As viewed from the General Theory of Relativity, light is a massless object which traverses the surface of space via the shortest path along that surface.
The Equivalency Principle states that an object operating under the Force of Gravity is equivalent to an object in an accelerated reference system. Generally, this is the starting point for the General Theory of Relativity.
One of the proofs, of the of Einstein’s General Theory of Relativity, is the fact that light passing near a massive object is deflected. The history of Einstein’s calculation and its verification is interesting, and it, along with some of the mathematics in calculating the deflection, can be found at mathpages.com/rr/s6-03/6-03.htm
Applying the Equivalency Principle, in the opposite direction to which it was intended, we should be able to infer that light does have inertial mass.
The reader might ask if there is any reason to believe that light or any other objects traveling close to the speed have any mass.
The answer is yes, maybe, kinda, sort of. With lots of equivocating.
Some years back Honeywell developed an unique guidance system. This was called a ring laser, and basically it consisted of three lasers aligned in a triangle. Laser light was reflected and/or refracted off mirrors at the corners of the triangles. By measuring the refracted light, the system could determine changes in direction.
Essentially, this system was the first commercial use of quantum mechanical principles.
Because of the minimal number of mechanical parts, Honeywell gave a life time guarantee of the system. As you may have already guessed, after a few years the guidance system started to fail. The failure was due to the fact that some of the atoms in the mirrors had been displaced. If light was simply being reflected or refracted this should not happen. If, instead, light had mass and some of this mass was interacting with the atoms nuclei then we could better understand the failure. There are other possibilities but they seem to me to be less likely.
To answer the question of whether or not there are massive objects traveling at or near the speed of light I will need to explain the physics project at the Sudan mine in northern Minnesota.
The University of Minnesota in collaboration with the Fermilab in Batavia Illinois and others designed an experiment to determine if Neutrinos could be experimentally determined to have inertial mass. There are actually two experiments in the underground mine. One is the Neutrino and the other is the dark matter experiment. Both are trying to find the so called missing mass of the Universe.
Basically Fermilabs sent Neutrinos through the Earth from Batavia to the Sudan under ground mine where they were monitored. The result, determined last year, was in fact that Nutrinos do have inertial mass. While I was visiting last year, I asked one of the physicists what the velocity of the Nutrinos was, expecting it to be something on the order of .9c. His answer surprised me. He said that at the current time they could not distinguish between the speed of light and the speed of Neutrinos. He sited evidence of a super nova some 10 plus (or was it 100 plus?) light years away where the Neutrinos arrival could not be distinguished from that of light.
Given conventional wisdom, experimental facts should conclude all the following conclusions to be false. You may want to quit now!!!
At this point I would like the reader to have, at least, an intuitive understanding of the relationship between energy and frequency. If you tie a rope to a poll and the start spinning that rope you should find that the faster you spin the rope (the more energy you apply) the more crests and valleys (the greater the frequency) will appear on that rope. A description of that motion, for a given energy in a single dimension, can be given, generally, by an equation of the form x=A(t)cos(pt/2π) + B(t)sin(pt/2π). Here p is the length of periodicity (where the motion starts to repeat itself and π is ~ 3.1416.
Light works that same way, and energy E is related to frequency f by the equation E=hf, where we are looking at a normalized monochromatic unit of light. For simplicity, you can just regard h as a constant that makes the equation work.
We also know that E=mc^2. For those people unfamiliar with the ^ notation, 2^3 is 2 raised to the third power or 2 cubed. You should also know that X^1/2 is the square root of X. The equation E=mc^2 means that E equals mass times the quantity c squared, where c is the speed of light. (At this point I do not mean to imply that the speed of light is fixed, only that the equation is valid and that c = 3 X 10^8 meters per second is the constant that makes the equation work).
Since hf=E and E=mc^2 we know that hf=mc^2. Dividing both sides by c^2 we get hf/(c^2) = m.
Now we have m in terms of f, but we can also get m in terms of wavelength using the relation f = c/l where l is the wavelength. Substituting c/l for f we get Equation A: m = h/lc.
This gives us a way to measure the mass of a particle of light.
Examples:
To compute the examples we need to know the following facts:
h = 6.63 x 10^-34 joule sec. And, because wavelengths are generally reported in angstrom units, that an angstrom unit = 10^-16 meters. And c which we will for the time being just consider a constant is given by c = 3 x 10^8 meters/sec
Red light has a wavelength of about 7000 angstrom units From Equation A:
m = (6.63x 10^-34)/(7000 X 10^-16)(3 x 10^8) = .3157 X 10^-29 = 3.157 X 10^-30 Kilograms.
Blue light has a wavelength of about 4000 angstrom units From Equation A:
m = (6.63x 10^-34)/(4000 X 10^-16)(3 x 10^8) = .5525 x 10^-29 = 5.525 X 10^-30 Kilograms.
