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0

As for "why", that is either simple or complex! The simple answer is that it is a consequence of the speed of light being constant for all observers. That means when travelling relative to each other they both measure time differently. This site provides a simple math explanation OTOH, time is measured differently in different gravitational fields, and that ...


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You don't have to be an astronaut traveling at near the speed of light to experience this effect. It can be measured quite precisely here on Earth and it has been measured both on elementary particles in accelerators and by flying atomic clocks around the world in planes. Humans are simply not used to it because the differences between the flow of time in ...


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As an example, http://en.wikipedia.org/wiki/Particle_decay, you can regard $\Delta t$ as the lifetime of the particle decayed.


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A fundamental postulate of QFT establishes that the theory admits a strongly continuous representation of (orthochronous proper) Poincaré group $\cal P$. A certain one-parameter subgroup of $\cal P$ describes time evolution (with respect to an inertial reference frame) which, as a consequence, turns out to be unitary since it is part of a larger unitary ...


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You people make this overly complicated. If the vacuum exists, time absolutely exists as well. For example, if the vacuum were to exist near a source of radiation, particles. They could enter the given vacuum as time passes by so time absolutely exists. also, time as a count of existence progressing, also must be identically throughout the universe to our ...


2

Rather than write something unintelligible, I'll quote from a page on cesium clocks. According to quantum theory, atoms can only exist in certain discrete ("quantized") energy states depending on what orbits about their nuclei are occupied by their electrons. Different transitions are possible; those in question refer to a change in the electron and ...


1

The traveling astronaut is younger. The situation is not reversible between both astronauts because the traveling astronaut is submitted to an effect which is similar to acceleration because he is following the curvature of space in the fourth dimension. The solution must consider the geometric/ topological constellation. And topologically, the traveling ...


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It's a unit conversion: $$ 1\,{\rm yr}=\frac{365\,{\rm days}}{\rm year}\times\frac{24\,{\rm hrs}}{1\,{\rm day}}\times\frac{3600\,{\rm sec}}{1\,{\rm hour}}=3.1556926\times10^7\,{\rm sec} $$ Since $3.1557$ is (somewhat) close to $\pi\sim3.1416$, we use the approximation you cite. Technically, the year is actually 365.25 days long, so using that gives a ...


0

time delation is very simple phenomenon.Time delation occurs because the speed of light is same for all observer in same media.so the rate of time experienced by the observer changes with respect to object moving near to speed of light because two events in space time having different time origin with respect to each other never coincide with each other.


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The acceleration is in the opposite direction from the initial velocity. If you are measuring positive upwards (by taking the initial velocity to be $+16.1$) the acceleration of gravity is negative. Your second answer is correct, but has one place too much accuracy as your data only goes to one place behind the decimal point.


1

When we think about the state of a hydrogen atom we instinctively think about the solutions to the time independant Schrodinger equation. These are the well known atomic orbitals. However for the time independant Schrodinger equation to apply the hydrogen atom must have existed unchanged for an infinite time and it will then continue to exist unchanged for ...


1

Yes, spin direction is also reversed. An "up" electron spin state simply becomes a "down" spin state. Electron spin remains fundamentally a type of angular momentum, even though the rules get a bit odder at the quantum scale. Incidentally, when time is reversed all particles also turn into their antimatter equivalents. So a bit of safety advise: Never, ever ...


0

Your statement: "Since the speed of light is constant, the light coming from the clock must travel a longer distance to reach my eye as it moves away. This would make time appear to slow down? If, on the other hand, the clock is moving towards me, the distance the light must travel to reach my eye becomes shorter and shorter, thus time would appear to speed ...


2

If you truly have a vector equation, then you really have three quadratic equations - one each for the X, Y and Z component. Let's write them: $$s_x = v_x \Delta t + \frac12 a_x (\Delta t)^2\\ s_y = v_y \Delta t + \frac12 a_y (\Delta t)^2\\ s_z = v_z \Delta t + \frac12 a_z (\Delta t)^2$$ If there is only one value of $\Delta t$, then this is an ...


