quantum mechanics - Is it possible to synthesize an EMW which is not quantized?

( I initially started to ask, "since according to Quantum-theory of light; the energy of a photon, depends only on the frequency of light-wave (E = h * nu), and no-mention of amplitude. So, does the energy of light doesn't depend upon amplitude?" But that question has already asked by >1 users , but I have 1 more question related to that )

Some of the other-answers suggests, the EMWs having same frequency but different amplitude, is a mixture-condition of photons.

Then, is-it really impossible to synthesize an electromagnetic wave of such an amplitude, so-that the energy is NOT an integral multiple of the basic photon's energy, using a dipole antenna and an oscillator ? like this? Fig-1, Electromagnetic-waves could be synthesized by creating an oscillating electric field in a dipole antenna. (image from http://www.kshitij-iitjee.com/production-of-electromagnetic-waves-by-an-antenna)

I think, in this-way, all-kinds of frequencies and amplitudes could be synthesized (at-least theoretically). Fig 2. EMWs with same frequency but varying amplitude. in Wave-2, I choose the value randomly, just to say, any-value could be placed arbitrarily. Neglect drawing-errors.

Then , what-thing restricting the antenna-system to produce an wave with such an amplitude?

Related question : Where is the amplitude of electromagnetic waves in the equation of energy of e/m waves? Where is the amplitude of electromagnetic waves in the equation of energy of e/m waves?

Answer

Possibly you are mixing different concepts together. A short answer is the following: indeed, for a single photon, the energy is proportional to its frequency. That is

\begin{align} E_{singlephoton}=h\nu. \end{align} However, for a wave package or an electromagnetic wave, the total energy is proportional to the photon number, $n_{photon}$, as well. That is \begin{align} E_{wave} = n_{photon}E_{singlephoton}=n_{photon}\cdot h\nu. \end{align} Therefore, the amplitude of an electromagnetic wave is proportional to both the photon number and frequency.

When you changing the envelope of the electromagnetic field, the power--which is the energy flow per unit time--is the vector product of the electrical fields and the magnetic fields integrated over the area you are interested in. That is $$P=\int \mathbf{E}\times \mathbf{B}\cdot d\mathbf{s},$$ where $\mathbf{E}$ and $\mathbf{B}$ are the electrical and magnetic field vectors respectively, and $d\mathbf{s}$ is the area element of the field. From the Maxwell equations, we know that \begin{align} \nabla\times \mathbf{E} &= -\frac{\partial\mathbf{B}}{\partial t}\\ \nabla\times \mathbf{H} &= \mathbf{J}+\frac{\partial\mathbf{D}}{\partial t} \end{align} with $\mathbf{D}$ the familiar displacement of the electrical field. Therefore, to generate the time-dependent and propagating electrical field as shown in your figures, you need to have a time-varying E-M field in some particular forms through the oscillation (changing spatial distribution of fields over time as shown in the Maxwell equations above) of the dipole field in the antenna. Roughly speaking, the oscillating frequency and the oscillation amplitude of the dipole of the antenna determine the energy and wave envelope of the propagating wave together. If the frequency is fixed, the larger the oscillation amplitude is, the more photons this antenna will generate in a unit time. You can change the oscillation frequency and amplitude of the dipole to generate arbitrary shapes of waves through some engineering techniques. A detailed analysis may take longer, but hopefully you get the main idea.

Since this is an exercise problem, I would let you to figure out the quantitative solution of the problem to check your understanding for yourself. For example, part (a) and (b) obviously have the same frequency but different amplitudes. You could use the ideas outlined above to solve the whole problem. Wish this helps.

material science - Why can one bend glass fiber?

Why can one bend glass fibers without breaking it, whereas glasses one comes across in real life is usually solid?

Is there also a good high-school level explanation of this?

supersymmetry - Mathematically: What is SUSY?

Wikipedia says:

In particle physics, supersymmetry (often abbreviated SUSY) is a symmetry that relates elementary particles of one spin to other particles that differ by half a unit of spin and are known as superpartners. In a theory with unbroken supersymmetry, for every type of boson there exists a corresponding type of fermion with the same mass and internal quantum numbers (other than spin), and vice-versa. There is only indirect evidence for the existence of supersymmetry [...]

I want a mathematical explanation of SUSY.

Answer

Mathematically, SUSY begins with the supersymmetry algebra, a Lie superalgebra, which is itself a special case of a more general class of algebras called graded Lie algebras. Of central importance is the supersymmetry algebra referred to as the super-Poincare algebra that extends the Poincare algebra to include supersymmetry "charges" and their anticommutators.

In the context of physics, one studies field theories, both classical and quantum, that exhibit invariance under some action of supersymmetry algebras on fields and Hilbert spaces of these theories. As a result, representations of supersymmetry algebras are especially important in physics.

I would highly recommend that you look at THIS set of notes written by Sohnius, one of the original supersymmetry masters and co-discoverers of THIS famous and important theorem which really motivates why supersymemtry is all the rage in physics. The notes talk about representations of supersymmetry algebras in a lot of detail, and the clarity of the prose is top-notch if you ask me.

Addendum. I almost forgot, you also hear the word "superspace" which is a construction that physicists use to, among other things, make constructing manifestly supersymmetric Lagrangians easier. The mathematics behind this is supermanifolds.