Other masses of note are:
the rest mass of an electron = 9.11 X 10^-31 Kilograms.
the rest mass of a proton = 1.67 X 10^-27 Kilograms.
You might, at first, object to the fact that light should weigh more than an electron, but the energy of light is measured at speeds of about 3 x 10^8 meters/sec. If electrons or protons were accelerated to near those speeds their measured mass would change by the Lorentzian factor of about 1/(1+(v/c)^2)^½, though I am getting ahead of myself here.
At this point I would like to consider the situation where a light particle is escaping a larger mass M, along the axis that connects the centers of these two objects. A classic analysis of this situation yields the following equation for the velocity of the light particle.
Equation B: v=(2/m)^½(T+(mMG)/x)^½ where T is the total energy of the system. This equation is derived from the fact that the total energy of the system T is equal to the kinetic energy K plus the potential energy V and the fact that the force of gravity is given by -mMG/x^2 in the book “Mechanics†by Kieth Symon second edition March 1964 page 37.
Bearing in mind that v is the speed of light, c, in this case we can write:
c = (2/m)^½(T+(mMG/x))^½ where m is the mass of light, M is the total mass of the emitting body(s), and G is the Gravitational constant.
What does this tell us about c?
- The largest value of c will be on it’s initial emission, because x will be small and the other terms are fixed.
- The value for c, during the initial time frame, will vary depending on the source of the emission. The speed of light emitted from a massive star will be larger than the speed of light emitted from our Sun.
- As a corollary to 2 the initial speed of light at or near the big bang would be very large.
We should also consider how big T actually is. Since the total energy is ½(m(v0^2)) + mMG/R0, where we assume that R0 is the radius of the Earth = 6.37 X 10^5 meters, M is the mass of the Earth = 5.983 x 10^24 Kilograms and G = 6.67 x 10^-11 , T = 1/2(3.157 X 10^-30)(3 X 10^8)^2 + (3.157 X 10^-30)(5.983 x 10^24)(6.67 x 10^-11)/6.37 X 10^5. Here I assumed that c = 3 X 10^8 though it is possible/probable that it could be larger). The initial Kinetic Energy portion = 14.21 X 10^-16 or K0 = 1.421 X 10^-15 or larger. The initial Potential Energy portion = 19.78 X 10^-22 or
V0 = 1.978 X 10^-21. Therefore T = 1.421 X 10^-15 + 1.978 X 10^-21. However, since I am not showing more than 4 significant decimals I can write T ~ 1.421 X 10^-15.
The curious thing about this is that T ~ 1/2E! I can only assume that the remaining energy does not play a factor in describing the motion of matter under gravity. (What does it do? Does it help multiple wavelengths be more cohesive?)
Additionally, the white light that we see from the Sun and stars is made up of the visible spectrum, why does it not separate? This is a simple classical problem.
From Equation A we can see that the different color lights should correspond to different masses, and from Equation B each of the colors of light should have their own velocities. For Newton this resulted in the unsolvable problem that the white light coming from the distant stars should not appear white because from Equation B each of the colors of light should have their own velocities.
Note: we should be careful about assumptions, because different frequencies/wavelengths of light will change the total energy T.
To bypass this problem we can simply write a given pulse of light as f(x.t) = Sum Ai(t)cos(t)+Bi(t)sin(t). Note Ai(t) and Bi(t) can be and generally are quantized and so we could write Ai(n,t) and Bi(m,t). Here we can simply calculate the mass by simply summing the component parts. I do not know why this wave does not simply fall apart into its different components. I suppose there could be some quantum argument to be made. It is also possible that the excess energy holds it together.
Perhaps the most compelling argument that the speed of light is constant is the Michelson-Morley experiment. This result has been verified a number of times.
One of the most important consequences of the Michelson-Morley experiment is the derivation of the Lorentzian transformations.
The following is a short sketch of that derivation. The following sketch is found in the book “Introduction to Special Relativity†by Robert Resnick and I shall highlight it by using a green font.
Consider two inertial reference frames S and S’. We will assume that S’ is moving with an uniform, constant, velocity v with respect to S. For simplicity sake we assume that the motion takes place solely along the x axis.
Now we let x,y,z, and t be the coordinates in the S system and x‘, y’, z’, and t’ be the coordinates in the S’ system.
Then we write:
x’ = a[size=75]11[/size][size=100]x[/size] + a[size=75]12[/size][size=100]y[/size] + a[size=75]13[/size][size=100]z[/size] + a[size=75]14[/size][size=100]t[/size]
y’ = a[size=75]21[/size][size=100]x[/size] + a[size=75]22[/size][size=100]y[/size] + a[size=75]23[/size][size=100]z [/size]+ a[size=75]24[/size][size=100]t [/size]
z’ = a[size=75]31[/size][size=100]x[/size] + a[size=75]32[/size][size=100]y[/size] + a[size=75]33[/size][size=100]z[/size] + a[size=75]34[/size][size=100]t[/size]
t’ = a[size=75]41[/size][size=100]x[/size] + a[size=75]42[/size][size=100]y [/size]+ a[size=75]43[/size][size=100]z[/size] + a[size=75]44[/size][size=100]t [/size]
We don’t use any odd functions like squares et cetera because we want the space around these two inertial reference frames to be smooth. Like wise we will eliminate the a[size=75]i2[/size] [size=100]and[/size] a[size=75]i3[/size] terms because [size=100]there[/size] is no motion between these systems in these dimensions.
