special relativity - Does $p=mc$ hold for photons?

Known that $E=hf$, $p=hf/c=h/\lambda$, then if $p=mc$, where $m$ is the (relativistic) mass, then $E=mc^2$ follows directly as an algebraic fact. Is this the case?

Answer

As you may know, photons do not have mass.

Relating relativistic momentum and relativistic energy, we get:

$E^2 = p^2c^2+(mc^2)^2$.

where $E$ is energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light.

As mass is zero, $E=pc$.

Now, we know that $E=hf$. Then we get the momentum for photon.

Note that there is a term called effective inertial mass. Photon does have it.

determinism - What are the principles of deterministic chaos?

I see in literature very different (and chaotic) descriptions of what is deterministic chaos.

Can you explain to me based in a type of formal definition, which principles need to be exactly fulfilled in order to justify a system's behavior is determistic chaos?

optics - Does light reflect if incident at exactly the critical angle?

A lot of textbooks and exam boards claim that light incident at exactly the critical angle is transmitted along the media boundary (i.e. at right-angles to the normal), but this seems to violate the principle of reversibility in classical physics. How would a photon or ray travelling in the reverse direction "know" when to enter the higher refracting medium? It can't know, so I conclude that such light is simply reflected?

Is this correct?

optics - Using complex numbers to represent waves

When talking about a plane wave of the form $$\vec E=\vec E_0\cos(\vec k \cdot \vec r-\omega t)$$ We can replace it by $$\vec E=\vec E_0\exp[i(\vec k \cdot \vec r-\omega t)]$$ so that it is easier for calculation and then only taking the real part as the physical quantity.

How can we be sure that the complex part of the wave equation will never become real and contribute to our calculations and let us arrive at an incorrect answer?

Answer

We have

\begin{align} E(r,t) &= E_0 \cos(kr - \omega t - \phi)\\ &=E_0\left(e^{i(kr-\omega t -\phi)} + e^{-i(kr - \omega t + \phi)} \right)\\ &= \tilde{E}_0 e^{i(kr-\omega t)} + \tilde{E}^* e^{-i(kr-\omega t)}\\ &= E^{(+)} + E^{(-)} \end{align}

Note I've set $\tilde{E}_0 = E_0 e^{-i\phi}$.

By (a particular choice of) convention, the first term is called the positive frequency term and the second term is called the negative frequency term.

As has been mentioned, if you are adding waves or performing other linear manipulations you can drop the negative frequency term and just work with the positive one, adding in the corresponding negative frequency term at the end of the calculation to recover a real final answer.

If you are confused, I recommend performing the manipulations I have shown above so that you have an expression in terms of complex exponentials rather than sines and cosines. However, instead of dropping the negative rotating part as is often (somewhat mysteriously) recommended in my courses and textbooks, just go ahead and keep it. You now have two terms to drag around instead of one but you will see that it is easier to perform manipulations on the complex expressions instead of sinusoidal expressions. You will also see that whenever you do something to the positive frequency term you basically do the complex conjugate thing to the negative frequency term. Then, if you insist, you can take the intuition you have built to understand that in some circumstances you can drop the negative frequency term so that you have less things to write down.

As a practice example try to answer the following question:

What is the amplitude and phase of the wave which is the sum of the two following waves:

\begin{align} E_1(r,t) &= E_1 \cos(kr-\omega t -\phi_1)\\ E_2(r,t) &= E_2 \sin(kr - \omega t + \phi_2) \end{align}

This can be solved in sinusoidal form using trig identities for sums and differences inside sines and cosines. It can also be solved using complex exponentials as described above. I recommend doing it both ways to see the differences.

Finally, to really put the nail in the coffin of this sinusoidal vs. exponential representation I recommend using the exponential formulas to derive the trigonometric identities needed to solve the above exercise. These trig identities can alternatively be derived from geometric considerations and drawing funny triangles and labeling the sides but I have a very hard time doing it that way. Once you get used to it, it is quite simple to prove them using the exponential representation as I hope you'll discover.

edit: Let me add a bit more to directly address your question: You ask how we know the complex part won't become real and influence the answer. What you should realize is that when you are working in the complex representation (after having ignored the negative frequency part) the complexity of the expression is actually critical to capture the phase of the wave. What you should in fact be concerned about is "how do we know that the negative frequency part will not develop a positive frequency part and affect the answer?" The answer is that if all of the operations are linear then it is only the coefficients of the exponentials which are affected, never the arguments of the exponentials which contain the phase and frequency terms. However, if you begin multiply waves (say you're mixing or homo/heterodyning signals or looking at other non-linear processes) you will see that terms pop up with different factors up in the exponential than you had originally. In this case I would not recommend dropping negative frequency terms as you could easily miss something.

