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0

You can always choose such coordinates that $g_{\mu \nu}$ is equal to the Minkowski metric $\eta_{\mu \nu}$ at some given point $p$, then the speed of light is just $c$. Everything else is like in SR: only the Lorentz subgroup of 4-d rotations (of the tangential space) keeps the interval $c^2 dt^2 - dx^2$ invariant. So, the usual concept of 'speed of light' ...


7

The answer to your questions is very nuanced for the most part. I'll start with the easy answers: Light does not experience time, neither does it experience no time, the most accurate statement I could probably make off-hand is that it experiences null time. Null time does not mean time has stopped, that would be zero time; null time means null time, null is ...


0

On your first question the answer should be here http://en.m.wikipedia.org/wiki/Metric_expansion_of_space, after the big bang only the distance between space is expanding but you should be able to understand on the link i have included, yes the speed of light remains constant but the expansion is causing it to have longer time than expected to arrive it is ...


2

Space is indeed expanding everywhere, and not only between galaxies. The reason we don't grow with it, is that the attraction between the electrons and the protons is strong enough to keep them bounded. You can look at it as if they always re-adjust their position to counter the expansion of space. This also applies to our solar system, our galaxy, and even ...


2

Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon. In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$ The four-momentum for ...


2

There is a difference between what "happens", described by the Lorentz Transformations, and what you see with incoming light beams, which is heavily influenced by differing light delay times from parts of the scene. There is a very good explanation in the second half of this essay: http://mathpages.com/rr/s2-05/2-05.htm, search for the phrase "Meanwhile, ...


1

The correct general equation is $E^2=m^2c^4+p^2c^2$, since $m=0$, $E=pc$. A photon's momentum is $p=\hbar k$, where k is the wavenumber ($k=\frac{2\pi}{\lambda}$). This would be consistent with $E=h\nu$ where $\nu$ is the frequency.


0

As is pointed out in the comments, it isn't possible for an object with a non-zero (rest) mass to travel precisely at the speed of light. However, it is possible to meaningfully talk about the behavior of a clock on a ship in the limit of the ship's (constant) speed approaching the speed of light. The equation for time dilation due to relative velocity is ...


0

No. At the speed of light special relativity predicts that in both reference frames the object will have the speed of light. So from outside of your ship you are moving at velocity c.


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It is important to think about what the velocity addition formula is meant to do. The velocity addition formula should be applied in situations of this type: When you (person $A$) see someone (person $B$) speeding by you with velocity $v$, and when this second person sees another person (person $C$, one can also substitute objects for people, of course) ...


3

This isn't definitive, since the answer is always undefined, but let's be cutesy. Let's let $u' = a*c$ and $v = -a*c$, where $a < 1$ Then, $$\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{ac-ac}{1- \frac{a^{2}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c-c}{-2a}\\ &= 0 ...


2

Your understanding is right, but you have to also realize that there is a rich variety of experimentally verified relativistic effects where the constant $c$ is occurring and this constant would suddenly have to be changed. Namely all the particles suffice the relation $$E = mc^2 = \sqrt{m_0 c^4 + p^2 c^2}$$ This relation is used e.g. to compute the mass ...


1

When travelling in a dielectric light isn't light. It intracts with the medium to form a composite system that has an effective mass and therefore travels slower than $c$. If the interaction is strong, as in a BEC, the interacting system can be described as a quasiparticle called a polariton. This isn't useful for weakly interacting systems like most ...


1

In GR you have to be careful to define exactly what you're asking, so let's define our experiment carefully. Take two observers stationary at a fixed distance from a spherically symmetric mass. We'll call the more distant observer Alfred and the nearer one Bert. Alfred shines a light of frequency $\nu_A$ towards Bert i.e. towards the spherical mass. When ...


0

Light moving away from a mass undergoes a gravitational redshift. Light moving toward a mass undergoes a gravitational blueshift.


2

While our definition of the meter is based on the speed of light, our definition of length is not. Naming and defining a unit of measurement is entirely different than defining a physical quantity. The most natural way to define distance (known by physicists as proper distance) is the measured spatial separation $\Delta r$ between two points in space. Rigid ...


0

Time is that which is measured by clocks. How clocks behave when they are being moved relative to each other is simply a collection of experimental facts. At no time do we need any light to do those experiments, and it totally doesn't matter to moved clocks if we are moving them during the day or the night. So what, exactly is your question? Is it why ...


1

"Most probably in reality there are some extremely complex laws and equations which makes this question more complicated." Not really. The equations are rather straightforward. Let's measure velocities in units of the speed of light and let's denote the velocity of $B$ as observed by $A$ as $v_{BA}$, the velocity of $A$ as observed by $C$ (the ...


0

Special relativity is often introduced to students using light clocks because this is a reasonably accessible way to understand that phenomena like time dilation and length contraction must occur. However you should not be mislead into thinking that we use light clocks to define special relativity. The fundamental principle of special relativity (and in fact ...


1

That light has a fixed velocity in vacuum comes from observations . In order to fit the data Lorenz transformation were imposed on the rigorous mathematical model for electromagnetism, Maxwell's equations. It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to ...


4

In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no". There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the ...


1

One aspect that comes to my mind is the concept of causality. Superluminal propagation would allow for a violation of this principle, creating various paradoxical phenomena.


0

For particles to go faster than the speed of light requires that Lorentz invariance must be broken. It is possible to formulate theories that violate Lorentz invariance, and such theories do indeed predict that particles going faster than the speed of light will emit radiation and lose energy as a result. This effect was invoked by Andrew Cohen and Sheldon ...


1

I'll make a few general comments first and then try to answer "What I am looking for is a clear and simple explanation on the exact purpose 'c square' serves in this equation." First off, while $E=mc^2$ is still widely used in the media, it is not widely used amongst physicists. That equation uses the concept of relativistic mass, which is more than a bit ...


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Electromagnetic radiation in a medium propagates according to the law $$ \mathbf E,\mathbf B \propto e^{\imath(\pm k_xx-\omega t)} $$ where $$ k_x^2 = \frac{n^2\omega^2}{c^2}\;. $$ The refractive index $n$ can also be complex, in which case its imaginary part describes the absorption of the EM wave in the medium. But the oscillating part is in any case $$ ...


-1

There is a philosophical question of what happens to your shadow, after it would travel faster than the speed of light. This is easily realized with the thought experiment of an object is moving around a point source of light. Given that the radial velocity of a rotating body is $\omega = vr$, then the shadow projected at a distance $r$ can then be ...


-1

This is basically about dispersion: EM waves with different frequencies travel at different speed in a medium because of interaction. Usually, as in a standard textbook experiment wherein a light beam is bent by a prism, higher frequency waves bends more. The more it bends, the more slowly it travels. So, lower frequency waves usually go faster. However, ...


6

There is a much better description here of Fizeau's nineteenth century experiment. Some of the key features that enabled Fizeau to succeed: A lens to collect the light from the source A collimating lens to prevent the light diverging during its journey A large diameter beam to minimise broadening of the beam by diffraction More lenses to focus the light ...


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Your shadow, surely, is not just the apparent darkening of a 2D region of a diffuse surface you happen to be standing in front of; it is the entire volume of space that your presence is preventing light from reaching, the extrusion of your silhouette from the light source out to infinity (for a point light source, or possibly to a point a finite distance ...


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Note: This is basically item 3 in jkel's answer. If you move at an appreciable fraction of the speed of light, then your shadow can appear to be "trailing" you, although it will always be "attached" to your feet if you're on flat ground. Suppose a person is moving in the direction shown below and that there are plane waves coming in from an angle. The ...


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Here is how you can "run faster than" (or at least, get away from) your shadow: you jump at sunset (I just realized 15 minutes after posting that this is the point that @jkej's answer made as possibility #2) Your shadow will detach from your feet, and it will "run away" from you. In the frame of reference of the shadow, you are running away from it. ...


18

Depending on exactly what you mean by away from your shadow, I can think of a number of methods: Position yourself in the shadow of some object larger than you. This would result in your shadow disappearing altogether; the ideal solution in my opinion, but perhaps considered cheating by the likes of Mr Elephant. Position yourself in such a way that your ...


4

This is borderline philosophical stuff, but... If you want to put some distance between you and your shadow, you have to fly. Actually, you have to put some distance between and the opaque surface right below you, so swimming in a tank would do the trick as well. This will disconnect you from your shadow. If you have a jetpack, or if you are in an ...


0

You're using cartoon physics. There's no "force carrying a photon". Photons will happily move on their own, without any external force being applied. This is basic Newtonian physics: F=0 <=> a=0 <=> dv/dt=0 <=> v(t) = v(0)


0

Your first idea would work if it was possible to accelerate a rocket to the speed of light rather than only being able to approach the speed of light. But a person on a rocket moving at $.99c$ relative to Earth will still look out his window and see photons zipping past at the speed of light, not at $.01c$. Since it would require infinite energy supplied to ...


