Academia Arena

Science in Uncertainty

Note: To many people, mathematics (a mere calculation-- an abstract intellectual activity that began in Greece in the sixth century BC) presents a significant barrier to their understanding of science. Certainly, mathematics has been the language of physics for four hundred years and more, and it is difficult to make progress in understanding the physical world without it.

Newtonian Laws Of Motion

If a force F acts on a particle of mass m₀ at rest and produces acceleration a in it, then the force is given by Newton’s second law (the law that describes the motion of bodies based on the conception of absolute space and time and held sway until Einstein’s discovery of special relativity -- postulated by Swiss mathematician and scientist Leonhard Eular after death of Sir Isaac Newton in 1736) which states that the body will accelerate, or change its speed, at a rate that is proportional to the force (For example, the acceleration is twice as great if the force is twice as great): F = m₀a. According to Newton’s First Law of Motion, every particle continues ‘in state of rest’ (v = 0, a=0) when no external force (F=0) acts on it. Under this condition the rest mass of the particle (a measure of quantity of matter in a particle; its inertia or resistance to acceleration in free space) becomes UNDEFINED.

m₀= F/a = 0/0

The equation

F = m³a / m₀² on rearranging lead to:

m= m₀ ^2/3 (F/a) ^1/3

Suppose no force acts on the particle (i.e., F = 0), then no acceleration is produced in the particle (i.e., a = 0). Under this condition: m = m₀^2/3 (0/0) ^1/3 i.e., m becomes UNDEFINED. There can be no bigger limitation than this (because m should be = m₀ under the condition: F = 0 and a = 0).

Newton’s third law of motion as stated in PHILOSOPHIAE NATURALIS PRINCIPIA MATHEMATICA (the most influential book ever written in physics – which rose Newton rapidly into public prominence – he was appointed president of the Royal Society and became the first scientist ever to be knighted):

“To every action there is always an equal and opposite reaction.”

Let us consider a boy is standing in front of wooden wall, holding a rubber ball and cloth ball of same mass in the hands. Let the wall is at the distance of 5 feet from the boy.

Let the boy kicks the rubber ball at the wall with some force F.

Action: Boy kicks the rubber ball at the wall from distance of 5 feet.

Reaction: The ball strikes the wall, and comes back to the boy i.e. travelling 5 feet. Now action and reaction is equal and opposite.

Let the same boy kicks the cloth ball at the wall with same force F.

Action: Boy kicks the cloth ball at the wall from distance of 5 feet.

Reaction: The ball strikes the wall, and comes back to the boy i.e. travelling 2.5 feet. Now action and reaction are not equal and opposite. In this case Newton’s third law of motion is completely violated.

Protein Ligand Binding

A protein in solution exists in two forms: bound and unbound. Depending on a specific protein’s affinity for ligand, a proportion of the protein may become bound to ligands, with the remainder being unbound. If the protein ligand binding is reversible, then a chemical equilibrium will exist between the bound and unbound states, such that:

P (metal) + L (ligand) ↔ PL (protein - ligand complex)

The dissociation constant for this reaction is,

K = [P] [L] / [PL]

In this equation [P] = [P] _T – [PL] and [L] = [L] _T – [PL] where [P] _T and [L] _T are the initial total concentrations of the protein and ligand, respectively. The dissociation constant K is a useful way to present the affinity of a protein for its ligand. This is because the number K quickly tells us the concentration of protein that is required to yield a significant amount of interaction with the target ligand. Specifically, when protein concentration equals K, the 50% of the target ligand will exist in the protein ligand complex and 50% of the ligand will remain in the free form [L]. (This holds true under conditions where protein is present in excess relative to ligand). Typically, proteins must display a K ≤ 1 x 10 ^–6 M for the interaction with their target ligand. When considering the K for proteins, smaller numbers mean better binding. The higher the K value the protein does not bind well to the ligand. At very high ligand concentrations all the protein will be in the form of PL such that

[P] = 0

If [P] = 0, then

K = 0

Since the binding constant K_B = 1/ K. Therefore: K_B = 1/0 i.e., the binding constant becomes UNDEFINED.

Using the equilibrium relationship K [PL] = [L] [P] and substituting,

[P] _T – [P] for [PL], [L] _T – [PL] for [L] and [P] _T – [PL] for [P] Gives:

K {[P] _T – [P]} = {[L] _T – [PL]} {[P] _T – [PL]}

K [P] _T – K [P] = [L] _T [P] _T – [PL] [L] _T – [PL] [P] _T + [PL] ² which on rearranging:

K [P] _T – [L] _T [P] _T + [PL] [P] _T = – [PL] [L] _T + [PL] ² + K [P]

[P]_T {K – [L] _T + [PL]} = [PL] {– [L] _T + [PL]} + K [P]

Further, if we substitute [L] _T = [PL] + [L]. Then we get

[P]_T {K – [PL] – [L] + [PL]} = [PL] {–[PL] – [L] + [PL]} + K [P]

[P]_T {K – [L]} = – [PL] [L] + K [P] which is the same as:

[P]_T {K – [L]} = K [P] – [PL] [L]

K – [L] = K {[P]/ [P] _T} – {[PL]/ [P] _T} [L]

Labeling [P] / [P] _T as F_FP (fraction of free protein) and [PL] / [P] _T as F_BP (fraction of bound protein) then above expression turn into

K – [L] = K F_FP – F_BP [L]

Any equation is valid only if LHS = RHS. Hence

If F_FP = F_BP=1, then the LHS = RHS, and the above Equation is true.

If F_FP = F_BP≠1, then the LHS ≠ RHS, and the above Equation is invalid.

Let us now check the validity of the condition

“F_FP = F_BP =1”.

As per the protein conservation law,

[P] _T = [PL] + [P]

From this it follows that

1= F_BP + F_FP

If we assume F_BP = F_FP =1, we get:

1 = 2

The condition F_FP = F_BP =1 is invalid, since 1 doesn’t = 2. In fact, the only way it can happen that K – [L] = K – [L] is if both F_FP = F_BP =1. Since F_FP = F_BP ≠ 1, Equation K – [L] = K F_FP – F_BP [L] does not therefore hold well.

Conclusion: Using the equilibrium relationship K [PL] = [L] [P] and substituting [P] _T – [P] for [PL], [L] _T – [PL] for [L], [P] _T – [PL] for [P] and simplifying we get the wrong result:

K – [L] = K F_FP – F_BP [L]

Considering the reaction: P + L ↔ PL the change in free energy is given by the equation:

ΔG = ΔG₀ + RT ln Q

where R is the gas constant (8.314 J / K / mol), T is the temperature in Kelvin scale, ln represents a logarithm to the base e, ΔG₀ is the Gibbs free energy change when all the reactants and products are in their standard state and Q is the reaction quotient or reaction function at any given time (Q = [PL] / [P] [L]). We may resort to thermodynamics and write for ΔG₀: ΔG₀ = − RT ln K_eq where K_eq is the equilibrium constant for the reaction. If K_eq is greater than 1, ln K_eq is positive, ΔG₀ is negative; so the forward reaction is favored. If K_eq is less than 1, ln K_eq is negative, ΔG₀ is positive; so the backward reaction is favored. It can be shown that

ΔG = − RT ln K_eq + RT ln Q

The dependence of the reaction rate on the concentrations of reacting substances is given by the Law of Mass Action (which was proposed by Cato Maximilian Guldberg and Peter Waage in 1864, based on the work of Claude Louis Berthollet’s ideas about reversible chemical reactions). This law states that the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants at any constant temperature at any given time.

Applying the law of mass action to the forward reaction:

v₁ = k₁ [P] [L] where k₁ is the rate constant of the forward reaction.

Applying the law of mass action to the backward reaction:

v₂ = k₂ [PL] where k₂ is the rate constant of the backward reaction.

Further, the ratio of v₁ / v₂ yields:

v₁ / v₂ = (k₁/ k₂) Q.

But equilibrium constant is the ratio of the rate constant of the forward reaction to the rate constant of the backward reaction. And consequently:

v₁ / v₂ = K_eq / Q.

On taking natural logarithms of above equation we get:

ln (v₁ / v₂) = ln K_eq – ln Q.

On multiplying by –RT on both sides, we obtain:

–RT ln (v₁ / v₂) = – RT ln K_eq + RT ln Q

Comparing Equations

ΔG = − RT ln K_eq + RT ln Q and

– RT ln (v₁ / v₂) = – RT ln K_eq + RT ln Q, the Gibbs free energy change is seen to be:

ΔG = −RT ln (v₁ / v₂)

or

ΔG = RT ln (v₂ / v₁).