Thus all we are left with is:
x’ = a[size=75]11[/size][size=100]x[/size] + a[size=75]14[/size][size=100]t[/size]
t’ = a[size=75]41[/size][size=100]x[/size] + a[size=75]44[/size]t
Since the light sphere will look to be the same in each reference frame, we know that:
x^2 + y^2 + z^2 = (ct)^2 [size=100]and[/size]
x’^2 + y’^2 + z’^2 = (ct’)^2
Substituting a[size=75]11[/size][size=100]x[/size] + a[size=75]14[/size][size=100]t[/size] for x’, and a[size=75]41[/size][size=100]x[/size] +a[size=75]44[/size][size=100]t[/size] for t’ we get:
(a[size=75]11[/size][size=100]x[/size] + a[size=75]14[/size][size=100]t[/size])^2 + y’^2 + z’^2 = (c(a[size=75]41[/size][size=100]x[/size] + a[size=75]44[/size]t))^2
Here we need to [size=100]simply[/size] a[size=75]11[/size][size=100]x[/size] + a[size=75]14[/size]t
To do this [size=100]we[/size] observe that at x’ = 0, 0 = x - vt. Rewriting, if 0 = a[size=75]11[/size][size=100]x[/size] + a[size=75]14[/size]t [size=100]then[/size] a[size=75]11[/size][size=100]x[/size] = -a[size=75]14[/size]t,
[size=100]or[/size] x = -a[size=75]14[/size][size=100]t[/size]/a[size=75]14[/size] [size=100]thus[/size] v = -a[size=75]14[/size][size=100]/[/size]a[size=75]11[/size].
Substituting [size=100]for[/size] v we get x’ = a[size=75]11[/size](x - vt). Therefore we can [size=100]write:[/size]
(a[size=75]11/size)^2 + y’^2 + z’^2 = (c(a[size=75]41[/size][size=100]x[/size] + a[size=75]44[/size]t))[1]2[/size]
By performing the squaring operations and rearranging the terms we get:
(a[size=75]11[/size][2]2[/size] - c^2a[size=75]41[/size]^2)[size=100]x[/size]^2 + y^2 + z^2 - 2(va[size=75]11[/size]^2 +[size=100]c[/size]^2a[size=75]41[/size][size=100]a[/size][size=75]44[/size])[size=100]xt[/size] = (c^2a[size=75]44[/size][3]2[/size] - v^2a[size=75]11[/size]^2)t^2.
We know that the light sphere must look the same in each system so the coefficients must be equal or in the xt term be 0.
Therefore [size=100]we[/size] get:
a[size=75]11[/size][4]2[/size] - c^2a[size=75]41[/size]^2 = [size=100]1[/size],
va[size=75]11[/size][5]2[/size] +c^2a[size=75]41[/size][size=100]a[/size][size=75]44[/size] = 0, [size=100]and[/size]
c^2a[size=75]44[/size][6]2[/size] - v^2a[size=75]11[/size]^2 = [size=100]c[/size]^2
A classic but annoying thing, which is common in Physics books, is to simply declare that the student can verify that a set of equations are the solutions to another set of equations. The problem is two fold. 1) you are forced to believe the book and 2) you do not know if they are the only solutions.
The solutions to these 3 equations while high school math, is still somewhat difficult and if you PM me I would be happy to provide the solution. The solutions are:
a11 = 1/(1 - (v/c)^2)^½
a41 = -(v/(c^2))/(1 - (v/c)^2)^½
a44 = 1/(1 - (v/c)^2)^½
By substituting for the aij we get the Lorentzian Transforms:
x’ = (x - vt)/(1 - (v/c)^2)^½
y’ = y
z’ = z
t’ = (t - (v/(c^2)x)/(1 - (v/c)^2)^½
These transforms are very important to the laws of Physics and essentially the whole of the Special Theory of Relativity is a direct consequence of these transforms. The important “Real Life†applications are to particle accelerators and radio active decay times for moving particles such as solar radiation.
Due to these transforms many people have concluded that mass = mass0/(1 - (v/c)^2)^½ where mass0 is the rest mass of an object.
Before I proceed I would like to get a feel for the force of gravity that would be affecting the Michelson-Morley experiment.
