Part 2. New definitions of Relativistic Energy and Momentum.
The new definition of mass obtained in Part 1, motivated us to look for its implications in the concept of Energy and Momentum. First of all, The definition of Kinetic Energy in its differential form, "dK" (scalar), is the dot product of an applied force "F" (vector) over a mass "m" by the differential of displacement "dr" (vector) produced by the effect of that force. Where the Force also equals the derivative of Momentum respect to time. In this concept of Force is where the concept of variability of mass is applied to produce the new concept of Energy.
Let's start by putting in equations what was said in words:
Observe that mass m is the constant Mo, the rest mass, divided by a Lorentz factor, where Mo is the same for any observer. Say, There are not other variable magnitude but the velocity inside the Lorentz factor. Thus, in this case the Lorentz factor is not independent of the integration and derivation processes. So,
Let's remember that in the earth movement, the starting point to make movement measurements was taken when all the following variables were zero: t = q = a = 0. Well, at that moment there was a velocity Vo, and also a mass with a value that is not the rest mass. It can be seen as the initial mass "mo". In this same way of reasoning we can see a final mass (which is identified as m). Then we have three different masses for the same body in its curvilinear movement: the initial mass "mo = Mo/(1-Vo²/c²)³/² ", the final mass "m = Mo/(1-v²/c²)³/² "and the rest mass "Mo".
With this in mind, let's continue. By integrating the differential of Kinetic Energy, we obtain the change between its final and initial state:
K - Ko = m.(2.v² - c²) - mo.(2.Vo² - c²).
By transforming this equation into a more convenient presentation:
K - Ko = m.v² + m.(v² - c²) - mo.Vo² - mo.(Vo² - c²).
K - Ko = m.v² - m.c².(1- v²/c²) - mo.Vo² + mo.c².(1 - Vo²/c²).
Reordering and substituting by the new mass' definition:
K - Ko = m.v² - mo.Vo² - m.c².(1- v²/c²) + mo.c².(1 - Vo²/c²).
K - Ko = Mo.[v²/(1-v²/c²)³/² - Vo²/(1-Vo²/c²)³/² - c²/(1-v²/c²)¹/² + c²/(1-Vo²/c²)¹/²]
Let's check obtained results for the case in which the velocity v is small respect to that of light. By expanding in binomial series all the terms at right, making resulting products and neglecting those divided by (c²) raised to any exponent, K - Ko reduces to:
By taking m ~ mo ~ Mo, change in Kinetic Energy reduces to the known Newtonian expression:
K - Ko ~ ½.m.v² - ½.m.Vo²
As we wished and expected!
Let's define the Kinetic Energy at a point as:
K = m.(2.v² - c²) + Mo.c²
Observe that for v = constant = Vo, the exact value of the kinetic energy for any particle is K = mo(2.Vo² - c²) + Mo.c². This result for Vo << c, reduces to mo.Vo/2, as it will be shown; and for Mo = 0, v = c, we obtain the photon's known relationship: K = m.c². This equation is currently used as a good approximation in experiments with very small non-zero-mass atomic particles.
With this definition, Kinetic Energy at low speed of masses (v << c) reduces to the consistent Newtonian expected result:
Let's check: If we do v = 0, then energy reduces to the internal energy, E = Mo.c², which is a consistent result!, and can be considered as if the mass at rest could be converted completely into pure energy, the same Einstein's conclusion.
Momentum is also redefined as:
p = m.v = Mo.v/(1-v²/c²)³/²
Substituting Mo by its expression in terms of m, in the equation of Energy, we obtain:
From where we can obtain the following expression for mass:
m = E/[2.c².(1-v²/c²)³/² - (c² - 2.v²)]
Which in the case of photon, whose velocity is c, this expression gives us the mass of the photon:
m = E/c²
and in the case of its momentum, by applying p = m.v,
p = v.E/[2.c².(1-v²/c²)³/² - (c² - 2.v²)]
for v = c:
p = E/c
All these results are consistent and known. (Einstein also obtained them, but ..)