Lastly, there is some discussion of these things on math.SE, see for example

https://math.stackexchange.com/questions/1204/why-are-superalgebras-so-important https://math.stackexchange.com/questions/51274/motivation-for-supermanifolds

homework and exercises - Confusion between thermal energy and heat

In the past, I usually misunderstood that thermal energy and heat are the same. However, some materials say that thermal energy is the intrinsic value of the system, including potential energy and kinetic energy, just as internal energy. Others say that it is just the average kinetic energy.

So, can anyone please clarify about thermal energy, heat, temperature and internal energy? Thanks a lot.

Answer

Expanding on the last point by @KevinZhou, temperature is most properly thought of as the "willingness" of the system to transfer heat. It is, in fact defined as

$\frac{1}{T} = \frac{\partial S}{\partial E}$

where $S$ is entropy and $E$ is internal energy. For a system consisting of two objects in thermal contact they will exchange energy such that they maximize the entropy of the system is maximized. So, if one can lose a little bit of entropy by giving up some energy and the other will increase its entropy by a large amount by taking in that energy then the energy will "flow" so that this happens. In other words, the system with a large $\partial S/\partial E$ "grabs" the energy from the one with small $\partial S/\partial E$. That is, energy flows from the one with high $T$ to the one with low $T$. The great thermal physics textbook by Schroeder makes the analogy that $E$ is "money", $S$ is "happiness" and $T$ is "generosity". The more generous system gives money to the less generous one to increase the overall happiness.

@KevinZhou says that temperature can be thought of as a measure of the amount of thermal energy. Most of the time you can get away with this but it isn't technically correct. We can form a relation between temperature and thermal energy for any system, but this isn't what temperature "is" (similarly $\sum F = ma$ means the sum of forces is equal to $ma$ not that the sum of forces "is" $ma$. Fundamentally the temperature is a measure of the system's willingness to transfer heat to other systems.

homework and exercises - Find angle of inclination when given mass, coefficient of friction, acceleration, and applied force

I have a problem that involves pushing an object with a:

• Mass of 80 kg


• Up a sloped plane (angle of incline is the unknown)


• With an applied force of 700N


• At an acceleration of 4.88 m/s^2


• With a coefficient of friction 0.07 acting against it

My goal is to find the angle that allows for that acceleration, using the 700N of force. I first flipped the axis to make the x-axis go up the inclined plane. So I can find the angle using the sum of the x-forces.

The sum of the x-forces (or m*a) = Applied force - (Weight force y component * coefficient of friction) - (Weight force of the x component).

I rewrote this as m*a = Applied - (coefficient of friction * w * cos theta) - (w * sin theta).

After trying for solve for the angle, I got stuck at (Applied - m*a) / w = (coefficient of friction * cos theta) + sin theta.

Is there a better way to solve this, without ending up with two different trig functions?

Answer

Construct an auxiliary right-angle triangle with angle $\alpha$, opposite side $\mu$ and adjacent side $1$. The hypotenuse is hence $\sqrt{1+\mu^2}$, and therefore

$$\mu \cos\theta+\sin\theta=\sqrt{1+\mu^2}\sin\alpha\cos\theta+\sqrt{1+\mu^2} \cos\alpha\sin\theta$$ $$=\sqrt{1+\mu^2}\sin(\theta+\alpha)$$ where $$\alpha=\tan^{-1}\mu$$

Can you take it from here?

electromagnetism - Measuring the spin of a single electron

Is it possible to measure the spin of a single electron? What papers have been published on answering this question? Would the measurement require a super sensitive SQUID, Superconductive Quantum Interference Device?

Answer

The spin of a single electron has been measured since the very first moment when the people understood that every electron possesses a spin. A Stern-Gerlach experiment - a magnetic field - is enough to measure the spin:

http://en.wikipedia.org/wiki/Stern-Gerlach_experiment

Electric field due to two opposite charges

We know that The net electric field due to two equal and oppsite charges is 0.

But let us consider a charge +Q in an isolated system. An electric field E will be emitted by it.

Similarly, a -Q charge will absorb an electric field E.

When both are in the same system then the magnitude field emitted by +Q and -Q should be added.

So can it be said that an electric field 2E is between +Q and -Q.

If an electric field E is generated by +Q towards -Q

And an electric field E is generated by -Q towards itself.

If both are in same direction, shouldn't both be aded to give 2E between them.

Of course the net will remain 0.

Answer

The net electric field due to two equal and oppsite charges is 0.

This is only true if the two charges are located in the exact same location. For example, a block of copper sitting on your lab bench contains an equal amount of electrons and protons, occupying the same volume of space, so the block of copper produces no net external electric field.

But if you separate the two charges from each other, they will produce a non-zero electric field everywhere in space. (This field will get very weak, but still non-zero, at locations much further from the charges than the distance between the charges)

It's actually easier to produce "zero net field" using two equal and same-signed charges.

For example, if I have two point charges with charge $+Q$ at locations $+x$ and $-x$ on the x-axis, then they will produce zero net field at the origin, since the field from one charge will be pointing right and the field from the other will be pointing left.

So can it be said that an electric field 2E is between +Q and -Q.

If an electric field E is generated by +Q towards -Q

And an electric field E is generated by -Q towards itself.

If both are in same direction, shouldn't both be aded to give 2E between them.

Exactly.

Of course the net will remain 0.

I'm not quite sure what you mean by this.

It is true in the sense of Gauss's Law. If you construct a closed surface around the two charges then the net electric flux through the surface will be 0.

But that's not the same as saying the "net field" is 0.

The net field is measured at a point in space, and is non-zero everywhere in your system. The net flux is measured over a surface, and can be zero if you construct the surface to contain the two charges.