At equilibrium: v₁ = v₂

ΔG = 0

Under this condition RT becomes UNDEFINED i.e.,

RT = 0 / 0

There can be no bigger limitation than this. RT cannot be undefined because R = 8.314 Joules per Kelvin per mole and T → undefined violates the third law of thermodynamics (which states that nothing can reach a state of absolute zero).

From the equation

ln (v₁ / v₂) = −ΔG / RT

it follows that

lnv₁ = −ΔG*₁ / RT + constant

lnv₂ = −ΔG*₂ / RT + constant

This splitting involves the assumption that reaction in the forward reaction depends only on the change ΔG*₁in Gibbs energy in going from the initial state to some intermediate state represented by the symbol *; similarly for the backward reaction there is a change ΔG*₂ in Gibbs energy in going from the product state to the intermediate state. For any reaction, we can therefore write

lnv = −ΔG* / RT + constant

Now, the Eyring approach assumes that we can assume a thermodynamic quasi-equilibrium can exist between reactants A & B and activated complex AB* (which is somewhat similar to a normal molecule with one important difference. It has one degree of vibration that is special. The AB* moves along this special vibrational mode to form product P (or to reform reactant A and B) at a certain course of time. If this is true, we can solve for constant (because at thermodynamic quasi-equilibrium ΔG* = 0 and v = v_eq, where v_eq = rate of reaction at thermodynamic quasi-equilibrium)

lnv_eq = 0 + constant

Substituting the value of constant, we get:

lnv = −ΔG* / RT + lnv_eq

or

v = v_eq e ^{−ΔG* / RT}

Since v = k_r C_A C_B (where k_r = rate constant for a given reaction (A + B → P), C_A & C_B = concentrations of reactants A & B). Therefore:

k_r C_A C_B = v_eq e ^{−ΔG* / RT}

By thermodynamics, we know that

∆G* = ∆G0 + RT ln(C*/ C_A C_B)

Where:

∆G* = Gibbs free energy of activation, ∆G0 = standard Gibbs free energy of activation, R = universal gas constant (8.314 J/K/mol), T = temperature in Kelvin and C* = concentration of activated complex AB*.

From this it follows that:

C_A C_B = C* e^{−ΔG* / RT} e^{−ΔG0 / RT}

Substituting the value of C_A C_B in the equation: k_r C_A C_B = v_eq e ^{−ΔG* / RT}, we get:

k_r C* = v_eq e ^{–ΔG0 / RT}

Since v_eq = k₂ C*_eq (where: k₂ = rate constant for product formation and C*_eq = concentration of AB* at thermodynamic quasi-equilibrium). Therefore:

k_r = k₂ (C*_eq /C*) e ^{–ΔG0 / RT}

But the expression in the existing literature of transition theory (which also widely referred to as activated complex theory -- has achieved widespread acceptance as a tool for the interpretation of chemical reaction rates − developed in 1935 by Eyring and by Evans and Polanyi) -- which pictures a reaction between A and B as proceeding through the formation of an activated complex, AB*, in a rapid pre-equilibrium – which falls apart by unimolecular decay into products, P, with a rate constant k₂) is:

k_r = k₂ e ^{–ΔG0 / RT}

k_r = k₂ (C*_eq /C*) e ^{–ΔG0 / RT} (rate equation when the thermodynamic quasi-equilibrium is not still attained between reactants A & B and activated complex AB*)

k_r = k₂ e ^{–ΔG0 / RT} (rate equation when the thermodynamic quasi-equilibrium is attained between reactants A & B and activated complex AB*)

Since e ^{–ΔG0 / RT} = K* (where K* is the equilibrium constant for the formation of activated complex). Therefore:

k_r = k₂ K*

Taking natural logarithm of the above equation we get:

lnk_r = lnk₂ + lnK*

Differentiating the above equation we get:

dlnk_r = dlnk₂ + dlnK*

which is the same as:

dlnk_r /dT = dlnk₂/dT + dlnK*/dT

Since:

dlnk_r /dT = E_a / RT²

dln K*/dT = ∆H*/ RT²

(where: E_a = energy of activation and ∆H* = standard enthalpy of activation).

