OP and MP is the altered longitutes.
You have that:
\bar {OM} ^2+ MP^2 = OP^2
That u can represent in complex numbers (isomorphic to vectors) this form:
OM+ MP i = OP
Well, really there are distances, but like there has pased one second, you can translate this to:
V_R + V_x i = c
Becouse 'c' is constant, Vx is the (virtual) component of the velocity into the black-box (and near the black hole) that causes a retard on the time... but velocity don't changes its constance.
But there's a problem: we have only 1 real module, and one virtual module (maybe the spacetime direction that changes the photon). But well, we can use hypercomplex numbers, clifford algebras, and the Hestenes ideas to make a generalization of the Einstein formulae:
x'= \frac{x}{\sqrt {1-(v/c)^2}}
how?
Well, like B. Russell said, u can draw spacetime in 2-D by using complex numbers. To draw spacetime in 3-D, u need hypercomplex numbers of Hamilton (or quaternions) that have one real value and 3 imaginary values. Forguet for a moment, the minkowsky space, and think only in a real value of the velocity, a virtual imaginary 3D space and try to draw it.
We will proceed this way:
In complex numbers, u know that module is the square of the 2 components. In the previous example. But u want to know the REAL velocity, not the VIRTUAL velocity. Is like this:
V_r + V_x i = c
V_r = \sqrt {(c^2- V_x^2)}
Now, with a hypercomplex velocity, you have that:
V_c=V_r+V_x i+V_y j+V_z k
With the hamilton algebra of hypercomplex numbers:
i^2=j^2=k^2=ijk=\sqrt {-1}
You want to know Vr, and this is:
V_r=V-V_x i-V_y j-V_z k
If Vcte=c, u have that:
Vr=c-V_x i-V_y j-V_z k
And his module is:
\sqrt {c^2-V_x^2-V_y^2-V_z^2}
If there's no change of direction, the REAL velocity is c, but if there is a change of direction (maybe becouse a spacetime alteration like a black hole, or others) Vr is alwais minor than c.
A vector (or multivector like this), his modulus is:
\frac {V_r+V_x i+V_y j+V_z k}{\sqrt {V_r^2+V_x^2+V_y^2+V_z^2}}
Now, with the drawing of the hypersphere, you know that
\frac {c^2} {\sqrt {c^2-V_x^2-V_y^2-V_z^2}}
And the velocity could be written like:
V= \frac {V_r+V_x i+V_y j+V_z k} {Vr^2+V_x^2+V_y^2+V_z^2} \frac {c^2}{\sqrt {c^2-V_x^2-V_y^2-V_z^2}}
What do u think about this?
Notes:
- Excerpt from "Pasos filosóficos hacia la unificación de la física. R. Aparicio".
- Source file (in spanish):
http://www.usuarios.lycos.es/Rufianin/constancia%20de%20la%20velocidad%20de%20la%20luz.pdf
I want to indicate a question on photons. Imagine that a photon goes in a direction and hits in a mirror to 45º. This photon is later observed like it has 90º respect its original direction.
Well, now imagine that this mirror is closed in a black box, and you don't know that's going to happen into this closed black box, but you see that the photon acts exactly like in first case. But someone says to you that, into the black box, there is a little BlackHole that has changed the direction on the photon.
It seems the same, but it's not, becouse in first case, is a reflection, but in second, is a change of the direction of the photon.
c is constant, and it can't have acceleration (neither normal or tangencial), but what has really happened is that photon has adquired a new direction by the action of the black hole (gravity atraction).
Well, now if you take one second, and you draw the Russell picture, u have this: first you draw the distance that you have seen that one object has
<br /> \begin{picture}(200,200)(0,0) <br /> \put(50,50){\textcolor{red}{\line(1,1){15}}} <br /> \put(50,50){\textcolor{black}{\line(0,1){50}}} <br /> \put(50,50){\textcolor{black}{\line(1,0){50}}} <br /> \put(50,51){\textcolor{green}{\line(1,0){15}}} <br /> \put(65,65){\textcolor{blue}{\line(0,-1){15}}} <br /> \put(50,50){\circle{100}}<br /> \end{picture}<br />
If the center is namened O, the green line is OM, the blue line is MP and the red line is OP.
Bertrand Russel says (the ABC of the relativity) that if you want measure a body that has coming to us, in a second it traces the distance OM. If we make a circle arround O of radius of the distance of light in one second, we can trace MP, perpendicular to MO, crossing the circle on P. Thus OP is the distance that light runs in one second, and the relationship within OP and OM is the relationship of the velocity of light with every one of they. The relationship on OP and MP is the altered longitutes.
You have that:
\bar {OM} ^2+ MP^2 = OP^2
That u can represent in complex numbers (isomorphic to vectors) this form:
OM+ MP i = OP
Well, really there are distances, but like there has pased one second, you can translate this to:
V_R + V_x i = c
Becouse 'c' is constant, Vx is the (virtual) component of the velocity into the black-box (and near the black hole) that causes a retard on the time... but velocity don't changes its constance.
But there's a problem: we have only 1 real module, and one virtual module (maybe the spacetime direction that changes the photon). But well, we can use hypercomplex numbers, clifford algebras, and the Hestenes ideas to make a generalization of the Einstein formulae:
x'= \frac{x}{\sqrt {1-(v/c)^2}}
how?
Well, like B. Russell said, u can draw spacetime in 2-D by using complex numbers. To draw spacetime in 3-D, u need hypercomplex numbers of Hamilton (or quaternions) that have one real value and 3 imaginary values. Forguet for a moment, the minkowsky space, and think only in a real value of the velocity, a virtual imaginary 3D space and try to draw it.
We will proceed this way:
In complex numbers, u know that module is the square of the 2 components. In the previous example. But u want to know the REAL velocity, not the VIRTUAL velocity. Is like this:
V_r + V_x i = c
V_r = \sqrt {(c^2- V_x^2)}
Now, with a hypercomplex velocity, you have that:
V_c=V_r+V_x i+V_y j+V_z k
With the hamilton algebra of hypercomplex numbers:
i^2=j^2=k^2=ijk=\sqrt {-1}
You want to know Vr, and this is:
V_r=V-V_x i-V_y j-V_z k
If Vcte=c, u have that:
Vr=c-V_x i-V_y j-V_z k
And his module is:
\sqrt {c^2-V_x^2-V_y^2-V_z^2}
If there's no change of direction, the REAL velocity is c, but if there is a change of direction (maybe becouse a spacetime alteration like a black hole, or others) Vr is alwais minor than c.
A vector (or multivector like this), his modulus is:
\frac {V_r+V_x i+V_y j+V_z k}{\sqrt {V_r^2+V_x^2+V_y^2+V_z^2}}
Now, with the drawing of the hypersphere, you know that
\frac {c^2} {\sqrt {c^2-V_x^2-V_y^2-V_z^2}}
And the velocity could be written like:
V= \frac {V_r+V_x i+V_y j+V_z k} {Vr^2+V_x^2+V_y^2+V_z^2} \frac {c^2}{\sqrt {c^2-V_x^2-V_y^2-V_z^2}}
What do u think about this?
Notes:
- Excerpt from "Pasos filosóficos hacia la unificación de la física. R. Aparicio".
- Source file (in spanish):
http://www.usuarios.lycos.es/Rufianin/constancia%20de%20la%20velocidad%20de%20la%20luz.pdf
Last edited:
