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## New answers tagged special-relativity

3

The basis is still $\{|\boldsymbol r\rangle\}$. The abstract Schrödinger equation is $$i\frac{\mathrm d}{\mathrm dt}|\psi\rangle=H|\psi\rangle$$ where $|\psi\rangle$ is a set of four kets, (with a slight abuse of notation) $$|\psi\rangle=\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\\|\psi_3\rangle\\|\psi_4\rangle\end{pmatrix}$$ Time is still a ...

0

It depends on whether the current is carried by a conductor or is in free space (an electron beam). In the case of an electron beam, the current will appear to have reversed in direction if you travel faster than the charge carriers, even without relativistic effects. This web page does the transformation roughly like you have attempted, using a charge ...

0

Ok, lets' say that at a distance from the earth, 3 spaceships were constructed. There were two small ships, and one long mother ship. Each ship was programmed to accelerate in the same manner, such that a specific G-force curve is followed as motion begins, and this is continued onward up until the intended target velocity is achieved. Each ship is sent ...

1

Reichenbach's original volume, "Axiomatization of the Theory of Relativity", appeared in 1924. It is one of a long string of works that periodically rediscover and/or explore the issue of non-Einstein synchronization in Special Relativity. See for instance this review on "Synchronization Gauges and the Principles of Special Relativity" and refs. therein ...

0

Timelike is when an event is inside the lightcone (as you have mentioned) and as a result, one event CAN affect the other event (there can exist a causality between the two events. E.g. lets say there are two events, where I shoot a laser and another event where someone gets hit by a laser. If they are timelike seperated then the laser that hit the dud could ...

-3

Is photon fixed in space time? Photon is not fixed to a point in space time, it moves as light at speed c relative to us. It covers 0 distance in in zero time relative to itself. In Last question I said, "They don't experience time, that is they are stationary with respect to time axis". The time axis is a virtual axis. It has no physical significance. ...

2

This answer has been hinted at in the others, but it's worth stating their collective knowledge as a succinct one liner that every physicist should know: Electric and Magnetic force only make sense in the light of special relativity if they are unified because if they were thought of as separate entities, then relatively moving observers would reach ...

5

The spacetime interval is a relativistic invariant, and is proportional to the travelers proper time. So in a since you are traveling one second per second, per your own wrist-watch. Every other measurement would be the speed of some other inertial reference system, measured with your clock. Let $s^2 = x^2 + y^2 +z^2- (ct)^2$, where $x$, $y$, $z$ are ...

-2

I don't know much about the math, but as speed increases, energy increases, therefore mass increases, Mass affects spacetime. If you bend spacetime two lines that are parallel can intersect. Think of the acceleration(Jim) as a squiggly line because of the effects on spacetime and John as a straight line. Jim branched off and branched back on when he left and ...

2

Photon experience infinite time dilation and hence, time is stationary for it. Does photon experience time  photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any ...

1

Also, the fact that $\Delta t' \rightarrow 0$ if we formally let $v \rightarrow c$ can be interpreted as saying that no time at all passes for a particle moving at the speed of light. Photons cannot "age" or in any other way change over time.

0

The question is whether time is an operator in the sense of ${\hat T}|t\rangle~=~t|t\rangle$. This at first glance would seem to make sense because we do have a position operator ${\hat X}|x\rangle~=~x|x\rangle$. However, this does not work. This is a subtle question in many ways. Quantum mechanics is unitary. Consider a state vector $|\psi(t)\rangle$ ...

5

Is time made into one observable? No. It is known that an operator $T$ that satisfies $[H,T]=i\hbar$ is either self-adjoing and $H$ unbounded below or anti-self-adjoint. Therefore, the theory is either intrinsically flawed (arbitrary negative energy) or $T$ is not observable (anti-self-adjoint $\Rightarrow$ imaginary eigenvalues).

0

The theorem that time is not an observable is quite general, Unruh W., Wald R. prove this in "Time and the interpretation of canonical quantum gravity, Physical Review D Volume 40 issue 8 1989" in the following form: "... in the context of ordinary Schrodinger quantum mechanics, no dynamical variable in a system with Hamiltonian bounded from below can act ...

0

In that sense, in Relativistic Quantum Mechanics based on Dirac's Equation, is time made into one observable? No, but it is treated on equal footing to the space coordinates. Dirac's equation is $$(i \hbar \gamma^\mu \partial_\mu - mc )\psi$$ where $\psi$ is not the classical wavefunction, but a four-component spinor. The components of the spinor are ...

6

Several answers have given a physical explanation as to why electric and magnetic forces are tightly coupled, and why you can't develop independent theories of "just electric" and "just magnetic" fields. Your subquestions (especially #1) make me think you're looking for some kind of symmetry. It turns out, there's a really nice one! All the asymmetry ...

3

The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration. The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of $\lambda \$ and some non-zero mass per unit length of $\rho \$ separated by ...

