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0

The observable universe radius is unique to any given point. By moving at all, we shift the center of our unique observable universes, causing the boundaries to shift.


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Other answers have effectively, or actually, said when you're dealing with $0.1 c$ or above is when you should be considering relativistic effects. You can see this by just slightly rearranging the $\frac{u+v}{1+ uv/c^2}$ formula: $$\frac{u+v}{1+ uv/c^2}=\frac{u+v}{1+ (\frac{u}{c})(\frac{v}{c})}$$ And from there you can see while both $u$ and $v$ are $\lt ...


1

In a comment Kyle points out the question Does strong magnetic field cause time dilation?, which is closely related to your question. However it isn't a duplicate because you are specifically asking whether experiments have been done, not whether the effect theoretically exists. Theoretically we would expect a magnetic fied to contribute to the curvature of ...


2

Energy is not a Lorentz invariant quantity, it is the zero-th component of the four-vector. Only proper orthochronous Lorentz transformations preserve the sign of the zeroth component, so if the energy is positive in one frame, a non-orthochronous Lorentz transformation would yield a frame in which the energy is negative. But we usually only allow the ...


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we can see negative energy solution as anti particle travelling in the -p direction in momentum space.creation operator have coefficents e^-ipx so it will create anti particle in the -p direction. here p is a four vector.four vector can be negative or positive.so we have solution in the positive p direction and solution in the - ve p direction.


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Maybe a chemists perspective could be useful to understand why we take both energies in quantum mechanics? It's not an elegant answer but is easy to rationalise! In computational chemistry we use the variational principle to compute the orbital mixing coefficients of molecular orbitals from an atomic orbital basis set. In doing so along the way we end up ...


0

There are two 'types' of velocities, relative velocities and mutual velocities. The relative velocity between object A and B is the velocity of A (say) as seen by B, to find relative velocities you use the Lorentz-velocity addition formula. The mutual velocity of two objects is the rate of change of the distance between them as seen by an observer another ...


1

When does a object go fast [enough to require the use of $\frac{u+v}{1+ uv/c^2}$]? When you are approaching light speed. How fast do you need to go to be "approaching" light speed? That depends on the precision. The key is that $\frac{u+v}{1+ uv/c^2}$ always works. However, it's not exactly simple, so generally we like to use an approximation: ...


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I'm usually a little more specific with my students. Let's consider a car traveling down the highway at $\rm 30\,m/s \approx 60\,mph$. Then the denominator in the relativistic formula is something like $$ 1 + \left(\frac vc\right)^2 = 1 + \left( \frac{30\rm\,m/s}{3\times10^8\rm\,m/s} \right)^2 = 1 + 10^{-14} $$ In your car, then, the difference between ...


9

Actually, regardless of the velocity of the objects in concern, the 'velocity addition' formula is always $\frac{u+v}{1+uv/c^2}$. There is no transition point where the formula changes from $u+v$ to the special relativity one. Its just that the difference you get in both the formulas at 'low velocities' is very very negligible. $c^2$, in $m^2/s^2$, is ...


51

For simplicity, consider the case $u=v$. The "slow" formula is then $2u$ and the "fast" formula is $\frac{2u}{1+(u/c)^2}$. In the plot you can see these results in units of $c$. The "slow" formula (red/dashed) is always wrong for $u\ne0$, but it is good enough [close enough to the "fast" formula (blue/solid)] for small $u/c$. The cutoff you choose depends on ...


2

The "fast" formula is always the correct one. A "fast" speed is one that is comparable to the speed of light. However, when both the speeds involved are much smaller than the speed of light, the "slow" formula is a very good approximation.


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In a rotating reference frame, the coordinate velocity of an object can exceed $c$. However, this doesn't mean that they're moving "faster than light". If we were to look at the light-cones at these distant locations, we would see that the four-velocities of these objects are still confined within the light-cones at those locations. To put this another ...


0

Although the most common answer to questions like that is "relative to what?", there are possible side-effects. If I recall correctly, the earth's velocity relative to the cosmic background radiation is on the order of a mere 600km/s. If we were to be travelling arbitrarily close to the speed of light as compared to our current motion, part of this ...


-1

Consider two inertial frames [...] [...] its origin strikes the origin of [the other] Let's give these two distinct participants (which are both called "origin" of their respective inertial frames) some more distinctive short names, for reference below; say $\mathsf J$ and $\mathsf P$. The event of the two "origins", participants $\mathsf J$ and ...


1

I drew the spacetime diagrams for you. On the l.h.s. you may see two simultaneous events in the unprimed (x,t) frame. The axes of a frame going in the positive direction (the primed frame) should be drawn into the unprimed as I have done it. You can find the new space coordinates by drawing a straight line parallel to the new time axis through the event. ...


3

This effect is called relativity of simultaneity. It means that two observers need not agree on the simultaneity of two events, or on their temporal order. This effect depends critically on whether the events are spacelike separated (i.e. $\Delta s^2=-c^2\Delta t^2+\Delta x^2>0$) or timelike separated (i.e. $\Delta s^2=-c^2\Delta t^2+\Delta x^2<0$). ...


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This is an area rife with potential misunderstanding, so we need to be absolutely clear what we mean. Suppose I take a ruler and a clock and I use rulers to mark out $x, y, z$ axes in space and the clock to note the positions of events in time. Assuming spacetime is flat, I now have a universal coordinate frame that everyone who is stationary relative to me ...


4

What you're missing is that the speed of light is not constant. There's this modern-day myth that says "Einstein told us that the speed of light is constant". But search the Einstein digital papers on "speed of light" or "velocity of light" for examples like this: The speed of light is spatially variable. And that isn't some discarded idea from 1911, see ...


1

Just write out the product $$u_\alpha u^\alpha = g_{\alpha\beta}u^\alpha u^\beta= g_{00}u^0 u^0\ ,$$ since $u^i=0\ ,\ i=\{1,2,3\}$. Then imposing $$u_\alpha u^\alpha = -1 \quad \Rightarrow \quad g_{00}=-1 \ .$$ I think the rest you can do on your own :)


2

$p^2 = m^2$ is the definition (up to a minus sign) of the mass of a momentum eigenstate. He derived that the same quantity (the expectation value of $[P^2,D]$ w.r.t. $\ket{p}$) equals $0$ and $2\mathrm{i}m^2$, so $m^2 = 0$. The $s$ is the scale parameter of the scale transformation induced by $D$, and it is any real number, so, starting from a given state ...


1

The cameras are at the same place, because we ignore the small distance between the cameras, right? When the cameras are at the same place, then the same information arrives at the same time at both cameras. The person in the train picture will look the same age as the person in the ground picture.


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In general relativity, the Minkowski metric plays a privileged role because it is the unique asymptotically flat solution to the vacuum Einstein equations that has zero ADM energy. The positive energy theorem in general relativity says all asymptotically flat spacetimes satisfying the dominant energy condition have non-negative ADM energy. Thus, one can ...


0

It is not possible to find a frame of reference where a photon is at rest. I will argument in two different ways: 1. Maxwell equations and electromagnetic argument: From Maxwell it is expected that electromagnetic disturbances propagate in vacuum at a constant speed c~299792458 m/s which is the maximum speed for the propagation of electromagnetic ...


4

It helps to write the full action: $$S = \int \frac{-mc^2}{\gamma}dt - \int U dt $$ The first term can be put in a much better form by noting that $d\tau = \frac{dt}{\gamma}$ represents the proper time for the particle. The action is then: $$S = -mc^2\int d\tau - \int U dt$$ The first term is Lorentz invariant, being only the distance between two points ...


0

You are using the wrong formula altogether, you need to be using relativistic Doppler effect rather then time dilation. The formula you are using ignores the fact that each successive signal sent by the person on the earth has to travel a further distance then the one before. The proper time is the time as measured in the earth frame. If compensate for ...


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I have made some videos of the twin paradox from the traveller's point of view (i.e. first person) here. I'll post the channel notes here to add a bit of substance, I would appreciate feedback and corrections. The twin "paradox" (in quotes because it is NOT a real paradox!) is a valuable learning tool for Special Relativity: ...


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The feed would have to be transmitted via electromagnetic signals, so the effects would be the same as if the person in the spaceship was trying to look back at earth normally. It would seem to the spaceman that the time on earth was running slower, and similarly, the video stream would arrive more slowly than if he had just been on earth. It would be like ...


1

The observer sees you travel from A to B - a distance that, in his frame of reference, is greater than one light year. He sees that you take more than a year. He concludes you are traveling at less than the speed of light. You, traveling so fast, "see" a much shorter distance (this is the concept of length contraction) $L' = L_0/\gamma$ where $\gamma = ...


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In the original paradox, there are two events of note: the front of the train exits the tunnel, and the rear of the train enters the tunnel. Call these events A and B, respectively. Because these two events are spacelike separated, the two observers (tunnel-based and train-based can disagree on the order in which they occurred. According to the tunnel ...


1

The Lorentz transformations are used to transform between different inertial frames. For example if you and I are in relative motion then the Lorentz transformations convert the positions of spacetime points in my rest frame to the positions of spacetime points in your rest frame. However anything travelling at the speed of light has no rest frame, so the ...


2

If we restrict ourselves to special relativity then the form of the Minkowski metric is an assumption. You can argue whether it is derived from the Einstein postulates or whether the Einstein postulates are derived from it, but this is really a philosophical nicety as you end up having to make equivalent assumptions either way. If you consider general ...


