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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 ...


2

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 ...


9

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 ? ...


49

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 ...


1

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 ...


0

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' ...


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 ...


3

The unified formula used in General Relativity is $$d\tau=\sqrt{\sum_{\mu=0}^3\sum_{\nu=0}^3 g_{\mu\nu}dx^\mu dx^\nu},$$ which by Einstein's notation (summation over doubly repeating indices is implicit) is also written as $$d\tau=\sqrt{g_{\mu\nu}dx^\mu dx^\nu}.$$ Here $d\tau$ is the proper time "felt" or measured by particle moving in the spacetime, for ...


2

The time dilation due to velocity and due to spacetime curvature can't be separated. Both are derived from the metric. There isn't a general formula for this because it depends on the metric in question. For example in my answer to the question A clock in freefall I calculate the time dilation for an observer falling from infinity towards a black hole, but ...


0

Okay, let's say you have a friend staying at home measuring the time and observing you with a telescope - the whole situation will be described from his point of view. Namely, we will use his measure of distance $x$ and the time $t$ he measures with the stopwatch in his hand. Special relativity tells us that he has a means of calculating your "proper ...


3

How a unit is chosen to be defined depends in large part on how precisely the unit can be reproduced based on that definition. Two different atomic clocks built using the best currently possible methods will produce almost exactly the same answer for how long a second is, to within about 1 part in $10^{14}$. The second is defined in terms of a property of ...


2

The webpage you were looking at is run by H. D. Zeh. So if you want to find out what he's talking about the best way to find out is to look up some of his papers, such as: http://arxiv.org/abs/1012.4708. He describes decoherence in quantum gravity in Section 5 of the paper. The basic answer is as follows. The Wheeler-DeWitt equation is time-independent in ...


1

I'd like to add something to these answers. In the classical mechanics, we cannot distinguish a moving body from the body at rest, if we look at it at any particular instant. So, we have to add some hidden information to the picture, that is instantaneous velocity. But that's what physics only knew in the 19th century. In 20th century physics, there have ...


0

The answer is more simple than you think. Time is that, which is measured by (technologically suitable) clocks. Physical theories will simply tell you how clocks behave under certain conditions. This is purely descriptive. There is not a single physical theory out there, that gives a microscopic description of time, although the similarity of time with ...


1

Okay, I am going to try and give this a shot, but this is most probably not going to be a decisive answer. Let us operate with the term event time and duration and consider only special relativity (SR). The conclusions of general relativity should be the same for reasonable space-times. (e.g. without closed time-like curves etc.) We expect event time to ...


1

Duration is certainly a more physical concept than time. Duration is something you may measure between timelike separated events while time is always something you compute by adding up duration measurements + an arbitrary constant to fix the origin. Duration is experimental and relational while time (e.g. GPS time) is an abstract a posteriori ...


4

At a "frozen" instant of time, the arrow may not be moving - but this is a tautology, since movement is something that requires time. However, even in that frozen instant the arrow does have a velocity (instantaneous velocity, if you will). Imagine that time is a series of huge number of discrete frames (or instead imagine that it is continuous, and that we ...


1

Take a landscape. It can be modeled by a function f(x,y,z). If all the derivatives, df/dx, df/dy, df/dz are zero, the landscape is flat to infinity and nothing interesting exists in the landscape. If one of the derivatives is different than zero, then we perceive a shape, and generally a landscape has a shape. As an example, suppose that we have a cone ...


2

What makes heat move from hot to cold? Entropy. How can you calculate entropy microscopically? Start counting states! What makes the universe change irreversibly from yesterday to tomorrow? Start counting states!


0

Planes of simultaneity in special relativity don't really mean much of anything. The real physical structure of spacetime is in the light cones. The takeaway from "relativity of simultaneity" is not that there are "different time orderings for different observers", but rather that there is no meaningful time ordering for spacelike separated events. They ...


1

Even though the forces started at different times, is there any displacement of the metal box in any of the situations? Or is there any movement at all but is the net displacement zero? Sure. If you think of each force as causing an acceleration, the first one begins an acceleration in one direction, the second an acceleration in the other (or a ...


11

What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ...


10

It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ...


1

"How do we know that clocks slow down relative to each other?" Experimentally. This has been observed many times in the lab. The same answer is true for ANYTHING in physics and science in general. We only know that it is true, because we have experimental evidence for it.


0

No, since by the principle of relativity: A body in constant velocity motion cannot determine whether it is in motion in a certain direction or whether everything else is in motion in the other direction. No physical experiment can determine this hence for all purposes a body in motion will simply claim that it is at rest while the other body is in motion. ...


2

Yes, the two are intimately related. One way, as in QMechanic's answer, is via Wick rotations, but in general there is a lot more freedom once you allow integration contours to go over into the complex plane. In my area, strong field physics, the use of complex time to understand tunnelling problems is everyday bread and butter for many people, and it is the ...



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