Therefore:

E_a/ RT² = dlnk₂/dT + ∆H*/ RT²

It is experimentally observed that for reactions in solution,

E_a = ∆H*

Hence,

dlnk₂/dT = 0

Since k₂ = (κ k_BT/h) where κ is the transmission coefficient (i.e., the fraction of activated complex crossing forward to yield the products), k_B and h are the Boltzmann’s constant and Planck’s constant respectively, T is the temperature in kelvin.

Therefore:

dlnκ /dT + dlnT/dT = 0

or

dlnκ = − dlnT

Integrating over dlnk from κ₁ to κ₂, and over dlnT from T₁ to T₂:

ln (κ₁ / κ₂) = ln (T₂ / T₁)

Taking ln ⁻¹ on both sides we get:

(κ₁ / κ₂) = (T₂ / T₁)

Which means: κ₁ is proportional to 1/ T₁ and κ₂ is proportional to 1/ T₂.

In general, κ is proportional to 1/ T which means: higher the temperature, lower the value of transmission coefficient. Lower the value of transmission coefficient, the fraction of the concentration of activated complex crossing forward to yield the products will be less. Lesser the concentration of activated complex crossing forward to yield the products, slower is the rate of reaction.

Conclusion: with the increase in temperature, the rate of reaction decreases.

Experimental Observation: The rate of reaction always increases with temperature. But in the case of enzyme catalyzed reactions, the rate increases with temperature up to certain level (corresponding to optimum temperature) after which the rate decreases with the increase in temperature.

Note: In the absence of information to the contrary, κ is assumed to be about 1. κ =1 implies no activated complex reverts back to the reactants (i.e., the activated complex always proceeds to products and never reverts back to reactants) and this assumption nullifies the description of equilibrium between the activated complex and the reactants and invalidates the quasi or rapid pre-equilibrium assumption.

Compton Effect

In physics, we define the kinetic energy of an object to be equal to the work done by an external impulse to increase velocity of the object from zero to some value v. That is,

KE = J × v

Impulse applied to an object produces an equivalent change in its linear momentum. The impulse J may be expressed in a simpler form:

J = ∆p = p₂ − p₁

where p₂ = final momentum of the object = mv and p₁ = initial momentum of the object = 0 (assuming that the object was initially at rest).

Impulse = mv

KE = mv²

In relativistic mechanics, we define the total energy of a particle to be equal to the sum of its rest mass energy and kinetic energy. That is, Total energy = rest energy + kinetic energy

mc² = m₀c² + KE

Solving KE = mv² we get:

m = m₀/ (1− v²/c²)

But according to Albert Einstein’s law of variation of mass with velocity,

m = m₀ / (1− v²/c²) ^½

m = m₀/ (1− v²/c²) ^½ implies transverse mass

m = m₀/ (1− v²/c²) ^3/2 implies longitudinal mass

m = m₀/ (1− v²/c²) →?

But according to the above equations: When v=c, m approaches infinity and if v>c, then m becomes imaginary i.e., these equations restrict body to move with speed equal or more than c.

Draft of letter from Bohr to Heisenberg, never sent.

In the handwriting of Niels Bohr's assistant, Aage Petersen.

Undated, but written after the first publication, in 1957, of the Danish translation of Robert Jungk, Heller als Tausend Sonnen, the first edition of Jungk's book to contain Heisenberg's letter

Dear Heisenberg,

I have seen a book, “Stærkere end tusind sole” [“Brighter than a thousand suns”] by Robert Jungk, recently published in Danish, and I think that I owe it to you to tell you that I am greatly amazed to see how much your memory has deceived you in your letter to the author of the book, excerpts of which are printed in the Danish edition [1957].

