# Tag Info

11

The electric and magnetic fields of a single photon in a box are in fact very important and interesting. If you fix the size of the box, then yes, you can define the peak magnetic or electric field value. It's a concept that comes up in cavity QED, and was important to Serge Haroche's Nobel Prize this year (along with a number of other researchers). In that ...

11

This is a reasonable question to ask, but the answer is probably not what you're expecting: the electric and magnetic fields don't have well-defined values in a state with a fixed number of photons. The electric and magnetic field operators do not commute with the number operator which counts photons. (They can't, because they are components of the ...

7

In the beginning, let's investigate the pair production. We know from relativity that mass can be equivalent to energy and if we set the most important physical constants to one $\hbar = c = \varepsilon = 1$, then we have the following relation (assuming that we produce electron-positron pair): $$E = \omega = \gamma_1 m_{e^-} + \gamma_2 m_{e^+}$$ Where ...

7

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 ...

7

Well, if the mode is speed-questioning, I'll attempt speed-answering: If photon will travel for million years without collisions, what subtle effects can be accumulated? The wave packet keeps expanding (or your uncertainty about location, take your pick), and the frequency drops due to space expansion. Nothing else that anyone knows about. Gravity ...

6

Starting with the Lagrangian for a massive $U(1)$ vector boson $A_\mu$ which like you said has 3 DOF: $\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 A^\mu A_\mu$ now if we change variables to $A_\mu\rightarrow A_\mu - \partial_\mu \theta$ and we have (Note that $F^{\mu \nu}$ and hence $F_{\mu \nu}F^{\mu \nu}$ is invariant under this ...

5

The answer by @gns-ank covers the kinematics of why. Below I tackle the Why does a photon "split" into an electron and positron, and not just bounce off the nucleus in your comment to his answer. In general physics can answer "why" in a nested way, like russian dolls. In the end, the kernel answer is "because it does". In this case though we are in ...

5

This is one of those cases when the ends do not justify the means... Just because you get a result that is true, from laws that aren't supposed hold in that situation, doesn't mean that the laws can be used there. If you're asking about whether there is a deeper connection between using $\frac{1}{2}mv^2$ and getting the radius, to the best of my knowledge, ...

5

The short answer is: no, $\frac{1}{2}mv^2$ is never valid for photons. A photon's energy is given by $$E = h f = \hbar \omega = \frac{h c}{\lambda}$$ always. The derivation of the Schwarzschild radius you mention is an incorrect one that happens to give the right answer accidentally. The correct derivation requires general relativity.

4

The relation $E=mc^2$ only works for particles at rest, which is evidently not the case for photons. In the general case, the relation is $$E^2=m^2c^4+p^2c^2$$ for a particle with momentum $p$. (Note, though that the momentum is not necessarily $p=mv$ as in the newtonian case! See for instance If photons have no mass, how can they have momentum?) For a ...

4

There is a simple way to understand the massive electrodynamics Lagrangian and limit, which is the Stueckelberg (Affine Higgs) mechanism. This is matematically equivalent to DJBunk's answer, but it is slightly more intuitive physically. Consider an Abelian Higgs model, with a massless electrodynamic vector potential $A$ and a scalar field with a $\phi^4$ ...

4

How is it possible to detect a single photon without making any change to it? In general, if you have detected a photon in your experimental apparatus, you have changed it drastically. It may have disappeared completely, as in this bubble chamber picture: The colored diagram shows the photon in the picture that has materialized into an electron ...

4

Here are real events relating to the last page of the pdf link you gave: Fig.1 This bubble chamber picture shows some electromagnetic events such as pair creation or materialization of high energy photon into an electron-positron pair (green tracks), the Compton effect (red tracks), the emission of electromagnetic radiation by accelerating charges ...