Demonstration that Einstein's definition of mass is wrong.
Let's do the following exercise in order to check the correctness of our procedure (or that of Einstein): Let's suppose that we wish to know the maximum speed developed by a particle with no rest mass (we know that it cannot reach the speed of light, because at this speed mass should take an infinite value (in Einstein's approach or in ours).
One way to do this, is by looking for a maximum of the Kinetic Energy of such mass, and finding out which is the velocity that allows this maximum.Taking the derivative of Kinetic Energy respect to the time and equalling it to zero, we have:
But, on the other hand, at the begining of this Part 2, we had found that dm = 3.m.v.dv/(c² - v²). From vectorial calculus v.dv = v.dv. By equalling both dm/m, we can put:
dm/m = 3.v.dv/(c² - v²) = -4.v.dv/(2.v² - c²)
From here, we obtain the value of velocity that maximizes Kinetic Energy (actually, it is a minimum):
v =± j.c/(2)¹/²
This result is equally obtained starting from dK = m.dv.v + v.v.dm = 0 ==> dm/m = -dv/v
Let's take a while to analize this result. The solutions to this problem are imaginary ones. The real solution, is the value of v = c, because if v>c mass is also imaginary, and this have no sense for us. Then, the actual solution for this problem are the real limits of the velocity of mass, that comes from our proper definition of mass: v =± c.
Now, Let's do the same exercise, but using Einstein's definition of mass:
(If we were taken the derivative of Einstein's definition of Kinetic Energy, m.c² - Mo.c², and equaled to zero, we were obtained the following strange result: c².dm/dt = 0, which obliges to be null one of the two factors!)
This is a wrong and absurd result. It does not admit any logical reasoning. The only way to obtain this kind of result is that there is something wrong in the only new definition we have used: the Einstein's definition of mass (or the Einstein's definition of kinetic energy, which comes out from the Einstein's definition of mass). Of course, also we have used the general definition of Kinetic Energy, dK = m.v.dv + v.v.dm, but this is a correct definition in physics. This subtile result shows the incorrectness of Einstein's definition of mass.
Because Einstein's definitions of Energy and Momentum are a direct consecuence of his definition of mass, they must also be wrong.
Relativistic relationships between Physical Magnitudes.
Now, we are going to do some discussion about the characteristics of the relativistic relationships between physical magnitudes that come from the new definitions of mass, energy, momentum, etc., when they are measured relative to the center of mass of the system . In Part 1, we had found that the Lorentz transformation between the differentials of radius measured by the moving observer and by the fixed one, when they make measurements relative to the center of mass of the system for particles with non zero mass rest, were the same obtained for the proper radius. One conclusion that we can infere from this is: that in general, it expands or compacts a variable physical magnitude times the same factor, independently as it is a differential or the proper magnitude, or, this is not true, and it is only valid for the case of radius, and that is it. But, Let's assume that the Lorentz Transformation, for variable physical magnitudes, will be the same as for the differentials, as also for the magnitudes themselves . If this is not so, somewhere we will find the inconsistency. In the case of the rest mass of a body, Mo, both observers measure the same magnitude, say this magnitude can not be considered as a variable physical magnitude, but as a constant.
Nevertheless, there are some examples that supports our assumption.
We know that r' = r/(1 - v²/c²)¹/², and that dr' = dr/(1 - v²/c²)¹/². Then, also we can put that:
d²r' = d²r/(1 - v²/c²)¹/².
And also that: d²r'/(dt')² = (d²r/dt²)/(1 - v²/c²)³/².
On the other hand, because angle is invariant w' = dq/dt' = dq/[dt(1 - v²/c²)¹/²], we also can put that:
d²q/(dt')² = (d²q/dt²)/(1 - v²/c²)
Or that dw'/dt' = (d²q/dt²)/(1 - v²/c²) = (dw/dt)/(1 - v²/c²)
Which implies that dw' = dw/(1 - v²/c²)¹/², where the factor is the same as that of w (say, w' = w/(1 - v²/c²)¹/²)
In the expression of the acceleration, which comes from v = dr/dt.Ur + r.dq/dt.Uq:
With the second term, as we will see, it is not so obvious that dm' = dm.(1 - v²/c²)³/², but it is a necessity (in agreement with our assumption). Let's see:
The previous used expression at the begining of this Part, for dm, was:
dm = 3.m.v.dv/(c² -v²)
If you think over this relationship, in order to make the Force invariant, you are pushed to establish that the expression v.dv/(c²-v²) must be invariant, which leads to (c²-v'²)=(c²-v²)/(1 - v²/c²)² in order to arrive at dm' = dm.(1 - v²/c²)³/²:
So, let's establish that this is so (Reality says last word!). Thus, the previous relationship becomes:
This way of reasoning will allow us to establish which are the Lorentz transformations, or factors, that applies for each physical magnitude. If the magnitude has no factor, between what is measured by the moving and what is measured by the fixed observer (in other words, factor is unity), then we will say that this magnitude is invariant to the Lorentz transformation, or that its measurement is the same, independent of the observer. The contrary result indicates that it is not invariant. Remember, Lorentz factors are unique for any magnitude and they come from either the equations or the relationships. Let's start with this in mind.
Electromagnetic Force:(Invariant, because it is a force)
F = q.(e + vxB)
Let's discuss it. Charge q seems not to be influenced by its velocity. Let's assume that it is invariant to Lorentz Transformation. Thus, Electric Field, e, must be invariant either, in order to preserve the invariance of Force. But we know that velocity is not ( v' = v/(1 - v²/c²). Then, the Magnetic Field Density will not be invariant either. So, the following must be met: B' = B.(1 - v²/c²), in order to make invariant the Force, as it was previously established.
Let's check the assumptions. In the equation of the electric Potential, we have that it is not invariant (under the assumption that Electric Field has been considered invariant):
The Energy of the Electric Field, where q has been assumed invariant, is given by:
dE' = q'.dV' = q.dV/(1 - v²/c²)¹/² ==> dE' = dE/(1 - v²/c²)¹/²
This result was the expected, because we had obtained it before (see the third result of relativistic relationships).
A charge located in an uniform magnetic field, which moves describing a circular path, by aplying classical analysis (which is dimensionally consistent), angular velocity is given by:
In Faraday's electromagnetical equation, we have that the electric voltage is given by:
V' = - df'/dt' = - (d/dt') Surface Integral of B'.ds'
Because of distances are related by the following Lorentz transformation: 1/(1 - v²/c²)¹/², then surfaces would be the square of this transformation: 1/(1 - v²/c²). So,
==> V' = V/(1 - v²/c²)¹/²
Which is another consistent output, because this result ratifies what had been obtained before under the assumption of the invariance of the charge and Electric Field.
Another check: We know that Maxwell's equations holds in any reference system. Let's see if our results support this:
Closed Line integral of e'.dl' = - (d/dt') Surface Integral of B'.ds' ==>
= Closed Line integral of e.dl/(1 - v²/c²)¹/² = - [1/(1 - v²/c²)¹/²].(d/dt).Surface Integral of B.ds.
And this is true, because we have seen that B'.ds' = B.ds, and both sides of the equation are divided by the same Lorentz Transformation. Say, we obtained that the equation meets in any reference system as it was expected.
By manipulating relationships and obtaining:
D' = (1 - v²/c²).D J' = (1 - v²/c²)¹/².J H' =H
It can be shown that with the other three Maxwell's equations we arrive at the same result (that the equations hold in any reference system). The same can be done with Pointing theorem, etc.
All things are consistent!. Which confirms the used procedure and taken assumptions.
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