Physics Forums FAQ

[ZapperZ]

This thread contains simple responses to the most frequently-asked questions in the Physics sub-forums. It is a continually evolving thread, and more FAQ's will be added.

For more in-depth coverage of Physics FAQ's, refer to

http://math.ucr.edu/home/baez/physics/

Please PM me all questions, comments, corrections, etc.

Zz.

[ZapperZ]

WHY DON’T ELECTRONS CRASH INTO THE NUCLEUS IN ATOMS?

Contributed by Marlon and edited by ZapperZ

If one describes atoms using only the Coulomb forces, the electron and the nucleus will attract each other and no stable atoms could exist. Obviously this is not the case. Niels Bohr was the first (1913) to propose a better model, which consisted of electrons moving around the nucleus in circular orbits. Each orbit corresponds to a certain discrete energy level. This model is based upon the quantisation of the angular momentum.

Unfortunately, electrons moving in a circular orbit have an acceleration due to the centripetal force. In classical electromagnetic theory, an accelerated charged particle must emit EM-radiation due to energy conservation. Hence, the electron would lose energy and spiral down towards the nucleus. Again stable atoms could not exist. What is wrong now?

It turns out that the picture of electrons moving in circular orbits around the nucleus isn’t correct either(*). The solution here is the implementation of Quantum Mechanics via the Schrödinger Equation and the concept of wavefunction. By applying such formalism, the “electron” occupies a volume of space simultaneously, so that it is “smeared” in a particular geometry around the nucleus. While there are no more “orbits”, we do use the term “orbitals” to indicate the shape of such geometry. However, this term should not be confused to mean an orbiting electron similar to our planets in the solar system. By describing the system in terms of the QM wavefunction, it creates stable states for the nucleus+electrons system that matches very well with experimental observation of standard atomic spectra.

Since there are no more “orbits” in the conventional sense, the problem of electrons radiating due to an accelerated motion is no longer meaningful. It explains why we have stable atoms.

To read more in detail: http://scienceworld.wolfram.com/physics/HydrogenAtom.html


(*) By saying that “an electron orbits the nucleus”, one is already implicitly assuming that one can track the position and momentum of that electron in an atom over a period of time. We have no such ability, and for those who know a bit about the Heisenberg Uncertainty Principle (HUP), one can already tell that such a statement violates this principle.

[ZapperZ]

IS LIGHT A WAVE OR A PARTICLE?

Contributed by Marlon and ZapperZ.

In our ordinary world, “wave” and “particle” behavior are two different and opposite characteristics. It is difficult for us to think that they can be one of the same. Is light a particle or a wave? The simple, naïve answer to that is “both” or “neither”.

Light, or photon, was never defined as a “particle” the way we normally define a particle. Light is not defined to have a definite boundary in space like a ping-pong ball, or a grain of sand. Instead, light is defined as having quanta of energy. So the discreteness is not defined as discrete object in space, but rather in the energy it can carry. Already, this is not your regular “particle”, and should not be confused as such.

Secondly, in quantum mechanics, the description and properties of light has only ONE, single, consistent formulation, not two. This formulation (be it via the ordinary Schrodinger equation, or the more complex Quantum Electrodynamics or QED), describes ALL characteristics of light – both the wave-like behavior and the particle-like behavior. Unlike classical physics, quantum mechanics does not need to switch gears to describe the wave-like and particle-like observations. This is all accomplished by one consistent theory.

So there is no duality – at least not within quantum mechanics. We still use the “duality” description of light when we try to describe light to laymen because wave and particle are behavior most people are familiar with. However, it doesn’t mean that in physics, or in the working of physicists, such a duality has any significance.

[ZapperZ]

Do Photons Move Slower in a Solid Medium?

Contributed by ZapperZ. Edited and corrected by Gokul43201 and inha

This question appears often because it has been shown that in a normal, dispersive solid such as glass, the speed of light is slower than it is in vacuum. This FAQ will strictly deal with that scenario only and will not address light transport in anomolous medium, atomic vapor, metals, etc., and will only consider light within the visible range.

The process of describing light transport via the quantum mechanical description isn't trivial. The use of photons to explain such process involves the understanding of not just the properties of photons, but also the quantum mechanical properties of the material itself (something one learns in Solid State Physics). So this explanation will attempt to only provide a very general and rough idea of the process.

A common explanation that has been provided is that a photon moving through the material still moves at the speed of c, but when it encounters the atom of the material, it is absorbed by the atom via an atomic transition. After a very slight delay, a photon is then re-emitted. This explanation is incorrect and inconsistent with empirical observations. If this is what actually occurs, then the absorption spectrum will be discrete because atoms have only discrete energy states. Yet, in glass for example, we see almost the whole visible spectrum being transmitted with no discrete disruption in the measured speed. In fact, the index of refraction (which reflects the speed of light through that medium) varies continuously, rather than abruptly, with the frequency of light.

Secondly, if that assertion is true, then the index of refraction would ONLY depend on the type of atom in the material, and nothing else, since the atom is responsible for the absorption of the photon. Again, if this is true, then we see a problem when we apply this to carbon, let's say. The index of refraction of graphite and diamond are different from each other. Yet, both are made up of carbon atoms. In fact, if we look at graphite alone, the index of refraction is different along different crystal directions. Obviously, materials with identical atoms can have different index of refraction. So it points to the evidence that it may have nothing to do with an "atomic transition".

When atoms and molecules form a solid, they start to lose most of their individual identity and form a "collective behavior" with other atoms. It is as the result of this collective behavior that one obtains a metal, insulator, semiconductor, etc. Almost all of the properties of solids that we are familiar with are the results of the collective properties of the solid as a whole, not the properties of the individual atoms. The same applies to how a photon moves through a solid.

A solid has a network of ions and electrons fixed in a "lattice". Think of this as a network of balls connected to each other by springs. Because of this, they have what is known as "collective vibrational modes", often called phonons. These are quanta of lattice vibrations, similar to photons being the quanta of EM radiation. It is these vibrational modes that can absorb a photon. So when a photon encounters a solid, and it can interact with an available phonon mode (i.e. something similar to a resonance condition), this photon can be absorbed by the solid and then converted to heat (it is the energy of these vibrations or phonons that we commonly refer to as heat). The solid is then opaque to this particular photon (i.e. at that frequency). Now, unlike the atomic orbitals, the phonon spectrum can be broad and continuous over a large frequency range. That is why all materials have a "bandwidth" of transmission or absorption. The width here depends on how wide the phonon spectrum is.

On the other hand, if a photon has an energy beyond the phonon spectrum, then while it can still cause a disturbance of the lattice ions, the solid cannot sustain this vibration, because the phonon mode isn't available. This is similar to trying to oscillate something at a different frequency than the resonance frequency. So the lattice does not absorb this photon and it is re-emitted but with a very slight delay. This, naively, is the origin of the apparent slowdown of the light speed in the material. The emitted photon may encounter other lattice ions as it makes its way through the material and this accumulate the delay.

Moral of the story: the properties of a solid that we are familiar with have more to do with the "collective" behavior of a large number of atoms interacting with each other. In most cases, these do not reflect the properties of the individual, isolated atoms.

[ZapperZ]

Contributed by vanesch.

The short answer is: no, there's no violation.

The longer answer can be this: Given the probabilistic aspect of quantum theory, what do we mean now, by "conservation of energy" ?

In quantum theory, it can be expressed in two different ways. The first way is this: A state with a precisely known energy will always keep this energy. The reason is that a state with a precisely known energy is an eigenstate of the Hamiltonian, and that's a stationary state under unitary evolution: so it remains (up to a phase factor) itself.

The second way goes as follows: for a given state, look at its EXPECTATION VALUE of the energy <psi | H | psi>. This is the statistical average over many trials of measuring the energy. Well, this expectation value is to remain the same during time evolution. It simply means that the energy value was not a well-defined quantity (only its expectation value was), and hence we cannot talk about violation of its conservation, given that it wasn't fixed initially.

And where does the time-energy uncertainty relationship come in ?
It tells you esentially that *in order to perform an energy measurement with precision dE*, you will need to measure (to have your measurement apparatus interact with) the system for a time of at least dt.

