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0

I don't think there is an analogous statement for spacelike intervals. The statement "The centre of mass follows a geodesic" is an immediate consequence of global momentum conservation. This is in turn follows from local momentum conservation. \begin{equation} \partial_\mu T^{\mu \alpha} = 0 \end{equation} where $T^{\alpha \beta}$ is the stress-energy ...


0

Under a Lorentz transformation $$ d^4 p \to | \det \Lambda | d^4 p = d^4 p $$ $$ p^2 \to (\Lambda p)^2 = p^2 \implies \delta(p^2 - m^2) \to \delta(p^2 - m^2) $$ and $$ dp^0 \text{sign}~p^0 \to dp^0 \text{sign}~p^0 \implies dp^0 \theta(p^0) \to dp^0 \theta(p^0) $$ Thus, the LHS is Lorentz invariant. Therefore the RHS is Lorentz invariant. BTW, a general ...


0

There are no the transverse Doppler effect in your two cases. Because there is always a classical Doppler shift when the distance between the source and the observer changes with time. See: Investigations on the Theory of the Transverse Doppler effect blog.sciencenet.cn/blog-267101-748804.html 2013-12-11.


0

If we say that the spaceship and earth are very close to each other, let us then imagine that the spaceship, right as it passes by earth, shoots a photon out at $45^{\circ}$ from its direction of travel. The simplest way to attack this problem will be with the transformation of velocities in the x-direction. From our known information, we can determine ...


0

I believe all things in the universe are always in motion. We stand on an earth that rotates, wobbles, orbits, expands and contracts, all while following a sun which orbits the milky way, waves up and down during this orbit, and does so within a milky way which itself is moving relative to the rest of the universe. I believe the sun is moving relative to ...


0

Apart from the issues of $SU(2)$ vs. $SL(2,\mathbb{C})$, complexification, etc, which is covered e.g. in this Phys.SE post and links therein, it seems OP's core question (v2) is essentially the following question. How does the Lie algebra generator $J_3=J^L_3+J^R_3$ act on a vector space $V_L\otimes V_R$ for a tensor representation? The short answer ...


0

When the pseudoscalar invariant $\vec E \cdot \vec B$ is zero, we have three cases. If $E^2<c^2B^2$ then you can switch to a frame moving with speed $E/B$ in a direction mutually orthogonal to $\vec E$ and $\vec B$ where there is no electric field in the new frame. Solve in the new frame. Then bring it back to the original frame. If $E^2>c^2B^2$ then ...


4

Because the photon definitely has energy, it must have a four-momentum vector, but it must be defined differently from mU because the proper time, $\tau$, along its worldline is zero. $$d\tau= dt\sqrt{1-v^2/c^2}$$ The photon four-momentum vector is defined to be ...


0

Note that $\nabla_\mu (\rho u^\mu)=0$ is not correct (and not the continuity equation). Keep in mind that (based on your definition of the stress tensor) $\rho$ is the energy density, and the conserved energy current is $T_{0\mu}$. The relativistic Euler (momentum conservation) equation is $$ D u_\mu = -\frac{1}{\rho+P}\nabla_\mu^\perp P $$ where ...


0

Nothing is wrong. If in your experiment, the tape is placed parallel to the rails in such a way that when the train passes (A), it grabs the tape and it "sticks" along the train "instantaneously", then the following happens: A person in terrain, before (A) sees a static normal-size tape and a contracted train approaching fast, after (A) he/she sees the ...


0

Gravity can change frequency. A light beam going towards a massive body is blue-shifted by the gravitational field. If it is escaping a gravitational body, then it is red-shifted.


2

The photon is an elementary particle. There are two ways do measure the frequency and therefore the energy of the photon since its energy E=h*nu . 1) using a diffraction grating which analyses the wavelengths in a beam of light , as below: This is the spectrum of iron. Each line is composed of zillions of photons with that frequency. If one sent one ...


0

In regards to your question: Can the frequency of light change in propagation from one media to another, the answer is no. I found a previous response to a similar question that might help you: Think of it like this: At the boundary/interface of the medium, the number of waves you send is the number of waves you receive, at the other side, almost ...


5

A force is defined as a change in momentum over time. In the Newtonian limit, this means a mass times an acceleration. But when dealing with things like photons, the formal definition is applied. Photons have no mass, only momentum. Therefore, if a force is applied to them, their momentum can be changed. This can happen in two important ways, a force can ...


0

Why is it important to incorporate both electric and magnetic forces into one single expression? Because the Lorentz force is "due to electromagnetic fields". I know people talk about electric fields and magnetic fields, but see Wikipedia: "Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole ...


3

Well, you could treat them separatedly, via two equations, say $$\mathbf{F}_\text{elec}=q\mathbf{E}$$ $$\mathbf{F}_\text{mag}=q\mathbf{v}\times\mathbf{B}$$ but since Newton's second law holds, in presence of an electric field and a magnetic field, the total force will be the sum of both, that is, the Lorentz force. I would say is just as simply as that.


3

The word "relativistic" means "compatible with principles of Special Relativity". This usually implies that we can no longer use the "classical" picture of universal stationary space and time. Instead we talk about 4-dimensional space-time. The word "quantum" means compatible with principles of quantum mechanics. You can look them up on wikipedia etc. But ...


4

Build up four-momentum from the mass and the four-velocity: $$m \vec u = m \gamma (\vec e_t + \vec v)$$ But four-momentum can always be decomposed according to energy and momentum: $$m \vec u = E \vec e_t + \vec p$$ Equate the $t$-components. Done.


1

Assuming tentatively that all velocities and notions of simultaneity are in the frame of reference of A, i.e., the events C/D passes A/B occur simultaneously with the event B reaches Z in the frame of reference of A, and C/D are moving past A/B at speeds 0.6c as measured in the frame of reference of A. Then the time will be offset by 2 years. If A and B ...


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.


1

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 ...


-3

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.


0

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: ...


29

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 ...


55

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.


3

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 ...


0

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. ...


4

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. $\Delta s^2=-c^2\Delta t^2+\Delta x^2<0$). ...


10

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.


1

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

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: ...


-1

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 = ...


0

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 ...



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