3

This process is the result of the cooperation of two theories of nature: (i) Special relativity: This is a huge topic to study but we shall only need a small part of it, and perhaps the most famous one, which tells us this $E=mc^2$. This equation shows us that matter and energy are equivalent and interchangeable. For example, if an amount of energy $E = ... 3 Kind of as an expansion on what drake said, this can be explained in several ways. For example: In electromagnetism, we know that Maxwell's equations govern electromagnetic radiation. From Maxwell's equations you can derive the EM wave equation $$\frac{\partial^2\vec{E}}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2}$$ (and the same ... 3 When atoms jiggle, their charged constituens get accelerated, and accelerated charges radiate electromagnetically (keep in mind that like most of physics, this is a lie-to-children). The (idealized) distribution of electromagnetic radiation is given by Planck's law, which cannot be derived from the classical picture I gave above because of the so-called ... 3 According to Special Relativity the relativistic energy for a particle is:$E^2= m^2c^4+p^2c^2$The invariant quantity under relativistic transformations is the rest mass$m$of the particle. For a photon$m=0$Using some simple algebra it is found$E=pc$for a photon. You will see this preserves the frequency and energy relationship. The error in the ... 3 This is really just an extension to JKL's answer since I wanted to pick up on his point about the microwave background, but first it's worth mentioning that although individual photons do not have a temperature EM radiation can be assigned a temperature. The EM radiation emitted by an object has a spectrum that depends on its temperature through Planck's ... 2 The deflection in each cell is twice that which it would be for a Newtonian particle coming in with velocity c. You apply the transverse force to change the direction, multiplying by 2 so as to account for the GR space-space parts of the metric tensor. This assumes that the matter making the gravity is nonrelativistic (so that you are justified in using a ... 2 The following improvement of your statements eliminates the apparent contradiction: The electromagentic field is the fundamental entity. Charges (electrons, positrons, nuclei) are accompanied by (''emit'') an electromagnetic field - a soft virtual photon cloud in terms of QED. Photons are elementary excitations of the quantum electromagnetic field. They ... 2 Remember, electromagnetic field is a distribution of electromagnetic force, not charge. Photon bosons are quantum of this field. So, they are force carriers.. not charge carriers. Only force is exchanged with these messenger particles. Based on this interaction, we determine charge of electrons etc involved. That's it! 2 In purely theoretical models, SUSY may be completely unbroken in which case photinos would be massless – and all particles would have the same mass as their superpartners. In the real world, SUSY has to be broken and a photino must consequently be massive (it is infinitely unlikely that the mass agreement will survive for any pair if SUSY is broken), ... 2 You should have a look at How does a photon travel through glass?. This is really a duplicate of your question but I haven't voted to close your question because I'm guessing you're asking for a fairly basic answer. You're thinking of the photon as little bullets fired from some point and hitting some other point, but this is only a partial description of ... 2 It's a non-trivial problem, which also involves how you define a photon in a medium - as a interacting particle and treating excitation of medium separately, or as a "dressed particle", including the interaction. From Abraham–Minkowski controversy Wikipedia page: The Abraham–Minkowski controversy is a physics debate concerning electromagnetic momentum ... 2 The central point of the question is somewhat ambiguous, but here is an effort to answer it. I am sorry in advance if I have misunderstood it. Does light/photons travel? The question whether light travels from place A to place B or not, can be answered mainly by experience and experiment/observation. When you hold a torch in the dark and you aim it at ... 2 The photons themselves do not have temperature as such. However, photons do contribute to the temperature of objects since they carry energy. A very good example is the microwave background radiation which is known to contribute a temperature to the universe at about 3K. One can work out the frequency of these photons using the basic relation$k_BT_{mwb}=hf$... 2 The act of measurement causes a quantum system to collapse into an eigenstate of the operator associated with the measurement. So unless the photon wave function is in an eigenstate of the detection operator, it will be changed. Unless a system is in an eigenstate of the hamiltonian it will not have a definite energy. If one measures the energy of the ... 2 The frequency of light is not a property of many photons but a property of a single photon. (This is also strictly inaccurate, since we should think about a field with ripples in it; we then call a little clump of waves a photon.) Anyway, let's imagine a photon/wavepacket as having a typical wavelength$\lambda\$ given by the peak-peak spacing in this ...

2

First, of course there's no perfect mirror. But let's assume there was one. Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever. But what if it does lose energy ...

2

They do not need to necessarily collide like balls. I guess the picture in your book is illustrative. Conservation laws apply to any kind of interaction between them. Note that details of the collision are not even provided in the question but you still can calculate the answer. The detailed theory of photon-electron interactions is called Quantum ...

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