0

A photon is fired across the width of a space ship a distance of one unit “ud”. The space ships velocity v relative to a fixed observer is 0.8*c The observer sees the photon travel a distance of cT and knowing the width to be one unit of distance calculates that 1ud = T*(c^2-v^2)^0.5 1ud = Tc*(1-(v/c)^2)^0.5 1ud = 0.6 Tc A photon is now fired forward in the ...


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In special relativity there is a distinction between 'experiencing events' and the concept of an observer: Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from In ...


1

You've written a vector equation, but any solution involving numbers has to involve one coordinate at a time, or what amounts to the same thing, three simultaneous equations. For simplicity, let's assume a one-dimensional version. All of the displacements, velocities, and the acceleration point in the same direction: $$s = v_i\Delta t + 1/2\, a\,(\Delta ...


1

No matter in which direction the clock moves (away or towards you) , the time will slow down in the clock. Time will always dilate as the clock moves faster and faster, but will be apparent to a human eye only once it reaches speeds close to the light speed. This is not taking into effect the doppler shift which is merely the increase/decrease in the ...


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Relativity is not needed: If you replace the clock with a strong laser which fires a ray of light every second with an atomic clock, you know that the ticks will not slow down because every tick will be followed by another after a second from the laser's perspecive . But how can you arrange that with the fact that the light move further and further away ? ...


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Analyzing one moving clock from the perspective of one stationary person will be inadequate to derive special relativity from. With just that set-up, you aren't actually using the key fact that the speed of light is the same for all observers – all you're actually using is just the fact that the speed of light is finite. With just taking into account that ...


1

In relativity the notion of simultaneity is relative to the obeserver. While one observer (the one "standing") see all the rays of light arrive at the same time, another observer (the one going near the speed of light) will see one ray arriving before another. The paradox here is you think simultaneity can be defined in an absolute way independent of the ...


1

In this case, the eigenstates of the Hamiltonian are not useful to solve the problem, and one has to work with the Schrödinger equation directly: $$ i\hbar \, \partial_t \psi(x,t)=\frac{-\hbar^2}{2m}\partial_x^2\psi(x,t)+P\,t\,\psi(x,t) $$ Using a Fourier transform in the variable $x$ you can show that the general solution is $$ \psi(x,t) = ...


0

For a particle in a time dependant external potential any Hermitian the Hamiltonian will still have a complete set of eigenstates, and the corresponding eigenvalues will still be the allowed energies of the system. These energies and eigenstates will however generally be time dependant and this means that they are of far less use for solving the TDSE. For ...


0

The idea of "relative to a photon" implies that a photon has a rest frame. A rest frame of some object is a reference frame in which the object's velocity is zero. One of the key axioms of special relativity is that light moves at c in all reference frames. The rest frame of a photon would require the photon to be at rest (velocity=0) and moving at c ...


1

Because we are not. Photons do not have a frame of reference and therefore we do not move at the speed of light with respect to them. This is also the reason why it does not make sense to ask questions like "What does a photon see?" or "How fast does time pass for a photon?".


0

I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension of space-time at 1 second per second as time goes forward. This is wrong. What troubles me here, is the fact that it is traveling through space-time ...


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The answers above are all correct as far as they go, but the next step is even more stunning. If I am A, the object at rest, then I am moving through spacetime at the speed of light. My friend B gets on a spaceship and rockets away at a high constant velocity and to me his time appears to slow down because of his velocity through space. But if my fried B ...


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Yes, because what slows down clocks is acceleration, whether by gravity or by centrifuge, if you like. It doesn't matter how fast the airplane is, but how high it is, because gravity is stronger at lower elevations.


-1

. . . . $$To \enspace begin,\qquad L^2=L'^2+b^2\qquad \qquad and\qquad \qquad c^2=a^2+v^2 \qquad $$ $$Thus\qquad L'^2/L^2+b^2/L'^2=1\qquad and\qquad a^2/c^2+v^2/c^2=1$$ $$ \enspace \qquad \qquad And\enspace \qquad \qquad x^o=y^o, \qquad \qquad Thus\enspace \quad L'^2/L^2+v^2/c^2=1 \qquad \qquad $$ $$\rightarrow \quad L'^2/L^2=1-v^2/c^2\quad \rightarrow ...


1

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' ...


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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 ...



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