Gravitational constant G = 6.67 x 10^-11 Newton meter^2/Kg^2. Mass of the Earth = 5.983 x 10^24 Kilograms. The radius of the Earth is 6.37 X 10^5 meters. Then the Force of Gravity on a particle of red light is:
(3.157 X 10^-30)(5.983 X 10^24)(6.67 X 10^-11)/(6.37 X 10^5)^2 =126.0 X 10^-17/40.58 X 10^10
Thus the Force of Gravity on Red light at the Earth‘s surface = 3.105 X 10^-27
Newtons. This is very small but still measurable.
If light does have mass, then the Michelson-Morley experiment would be wrong, and the speed of light traveling away from the Earth should be different than the light traveling towards the Earth. The Earth could be interchangeable with any massive object such as the Sun or a black hole.
At this point I feel logically compelled to find a reasonable altnerative to the Lorentzian Transforms.
Personally, I have always felt that systems that measure things, need to have things that can be measured. Therefore, in the case of S and S’ we need to have some massive object in each system to measure. So I will assume that S has a mass designated by M. Likewise S’ needs to have a mass of m’.
To replicate this derivation of the Lorentzian Transform operating under the force of gravity, first I will need to describe a light sphere which is deformed by mass M.
Consider the following picture:

The motion of a particle operating under the force of gravity will take place in a single plane. This means that I can use the simpler diagram that follows to describe this motion

To obtain the equation for the initial light sphere I will start with the equation for the change in the radial velocity given in Mechanics by Kieth Symon
Equation C: dR/dt = (2/m)^½(E + Gmm/R - L^2/R^2)^½ where L is the fixed angular momentum.
In general I would like to check out a little more thoroughly what this light pulse looks like.
The math and logic in blue are mine so the reader should be skeptical
To check out the velocity along the x axis we note that because x = Rcosθ. dx/dt = (dR/dt)cosθ + (Rdθ/dt)d(cosθ)/dθ. Here θ is the current angle between the x-y axis and the radius to that point from the emission source. Then, since dθ/dt is 0, the angular momentum L is a constant . Now we can write, dx/dt = (dR/dt)cosθ + RLd(cosθ)/dθ.
Since d(cosθ)/dθ = -sinθ, we can write dx/dt = (dR/dt)cosθ - RLsinθ Substituting (2/m)^½(E + Gmm/R - (L^2)/R^2)^½ for dR/dt we get:
dx/dt = (2/m)^½(E + Gmm/R - (L^2)/R^2)^½cosθ - RLsinθ
Now we can notice that as θ decreases sinθ goes to 0 which will minimize the term RL and thus we will be subtracting a minimal amount and thus increasing dx/dt (the velocity). Similarily we are increasing the term (2/m)^½(E + Gmm/R - (L^2)/R^2)^½cosθ because cosθ will increase to 1 as θ goes to 0.
Therefore the velocity in the x dimension increases as the initial angle of θ is decreased from 90 degrees to 0.
Note: I would like to know a lot more about what the light pulse looks like.
In an effort to find R as a function of t we start with Equation C and multiply both sides by dt and divide by (E + Gmm/R - (L^2)/R^2)^½ to get:
dR = ((2/m)^½)dt
(E + GmM/R - ((L^2)/2mR^2)^½
Integrating both sides we get:
∫ __dR = (2/m)^½ t
(E + GmM/R - ((L^2)/2mR^2)^½
We know from the CRC Handbook of Chemistry and Physics 47 Edition 1966-1967 that the integration formula 203 on page A195 gives us:
∫ RdR___ = ((X)^½)/c - b/2c times ∫ dR/X^1/2
(X)^½
From formula 193 we know that ∫ dR/(X)^½ = (1/(c^½))ln((X)^½ +R(c^½) + b/2(c^½).
To make ___dR look like
(E + GmM/R - (L^2)/2mR^2)^½
RdR I need to multiply the
(X)^1/2
denominator by (R^2/R^2)^1/2 to get:
_____dR = __dR
(R^2/R^2)^1/2(E + GmM/R - (L^2)/2mR^2)^½ (1/R^2)^½(ER^2 + GmMR - (L^2)/2m)^1/2
Now we can write:
dR_ = RdR_
(E + GmM/R - (L^2)/2mR^2)^½ (ER^2 + GmMR - (L^2)/2m)^1/2
Here we set a = -(L^2)/2m, and b = GmM, and c = E
Thus we have:
∫ RdR = (X^½)/c - b/c(1/(c^½))ln((X)^½ +R(c^½) + b/2(c^½).
(X)^½
Now we can write:
t = (m/2)^½((X^½)/c - (b/c)(1/c^½)(ln(X^½ + Rc^½ + b/(2c^½))))
Where X = a + bR + cR^2 and a = -(L^2)/2m, and b = GmM, and c = E.
The obvious problem here is that I have found the time coordinate in terms of R and not the other way around.
Now to find the equation for θ.