Personally, I remember every word of our conversations, which took place on a background of extreme sorrow and tension for us here in Denmark. In particular, it made a strong impression both on Margrethe and me, and on everyone at the Institute that the two of you spoke to, that you and Weizsäcker expressed your definite conviction that Germany would win and that it was therefore quite foolish for us to maintain the hope of a different outcome of the war and to be reticent as regards all German offers of cooperation. I also remember quite clearly our conversation in my room at the Institute, where in vague terms you spoke in a manner that could only give me the firm impression that, under your leadership, everything was being done in Germany to develop atomic weapons and that you said that there was no need to talk about details since you were completely familiar with them and had spent the past two years working more or less exclusively on such preparations. I listened to this without speaking since [a] great matter for mankind was at issue in which, despite our personal friendship, we had to be regarded as representatives of two sides engaged in mortal combat. That my silence and gravity, as you write in the letter, could be taken as an expression of shock at your reports that it was possible to make an atomic bomb is a quite peculiar misunderstanding, which must be due to the great tension in your own mind. From the day three years earlier when I realized that slow neutrons could only cause fission in Uranium 235 and not 238, it was of course obvious to me that a bomb with certain effect could be produced by separating the uraniums. In June 1939 I had even given a public lecture in Birmingham about uranium fission, where I talked about the effects of such a bomb but of course added that the technical preparations would be so large that one did not know how soon they could be overcome. If anything in my behaviour could be interpreted as shock, it did not derive from such reports but rather from the news, as I had to understand it, that Germany was participating vigorously in a race to be the first with atomic weapons.

Besides, at the time I knew nothing about how far one had already come in England and America, which I learned only the following year when I was able to go to England after being informed that the German occupation force in Denmark had made preparations for my arrest.

All this is of course just a rendition of what I remember clearly from our conversations, which subsequently were naturally the subject of thorough discussions at the Institute and with other trusted friends in Denmark. It is quite another matter that, at that time and ever since, I have always had the definite impression that you and Weizsäcker had arranged the symposium at the German Institute, in which I did not take part myself as a matter of principle, and the visit to us in order to assure yourselves that we suffered no harm and to try in every way to help us in our dangerous situation.

This letter is essentially just between the two of us, but because of the stir the book has already caused in Danish newspapers, I have thought it appropriate to relate the contents of the letter in confidence to the head of the Danish Foreign Office and to Ambassador Duckwitz.

Compton Effect-- An effect published in the Physical Review that explained the x-ray shift by attributing particle-like momentum to light quanta – discovered by American physicist Arthur Compton in early 1920s at Washington University in St. Louis, which amply confirmed the particle behavior of photons at a time when the corpuscular nature of light suggested by photoelectric effect was still being debated. This effect is suggested that when an x-ray quantum of energy hυ and a momentum h/ λ interacts with an electron in an atom, which is treated as being at rest with momentum = 0 and energy equal to its rest energy, m₀c². The symbols h, υ, and λ are the standard symbols used for Planck’s constant, the photon’s frequency, its wavelength, and m₀ is the rest mass of the electron. In the interaction, the x- ray photon is scattered in the direction at an angle θ with respect to the photon’s incoming path with momentum h/ λ_s and energy hυ_s. The electron is scattered in the direction at an angle φ with respect to the photon’s incoming path with momentum mv and energy mc² (where m is the total mass of the electron after the interaction). The phenomenon of Compton scattering may be analyzed as an elastic collision of a photon with a free electron using relativistic mechanics. Since the energy of the photons (661. 6 keV) is much greater than the binding energy of electrons (the most tightly bound electrons have a binding energy less than 1 keV), the electrons which scatter the photons may be considered free electrons. Because energy and momentum must be conserved in an elastic collision, we can obtain the formula for the wavelength of the scattered photon, λ_s as a function of scattering angle θ: λ_s = {(h/m₀c) × (1− cosθ) + λ} where λ is the wavelength of the incident photon, c is the speed of light in vacuum and (h/m₀c) is λ_C the Compton wavelength of the electron (which characterizes the length scale at which the wave property of an electron starts to show up. In an interaction that is characterized by a length scale larger than the Compton wavelength, electron behaves classically (i.e., no observation of wave nature). For interactions that occur at a length scale comparable than the Compton wavelength, the wave nature of the electron begins to take over from classical physics).

λ_s = λ_C (1− cosθ) + λ

λ_C = (λ_s – λ) / (1− cosθ)

It has been experimentally observed that for θ = 0^o there is no change in wavelength of the incident photon (i.e., λ_s = λ). Under this condition the Compton wavelength of the electron (which is = 2.42 × 10 ^–12 m) becomes undefined i.e.,

λ_C = 0/0.

The rate of transfer of photon energy to the electron i.e., − (dE/dt), is given by the relation: − (dE/dt) = hυ², where E = hυ. But υ = c/λ. Therefore:

dλ= c × dt

Integrating over dλ from λ (the wavelength of the incident photon) to λ_s (the wavelength of the scattered photon), and over dt from zero to t:

(λ_s − λ) = c × t

Since λ_s− λ = h/m₀c × (1− cosθ) – which Arthur Compton derived in his paper “A Quantum Theory of the Scattering of x-rays by Light Elements” by assuming that each scattered x-ray photon interacted with only one electron. Therefore:

t = h/m₀c² × (1− cosθ)

For θ = 0^o: t = 0 (i.e., scattering process is instantaneous at θ = 0^o). Under this condition h/m₀c² becomes undefined i.e., h/m₀c² = 0/0.

Velocities of recoil of the scattering electrons have not been experimentally determined. This is probably because the electrons which recoil in the process of the scattering of x-ray photons have not been observed. However, velocity of recoil of the scattering electrons can be calculated using the

• 
  Law of Conservation of Energy.
  
• 
  Law of Conservation of Momentum.
  

The conservation of energy merely equates the sum of energies before and after scattering i.e., the energy of the x-ray photon, hυ, and the rest energy of the electron, m₀c², before scattering is equal to the energy of the scattered x-ray photon, hυ_s, and the total energy of the electron, mc², after scattering i.e.,

hυ + m₀c² = hυ_s + mc²

or

(hυ – hυ_s) = mc² – m₀c²

But according to law of variation of mass with velocity (which states that mass and energy are “only different expressions of the same thing,” even though mass is a relativistic invariant, i.e., a four-dimensional scalar, while energy is the fourth component of a four-dimensional vector),

mc² = m₀c² / (1− v²/c²) ^½

Therefore:

(hυ – hυ_s) = m₀c² {1 / (1− v²/c²) ^½ − 1}

For θ = 90^o

hυ = 28.072 × 10⁻³⁶ Joules, hυ_s = 27.226 × 10⁻³⁶ Joules

Therefore:

(28.072 × 10⁻³⁶ – 27.226 × 10⁻³⁶) = m₀c² {1 / (1− v²/c²) ^½ − 1}

(28.072 × 10⁻³⁶ – 27.226 × 10⁻³⁶) = 81.9 × 10⁻¹⁵ × {1 / (1− v²/c²) ^½ − 1}

(28.072 − 27.226) × 10⁻³⁶ = 81.9 × 10⁻¹⁵ × {1/ (1− v²/c²) ^½ − 1}

(0.846 × 10⁻³⁶ / 81.9 × 10⁻¹⁵) + 1 = 1/ (1− v²/c²) ^½

[1.0329 × 10 ⁻²³ + 1] = 1/ (1− v²/c²) ^½

Since: 1.0329 × 10 ⁻²³<<<< 1. Therefore: [1.0329 × 10 ⁻²³ + 1] ≈ 1

1 = 1/ (1− v²/c²) ^½

From this it follows that

v = 0 (illogical and meaningless result because v = 0.04 c – which is shown below).

The principle of the conservation of momentum accordingly demands that the momentum of recoil of the scattering electron shall equal the vector difference between the momenta of these photons. The momentum of the electron, p_e= m₀cv/ (c² − v²) ^½, is thus given by the relation

m₀² c²v²/ (c² − v²) = p² + p_s² − 2pp_s cosθ

Solving p ² = (h / λ) ² = 87.553 × 10 ^{− 48} J²s²/m², p_s ² = (h / λ_s) ² = 82.355 × 10 ^{− 48} J²s²/m² and θ = 90^o, we get:

m₀² c²v²/ (c² − v²) = (p² + p_s²)

m₀² c²v²/ (c² − v²) = (87.553 + 82.355) × 10 ^{− 48}

m_e² c²v²/ (c² − v²) = 169.908 × 10 ^{− 48} J²s²/m²

But m₀²c² = 745.29×10 ⁻⁴⁶ J². Therefore:

v²/ (c² − v²) = (169.908 × 10 ⁻⁴⁸ / 745.29×10 ⁻⁴⁶) = 2.279 × 10⁻³

v² = 2.279 × 10⁻³c² − 2.279 × 10⁻³v²

v² (1 + 2.279 × 10⁻³) = 2.279 × 10⁻³c²

From this it follows that

v = 0.04c