# Tag Info

21

If (and that's a big if) tomorrow we had a $70\sigma$ detection in a repeatable experiment of a particle that travelled faster than $c$, then one of several things would be true. 1) We would be forced to conclude that $c$ is not, in fact, the limiting speed of information transfer; everything based on this assumption would have to be scrapped (pretty much ...

18

When an electron absorbs a photon, it remains an electron and the photon disappears. The electron energy and momentum are altered to account for the energy and momentum the photon was carrying. For a free electron, it will not be possible to balance energy and momentum simultaneously. There will have to be another interaction to make that work. If the ...

14

The equation you are mentioning is the gravitation force derived by Newton. This force doesn't apply to particles such as photons for two reasons: Photons are too small, and you can't use Newtonian physics to describe their properties. Photons travel too fast (their velocity is the speed of light) and at such a velocity Newtonian mechanics cannot be ...

8

As Ross pointed out, two scenarios are possible: free electron / electron as part of an atom. They're treated in two totally different ways. Free electron: free electrons can't really "absorb" photons. They can collide with them, and some things can happen (this, for instance). Those types of collisions are described by QED and there are a bunch of ...

6

That's because the relation $p=\gamma mv$ doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ...

6

Yes - you can even propel spaceships with it - Solar Sail Although Solar radiation pressure at the Earth is around 9E-6N/m2 while the thrust from a Saturn V rocket is 34 MN, so you would need a solar sail something like 2000Km on a side to get the same acceleration

5

But what if tomorrow we happen to observe a particle X that travels with a speed V>c? We would have made the first observation of a tachyon. In special relativity, a faster-than-light particle would have space-like four-momentum, in contrast to ordinary particles that have time-like four-momentum. It would also have imaginary mass. Being ...

5

You can understand the expression by attempting the limit for $v\rightarrow c$ and $m\rightarrow0$. Notice that $\gamma\rightarrow\infty$ when $v\rightarrow c$. Therefore $m v \gamma$ is an undetermination of the form $0\cdot\infty$. From your expressions, you cannot say that $E=0$. The limit does not exist, and this implies that this expression is not valid ...

5

Your logic is ultimately wrong because that equation doesn't reveal the true nature of gravity. According to general relativity, objects themselves bend space-time. Imagine space like a rubber sheet. If you stretch it and place a mass in the middle and roll a ping-pong ball past the mass, it will curve towards the mass. Similarly when space-time is being ...

4

There is no way to be 100% sure, but we can put upper limits on the mass. Massless particles don't have a rest frame, so it doesn't make sense to talk about time dilation in the photon's frame. A massive photon would have a rest frame, so you could eventually catch up to it and move alongside it. List of experimental limits on photon mass more ...

4

The problem with your idea is that each time the light reflects off the mirror it transfers some of its energy to the mirror (to increase the mirror's kinetic energy) and is red shifted as a result. So the thrust would fade as the light red shifts away to nothing. For obvious reasons the light can only transfer as much energy to the mirrors (in the form of ...

3

Is the photon physically oscillating through space as it travels? I wouldnt imagine so. Which periodic occurrence is referred to when one talks about the frequency of a particle? No the photon is not oscillating through space. It is an elementary particle of the standard model which is the quantum mechanical description of most of our experimental ...

3

The easiest way to see frequencies is in interference. Imagine you have waves coming towards a wall. Imagine too that the frequency of the waves is way higher that what you can see. You cannot directly observe the waves, but you will see that the wall is wet a few centimetres over the surface. Now, instead of one wave, you have two coming from different ...

3

Ah, all that talk about curved space-time. Well, there is a simpler argument. The fundamental axiom of general theory of relativity, "principle of equivalence", says: The effect of a homogeneous gravitational field is equivalent to that of a reference frame in uniform acceleration in the direction opposite to that of the gravitational field. All that ...