From Mechanics by Kieth Symon, we get θ = θ0 +
∫ L dt
mR^2
But if we write dt = dR(dt/dR), then θ = θ0 +
∫ L dt becomes
mR^2
θ = θ0 + ∫ _____LdR
(mR^2)(2/m)^½(E + GmM/R - (L^2)/2mR^2)^½
Bringing out the constant terms and using formula 215 of the CRC Handbook of Chemistry and Physics of we get
θ = θ0 + L/(2/m^3)^½ times ∫ dR we get:
R^2(X)^½
θ = θ0 + L/(2/m^3)^½ 1/(-a^½)arcsin((bR + 2a)/(R(b^2 - 4ac)^½) where a = -(L^2)/2m, and b = GmM, and c = T. The reader should notice that a is negative thus insuring that the quantity is well defined.
Now we have t in terms of R and θ in terms of R. The classic solution, of course, is for R in terms of θ , and is generally written 1/R = m^2MG/(L^2) + Acos(θ - θ0) where A^2 = (m^2(GmM)^2/(L^2.) + (2mT/(L^2)). This equation results the motion of ellipses hyperbolas, and parabolas. I must say that you have to be impressed with that old geyser Newton, cause this stuff is not that easy.
I should mention before proceeding that the above mechanics are actually a failure in describing a specific astrological phenomenon. Mercury does not revolve about the Sun as an ellipse, this is part of the evidence for the superiority of Einstein’s General Theory of Relativity.
I would also like to note that the equation 1/R = m^2MG/(L^2) + Acos(θ - θ0) where A^2 = (m^2(GmM)^2/(L^2.) + (2mT/(L^2)) does not work for our particular problem. This is because when T is positive the conic section should be a parabola, but its directrix (the line used in the definition of the parabola) is at infinity.
At this point I need to find an equivalent to the Lorentzian Equation in order to have a genuinely meaningful theory. I also know that a light pulse given off by one system should look the same in another system traveling along the same radial axis. This should be true even though the pulse should be dampened by the force of gravity.
Consider the following picture:

Here we will start with the general equation for t given by:
t = (m/2)^½((X^½)/c - (b/c)(1/c^½)(ln(X^½ + Rc^½ + b/(2(c^½)))))
By trail and a lot of error I have found that it will be easier to work with the square of this equation. I failed with polar coordinates and cartisian coordinates when the equation was left in the first degree).
I will also fail here but you might like to follow the logic. If you are not interested in the logic you might skip ahead to:
Skip to here
So I will write:
2c/m(t^2) = X - 2^½(b/(c^½)ln(X^½ + Rc^½ + b/2(c^½)))) + (ln(X^½ + R(c^½) + b/(2(c^½))))^2
Where X = a + bR + cR^2 and a = -(L^2)/2m, and b = GmM, and c = T
(2c/m)(t^2) = (a + bR +c(R^2)) - (2b/(c^½))((a + bR +c(R^2))^½ln((a + bR +c(R^2))^½ + Rc^½ + b/(2(c^½))) + (ln((a + bR +c(x^2))^½ + Rc^½ + b/(2(c^½))))^2
Like the solution for the Lorentzian equations we will set
x’ = f11x + f12y +f13z + f14t
y’ = f21x + f22y +f23z + f24t
z’ = f31x + f32y +f33z + f34t
t’ = f41x + f42y +f43z + f44t
Only in this case fij are all functions and space can be squeezed because of the pull towards a massive object.
I will assume that the motion is a in single dimension (the x dimension) because the problem is probably already too difficult for me.
Since these pulses look the same in each system I will write:
(2c/m)(t‘^2) = (a + bR’ +c(R‘^2)) - (2b/(c^½))(a + bR’ +c(R‘^2))^½(ln((a + bR’ +c(R‘^2))^½ + R’c^½ + b/(2(c^½)))) + (ln((a + bR’ +c(R‘^2))^½ + R’c^½ + b/(2(c^½))))^2
Writing R’ = (x’ + R0) we get:
(2c/m)(t‘^2) = (a + b(x’ + R0) +c((x’ + R0)^2)) - (2b/(c^½))(a + b(x’ + R0) +c((x’ + R0)^2))^½(ln((a + b(x’ + R0) +c((x’ + R0)^2))^½ + (x’ + R0)c^½ + b/(2(c^½)))) + (ln((a + b(x’ + R0) +c((x’ + R0)^2))^½ + (x’ + R0)c^½ + b/(2(c^½))))^2
Again I will make the same simplifying assumptions that there is no motion in the directions away from the x axis, and thus y’ =y, and z’=z. Now we can write:
x’ = f11x + f14t
t’ = f41x + f44t
(f11x + f14t)
Now we will see if we can simplify x’ analogously to what was done in the derivation of the Lorentzian transform.
We know that if x’ = 0 then, 0 = f11x +f14t Or x = - f14/f11t. But since v is not constant we need to be a little more careful.