-1

1st Law of Motion Every object in a state of consistent motion tends to remain in that state of motion unless an external force applied to it. 2nd Law of Motion It is pertaining to the relationship between an object’s mass, its acceleration, and the applied force. In this law, the direction of the force vector is the same as the direction of the ...


-1

Why do i have the sense that this can have a meaningful physical interpretation (even if velocities are both $c$)? First lets assume that both velocities are $c$ or $1$ in the question's notation. Then the relativistic velocity addition formula simply gives: $1_A \oplus 1_B = 1_B \oplus 1_A = 1_{tot}$, correct? Yes, does the fact that this is mathematicaly ...


2

After a lot more searching, I have found the answer to my question! :D Below is a summary of the information I found. There is no specific webpage I can link to because I relied on sources who quoted other sources which no longer exist, but maybe this information can be useful to someone else someday. Most of what I learned comes from Professor Lou ...


4

Of the choices given, I would favor explanation #2. It doesn't require quantum physics; modeling atoms as ball-and-spring systems works pretty well. In his famous textbook for undergraduates Griffiths does this, and if you have some math training that would be a fine place to head for the details. I think #5/#6 are also, arguably, correct if you treat the ...


1

Due to the high speed of light, the light pollution would clear, at least in a practical sense, almost instantaneously from the area once all lights were shut off. One's eyes might need up to about 30 minutes to fully adjust to the dark, so it would take about that long for most of the stars to appear to "come out" to the naked eye.


2

if higher speed than the speed of light will be discovered, will scientists be able to adjust the special relativity to the new situation? I suppose so. Science has encountered paradigm-shifting discoveries many times, and has always come up with new ways to describe reality. If this does happen, it could be an adjustment to special relativity, or ...


2

From the OP's comments I have a feeling that the real issue is not the virtual particle, but an actual annihilation of a positron and electron into 2 real photons. Like here: And the question is, how can the system of 2 massive electrons accelerate into the speed of light, when accelerating to c requires infinite amount of energy. The answer has two parts: ...


3

In Feynman diagrams all four vectors conform with special relativity algebra, BUT the internal lines, even though they have the name "photon" are virtual, which means the mass can be different than zero, only the quantum numbers identify them as a photon, quark, electron, etc.. Feynman diagrams are an iconic representation one to one with the integrals and ...


3

it seems curious to me that a system can have a mass, and then accelerate to the speed of of a massless particle There is no paradox here. Special relativity doesn't preserve velocity - in fact the opposite - it allows one to move between frames of reference in which the velocity of objects is relative to the observer. [do the particles] accelerate ...


0

I am also puzzled by this question. I would like to pose it in an alternative fashion. The question posed by kaka is simple and clear but the answers are too complicated. Rational numbers and irrational numbers are mutually exclusive sets of numbers. Speed of light in vacuum has a constant numerical value,say in units of m/s. The question is does the ...


1

For the observer traveling faster than the speed of light the light of the laser must appear to move away from him at the speed of light. For the observer in the rest frame, the laser light travels slower than the person firing it. I conclude that time is moving backwards for the guy with the laser... But seriously - we are breaking laws of physics here. ...


1

Since $n=\sqrt{\varepsilon_r\mu_r}$ (the relative permeability $\mu_r$ being almost always $1$), and $v=\frac{c}{n}$, you can also write $I = \frac{nc\varepsilon_0}{2} E_0^2$. (We used the decomposition $\varepsilon=\varepsilon_r\varepsilon_0$.) So, the intensity depends linearly on the refractive index.


1

A more particle-based view of zakk and Moonraker's observation that photons experience zero time between emission and absorption is this: The postulated event that a massless particle (say a photon) could emit another massless particle (say another photon) in the retrograde direction — or in any direction for that matter — is always zero. A photon from its ...


4

I'm getting the sense that my fears were correct, it's physically a nonsensical situation. Applying a formula outside of the context in which it was derived will likely produce nonsensical results. In the derivation of the relativistic velocity addition formula, and using your notation, there is an object B with uniform velocity $w$ in some inertial ...


1

As pointed out by zakk, any proper time of objects moving at light speed would be zero. That means physically that there does not exist any point of time where you could perform the addition c + (-c). Example: A photon is emitted at A and absorbed at B. There is no point of time A < t < B where the photon is actually moving, the age of the photon at ...



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