# Tag Info

16

@Florin is absolutely right, but sometimes a picture is worth a thousand words. This website has multiple pictures and explanations about how the future light cone starts to point only to the inside of the black hole once you pass the event horizon. Here is one of the images: Time is vertical, the cylinder represents the event horizon and the cones are ...

13

The alien doesn't really see our future. He's still seeing our past, but a more recent past than he did before. Assuming that the alien is 100 light years away when he starts cycling then he is seeing what happened to us 100 years ago. If he "cycled" fast enough (i.e. at an appreciable percentage of the speed of light) so that he was now only 50 light ...

11

Since the Lorentz transformations are a consequence of the postulate of constancy of the speed of light, together with some homogeneity and parallel postulates, it is a little difficult to make precise the request for a Lorentz-transformation free demonstration. But I will interpret the question as asking for a synthetic proof of the addition of velocities. ...

11

I wish those who popularize science would stop talking about the speed of light. It just confuses people. Forget about speed. Think in terms of energy. You'd need an infinite amount of energy to get out of it. Infinite. Meaning, it's like dealing with the Mafia: no matter how much you're spending, it's still not enough. Think in terms of topology. From the ...

10

"Relativity" is actually a misleading word that Einstein didn't like. It doesn't mean "every vantage point is equivalent and it's all relative". It really means only inertial, non-accelerating vantage points are equivalent. You could think of it as, prior to relativity, people believed that there was an absolute position/speed to the universe. Special ...

10

The object you're talking about is called, in mathematics, a Clifford algebra. The case when the algebra is over the complex field in general has a significantly different structure from the case when the algebra is over the real field, which is important in Physics. In Physics, in the specific case of 4 dimensions, using the Minkowski metric as you have in ...

10

It's just funny. Note that your equation doesn't actually use any single general quaternion. You only use the $i,j,k$ imaginary units in an ad hoc way to get three minus signs whenever you need them. If you were using an actual quaternion $$q = t + xi + yj + zk,$$ then the only semi-natural real bilinear invariant you may construct out of it is $$q\bar q ... 10 It appears to me the issue is understanding momentum conservation. An even cruder example would be to shine a bright torch out the back of your vehicle. Even though the photons have no mass, wouldn't the vehicle move forward? You also refer to mass in this manner in the paraphrasing of Newton's third law "proportional opposite mass/acceleration ratio ... 10 Newtonian gravitation is just the statement that the gravitational force between two objects obeys an inverse-square distance law, is proportional to the masses and is directed along the line that joins them. As such, it implies that the interaction between the objects is transmitted instantaneously and it must be inconsistent with special relativity (SR). ... 10 Yes--- this is directly analogous to the following statement from geometry: The plane curve with everywhere constant curvature is the circle. You could prove this by integrating the condition of constant curvature (which is just as messy as in relativity), or by doing an arclength parametrization (which is clean, see below). But easiest of all is to note ... 8 Dear rubenb, yes, what your professor says is surely based on solid maths. The reason is that the 4-component Dirac spinor is actually composed of two separate 2-component pieces. The elementary "spinors" for 3+1 dimensions have two complex components. That results from the isomorphism between groups$$SL(2,C) \sim Spin (3,1). Note that both groups have 6 ...

8

There are solutions to Einstein's field equations, which have closed timelike curves. For example Godel's solution. Would that constitute time travel, if you can reach the same point on your world line in finite time? One objection may be that such solution do not describe the universe, but examples as the Tipler's cylinder suggest that at least in theory we ...

8

I endorse Ron's answer – it's the systematic way to proceed. The velocity $v/c$ may be written as $\tanh \eta$ where $\eta$, the rapidity or whatever, is the hyperbolic (Minkowski) counterpart of the (Euclidean) angle. The addition of velocities then boils down to an addition formula for $\tanh(\eta_1+\eta_2)$ because the rapidities just add additively. Let ...

7

There is a classic treatise on "Relativity, Thermodynamics and Cosmology" from R. Tolmann from the 1930s - it is still referenced in papers today. This generalises Thermodynamics to Special Relativity and then General Relativity. As a simple example the transformation law for Temperature is stated as: $T=\sqrt(1-v^2/c^2)T_0$ when changing to a Lorentz moving ...

7

You say that both twins are "exactly 20 years old." I assume you mean that they are both 20 years old at the same time. But part of the point of special relativity is that a phrase like "at the same time" means different things in different reference frames. To be specific, suppose that these two moments (Peter's birthday party and Michael's birthday ...

7

Kinematically, yes. In terms of describing the positions of objects, it is equivalent to say "A is accelerating away from B" and "B is accelerating away from A". However, it is an observed fact that the universe treats these two situations differently. A and B can check whether they feel artificial gravity in their reference frame. If so, it's ...

7

All basic laws of physics are frame-independent. They either exhibit Galilean (non-relativistic) or Lorentzian (relativistic) invariance. Examples are Newton's laws (Galilean), Maxwell's equations (Lorentzian), Navier-Stokes equations (Galilean), etc. A notable exception is formed by Schrödinger's equation which, upon closer inspection, can be fixed into ...

7

One way to think of a "moving shadow" is by following the last photon that was allowed through. In that case, the speed of a shadow is exactly the speed of light. On the other hand, you could also define the speed of a shadow as the speed of the boundary between dark and light. In that case there is no thing that's actually moving, so there's no bound on ...

6

One thing to be aware of is that the principle of relativity would not apply to this computer--rotating reference frames are not inertial, and therefore, will not be related to 'stationary' reference frames by simple Lorentz transformations. Also note that if there are any capacitors or anything along those lines in the computer, then they would be ...

6

It is mathematically possible to create some instances in which an object goes back in time relative to some observer. For example, simply going faster than light causes such an effect, but of course, speed of light is the limit for any massive object. While it is mathematically possible, there are many paradoxes caused by time travel to past, unless you ...

6

Well, this is a migrated question and it deserves an answer from a physicist. In my opinion, within the physics framework we have developed up to now, i.e. the totality of the accumulated theories which are based on experimental results, time travel in the sense of a human controlling his/her position in time as one can control it in space, is not possible. ...

6

No, the light cone does not depend on the frame in which it is viewed. The light cone is a collection of events that are lightlike-separated from $P$. This collection of points is the same in all reference frames because in special relativity the interval is invariant. If you swept out a light cone from $P$ by having a source at $P$ emit a spherical ...

6

Your confusion comes from the difference between special and general relativity. In special relativity, the space-time manifold is assumed to carry the structure of 4-dimensional Minkowski space, which has the nice property that it is canonically identified with its own tangent space at the origin (since it is a vector space). So in special relativity you ...

6

This issue is largely settled today--- you cannot go to another disconnected universe, but you are either trapped in the black hole, or reemerge in this universe. The reason is the no-information loss property. If material could go between disconnected universes, information about the state of this material would be permanently lost to the other universe. ...

6

Let me attempt a more "popular science" answer (Ron please be gentle with me!). In GR a geodesic is the path followed by a freely moving object. There's nothing especially complex about this; if you throw a stone (in a vacuum to avoid air resistance) it follows a geodesic. If the universe is simply connected you'd expect to be able to get anywhere and back ...

6

Intuition and perception (or the lack of there of) can be a big problem when you're trying to comprehend the implications of special/general relativity. You must understand that in everyday life which fuels our intuition is pretty slow. Most people don't move faster than $900 km/h$ or $250 m/s$. And that's a luxury for most, to travel by a fast jet. The ...

6

Yes, it is a one-dimensional subgroup generated by exponentiating an infinitesimal boost. Every one dimensional exponentiation of a generator forms an abelian group, because $e^{aG} e^{bG} = e^{(a+b)G}$, there is nothing to not commute. This result is the addition of velocities, you can explicity check that this is associative (it is always manifestly ...

6

where t_0 is the time it would take if traveling at c This is your problem. $t_0$ by the pilot, and $t$ the time the trip takes as seen by an observer moving with respect to the pilot at velocity $v$. From the stay at home observers point of view the trip takes $t = d/v$, but the pilot experiences proper time $t_0 = t/\gamma < t$. Unfortunately ...

6

You are at 'O' the object is moving from A,B and is further from you at A than it is at B (ie O-A is longer than O-B) Imagine it flared when it was at A, the light from that would be on it's way to you along A-O, meanwhile the object moves to B and flares again. That light left later but has a short distance to go to you. Imagine for example the two ...

6

(I will assume in my answer that people have read the discussion on the old question, linked to by the OP.) No, it is not like the aether. It is still true that locally, there is no preferred reference frame. You don't even really need to think about spacetime to see what is going on. Consider a two-dimensional plane, parametrised by $(x,y)$, and roll it ...

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