FULafiaPhysics

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We'll make FULafia best nuclear research centre in Africa - Mgt | The Nation Federal University of Lafia has procured ORTEC/AMETEK High Purity Germanium (HPGe) gamma/nuclear detector with its accessories such

Congratulations to the new Head of Department of Physics, Dr. Oladiran Johnson Abimbola.

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Dr. Oladiran Johnson Abimbola Federal University of Lafia located in Nasarawa State Nigeria, was established in February 2011 by the Federal Government of Nigeria. The University is one of the nine Universities that were created......

Deriving the Lorentz Transformation very simply:

For the relative orientation of the co-ordinate systems indicated in the figure, the x-axes of both systems pernumently coincide. In the present case we can divide the problem into parts by considering first only events which are localised on the x-axis. Any such event is represented with respect to the co-ordinate system K by the abscissa x and the time t, and with respect to the system K1 by the abscissa x' and the time t'. We require to find x' and t' when x and t are given.

A light-signal, which is proceeding along the positive axis of x, is transmitted according to the equation
x = ct
or
x - ct = 0 (1)

Since the same light-signal has to be transmitted relative to K’ with the velocity c, the propagation relative to the system K’ will be represented by the analogous formula
x' - ct' = 0 (2)

Those space-time points (events) which satisfy (x) must also satisfy (2). Obviously, this will be the case when the relation
(x' - ct') = λ (x - ct) (3)

is fulfilled in general, where λ indicates a constant; for, according to (3), the disappearance of (x - ct) involves the disappearance of (x' - ct').

If we apply quite similar considerations to light rays which are being transmitted along the negative x-axis, we obtain the condition
(x' + ct') = µ(x + ct) (4)

By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants a and b in place of the constants λ and µ, where
a = (λ + µ)/2
and
b = (λ - µ)/2
we obtain the equations
x’ = ax – bct
ct’ = act – bx (5)

We should thus have the solution of our problem, if the constants a and b were known. These result from the following discussion.
For the origin of K’ we have permanently x' = 0, and hence according to the first of the equations (5)
x = (bc/a)t

If we call v the velocity with which the origin of K’ is moving relative to K, we then have
v = bc/a (6)

The same value v can be obtained from equations (5), if we calculate the velocity of another point of K’ relative to K, or the velocity (directed towards the negative x-axis) of a point of K with respect to K'. In short, we can designate v as the relative velocity of the two systems.

Furthermore, the principle of relativity teaches us that, as judged from K, the length of a unit measuring-rod which is at rest with reference to K’ must be exactly the same as the length, as judged from K', of a unit measuring-rod which is at rest relative to K. In order to see how the points of the x-axis appear as viewed from K, we only require to take a " snapshot " of K’ from K; this means that we have to insert a particular value of t (time of K), e.g. t = 0. For this value of t we then obtain from the first of the equations (5)

x' = ax

Two points of the x'-axis which are separated by the distance Δx' = 1 when measured in the K’ system are thus separated in our instantaneous photograph by the distance
∆x = 1/a (7)

But if the snapshot be taken from K'(t' = 0), and if we eliminate t from the equations (5), taking into account the expression (6), we obtain
x’ = a(1 - v²/c²)x

From this we conclude that two points on the x-axis separated by the distance I (relative to K) will be represented on our snapshot by the distance
∆x’ = a(1 - v²/c²) (7a)

But from what has been said, the two snapshots must be identical; hence Δx in (7) must be equal to Δx' in (7a), so that we obtain
a = 1/(1 - v²/c²)

The equations (6) and (7b) determine the constants a and b. By inserting the values of these constants in (5), we obtain the first and the fourth of the equations given in the theory of special relativity.
x’ = (x -vt)/√(1 - v²/c²)
t’ = (t – vx/c²)/√(1 - v²/c²) (8)

Thus we have obtained the Lorentz transformation for events on the x-axis. It satisfies the condition

x'² - c²t'² = x² - c²t² (8a).

The extension of this result, to include events which take place outside the x-axis, is obtained by retaining equations (8) and supplementing them by the relations
y’ = y
z = z’ (9)

In this way we satisfy the postulate of the constancy of the velocity of light in vacuo for rays of light of arbitrary direction, both for the system K and for the system K'. This may be shown in the following manner.

We suppose a light-signal sent out from the origin of K at the time t = 0. It will be propagated according to the equation
r = √(x² + y² + z²)

or, if we square this equation, according to the equation

x² + y² + z² = c²t² = 0 (10)

It is required by the law of propagation of light, in conjunction with the postulate of relativity, that the transmission of the signal in question should take place — as judged from K1 — in accordance with the corresponding formula

r' = ct'
or,
x'² + y'² + z'² - c²t'² = 0 (10a).

In order that equation (10a) may be a consequence of equation (10), we must have

x'² + y'² + z'² - c²t'² = σ (x² + y² + z² - c²t²) (11)

Since equation (8a) must hold for points on the x-axis, we thus have σ = I. It is easily seen that the Lorentz transformation really satisfies equation (11) for σ = I; for (11) is a consequence of (8a) and (9), and hence also of (8) and (9). We have thus derived the Lorentz transformation.

The Lorentz transformation represented by (8) and (9) still requires to be generalised. Obviously, it is immaterial whether the axes of K’ be chosen so that they are spatially parallel to those of K. It is also not essential that the velocity of translation of K’ with respect to K should be in the direction of the x-axis. A simple consideration shows that we are able to construct the Lorentz transformation in this general sense from two kinds of transformations, viz. from Lorentz transformations in the special sense and from purely spatial transformations. which corresponds to the replacement of the rectangular co-ordinate system by a new system with its axes pointing in other directions.

Mathematically, we can characterise the generalised Lorentz transformation thus:

It expresses x', y', x', t', in terms of linear homogeneous functions of x, y, x, t, of such a kind that the relation

x'² + y'² + z'² - c²t'² = x² + y² + z² - c²t² (11a)

is satisficd identically. That is to say: If we substitute their expressions in x, y, x, t, in place of x', y', x', t', on the left-hand side, then the left-hand side of (11a) agrees with the right-hand side.

New Feynman Diagram! This diagram represents a neutral pion (which consists of either an up-antiup or down-antidown quark structure) decaying into two gamma particles. The angle in which they’re spread out represents the kinetic energy of the pion. The greater the angle, the lower the kinetic energy.

Quantum mechanics is a lot to handle at first. However, let's remember that there are postulates that guide the madness!

First Postulate:
At each instant the state of a physical system is represented by a ket |ψ⟩ in the space of states.

Second Postulate:
Every observable attribute of a physical system is described by an operator that acts on the kets that describe the system.

Third Postulate:
The only possible result of the measurement of an observable A is one of the eigenvalues of the corresponding operator Â.

Fourth Postulate:
When a measurement of an observable A is made on a generic state |ψ⟩, the probability of obtaining an eigenvalue a𝘯 is given by the square of the inner product of |ψ⟩ with the eigenstate |a𝘯⟩, |〈a𝘯|ψ|².

Fifth Postulate:
Immediately after the measurement of an observable A has yielded a value a𝘯, the state of the system is the normalized eigenstate |a𝘯⟩.

Sixth Postulate:
The time evolution of a quantum system preserves the normalization of the associated ket. The time evolution of the state of a quantum system is described by
|ψ(t)⟩ = Û(t, t₀)|ψ(t₀)⟩
for some unitary operator Û.

Brownian motion :
The erratic random movement of microscopic particles in a fluid, as a result of continuous bombardment from molecules of the surrounding medium.

The random movement of particles suspended in a fluid resulting from their bombardment by the fast moving atoms or molecules in the gas or liquid, and there is no preferred direction for these random oscillations.
In 1905, Einstein published a diffusion equation that predicts the average distance a pollen grain travels through a liquid in a given time. The equation is much more complex than the typical equation:

d² = y(T/γu).(t/N)

d - the distance the pollen grain travel during time t.
N - no. of atoms in a specific volume of the liquid.
y - a numerical constant.
T - Liquid's temperature.
γ - radius of the pollen grain.
u - Liquid's viscosity, it's resistance to motion.

Einstein's diffusion equation predicts the average distance pollen grains travel for any values of four parameters: grain size, time, and the liquid's temperature and viscosity.
Einstein's insight was to figure out how far they travel on average. This is called a 'random walk' or a ‘drunken sailor problem'. Einstein found it takes four times as long to travel twice as far when each step is taken in a random direction.

HEAT ENERGY can leap across a few hundred nanometres of a complete vacuum, thanks to a quantum mechanical phenomenon called the CASIMIR INTERACTION.

In a new study, University of California, Berkeley, researchers show that heat energy can travel through a complete vacuum thanks to invisible quantum fluctuations. In the experiment, the team placed two gold-coated silicon nitride membranes a few hundred nanometres apart inside a vacuum chamber. When they heated up one of the membranes, the other warmed up, too, even though there was nothing connecting the two membranes and negligible light energy passing between them.

➡️ Vibrations of atoms or molecules, which carry thermal energy, simply can't travel if there are no atoms or molecules around. But a new study by researchers at the University of California, Berkeley, shows how the weirdness of quantum mechanics can turn even this basic tenet of classical physics on its head.

🔹This [Casimir] interaction is only significant on very short length scales.🔸

➡️ "Even if you have empty space—no matter, no light—quantum mechanics says it cannot be truly empty. There are still some quantum field fluctuations in a vacuum," - said King Yan Fong, a postdoctoral scholar at UC Berkeley and the study's other first author. "These fluctuations give rise to a force that connects two objects, which is called the Casimir interaction. So, when one object heats up and starts shaking and oscillating, that motion can actually be transmitted to the other object across the vacuum because of these quantum fluctuations."

➡️ Though theorists have long speculated that the Casimir interaction could help molecular vibrations travel through empty space, proving it experimentally has been a major challenge. To do so, the team engineered extremely thin silicon nitride membranes, which they fabricated in a dust-free clean room, and then devised a way to precisely control and monitor their temperature.

➡️ They found that, by carefully selecting the size and design of the membranes, they could transfer the heat energy over a few hundred nanometers of vacuum. This distance was far enough that other possible modes of heat transfer were negligible—such as energy carried by electromagnetic radiation, which is how energy from the sun heats up Earth.

➡️ "Because molecular vibrations are also the basis of the sounds that we hear, this discovery hints that sounds can also travel through a vacuum", said Xiang Zhang, the professor of mechanical engineering at UC Berkeley who guided the study.

© University of California - Berkeley

NSPS CONFERENCE 2020

NSPS CONFERENCE 2020