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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 ...


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 ...


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, ...


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 ...


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 ...


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


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 ...


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 ...


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 ...


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, ...


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 ...


2

This is the plot of sunlight, red at ground level. Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power. As all light comes from the sun during daylight this should suffice. One can get the number of photons by ...


2

As an analogy, consider the photon that strike your face and reach my eyes, we say that that photon carries information about your face which then helps me to identify you, You are confusing the individual photons with the electromagnetic wave that is light, which is composed out of a zillion photons. but don't these photons collide midway with air ...


2

charge is an intrinsic property of any particle. we in principle cannot change the intrinsic property of any particle. photons are the carriers of electromagnetic interaction(action at a distance).


2

The photons do not get "stuck" at the event horizon, from their own reference frame they are still traveling at C. The event horizon is simply the point where gravity from the singularity is strong enough that the escape velocity exceeds C. Just above the event horizon, photons can still escape the black hole, and just below it they sink into the ...


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 ...


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 ...


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 ...


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.


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 ...


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

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 ...


1

Unfortunately, no, your calculation does not seem to be correct. Your calculation is based on Einsteins famous equation $E=mc^2$; however, this equation is actually only valid for objects at rest, while all experiments confirm that photons in a vacuum move with a constant speed of $2.99...\times10^8$~m/s. The equation Einstein gave for moving particles is ...


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 ...


1

Yes, this is actually often used in a spectroscopic technique called REMPI -- see the image on this wikipedia page https://en.wikipedia.org/wiki/Resonance-enhanced_multiphoton_ionization There are some important physics techniques that rely on interaction with two photons -- two photon spectroscopy (http://cua.mit.edu/8.421_S06/Chapter9.pdf). Some other ...


1

You are asking about wavelength, but your question is a problem not very well-posed. Then, let's say so: from a single photon we cannot get a conclusion about it characteristics. Only if we know the beam to which the photon belongs, can we know the wavelength (or wavelengths see below). Next, the wavelength is the result of preparation of the beam, it is ...


1

Yes, a photon is defined by its wavelength, which also directly relates to its energy. I'm not sure what you mean by interacting with other particles, but a simple glass prism is able to separate the different wavelengths of white light into its constituent bands. The wavelength of a photon of light is related to its energy by planck's constant. ...


1

Conservation of energy cannot be violated - so you cannot have a single interaction in which a photon of energy $E$ results in the emission of a photon of energy $E + \Delta E$ unless there is another source of energy. It is conceivable that an atom in an already excited state could result in such a phenomenon; and of course there is two-photon interaction ...


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 ...



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