3

by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ... 2 There is no position operator for photons, so photons do not have a spatial probability density. Associated with a photon (in a laser beam, say) one has only a probability density of hitting any given surface crossing the beam at a particular point of the surface. See Chapter B2: Photons and Electrons (and the entries ''Particle positions and the position ... 2 The abstract of that paper mentions the annihilation of the previously measure photons, where now, they aren't totally conserved, but not totally annihilated. Further down, they mention survival probability, and without access to the whole article, I can only conjecture, but I would go so far as to say that that is them implying that not ALL of their photons ... 2 One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric: $$ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 ... 2 In Newtons theory of gravity photons are not affected by gravity (created by masses). So your conclusion is correct. But in General Relativity the curves of free objects like test particles or photons (geodesics) are determined by the space-time geometry. The geometry is described by the metric which is given by the energy and mass distribution of the ... 2 You have misunderstood the nature of a singularity. The singularity at the centre of a black hole is not a point in space. Instead it is a place where spacetime becomes infinity curved, and it isn't possible to describe what happens there. Well, it's not possible using General Relativity, but we hope some future theory of quantum gravity will explain what ... 2 Newton's formula is an approximation of how "gravity really works". We actually still don't know how gravity really works, but we have vastly refined our understanding of it thanks to Einstein's general theory of relativity. Gravity is simply a measure of curvature of a 4-dimensional manifold we earthlings call space-time. Local concentrations of mass or ... 1 So the key point to understanding this problem is to understand that it is the modes that contain information about the physical parameters of your photons (such as the momentum or angular momentum), and quantization is just a description of excitation of those modes. For instance the canonical quantization of the plane-wave expansion which you've ... 1 you are correct in your thinking, because you are using newton's laws. However, newton's laws were not completely accurate. For circumstances like ones with photons and extreme gravity like black holes, you must use Einstein's general theory of relativity. These laws say that all particles follow the shortest path along spacetime, including photons. But ... 1 Consider a very distant supernova; for example, suppose that the photons of the explosion have to travel a billion lightyears to reach us. If these photons had different velocities, then these differences would cause an accumulating difference in their travel time. Even if their velocities would differ by as little as a billionth, then the fastest, most ... 1 In special relativity the energy is related to mass and momentum by E^2 = (pc)^2 + (mc^2)^2, where p is the momentum. m here is the rest mass of the particle, so for the photons case there is only energy from the momentum. The E = mc^2 you are likely familiar with ignores the momentum term, and hence only involves the rest mass. Photons are the ... 1 I assume you don't mean the speed of light, but you are essentially asking: Will light escape that strong gravitational pull? If this is your question then first: Direct quote from wikipedia -> "An object whose radius is smaller than its Schwarzschild radius (r_s = \frac{2GM}{c^2}) is called a black hole. The surface at the Schwarzschild radius acts as ... 1 The speed of light is a constant regardless of where or when you measure it. The speed of the light as it leaves the star will be c=299792458\frac{m}{s}. The speed of the same light far from the star will also be c. Instead of slowing down like newtonian objects, the light will instead lose energy as it attempts to leave the star. This will correspond ... 1 In undergraduate physics courses (c.f. MIT 8.03 course on waves), electromagnetic radiation is introduced as a pair of time-dependent oscillating magnetic and electric fields which satisfy the electromagnetic wave equation, as derived by Maxwell:$$\frac{1}{c^2}\frac{\partial^2 E(x,t)}{\partial t} - \frac{\partial^2 E(x,t)}{\partial x^2} = 0$\$ ...

1

There can be a pulse made up of many photons whose temporal length is a femtosecond, or there could be a single photo - one energy quantum, with a single wavelength. So, if you know the spectral bandwidth of your short-pulse signal, compare that with the spectral transmittance of your beam splitter. I'd suggest that unless you're right on the cutoff edge ...

1

the simple answer is that a quantum state has several variables (degrees of freedom), so if you measure only one of them and leave the others unchanged, then you detect the photon , change its state but do not destroy it completely. this is what they say in introduction Second, nondestructive detection can serve as a herald that signals the presence ...

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