1

Physics explains how the universe works. It models the behavior of the universe as mathematical laws. The only help it can offer why one law is true is that it can be verified by experiment, or that it can be derived from another law. As in mathematics, this leads back to more fundamental laws until you reach "axioms." At this point, why can only be answered ...


1

the forward land delegate IS approaching the light and the backwardland delegate IS moving away from it Are they? Or are the delegates sitting perfectly still, and the Earth spinning quickly beneath them? Of course we have a convention that the Earth is stationary and the train moves across it, but we also know that the Earth is not stationary - it ...


2

Simply put, relativistic speeds cause for events previously thought of as simultaneous to no longer be simultaneous if the velocity of the reference frame of the event changes relative to the defined observer. The best way to wrap your head around this is to pictorially trace what is happening in space time. The case you describe is v>0 Think of v in ...


3

As it happens, you are absolutely correct. The velocities we encounter in everyday life are 3D velocities that are vectors defined as: $$ \vec{v} = \left(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right) $$ In special relativity we use a 4D velocity called the four-velocity, and this is a four-vector defined as: $$ \vec{v} = \left(c\frac{dt}{d\tau}, ...


2

Time dilation is a result of a fundamental symmetry of the universe. Start with good old Newtonian motion. Suppose I see some object move a distance $(dx, dy, dz)$, then Pythogoras' theorem tells me that the total distance it has moved, $ds$, is given by: $$ ds^2 = dx^2 + dy^2 + dz^2 \tag{1} $$ Now suppose you're using a coordinates rotated relative to ...


1

I think the answer to your question of are we able to tell why it happens is because we are able to predict, measure and mathematically demostrate time dilation.


-1

If the speed of light is constant no matter the speed of the source Actually, the speed of light isn't constant. It varies with gravitational potential, see Einstein talking about that here. However in gravity-free space the speed of light is constant. In addition you measure it to be constant regardless of your motion through space. That's because of the ...


0

It does not matter what your velocity is relative to a photon - the photon will always move at a fixed speed from your point of view. This is the 2nd postulate of special relativity - the speed of light in a vacuum has the same value (it's invariant) for all non-accelerating observers (in a medium of uniform density, the speed would be different from its ...


0

Einstein's postulate is that the relative speed between the observer and the source does not affect the speed the observer sees the photon moving, as long as the observer isn't accelerating, that is. This basically means there is no such thing as absolute 0 speed. Speed of light is your new and only constant friend.


1

We'll consider the relevant terms in the Schwarzschild metric, and substitute $d\phi = \frac{v}{r} dt$ with $\theta = \pi/2$ to get (assuming circular orbits): $$d\tau^2 = \left(1-\frac{r_s}{r}\right)dt^2 - \frac{v^2}{c^2}dt^2$$ where $\tau$ is the proper time (time as measured by the object), and $v$ is the orbital speed. The relevant quantity to measure ...


1

The speed of light in a vacuum, or c, is 299,792,458 meters per second. Any other medium will slow light down. For example light takes about 40% more time to go through glass than vacuum. But media that would increase the velocity of light would violate fundamental laws of physics and can therefore not exist.


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The Planck length isn't the smallest possible length, and the Planck time isn't the smallest possible time. As far as we know spacetime is continuous so velocities do not have to be an integral number of Planck lengths divided by an integral number of Planck times. The Planck length is the smallest length that can be measured, but the reason we can't ...


0

Now obviously, the moving magnet produces a circulating electric field around it: if you draw any loop in space, and the magnet goes through it, the flux through the loop changes in time, producing an emf and so forth, according to Faraday's law. This is the second answer I give here and the short answer is again No. A moving magnet doesn't produce in ...


0

First, let's look at terminology (which is important when you've learnt a particular phrasing of the laws). A time-varying magnetic field is when a magnetic field at a point is changing in time. A space-varying magnetic field is when the magnetic field is varying in space. And that means a static magnet is sufficient to produce a space-varying magnetic ...


0

Yes, your intuition is correct: two different boosts do contain one rotation, and precisely two boosts along two orthogonal axes contain one rotation around the third orthogonal axis --- the most direct way to see that is by considering that the commutator of two different boosts is one rotation, and more completely the Lorentz algebra of rotations $R_{a}$ ...


-2

But can you say the same for a space varying magnetic field i.e would there be an electric field if you move a magnet simply in space in the absence of a conducting body? Clear answer, no. In the absence of a second body, containing electrons, protons or neutrons a permanent magnet does not induce a electric field. But this is right or not depending ...


1

You should have two boost generators. You have constructed one for boost in the $x$ direction, but there is also one for boost in $y$.


-1

You should remember that photons don't have any mass, so relation is $E=p/c$ for photons. Moreover, you have to count electrons mass when you apply energy equations. Electron mass will vanish, but it's important to take account of it. I mean, before the square root, you don't have the electron mass, you have the kinetic energy.



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