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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 ...


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 + ...


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 ...


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 ...


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


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 ...


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 ...


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

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 ...


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 ...


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 ...


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. ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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 ...


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 ...


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 ...



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