Here we know that t = (m/2)^½((X^½)/c - (b/c)(1/c^½)(ln(X^½ + Rc^½ + b/(2c^½)))) in general, and we can ASSUME that in the moving system c at the origin of the coordinate system is not equal to the original c because the initial velocities are different. Here we will set
t = (m/2)^½((X^½)/c’ - (b/c‘)(1/c‘^½)(ln(X^½ + Rc‘^½ + b/(2c‘^½)))). Because it will become awkward carrying this equation forward I will set
Q = (m/2)^½((X^½)/c’ - (b/c‘)(1/c‘^½)(ln(X^½ + (x+R0)c‘^½ + b/(2c‘^½))))
Where X =a + b(x+R0) + c((x+R0)^2). We should also note that because the motion of the S’ system is taking place on the x axis L =0. Therefore a = 0
Now if 0 = f11x + f14t then we know that (f11/f14)x+ t =0 Or t = -f11/f14x. Since t = Q at this point we must have Q = -f11/f14x. Now we can write:
x’ = f14(-Q + t). In order to get the proper form I will write Q’ = Q/x. The reader should note that I have introduced a dreaded singularity here; and will probably pay for it for the rest of my life. Then
x’ = f14(-Q’x + t).
Substituting we get:
(2c/m)((f41x + f44t)^2) = (a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2)) - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
We know that:
(f41x + f44t)^2 = (f41^2)(x^2) + 2f41f44xt + f44^2t^2
(f14(-Q’x + t) + R0)^2 = (f14^2)(-Q’x + t)^2 - 2R0f14(Q’x - t) + R0^2 and
f14^2(-Q’x + t)^2 - 2R0f14(Q’x - t) + R0^2 = (f14^2)(Q’^2)(x^2) + 2(f14^2)(-Q’x)t + (f14^2)t^2 - 2R0f14(Q’x - t) + R0^2
Substituting we get:
(f14(-Q’x + t) + R0)^2 = (f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2
Now we have:
(2c/m)((f41x + f44t)^2) = (a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2)) - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c((f14(-Q’x + t) + R0)^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
Where
(f41x + f44t)^2 = (f41^2)(x^2) + 2f41f44xt + f44^2t^2
(f14(-Q’x + t) + R0)^2 = (f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2
Substituting for the Squared items we get:
(2c/m)((f41^2)(x^2) + 2f41f44xt + (f44^2)t^2) = (a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2)) - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
Leaving only the t^2 terms on the left side of the equation and doing some partial regrouping:
(2c/m)(+ 2f41f44xt + (f44^2)t^2) = (a + b(f14(-Q’x + t) + R0) - (2c/m)(f41^2)(x^2) -(2c/m)2f41f44xt + c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2)) - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
Moveing the c(f14^2)t^2 on the right side to the left side and grouping the x^2 and xt terms we get:
(2c/m)(f44^2)t^2 - c(f14^2)t^2 = (a + b(f14(-Q’x + t) + R0) - (2c/m)(f41^2)(x^2)+cf14^2)(Q’^2)(x^2) - (2c/m)2f41f44xt - c2(f14^2)Q’xt - c2R0f14Q’x + c2R0f14t + cR0^2 - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
Using the distributive property to take out the t^2, x^2, and xt terms we get:
((2c/m)(f44^2)- c(f14^2))t^2 = (a + b(f14(-Q’x + t) + R0) - ((2c/m)(f41^2) + cf14^2)(Q’^2))(x^2) - ((2c/m)2f41f44 - c2(f14^2)Q’)xt - c2R0f14Q’x + c2R0f14t + cR0^2 - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
Because we want the two light pulses in the different systems to look as similar as possible, (2c/m)(t^2) = (a + b(x + R0) +c((x + R0)^2)) - (2b/(c^½))(a + b(x + R0) +c((x + R0)^2))^½(ln((a + b(x + R0) +c((x + R0)^2))^½ + (x + R0)c^½ + b/(2(c^½)))) + (ln((a + b(x + R0) +c((x + R0)^2))^½ + (x + R0)c^½ + b/(2(c^½))))^2
we need to set:
2c/m(f44^2) - c(f14^2) = 2c/m The t^2 coefficient Equation 1
Now we have
((2c/m)(f41^2) + cf14^2)(Q’^2))(x^2) must be written in the form of c((x + R0)^2 or cx^2 +2cxR0 + c(R0^2)
We will therefore move the R0^2 term and R0x term in proximity to the x^2 tern to get:
((2c/m)(f44^2)- c(f14^2))t^2 = (a + b(f14(-Q’x + t) + R0) - ((2c/m)(f41^2) + cf14^2)(Q’^2))(x^2) - c2R0f14Q’x + cR0^2) - ((2c/m)2f41f44 - c2(f14^2)Q’)xt + c2R0f14t - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2
Now we have the terms ((2c/m)(f41^2) + cf14^2)(Q’^2))(x^2) - c2R0f14Q’x + cR0^2. To get what we want we add ((2c/m)(f41^2) + cf14^2)(Q’^2))2R0x + ((2c/m)(f41^2) + cf14^2)(Q’^2))R0^2.
((2c/m)(f44^2)- c(f14^2))t^2 = (a + b(f14(-Q’x + t) + R0) - ((2c/m)(f41^2) + cf14^2)(Q’^2))(x^2) + ((2c/m)(f41^2) + cf14^2)(Q’^2))2R0x + ((2c/m)(f41^2) + cf14^2)(Q’^2))R0^2 - c2R0f14Q’x - ((2c/m)(f41^2) + cf14^2)(Q’^2))2R0x + cR0^2 -((2c/m)(f41^2) + cf14^2)(Q’^2))R0^2) - ((2c/m)2f41f44 - c2(f14^2)Q’)xt + c2R0f14t - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2.
Now we can write:
((2c/m)(f44^2)- c(f14^2))t^2 = (a + b(f14(-Q’x + t) + R0) - ((2c/m)(f41^2) + cf14^2)(Q’^2))(x+R0)^2 -c2R0f14Q’x - ((2c/m)(f41^2) + cf14^2)(Q’^2))2R0x + cR0^2 - ((2c/m)(f41^2) + c(f14^2)(Q’^2))R0^2) - ((2c/m)2f41f44 - c2(f14^2)Q’)xt + c2R0f14t - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2.
Now ((2c/m)(f41^2) + cf14^2)(Q’^2)) = c The x^2 coefficient Equation 2
And ((2c/m)2f41f44 - c2(f14^2)Q’) = 0 The xt coefficient Equation 3
Rewriting equations 1, 2, & 3 we get:
2c/m(f44^2) - c(f14^2) = 2c/m
((2c/m)(f41^2) + cf14^2)(Q’^2)) = c
(2c/m)2f41f44 - c2(f14^2)Q’) = 0
These are the analogs to the equations that determined the Lorentzian Transforms.
The obvious problem is that, unless I am very lucky, the terms
-c2R0f14Q’x - ((2c/m)(f41^2) + cf14^2)(Q’^2))2R0x + cR0^2 - ((2c/m)(f41^2) + c(f14^2)(Q’^2))R0^2) - ((2c/m)2f41f44 - c2(f14^2)Q’)xt + c2R0f14t - (2b/(c^½))(a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½(ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½)))) + (ln((a + b(f14(-Q’x + t) + R0) +c(f14^2)(Q’^2)(x^2) - 2(f14^2)Q’xt + (f14^2)t^2 -2R0f14Q’x + 2R0f14t + R0^2))^½ + (f14(-Q’x + t) + R0)c^½ + b/(2(c^½))))^2,
except for ((2c/m)2f41f44 - c2(f14^2)Q’)xt will not translate properly.
But I am going to ignor this problem because the terms t^2, x^, and xt will (hopefully) dominate the (lnx)^2 and xlnx). To see this we can look at lim (x/(lnx)^2) as x goes to infinity. By L’Hospital’s Rule we get 1/(1/x)(2lnx)=x/2lnx by applying the rule a second time we get 1/(2/x) = x/2 which goes to infinity. xlnx is a minus term, and will only add in diminishing the additional terms. Anyway x/xlnx = 1/(lnx+ x/x) = 1/(lnx + 1) = 1/(1/x) = x. Therefore the additional terms should be dominated by t^2, x^2, and xt.
Now to solve the system of equations:
2c/m(f44^2) - c(f14^2) = 2c/m
((2c/m)(f41^2) + c(f14^2)(Q’^2)) = c
(2c/m)2f41f44 - c2(f14^2)Q’) = 0
First I will clean up these equations.
f44^2 -(m/2)(f14)^2 = 1. Equation 1
(2/m)(f41)^2 + (Q’^2)(f14^2) = 1. Equation 2
f41f44 - (m/2)Q’(f14^2) = 0. Equation 3
We want to eliminate the cross terms f41f44. To do this we will write an intermediate equation in the form of a(f41^2 ) + b(f41f44) + c(f44^2) = 0 which we could factor as (p1f41 +q1f44)(p2f41 + q2f44) = 0.
Now we will try to eliminate f14 from the equations above.
Q’(f44^2) - (m/2)Q‘(f14)^2 = Q’. Equation 1 times Q’
f41f44 - (m/2)Q’(f14^2) = 0. Equation 3 to be subtracted from Equation 1
Q’(f44^2) - f41f44 =Q’. Equation a
(m/2)(2/m)(f41)^2 + (m/2)(Q’^2)(f14^2) = m/2. Equation 2 multiplied by m/2
Q’f41f44 - (m/2)(Q’^2)(f14^2) = 0. Equation 3 (multiplied by Q’) to be added to Equation 2
f41^2 + Q’f41f44 =m/2. Equation b
In order to get the factoring terms = 0, we need to multiply Equation a by (m/2) and Equation b by Q‘. Thus we get:
(m/2)Q’(f44^2) - (m/2)f41f44 = (m/2)Q’
Q’f41^2 + (Q’^2)f41f44 =(m/2)Q’ Subtracting from the above equation we get:
(m/2)Q’(f44^2) - (m/2)f41f44 - (Q’^2)f41f44 - Q’(f41^2) = 0
Now we want to write this in terms of (p1f44 +q1f41)(p2f44 + q2f41) = 0.
