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0

Note that $\nabla_\mu (\rho u^\mu)=0$ is not correct (and not the continuity equation). Keep in mind that $\rho$ 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 $D=u^\mu\nabla_\mu$ and ...


1

I'll answer to 1 Maxwell equations are already relativistic, but - in a flat spacetime. You can write Maxwell equations for a general metric $ g_{\mu \nu} $ (The original Maxwell equations are formulated for a flat spacetime - $ g_{\mu \nu} = \eta _{\mu \nu}$ ). One way this can be done is by the following algorithm: Transform coordinates to a local ...


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

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 ...


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


4

A few quick clarifications: a particle cannot just annihilate. It disappears when it interacts with something else. The obvious example of this is an electron and positron annihilating to turn into two photons. Also, the total energy of a particle (this applies to electrons, positrons and photons) is given by: $$ E^2 = p^2c^2 + m^2c^4 $$ where $p$ is the ...


12

First, something we need to get out of the way: Kinetic energy as $\frac{1}{2} m v^2$ is not a precise formula; it is merely a good approximation for anything that is traveling slowly compared to the speed of light. In fact, more precisely, the energy is \begin{equation} E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + ...


1

Short answer: Yes, spacetime is relative. Einstein's relativity mixes space and time when transforming from one set of coordinates to another, however there is no preferred frame. This is part of the criteria known as Lorentz invariance. If you are given the physical results of an experiment, there is no way to determine which particular inertial frame it ...


2

There is a property of spacetime which is independent of frame of reference. The geometrical properties of the spacetime are described by the metric tensor, $ \eta _{\alpha \beta} =diag({-1,1,1,1})$ is SR (flat spacetime) or more generally $ g_{\alpha \beta} $ (any spacetime) in GR. This tensor specifies the distance between two infinitesimally close ...


0

Whichever clock switches it's rest frame in order to meet with the other clock will be the one with a lower count (younger twin). If both clocks switch their rest frame symmetrically, then both will be of the same age. Meaning, if an observer who was at rest with those clocks before they started accelerating, sees those clocks move away from him at Vrel_a ...


0

"What would determine which clock ticks faster?" If both clocks experience a uniform change in motion they will show the same time when they convene. Acceleration/deceleration changes a reference frame's time dilation factor. If these two clocks experienced different changes in motion (ex. one accelerates/decelerates differently than the other) the ...


1

Beginners to special relativity generally learn the equation for time dilation: $$ t' = \frac{t}{\gamma} \tag{1} $$ where $\gamma$ is the Lorentz factor: $$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$ For any $v > 0$ the Lorentz factor is greater than one and hence $t' < t$, which is what we mean by time dilation. And this is all very well, but ...


0

When you discuss simultaneity in special relativity you are talking about classical events (locations and times together) that are simultaneous in a frame. A frame extends outwards in space and time. When you change your speed you find out there is a totally different frame at which you are now at rest. Your old frame (which extended throughout space and ...


-2

Simultaneity of two events relies on the two events occurring at the same time and at the same place; it is in fact the bare minimum that a notion of simultaneity has to satisfy and is derived from the classical motion; moreover in this picture the older classical notion of simultaneity no longer even makes sense; ie it's frame dependent. The classical ...


1

How can you say that if you walk on the street with 1m/s you are simultaneous with a certain event A in Andromeda, and if you start running with 2m/s you become simultaneous with an event B many years in the future of A (that is an effect of A) in Andromeda, when there is no causality in the first place at the most fundamental level ? So isn't this ...


0

The answer to the main question is no. One gram of matter (electrons), will annihilate exactly one gram of antimatter (positrons), regardless of the speed with which they approach each other. If they collide with a speed other than zero, the energy due to the motion will also be added to the energy produced by the annihilation.


0

The key point here is the "weird" change in the values you measure depends on what frame you observe them in relative to the object. There is no experiment you can conduct to determine an absolute value for your speed. Speed only makes sense relative to an object. So if a near light speed object relative to us was heading towards us, from its perspective ...


-1

First of all Galilean Transformation laws are not valid in most cases. For the moving reference frames the special theory of relativity is more suitable. In fact it is the most suitable one given the experimental data. In special relativity one defines a so-called energy-momentum four vector, whose magnitude every observer in an inertial reference frame ...