To summarize: In sinusoidal representation the relevant information is contained in the amplitudes and phases of the sines and cosines and in the complex representation the relevant information is contained in the (complex) amplitude of the coefficients of the positive and negative frequency terms, bearing in mind that the information in the coefficient of the positive frequency term is redundant with the information in the negative frequency coefficient as those two terms are complex conjugates of each other.

astronomy - How can a Population III star be so massive?

How can a Population III star have a mass of several hundred solar masses? Normally the limit is about 100 solar masses.

Answer

I think there are really three questions that need to be answered for this to make sense:

1. is there a "normal" limit to how large a star can be?


2. how can population III stars form with such large masses?



3. how can population III stars retain their large masses?

An answer to the first question is tricky. We expect large stars to be rare, and the largest stars to be the rarest. On top of this, they'll lead the shortest lives. Getting observational constraints has thus been tricky. There might be a limit to the amount of mass that is available to turn into stars when they form. As for the "normal" limits on the masses of stars, most (as far as I know) involve around pulsational instability. But the recent discovery of massive stars in and near the cluster R136a suggests that stars with masses over 150 solar can form even in material that has a non-negligible metal content. So whether there is a "normal" limit is open question.

The second question is much better understood, thanks to a lot of numerical work. Tom Abel recently wrote an article for Physics Today that summarizes current understanding of pop III star formation. Basically, the smallest amount of gas unstable to collapse under its own gravity, the Jeans Mass, increases with temperature (like T^3/2). So the cooler the gas can become, the smaller the fragments we expect to see. What determines how cool the gas can become? The atoms and molecules that radiate within it, and whether this radiation can escape. In metal-polluted gas, various molecular and atomic lines allow the gas to cool to tens of K. In metal-free material, the most effective coolant (in terms of the low temperatures it can achieve) is molecular hydrogen, which will only cool to around 200 K. This is a higher temperature, so we expect more massive fragments. This is a gross simplification! The situation really involves complex dynamics, shock formation, and all sorts of other stuff. Even the question of whether or not molecular hydrogen can form is contested.

Finally, if a massive pop III star formed, would it keep its mass? We know that the some massive stars in the local universe, like Eta Carinae, are violent beasts. This kind of episodic, pulsational mass loss could be present in Pop III stars, but since such mass loss is so poorly understood, this is often ignored. More generally, we expect that the metals in the atmospheres of massive stars absorb enough of the radiation created inside the star to be driven away in a wind. Again, there aren't any metals in metal-free gas, so we expect this effect to be much smaller in Pop III stars.

So, we expect Pop III stars to be larger because there is more gas available, because the gas fragments less owing to its higher temperature, and because we don't think the stars lose as much mass as modern stars do. And, we aren't even sure that there's a limit on how massive stars can be in the first place!

visible light - What "happens" to the energy of a photon after it is absorbed?

The simple model of the colour of reflected light from objects (yes, colour perception is a function of the eye/brain) as I understand it is:

Firstly I will write what I understand happens - which may be the source of my misunderstanding.

Consider white light incident on a material.

1. 
   

   Photons of particular wavelength can be absorbed by an atom by causing an electron to jump from its "base" state to some higher energy level.

   
   


2. 
   

   If the remainder of the incident light is reflected or transmitted the colour of the material is whatever the eye/brain perceives to be the colour of white light less the frequencies absorbed.

   
   


3. 
   

   The material must absorb a range of frequencies otherwise all colours of reflected/transmitted light would appear white until observed through a spectroscope which would indicate individual frequencies missing from the white light spectrum - which would be too little of the whole spectrum to notice.

   
   

Which leads to my actual question ....

If an electron has been "excited" by absorbing a photon from the incident light, surely at some moment later in time it will fall to a lower energy level and re-emit the original frequency absorbed? Hence there will be no "missing" frequencies from the reflected / transmitted light and every object will appear to be white? (but of course this does not happen).

Answer

If you are considering a single isolated atom then it's true that the atom has no way of getting rid of the energy from the photon except by emitting another photon. However as soon as the atom is surrounded by other atoms there are various mechanisms for radiationless decay i.e. transferring the energy of the absorbed photon into channels that don't involve reradiating the photon.

In a gas the excited atom or molecule can collide with another atom/molecule and transfer the excitation energy into kinetic energy. This is known as collisional de-excitation (that Wikipedia article is for collisional excitation, but de-excitation is the same process in reverse).

In a solid the energy can be transferred to lattice vibrations, i.e, heat, which is generally known as quenching. In fact in most solids quenching is so efficient that almost no energy is reradiated as photons. Reradiation in fluorescence or phosphorescence is the exception rather than the norm.

optics - Why do we see an inverted image in a spoon when kept far from our face with concave side towards our face?

We see a virtual inverted image whereas in case of concave mirror we can see a virtual image which is erect.