0

The four vector $p$ is given as $(E,{\bf p})$ so that $$(p' -p)^2 = (E' - E)^2 - ({\bf p'} - {\bf p})^2 = (E'^2 - {\bf p'}^2) + (E^2 - {\bf p}^2)- 2E'E + 2 {\bf p'} \cdot {\bf p}$$ Then put $E'E \approx (m + \frac{{\bf p'}^2}{2m})(m + \frac{{\bf p}^2}{2m}) \approx 2m^2 + {\bf p'}^2 + {\bf p}^2$ (ignoring the term $\frac{{\bf p'}^2{\bf p}^2}{4 m^2}$). Using ...

19

The arguments from special relativity given in the other answers is correct. What is charge according to one observer is current according to another observer that is in relative motion to the first. But this is, from a historical perspective, somewhat backwards. This consideration is what led Einstein to develop special relativity -- the paper is called On ...

80

Consider this: A charged particle at rest creates an electric field, but no magnetic field. Now if you walk past the charge, it will be in motion from your point of view, that is, in your frame of reference. So your magnetometer will detect a magnetic field. But the charge is just sitting on the table. Nothing about the charge has changed. Evidently ...

4

Regarding 1) observe that there is a pattern in common - namely that there is some region (volume for Gauss and a surface for Ampere) and integral of the source on this region is equal to the integral of the field on the boundary. This is a striking similarity. 2) currents are nothing else than moving charges. So both fields are generated by charges. These ...

8

You are right, the electric field and the magnetic field are distinct fields that have different properties. The reason why they are still classified as the cause for the "electromagnetic force" are the following: In higher theories, like the field theory, the electric and the magnetic field are caused by the same gauge principles. There is just "one" ...

1

I am not sure why you are having the confusion. Your calculations are correct. The distance you measure in your reference frame are the actual distances, so you would measure 420 meters. The "clock" for the particle in your reference frame would be running slower so the particle would decay in 1.4 microsecond in your reference frame. In the particle's ...

2

The metric is usually defined this way because of the speed limit (c). Imagine that: If you are centered at the origin of a coordinate frame (x,y,z) a emit a signal, this signal would never travel a distance bigger than the distance the light would travel. So, $${|r|} ^{2} = {x}^{2} + {y}^{2} + {z}^{2}$$ And set it equal to the distance the light would ...

1

Invariants are useful, in general, because they represent something that all observers can agree upon. Relativity showed us that the concept of time-intervals, spatial-distances, and even sequences of events can be drastically different from different observers. So how can one observer 'relate' to another? I.e. how could I, standing still, figure out what ...

3

L. Motl offered quite a few arguments suggesting that complex numbers are required for quantum theory. Surprisingly, this is not quite correct, at least in some general and important cases, and I do not have in mind replacing complex numbers with pairs of real numbers. Schroedinger noted that, in the case of a scalar field interacting with electromagnetic ...

1

The spacetime interval invariance property allows us to, for example, compare the rate of time passing for two observers moving at relative velocities to each other. Although no observer in the universe is at complete rest, the interval is a benchmark for comparison of the physical effects of differences in velocity, or indeed location. Say one observer is ...

1

You need to distinguish between seeing and observing. These are technical words in relativity. And totally distinct things. Seeing always happens where you are. When there is a distant event, an observation is when you correct the seeing to compute when and where the distant event happened. So if Alice and Bob are at rest and 8 light years apart then when ...

4

You may redefine various quantities according to $p\to ip_{\rm yours}$ and things like that but that clearly doesn't change physics, just conventions. For example, the mixed-signature spacetime may indeed be emulated by having the ${+}{+}{+}{+}$ signature but with one (or three) components pure imaginary, and Einstein actually favored this convention at some ...

5

Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound $c_s$. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations ...

17

Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are ...