We know that p1p2(f44^2) + p1q2f41f44 + p2q1f41f44 + q1q2(f41^2) = (p1f44 +q1f41)(p2f44 + q2f41). So the question is what are p1, p2, q1, and q2?
We know that
p1p2 = m/2Q’
p1q2 + q1p2 = - ((m/2) + (Q’^2)) and
q1q2 = -Q‘.
If we set p1 = m/2 and p2 = Q’ then
m/2q2 + q1Q’ = -((m/2) + (Q’^2))
Set q1 = -Q’ and q2 = -1 Then
p1 = m/2
p2 = Q’
q1 = -Q’
q2 = -1
Now we want to verify that p1p2(f44^2) + p1q2f41f44 + p2q1f41f44 + q1q2(f41^2) =
(m/2)Q’(f44^2) - (m/2)f41f44 - (Q’^2)f41f44 - Q’(f41^2) = 0
p1p2f44^2 = [m/2][Q’]f44^2, p1q2f41f44 = [m/2][-1]f41f44, p2q1f41f44 = [Q’][[-Q’]f41f44, and
q1q2f41^2 = [-Q’][-1]f41^2
Now p1p2(f44^2) + p1q2f41f44 + p2q1f41f44 + q1q2(f41^2 =
[m/2][Q’]f44^2 + [m/2][-1]f41f44 + [Q’][[-Q’]f41f44 + [-Q’][-1]f41^2
Or p1p2(f44^2) + p1q2f41f44 + p2q1f41f44 + q1q2(f41^2 =
= (m/2)Q’f44^2 - m/2f41f44 - (Q’^2)f41f44 + Q’f41^2
Now because (p1f44 + q1f41)(p2f44 + q2f41) = 0, we can write:
((m/2)f44 - Q’f41)(Q’f44 - f41) = 0
Thus we have f44 = (2/m)Q’f41 OR f41 = Q’f44
If we let f41 = Q’f44 then by substituting into Equation a we get:
Q’(f44^2) - [Q’f44]f44 = Q’ Or
Q’(f44^2) - Q’(f44^2) = Q’.
Contradiction because Q’(f44^2) - Q’(f44^2) = 0 and Q’ is not always 0
If we let f44 = (2/m)Q’f41 then by substituting into Equation a we get:
Q’[(2/m)Q’f41]^2 - f41[(2/m)Q’f41] = Q’ Or
Q’(2/m)^2(Q’^2)(f41^2) - (2/m)Q’(f41^2) = Q’ Or
((2/m)^2(Q’^3) - (2/m)Q’)(f41^2) = Q’. Taking a Q’ out we get
((2/m)^2(Q’^2) - (2/m))(f41^2) = 1. Taking 2/m out we get
((2/m)(Q’^2) - 1)(f41^2) = (m/2) Or
f41^2 = (m/2)/((2/m)(Q’^2) - 1)
Substituting f44 = (2/m)Q’f41 into Equation b we get
f41^2 + Q’f41[(2/m)Q’f41] = m/2 Or
f41^2 + (2/m)(Q’^2)(f41^2) = m/2 Or
f41^2(1 + (2/m)(Q’^2)) = m/2 OR
f41^2 = (m/2)/(1 + (2/m)(Q’^2))
Now we have the two equations
f41^2 = (m/2)/((2/m)(Q’^2) - 1) And
f41^2 = (m/2)/(1 + (2/m)(Q’^2)) This will require
((2/m)(Q’^2) - 1) = (1 + (2/m)(Q’^2)) Or subtracting (2/m)(Q’^2) from both sides
-1 = 1. This is really not good!!!
My math is wrong, or the system of equations can not be solved!!!
Skip to here:
My hope was that the system of equations:
f44^2 -(m/2)(f14)^2 = 1. Equation 1
(2/m)(f41)^2 + (Q’^2)(f14^2) = 1. Equation 2
f41f44 - (m/2)Q’(f14^2) = 0. Equation 3
would, as b goes to 0, look similar to the system that determined the Lorentzian transforms:
a11 = 1/(1 - (v/c)^2)^½
a41 = -(v/(c^2))/(1 - (v/c)^2)^½
a44 = 1/(1 - (v/c)^2)^½
And, with even more luck, I could account for the problems associated with:
m = m[size=75]0[/size]/(1 - (v/c)[7]2[/size])^½
This approach also avoids the problems with the arbitary parameters used in the General Theory of Relativity.
In general a surface in differential geometry (the general theory) is described by (f1(u,v),f2(u,v),f3(u,v)). Here the parameters u and v need have no significance to the surface being studied.
Any way I feel like a complete fool, but nonetheless will continue the search.
