# Tag Info

22

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 ...

21

Yes, the photons actually reach you, like rain falling on you, not like watching rain from a distance. When you see a star, photons from the star actually enter your eye. In for example rods of your eye, the photon causes a molecule of retinal to react by change from cis to trans isomer.

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 ...

7

A photon is emitted by a star, travels in a straight line* and enters your eye when you look at that star. You see the star continuously shining because photons are continuously being emitted, so you can't actually tell when each photon enters your eye. It's like a constant flow of water as opposed to dripping. You can't see photons in the same way you can ...

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

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

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

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 ...

5

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 ...

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 ...

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 $v ... 3 Is the photon physically oscillating through space as it travels? I wouldnt imagine so. Which periodic occurrence is referred to when one talks about the frequency of a particle? No the photon is not oscillating through space. It is an elementary particle of the standard model which is the quantum mechanical description of most of our experimental ... 3 The easiest way to see frequencies is in interference. Imagine you have waves coming towards a wall. Imagine too that the frequency of the waves is way higher that what you can see. You cannot directly observe the waves, but you will see that the wall is wet a few centimetres over the surface. Now, instead of one wave, you have two coming from different ... 3 Ah, all that talk about curved space-time. Well, there is a simpler argument. The fundamental axiom of general theory of relativity, "principle of equivalence", says: The effect of a homogeneous gravitational field is equivalent to that of a reference frame in uniform acceleration in the direction opposite to that of the gravitational field. All that ... 2 One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric:$\$ 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

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 ...

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

A "normal, healthy" human eye has two types of light-sensitive cells in the retina: rods ("color blind", but capable of sensing low light levels) and cones: cells that are sensitive to different bands. See this figure for their relative sensitivity (from http://hyperphysics.phy-astr.gsu.edu/hbase/vision/colcon.html) When you look at a spectrum of light, ...

2

I know that when I observe rain, I can both observe it from a distance but could also be immersed in it as well if in the path of that rain. But with distant starlight, are we just observing it or are the photons actually reaching and penetrating the earth around us? When you're immersed in rain, you interact with it directly: you get wet. That's ...

2

In the presence of strong gravitational sources, such as black holes, photons will experience redshift. This means that with respect to an observer in a region of a weaker gravitational field, they lose energy. This is equivalent to the statement that their frequency decreases and their wavelength increases. This is a consequence of time dilation. The ...

1

Light of different color is understood as a wave of different frequency of oscillation. Matter in surfaces has characteristic behaviour for each of these frequencies, based on its chemical composition: electrons in the surface oscillate and absorb a lot of light when its frequency is around one of resonance frequencies of the matter; and the electrons ...

1

In my answer to this question: (What is the sun's spectral series?), I give a very detailed answer about why mixing colours of light produces other colours and how it is purely a result of biology and evolution. I also delve a bit into the structure of the human eye and why, in fact, only three colours are necessary to reproduce all of the colours we can ...

1

The cells in our retina that detect by frequency (read: colour) detect most strongly in three slightly different bands we know as Red, Green and Blue. To make a slight correction I would say an incadescent bulb is quite far from white, so I would rather proceed talking about sunlight on a clear day. The reason why sunlight appears as white as say a white ...

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible