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3

We don't believe this is possible. The justification for this belief is nothing less and nothing more than experimental observation. We have never observed a process where an effect comes before its cause, so we simply reason inductively to establish a postulate that the preferred order of events in physical processes is always the same, for any observer. ...


0

Two things, both simple. It is not that complicated,physics assumes causality and insures that all its laws obey it. Anything else at this point is speculation. 1) Special relativity was built based in the speed of light being invariant in all inertial coordinate frames. Events inside the null cones of each other (one past, one future, maybe both at apex) ...


2

Causality in physics is not defined in a satisfactory way. The meaning of cause and effect has been debated for centuries. In the end, at the microscopic scale (with the exception of a few processes) the laws of physics are reversible, and the best you can do is define if two events could have influence on each other. Everything are particles that collide, ...


1

Causality and chronological order of the events are very interesting notions - especially when considered under Relativity. I don't yet know much about GR rigorously, so I will not touch that part. (But I have heard that things become exceedingly interesting in that regime!) It can be shown with the help of Lorentz Transformations that if the square of the ...


9

The notion of "Causality" is notoriously subtle, something philosophers still struggle with and is probably impossible to define rigorously. A search of something like the Stanford Encyclopedia of Philosophy will turn up a dizzying array of articles that all show this notion tests the intellect of very many highly capable and insightful people. But from the ...


1

I think what this means is that: if two distinct events $a$ and $b$ are separated by a timelike or null future-directed curve then all observers agree on this (not just inertial ones); and in this case there is no such curve which also goes from $b$ to $a$ (which all observers also agree on); and if there is no such curve as that specified in (1), then all ...


2

The answer is that it has nothing to do with light, c, black holes, event horizons or relativity. It is simply escape velocity. We know that two bodies attract each other with a force $f = G\frac{M m}{d^2}$ and we know that something is in a stable orbit when its potential energy $PE = -GMm/d$ matches its kinetic energy $KE = m{v^2}/2 $. Solve these two for ...


2

The explanation I like is thus: In GR, all things, from planets to photons, travel in straight lines through curved space bent by mass. Black holes bend and distort spacetime so severely that the curvature captures the photon. Scale things down and it behaves much the same way passing asteroids can be captured by a star. For us, the speed of the asteroid(...


1

The picture I always liked is for an observer free-falling into the black hole, when they're just outside of the event horizon, it looks like the event horizon is propagating outward at nearly the speed of light. After the observer falls just inside, the event horizon now looks like it's propagating outward at greater than the speed of light.


-3

The short answer is the speed of light is constant until its not. Its constant until it runs into a wall and its constant until effected by gravity. The extreme gravity of a black hole will deflect the path of the photons more and more until finally at the event horizon all deflection is toward the black hole.


33

Suppose you are floating in a river, and you have with you a model boat, called the SS Lightray, that can do 3 m/sec through the water. When you set the boat travelling upstream as far as you're concerned it is doing 3 m/sec. But I'm standing on the bank watching the river flowing at 1 m/sec, so when I look at your boat I see it travelling at a net speed of ...


6

To explain it in layman's terms, without using advanced concepts: Space is warped in the "inside" of the black hole (that is, under the event horizon) so much, that it behaves completely different than what we perceive hear on Earth. The "outward direction" simply does not exist. For example, here on Earth, we can go in the three spacial dimensions in both ...


-3

Isn't the saying "The speed of light is constant 'in a vacuum'"? The 'in a vacuum' bit is commonly missed off (in the same way, "'The love of' money is the root of all evil", is often commonly misquoted). If a black hole isn't a vacuum (because at some point, particles are close enough together it is no longer a vacuum), then the speed of light slows ...


66

The speed $c$ that is constant is so when measured locally relative to a freefalling frame (i.e. one for which all points follow spacefime geodesics wrt to the metric $g$). Local means that the frame's extent must be "small" enough that it can be thought of as flat: think of this as zooming in on the spacetime manifold, which is a smooth object, with enough ...


2

The Thorne time machine, a wormhole with one opening accelerated or Lorentz boosted outwards and then conversely brought back, does not permit time travel prior to the Cauchy horizon. This is the point where the time machine is "turned on." This Cauchy horizon has in regions of spacetime prior to its formation a set of curves winding through the wormhole ...


0

It is not clear to me exactly what you are asking for : a derivation such as given by Lawrence B Crowell, or a recommended reading list either for the mechanics of macroscopic bodies or for the microscopic properties of materials. Your question "how does it know?" seems rather a naive one for an electrical engineer to be asking, because you must have asked ...


2

Think of a single particle with mass $m$ at $\vec r$ from a coordinate origin. Now suppose this body is in circular motion with angular velocity $\omega$ around this origin. The momentum of the particle is then $$ \vec p = m{\vec v} = m\frac{d\vec r}{dt}. $$ The velocity is then $\vec v = \vec\omega\times\vec r$. Now let us consider the angular momentum of ...


3

A manifold $M$ is geodesically complete (i.e. has the property of "geodesic completeness") if for every point $P\in M$ and every direction $dx^\mu$ from that point, the infinitesimal line interval $dx^\mu$ away from $P$ may be transported by parallel transport by any amount $K\in(-\infty,+\infty)$ in both directions, i.e. if the geodesics (maximally straight ...



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