On photons in perpendicular directions

by Raparicio
Tags: directions, perpendicular, photons
Raparicio is offline
Apr3-05, 02:53 PM
P: 112
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 blackhole (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


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:

[tex] \bar {OM} ^2+ MP^2 = OP^2 [/tex]

That u can represent in complex numbers (isomorphic to vectors) this form:

[tex] OM+ MP i = OP [/tex]

Well, really there are distances, but like there has pased one second, you can translate this to:

[tex] V_R + V_x i = c [/tex]

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:

[tex] x'= \frac{x}{\sqrt {1-(v/c)^2}} [/tex]


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:

[tex] V_r + V_x i = c [/tex]

[tex] V_r = \sqrt {(c^2- V_x^2)} [/tex]

Now, with a hypercomplex velocity, you have that:

[tex] V_c=V_r+V_x i+V_y j+V_z k [/tex]

With the hamilton algebra of hypercomplex numbers:

[tex]i^2=j^2=k^2=ijk=\sqrt {-1} [/tex]

You want to know Vr, and this is:

[tex] V_r=V-V_x i-V_y j-V_z k [/tex]

If Vcte=c, u have that:

[tex] Vr=c-V_x i-V_y j-V_z k [/tex]

And his module is:

[tex] \sqrt {c^2-V_x^2-V_y^2-V_z^2} [/tex]

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 blackhole, or others) Vr is alwais minor than c.

A vector (or multivector like this), his modulus is:

[tex] \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}} [/tex]

Now, with the drawing of the hypersphere, you know that

[tex] \frac {c^2} {\sqrt {c^2-V_x^2-V_y^2-V_z^2}} [/tex]

And the velocity could be written like:

[tex]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}} [/tex]

What do u think about this?


- Excerpt from "Pasos filosóficos hacia la unificación de la física. R. Aparicio".
- Source file (in spanish):
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Raparicio is offline
Apr5-05, 07:39 PM
P: 112
Dextercioby??? Where are u?
Raparicio is offline
Apr12-05, 02:15 PM
P: 112
Iff you try to unify quantum mechanics with relativity, you need formula that have "curves". The above formula, has a direct implication with Pin(k) an Spin(k) throught the Clifford Algebras... and a very interesting applications.

Are not responses becouse it's not clear, or becouse it's not of interest???


sweetser is offline
May10-05, 08:42 PM
sweetser's Avatar
P: 361

On photons in perpendicular directions

Hello Raparicio:

In special relativity, there are two important things to define precisely: what is invariant and how things change under a change in the inertial reference frame. Neither issue is addressed in this thread. The first statement about velocity (VR +Vxi=c) looks wrong to me. Velocity is a 4-vector that gets contracted with its dual vector (or contravariant vector with the covector) to make the invariant c^2. Both the velocity and its dual point in exactly the same direction. The 4-vector has 4 basis vectors: 1, i, j, and k. These are effectively one quaternion or equivalently three complex numbers that share the same real (1, i), (1, j), and (1,k).

It is my own personal quirk, but for me there cannot be _another_ i that is like the i, j, k of above except that this new i commutes with everything else. The new i behaves exactly like the i of (1, i) and the j of (1, j) and the k of (1, k). It is understood that this new i is helpful, fun, and widely used, but the unexplained redundancy is unacceptable for me (Nature is elegant).

You must show that the addition of velocity formula gives exactly the same result as is seen in special relativity since that formula has been confirmed in countless high energy physics experiments. It looks like there are too many v's for it to work out.


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