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-1

$$ \frac{\eta^{\mu\nu}}{\Box} $$


0

If we write $A_\mu(x)=\varepsilon_\mu(p)e^{ipx}$, the polarization vector should satisfy $\varepsilon_\mu p^\mu=0$, which is a Lorentiz-invariant relation, and is necessary to make sure that we have an irreducible representation of the Lorentz group (actually, the little group that leaves the momentum invariant). This knocks down the number of D.O.F to 3. ...


3

The longitudinal mode decouples from all physical processes as a consequence of gauge invariance, which in turn forces the Ward identity $$ k^\mu \mathcal{M}_\mu = 0$$ where the S-matrix element decomposition $\mathcal{M}^\mu$ is obtained from the polarization vector $\epsilon^\mu(k)$ by $\mathcal{M} = \epsilon^\mu(k) \mathcal{M}_\mu$. This decoupling (and, ...


1

Punk Physicist's answer is spot on. But I'd like to add a little to his/her last two paragraphs, in particular, a description of what it is that you see in an interference pattern. You cannot define a position observable, but you can of course define the state of the second quantized field. Moreover you can describe the probability amplitude for a photon to ...


2

There's an old argument by Newton and Wigner, that the photon as a massless particle can't have a position operator and therefore no position space wave function. The paper you're thinking of is T. Newton and E. Wigner, "Localized States for Elementary Systems," Rev. Mod. Phys. 21, 400–406 (1949) doi:10.1103/RevModPhys.21.400. Photons are ...


0

Wouldn't the photons or waves of EMF just fly away into space and be lost (the energy would be lost, not stored) One must distinguish between electromagnetic waves and, e.g., static electric and magnetic fields. Essentially, ('real') photons are associated with electromagnetic radiation (radio, light, x-rays, gamma rays). Electromagnetic waves ...


2

Your argument that the energy should radiate away would be true if your inductor were a good antenna, in which case it would be a bad inductor! The problem is an impedance mismatch: The inductor produces a magnetic field (which stores the energy you inquire about), but little electric field. That is the wrong ratio, or impedance, to couple to the vacuum ...


0

A moved charge - in this case an electron - which is accelerated in a circle (the inductive coil) will induce a magnetic field. How does the electron induce the magnetic field? The electron has a magnetic moment and it spins. The movement of the electron in the coil align the magnetic moment and all moved electrons induce the common magnetic field of the ...


1

suppose it is possible to accelerate matter at speed of light By this you must mean suppose that relativistic mechanics is, at its root, wrong. What will the time reflects on these two clocks? Since you've stipulated that relativistic mechanics is wrong, which incorrect, non-relativistic mechanics would you like to apply to this problem?


1

Now, what I'm wondering is this- do we have a single model/theory whose equations accurately describe the behavior of light in all scenarios? Yes, for all intents and purposes: QED. Clearly, we can not say that QED works at all scenarios (e.g., very tiny length scales and very high energy scales), since we can not test it in all scenarios. But, as ...


0

There used to be two models: wave and particle. QM mechanics does not pick particle over wave. QM actually picks neither and says either (all 3 interpretations are equally valid) it is both at the same time or it is something we can not yet understand.


-3

yes, we use a single model, called quantum mechanics. For interpretations of quantum mechanics please search the web.


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@dmckee is correct: the positronium atom indeed does have spdf energy levels. What threw me off was missing the fact that the s orbital does have a finite probability of 'collision' between the antiparticles and thus their annihilation. I was viewing the s sate as a Bohr 'orbital' where the electron could not spiral into the positron because it could not ...


2

The grand canonical ensemble allows the number of particles of your system to fluctuate but makes the assumption that it is constant amongst the reservoir and system combined i.e. $$ n_{res} + n_{sys} = const $$ For the case of photons this is not true.


0

Instead of the word light it would be better to use the word electromagnetism. Newton and Young were fascinated about the decomposition of white light into it colors and about fringes behind edges. Since Maxwell it was obvious that light was only a small part of the electromagnetic spectrum. Later were discovered the weak and the strong nuclear power. But in ...


0

Light exists in nature therefore physicists were bound to be interested in it at some point. But light is special in a sense. All we see, we see with light. Our eyes are sensitive to electromagnetic fields which make up this light. Furthermore, the main force which keeps the stars together is electromagnetism, which is the interaction of matter and light ...


-1

Light is a rare phenomenon as it has no mass. Most particles have mass and therefore cannot act in the way that light is, which is to move at the speed of light. At this speed, due to relativity light moves at the same speed regardless of your perspective; no matter how fast you are able to move you can never catch up to light, and even as you try harder to ...


10

Weinberg is right. The issue here is with the usual interpretation of the wavefunction as an amplitude density. This implies being able to localize the particle in an arbitrarily small region. However, it is not possible to localize photons (or any massless particles with spin, for that matter). The reason for this is the careful definition of what ...


5

Wave function for $N$ particles is a scalar function of $3N$ spatial coordinates. Wave function for $N$ photons, if named so at all, should thus also be a scalar function of $3N$ spatial coordinates. EM field, on the other hand, is a vector function of 3 spatial coordinates. Any chosen wave function is non-unique; there are many different wave functions, ...


0

What is Magnetic Field Made of (Not the Dipole which Cause but the Magnetic field lines)? No one knows. Electric and Magnetic fields are now the starting point (i.e. Natural Laws) in physics. No one knows what lies underneath them We just know Nature obeys these Laws. They are the starting point in building a physical theory. We assume Newton's third Law ...


4

I believe Weinberg is trying to make a distinction between two viewpoints of quantum particles, one historical (although we still use it when thinking about and teaching non-relativistic quantum mechanics) and one modern. In non-relativistic quantum mechanics, we typically start by assuming the existence of "particles" familiar to us from classical ...


0

Of course the photon will have an amount of energy and that is radiation energy which if defined by Planck's equation. Radiation energy is also related directly to wavelength,frequency and wavelength number.The higher the energy, the less is the wavelength ( all this from Planck's equation). What Heisenberg equation is about, does not include the energy.It ...


0

Although there is already an accepted answer, I want to give some further ideas: While the Bragg-peak is the "perfect" solution to get a high percentage of effective dose to the desired tissue, the exact positioning of this peak in the tumor tissue can be challenging due to misalignments of CT/MRI data and the application gantry and furthermore due to ...


0

In short, no, the individual photons are not actually being split into two new photons of lesser energy. It is true that in the classical interpretation of light, an ensemble of many distinct waves of photons, will create destructive and constructive interference patterns ( e.g. Young's Double Slit Experiment ). It is even more intriguing that these ...


0

Reflection like transmission are active process involving the electrons around the atoms interacting with the electric field of the incident light. If the intensity is too high, heating (vigorous all molecule movements) will result and the specimen is damaged. If the frequency is too high, the molecules are disfigured and the specimen is simply pierced ...


-1

I am going to answer the question in the simplest terms I know. Assume you have only one hydrogen atom. If a photon of the "correct" frequency "hits" the atom, its electron will jump to a higher excited state. What this means, is that the energy of the photon (the photon itself, since it has no mass) is used (goes into) to make the electron move to a ...


0

Photon has spin 1 :) I am serious. Check it if you don't believe me. The physical meaning of the photon propagator $\Delta_{\mu \nu}$ is the following: $$ \Delta_{\mu \nu} (x-y) = \left< A_{\mu}(x) A_{\nu}(y) \right>, $$ where $A_{\mu}$ is the electromagnetic potential form and $\left< ... \right>$ is a shorthand for the vacuum expectation ...


0

[...] since photon is spin-zero gauge boson. Why does it have gμν ? Does this somehow about Ward identity? Photon is not "spin-zero gauge boson". It is "spin one gauge boson".


0

Photon is a (one) quantum of electromagnetic radiation/light, ok? When it has low energy, means large wavelength, means low frequency, we know it as radio waves. Some higher energy, means higher frequency, means lower wavelength, we see it as light. At, even higher energy it is X-rays. Let us stop here. The formula for single photon's energy is ...


0

$eU$ is the energy of a charge $e$ after it's accelerated by a potential difference (voltage) $U$. In this case, yes, $e$ is the elementary charge. $hf$ is the energy of a photon (X-ray or otherwise) with frequency $f$. If we assume that all the energy of the accelerated charge gets converted into the energy of a photon, and here this will be the maximum ...


1

You only have a problem if you start without any tangential velocity and have no reflection (or you collide with a celestial body before you have a chance to evade it). Let us say you start in a stable orbit. Your ship has a bow and stern, in the direction of your velocity. If you can reflect the light towards your stern, it will accelerate you and you get ...


2

Some practical information first - the charged particle beam passing through the matter suffers from energy (and also angular) straggling. That means that even if an ideally monoenergetic beam is used, there will be always a finite volume with Bragg peak losses. The bigger initial energy, the bigger is the volume. Protons stopping at 40 mm have straggling ...


0

A photon with 0 frequency wouldn't interact with anything, so it is impossible to know if it exists.


3

We can represent a monochromatic electromagnetic wave by one of its fields, the $\vec E$ or $\vec B$ field (or $\vec H$ in the case of the diagram further down). For example, we can write $$\vec E (\vec r, t) = \vec E_0 e^{i(\vec k \cdot \vec r - \omega t)}$$ where the relevant part for this question is $\omega$, the angular velocity. When the frequency of a ...


1

It will help if your read the answer by Motl in a similar question Virtual photon description of B and E fields. Is it possible to create a beam of light with frequency of 0? It will not be light, as you approach 0, it will be a radio wave, i.e. it will be generated by an antenna. Think of starting with a very low frequency and keep on diminishing it ...


4

The uncertainty principle limits our ability to determine the wavelength of a particle with infinite precision. At the same time, there is no fundamental reason why any two photons (even if generated by exactly the same process) should produce exactly the same wavelength; however, you can be sure that there will be plenty that are the same within the limits ...


-1

Surely you'd agree that all electrons and protons are $exactly$ the same (indistinguishable). Now consider a regular Hydrogen-1 isotope: A proton and an electron bound together. There's definitely more than just one of these atoms in the observable universe. Well then consider an electron floating around the $n=2$ shell of a Hydrogen-1 isotope, then ...


0

It is confusing because they say superposition when they mean entanglement. The yellow beam containing the information about the cutout got it from an entangled red beam which made contact with the cutout. In fact, National Geographic has a much better article: ...


1

I believe that Paul's (http://www.mathpages.com/home/kmath210/kmath210.htm) source answers it brilliantly. "The main bunch of riders - may be moving at a constant speed. But within the bunch an individual rider may moving more slowly, dropping back for a rest or a drink." Coupled with Mr. Witthoft's response "What you're missing, Dirk, is that there is no ...


-7

Once emitted photons are indivisible units. When a photon hits an electron moving in the same direction,the photon will be absorpted partially and the electron emits an other photon with lower energy. This happens for example at linear particle accelerators. In both cases energy from the photon goes over to the electron and the electron moves faster. If a ...


5

How are photons created? An accelerating charged particle generates photons tangentially as well as a decelarating one. Where do these photons come from? From the energy carried by the electron. In this sense photons are just a packet of energy which is associated with the electromagnetic field. This type of interactions of electrons and ions with fields ...


18

When you turn on a lightbulb, you easily create many photons. They can go away just as easily. That's because they are bosons and they have no charge. Think of waves on a pond. Where do they "come from" when you throw a stone in? Where do they go when they dissipate? That's actually a very good analogy in some ways because the math that describes ...


0

are there things called infrared photons, or ultraviolet photons etc, as there are infrared waves, or ultraviolet waves? Yes, absolutely. Certain aspects of the photon notion can be thought of as even more like a classical wave than even the above. One photon alone can be in a pure quantum superposition of energy eigenstates, or polarization ...


4

A photon, unlike some other particles, has no number that must be conserved, thus when absorbed all of the energy present goes into exciting the particle which absorbed it, allowing no laws to be broken. This is due to Noether's Theorem. http://en.m.wikipedia.org/wiki/Noether%27s_theorem


52

Well, the answer you usually get is half right. They do disappear (more on this in a second). I'd hesitate to say they turn into "heat energy," both because we don't use the term "heat" that way in a technical sense and because most of the time we like to talk about atoms absorbing photons. In this case the energy of the photon becomes potential energy of ...


1

In reply to various points you make in the question Edit 4 - with sound - the energy of wave that you are talking about that can be increased by focusing is equivalent to he intensity of light that we can increase by focusing with a lens. To make an X-ray we need to change the frequency of light - focusing sound waves does not change their frequency - does ...


-2

About quantum mechanics connection through the USM www.kanevuniverse.com and is it flowing towards the general relativity? First of all congratulation to Brian about very interesting point about quantum mechanics and general relativity compatibility! First of all let me begin with this: Now let talk a little about the last episodes (there is talking about ...


3

Point the light source up or down in a gravitational field and let Einstein take care of you. Of course, if are forced to use the Earth as the source of your gravitational field you are going to need a pretty sensitive tool to measure the change in frequency. Luckily for us there is a set of of high precision sources in orbit and you can buy the ...


4

To my knowledge, other than red/blue shifting light, the frequency cannot be changed after it's emission. However you may be able to somehow set up some other material which would emit x-rays when infrared light is absorbed. Due to the nature of light and how it acts like a particle when interacting with particles so the energy gained doesn't accumulate when ...


0

Not "in thin air." The problem (well, a problem) is that if you take an infrared photon and have it become an X-ray photon, your new photon has greatly increased energy, which is obviously not something that just happens. However, there are effects that can sort of do this. The key is that if you want to make one 400nm photon you need two 800nm ones to ...



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