So this means that when you are discussing about a system for a time less than dt, that there is no visible difference between a stationary state with precisely energy E, or with a superposition of stationary states of which the energy eigenvalues lie within dE of E. Indeed, below a time dt, the unitary evolution equation (Schroedinger equation integrated) will not have altered significantly the phases between these contributions as can easily be verified (each term taking a factor exp(-i E t / hbar) ).
So you're not able to find out the difference between the two situations, and hence you cannot know whether the system is in such a superposition, or in a precise energy eigenstate. As such, the uncertainty is a matter of uncertainty on the INITIAL condition (was the system in a pure energy state or not ?) or of the energy transfer during the measurement interaction between apparatus and system. It is not a question of "stealing energy from nowhere" or something of the kind.

There are two typical cases:

1) the system was "created" in a time dt. This means that during its "creation interaction" one cannot be sure that it was in a pure energy eigenstate: it could be created in a superposition of eigenstates with eigenvalues spread over dE. So you measuring (precisely) the energy value just means you selected out one of the possible eigenstates of which the system was in a superposition: no violation of conservation of energy. This is often the case with "particle resonances" or other short-lived phenomena.

2) The system is *prepared* in a precise energy eigenstate, and you quickly measure, during time dt. In this case, it can be shown that the interaction between the system and the measurement apparatus can give rise to a transfer of energy of order of dE. So reading again another value (within dE) of the energy is then just part of the "perturbation" introduced by the energy measurement apparatus. Again no violation of energy conservation.

Finally, in the long run, the *expectation value* will be recovered, as the average of a great many number of measurements. So there will never be a net gain or loss of energy.

[ZapperZ]

Do Photons Have Mass?

Contributed by jtbell, Hootenanny, and ZapperZ.

The quick answer: NO.

However, this is where it gets a bit confusing for most people. This is because in physics, there are several ways to define and measure a quantity that we call "mass". Now, it doesn't create any confusion among physicists because we tend to know in what context such a quantity is defined. However, for the general public trying to decipher scientific papers and presentation, this is a trap that many fall into and can be the source of many confusion.

In physics, the most important definition of a bare mass (we are not going to deal with effective mass that is a part of solid state/condensed matter physics) is what is known as the invariant mass. Invariant mass (m_0) (aka rest mass, proper mass or intrinsic mass) is independent of reference frame. In other words, an object’s invariant mass has the same value no matter who is observing the object and no matter what their velocity is relative to the object. The invariant mass of a particle is defined as the total energy of the particle measured in the particle’s rest frame divided by the speed of light squared. More generally, the invariant mass is defined as

m_0 = \sqrt{\frac{E^2}{c^4} - \frac{p^2}{c^2}}

Now for a photon, this is zero since E = pc. In many aspect, this is all that we need to know. In physics, something that is invariant after some operation is very desirable.



But Photons Have Energy. By E = mc^2, Doesn't This Mean That They Have A Mass?

The equation above was derived from this expression:

E^2 = (pc)^2 + (m_0 c^2)^2

A photon can still have zero invariant mass (m_0), and can still have energy. There's nothing inconsistent here. All of the photon's energy is in the term pc. Some people would say that this is the photon's "inertial mass", since it is similar to the inertia that one feels when trying to stop a moving mass. This may or may not be useful to consider. However, it certainly should not be confused with the concept of the ordinary mass that most people are familiar with.

There are, of course, other forms of mass. Most commonly used terminology is something called "relativistic mass". This mass is defined as

m = \gamma m_0

where

\gamma = \frac{1}{\sqrt{1-\beta^2}}[/itex]

[tex]\beta = \frac{v}{c}

This relativistic mass is what most people attribute to the "gain in mass" of particle moving at relativistic speeds. However, one needs to be aware that in professional circles, such concept is very seldom used. One very seldom hears this when one attends a high energy physics seminar, for example, or read a particle collider experiment paper. This is because in citing a relativistic mass, one must also cite the speed of the particle with respect to what reference frame. This is cumbersome and unnecessary especially when the invariant mass would have been clearer (that's why we love invariant anything).

You can read more about this at the Usenet FAQ (http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html).

[Hootenanny]

Written by Hootenanny (http://www.physicsforums.com/member.php?u=44656). Edited by berkeman (http://www.physicsforums.com/member.php?u=8921) and Kurdt (http://www.physicsforums.com/member.php?u=6034). Feel free to contact me with any questions, comments or corrections.

What is Heat?

One of the most frustrating misconceptions in Thermodynamics centres on the actual definition of heat. Many science/engineering students often refer to a body as to be possessing heat, but as we shall see, to do so is completely nonsensical.

We start by examining the first law and defining each of it’s terms. The first law is simply a statement of the principle of conservation of energy and is often stated thus;

\Delta U = Q + W

Where \Delta U is the change in internal energy, Q is the heat added to the system and W is the work done on the system. So what do all these terms actually mean?

Internal EnergyWe define the internal energy as the energy associated with the microscopic energies of system, that is with the energy associated with the random motion of the molecules within a system. So for a general fluid, the internal energy of a system is the sum of the translational kinetic energies, the rotational kinetic energies, the vibrational kinetic energies and the potential energies of all the molecules in that system. The internal energy of a system is often erroneously referred to as the heat of a system and we shall see why this is incorrect later. One important point to note here is that the internal energy is a state variable, that is, the change in internal energy between any two states is independent of the path taken.

WorkWell, if you're reading this I assume that you know the definition of work; in thermodynamics work is usually associated with a transfer of energy into or out of a system. An example of work specific to thermodynamics would be the application of a force to a piston, which would then compress the gas within the cylinder, thus doing work on the gas. Since work is being done on the gas the W term in our expression would be positive. If we assume that the walls of the cylinder are adiabatic (no heat transfer) then all the work done would be converted to internal energy. Suppose that after we have compressed the piston, we release it. Intuitively, we would expect the piston to recoil back, and this is exactly what happens; the gas expands and does [an equal amount of] work on the piston against atmospheric pressure. In this case, since it is the gas that is doing work, our W term would be negative.

HeatSo we have defined the internal energy of a system and we can quantify the work done on the system, but what about heat? First let us examine temperature. One useful definition of temperature is often called kinetic temperature and is derived from kinetic theory. Using kinetic theory the temperature of a system is taken to be a measure of the average translational kinetic energy associated with the random motion of the molecules with the system. It should be noted that although related to internal energy, temperature is not directly proportional to internal energy since internal energy also involves the rotational and vibrational kinetic energies and the potential energies of the constituent molecules.

So, we come to the definition of heat. If we examine the first law, we can see that we can increase the internal energy of a system either by doing work on it, or adding heat to it. Consider a piston and a cylinder filled with gas, we can increase the internal energy of the system by either compressing the gas by applying a force to the piston (work) or by fixing the piston and placing the cylinder in a flame (heat). We can compress and heat the gas in such a way that after the operation all the macroscopic properties (pressure, volume & temperature) are identical, that is the two cylinders are in identical states. Suppose we take two identical cylinders (but not necessarily in identical initial states) filled with a gas at 373K, one of which we compress and the other of which we heat such that both cylinders are at 473K and all their macroscopic quantities are identical, that is the final states of the two cylinders are identical. If we were to now examine the final states of the two cylinders, we have no way of knowing which was compressed and which was heated; the only conclusion we can draw is that their internal energies have increased. In this way we can consider heat as the microscopic analogy of work (macroscopic). I therefore, offer you a formal definition of heat:
"Heat is the non-mechanical exchange of energy between the system and surroundings as a result of a difference in temperature"

Both work and heat can be considered as methods of transferring energy within or between systems. It should now be apparent why the statement "a body posses heat" is nonsensical. To say that a body posses heat is analogous to stating that a "body has work", which you must agree is utter rubbish. Rather, one transfers energy to a body by doing work on that body and one transfers energy to a body by heating or adding heat to that body. Similarly, it is incorrect to sate the a body's heat has increased, rather it's internal energy has increased.

A note about Thermal EnergySome texts make use of the term "thermal energy" when discussing the "translational kinetic energy" of the molecules, I personally find that the term "thermal energy" only serves to confuse discussions further.

Further Reading
Thermal Physics, 2nd Edition, C.B.P. Finn
Heat@Hyperphysics (http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/heat.html#c1)
Internal Energy@Hyperphysics (http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/inteng.html#c2)

[jtbell]

In general relativity, gravitation is a manifestation of the curvature of spacetime. The motion of all objects is affected by this curvature, regardless of whether they have mass or not. Light follows geodesic paths in spacetime, which are straight lines in flat spacetime, and curved paths in curved spacetime.

Note that by "mass" above I mean "invariant mass" as discussed in post #6 above, because it is the invariant mass that is zero for a photon. If you prefer to think in terms of "relativistic mass" (which is related to energy via E = m_{rel} c^2, note that all photons follow the same geodesics, regardless of their energy.