0

It would be more correct to state Newton's second law as following: $$\vec F = \frac{d \vec p}{dt}$$ This holds in relativistic mechanics too: $$\vec F = \frac{d \vec p}{dt} = \frac{d (\gamma m \vec v)}{dt} = \gamma^3 m \vec a_{//}+\gamma m \vec a_{\perp}$$ Where $\vec a_{//}$ is the component of the acceleration which is parallel to $\vec v$ and $\vec ... 0 Skipping the normal caveats about the obsolescence of relativistic mass,$F = ma$still doesn't work. The more proper expression for force is the rate of change of momentum. So, \begin{array}{ll} F &= \frac{dp}{dt} \\ &= \frac{d(\gamma m_0v)}{dt} \\ &= m_0\left(\gamma\frac{dv}{dt} + v\frac{d\gamma}{dt} \right)\\ &= m_0\left(\gamma a + ... -1 No, because $$F = \frac{dp}{dt}$$ (p is momentum) $$\rightarrow F = \frac{d \ m \times v}{dt}$$ $$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$ cross-product rule $$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$ $$\rightarrow F = \frac{d\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}}{dt} + \frac{dv}{dt}$$ which gives the correct relation between Force, mass and ... 3 As is written here the two remaining equations follow from the Bianchi identity which says that the anti-symmetrized derivative is zero, ie. $$\partial_{[a} F_{bc]} = \partial_{a} F_{bc}+\partial_{b} F_{ca}+\partial_{c} F_{ab} = 0$$ (remember the$F_{\mu\nu}$is antisymmetric itself!) 0 I believe$F^{\mu\nu}$is better known as the Faraday tensor; Maxwell tensor is only the spatial part of it, 3x3 matrix. The argument goes like this:$F$is a four-tensor, so if any quantity is constructed from it by multiplication of its components, this new quantity also has to transform as a tensor. The construct $$... 0 Proper time is a directed scalar. As of this writing, there is no mathematical terminology or analysis technique that does justice to defining it, including and especially as proper time was defined by Minkowski, hyperbolic / rotationally bound to something else defined as space, which is the same directed scalar. For a directed virtual or real EM wave of ... 4 How does a scalar quantity transforms under a Lorentz transformation? I will show you here how this works for a vector (contravariant) and a covector. The transformation rule for these objects are$$u^a = { \partial x^a \over \partial x' ^b} u' ^b u_a = { \partial x'^b \over \partial x^a}u' _b $$Multiplying the two:$$ u^{a} u_{a} = { \partial ... 2 If I'm reading the question (v1) right, you present a paradox in paragraph 2 (commonly called Bell's spaceship paradox) and then try to resolve it in the next few paragraphs. Your resolution doesn't make sense, as pointed out in the comments. The mistake is that$u_2'$is not zero, by the relativity of simultaneity: different observers will disagree on the ... 3 The traveller doesn't see the entire universe the same way. He sees everything in front of him extremely blue shifted, with all of the future from that direction happening in five minutes, everything behind him extremely red shifted and essentially coming to a standstill. Events close to his trajectory will go from blue to red extremely quickly and he can ... 0 In 3 dimensions you can use the cross product to get an appropriate rotation axis. If$\hat{x}$and$\hat{n}$are non-parallel 3-dimensional unit vectors then$\vec{s}=\hat{n}\times \hat{x}$is non-zero. Since$\vec{s}$is orthogonal to both$\hat{n}$and$\hat{x}$there is some rotation about$\vec{s}$that takes$\hat{n}$to$\hat{x}$. You can get the ... 1 Let$\theta>0$denote the angle between the$\hat{\mathbf x}$and$\hat{\mathbf n}$. Notice that $$\hat{\mathbf u} = \frac{\hat{\mathbf x}\times\hat{\mathbf n}}{\sin\theta}$$ Is a unit vector perpendicular to both$\hat{\mathbf x}$and$\hat{\mathbf n}$. The desired rotation is a right-handed rotation around$\hat{\mathbf u}$by the angle$\theta$. ... 1 A body with finite dimensions will always contract in the direction of motion. A square sheet will contract in either direction by a factor of$L = L_0/\gamma$where$\gamma = 1/\sqrt{1-\frac{v^2}{c^2}}$. A sheet moving in both directions will experience contraction in both dimensions. Length contraction is motivated, as you say, by a need to keep the speed ... 0 I think you misunderstood something.Length contraction in a moving body,does not occur in the perpendicular direction of its velocity (not length as you mentioned). Length contraction occurs in the direction of its motion. 0 All the quantities are scalars ($\Box=\partial^\mu\partial_\nu$,$\phi$and$\rho$) and so$\Box\phi=4\pi\rho$is clearly Lorentz invariant. 1 Mandelstam$t = (p_1 -p_1')^2$corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum. 3 Temperature can be thought of as the vibration or oscillation of individual particles. More the vibration, more the temperature. The frames velocity is just the velocity of its mean position, as the vibration is independent of the frame velocity, so is the temperature. -4 Mikael Fremling wrote: “If the speed of light would be infinite, we would not have light at all.” With nice, logical, rational, reasonable explanations after that. So, photon would not exist. http://en.wikipedia.org/wiki/Two-photon_physics http://www.natureworldnews.com/articles/4178/20130926/new-state-matter-created-fusing-photons.htm ... 3 This is a subtle and somewhat complicated question, but I think the basic answer is no''. 1) The relativistic Boltzmann equation is $$p^\mu\partial_\mu f = C[f]$$ which has the same structure as the non-relativistic Boltzmann equation. This equation can be used to derive relativistic Fokker-Planck equations. One example is the Landau collision term, ... 1 It's the time coordinate of an event in the$S$frame (the coordinate frame you're moving with respect to), then a fixed time. Important remark: the equation that you are using does not give the decrease (not inflation) in length of an object as measured by an observer who's moving with respect to the$S$frame, but the$x\$ coordinate transformation between ...

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