1

When considering relativistic speeds, the notion of "particle & anti-particle" somewhat blurs. The correct treatment of a relativistic free electron for example is given by the Dirac Equation, which relates Dirac Spinors. A spinor is something like a 4-vector, describing the wave function of our electron. In it's rest frame, two of the spinor's ...


1

Here is a sketch of where it comes from. First just consider the perfect fluid terms and note the thermodynamic relation $$ \rho + p = \mu n + T s, $$ where $T$ and $s$ are temperature and entropy, $\mu$ and $n$ are a chemical potential and number density. We also have a relation for derivatives of $p$ $$ dp = n d\mu + s dT. $$ Now if you take the ...


11

Each particle only annihilates its exact antiparticle. Electrons annihilate positrons. A blue up quark annihilates an anti-blue anti-up quark. A muon annihilates an anti-muon. The thing about anti-matter is that it postulates an exact opposite of every particular particle type (except for things like photons that are their own antiparticles). It's about ...


33

A sophisticated, yet easy way to see that this the answer must be "No." is to recall that velocity is relative — that there is no absolute notion of velocity. You said the matter was moving and the antimatter still, but that point of view (AKA frame of reference) is not privileged in any way. An observer at rest with respect to the matter has just as much ...


13

The particle-antiparticle annihilation is on a per-particle basis. One electron annihilates on positron. One up quark annihilates one anti-up quark. One down quark annihilates one anti-down quark. Moving at relativistic speeds doesn't change the number of particles. For that matter, you could annihilate an electron with an anti-muon, since an electron and a ...


2

Simultaneity is real in precisely the same sense that the component values of a vector or tensor are real: your instruments will certainly measure certain things as co-incident and those measurements are real and instruments can measure vector components. But these events are not needfully co-incident in other inertial frames, just as the component values of ...


2

An observer at any given location can only define events to be simultaneous in retrospect, after there's been time for the light from each event to reach them. Simultaneity in each inertial frame is defined based on the assumption that light travels at a uniform speed in that frame (the second postulate of special relativity), so for example if at the stroke ...


1

It's a paradox about relativity. How can you know that simultaneity is real? To keep things simple, we can define simultaneity of two events for a particular observer as light (or information) reaching the observer at the same time. According to this definition, simultaneity is a "real" phenomenon. A layer of complexity (that doesn't much affect the ...


2

Yes, you're quite correct that in the example you describe Andromeda can be moving faster than the speed of light. But that's perfectly in accord with special relativity. It's just that the rule that nothing can move faster than the speed of light is true only for unaccelerated motion. Although beginners are (usually) taught special relativity using ...


2

Suppose Andromeda is at rest relative to earth. You are on earth, holding your ruler, which just touches Andromeda. Now you instantly start traveling toward Andromeda at a high speed. Your ruler, of course, travels with you, still pointing toward Andromeda. Your journey just began an instant ago, so neither you nor your ruler has yet moved appreciably. ...


1

The high speed of travel causes both time dilation and space contraction, so the numbers add up from both frames of reference. For the astronaut traveling at near lightspeed, the space between him and Andromeda is contracted; at a very high speed, Andromeda would appear to be only 25ly from the Milky Way, but since the astronaut is traveling slower than ...


2

Which is more fundamental is probably a meaningless question, but you can think of the geometric notion of a manifold (the mathematical abstraction of spacetime) as being more general than Lorentzian manifolds, i.e. ones whose metric is locally like that given in John Rennie's Answer. So one could in principle conceive of universes that were described by ...


0

fundamental is a choice dependent term, its what one considers something to be more basic. spacetime is a consequence of both constancy of c + invariance of physical laws, which describes how the general nature of laws should be, which is called Lorentz invariance. spacetime is one consequence of Lorentz invariance, there are others like energy and momentum ...


2

It is an artificial distinction to say one is more fundamental than the other. The geometry of flat spacetime is given by the Minkowski metric: $$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$ and this is fundamental in the sense that all of special relativity is described by this equation. But it is also fundamental that the parameter $c$ in the equation (which ...



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