Janus

Of course it is the same space ship! The oval to the left is the spaceship at the instant when the light hits the mirror. The oval to the right is the Same spaceship at the instant when the light returns to the center. This is as seen by an observer outside of the ship, watching it go by at v. (to keep the image uncluttered, I omitted the spaceship at the instant the light is emitted from the center from the image) To someone in the ship, the light simply travels out at angle alpha (wrt the relative motion) for a distance Y, hits the mirror and returns along the same path to the center. Since the light returns to the center of the ship in S' (the frame of the ship), it must also return to the center in frame S (the frame in which the ship is moving at v) Thus in S' the light follows the path shown in my images. Alpha is the angle the light is launched as measured by someone traveing with the ship. Y is the radius as measured by someone traveling with the ship The length of the ship is only contracted along the direction of motion when measured by someone to which the ship has a relative motion. The ship has zero relative motion wrt to someone traveling with the ship, thus it has no length contraction for that person. To say otherwise shows a complete and utter lack of understanding of what Special Relativity is about. To refute Relativity you must first understand it. The simple fact is that Relativity is a completely self-consistant theory. It contains no mathematical or logical errors. It makes numerous predictions which have held up under every real experimental test put to it. No thought experiment you can ever dream up will ever refute it. If you think you've come up with one, all it means is that you have made an assumption that is contradictory to one of the postulates of Relativity. All you've done in this case is show that Relativity is not compatible with your assumption. This in of itself provides nothing in the way of proving which is correct, your assumption or Relatiivity. The only way to refute Relativity is to produce results from a Actual Physical Experiment that gives results that differ from that which Relativity predicts.

In order to examine this claim that light emmited at different angles somehow changes the results let's try the the folowing exercise. Let's call the frame of the sphere frame S' and the frame in which the sphere moves at v, S In S the light travels a distance of ct1 while our sphere travels a distance of vt1. Now considering image 1: Assuming the light is emitted at at an angle of alpha from the center of the sphere as measured in S' , then the point at which the light hits the sphere is as measured in S is: [math] ( Y \sin \alpha , vt_1+ \gamma Y \cos \alpha ) [/math] where Y is the radius of the sphere as measured in S'. and that the center of the sphere at the instant of impact is at (0,0) [math]\gamma = \sqrt{1- \frac{v^2}{c^2}}[/math] The gamma factor is needed because the sphere is length contracted in S Thus: [math]c^2t_1^2 = Y^2 \sin^2 \alpha + ( vt_1 + \gamma Y \cos \alpha)^2 [/math] [math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \gamma^2 Y^2 \cos^2 \alpha [/math] [math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( 1- \frac{v^2}{c^2} \right)Y^2 \cos^2 \alpha [/math] [math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( \cos^2 - \frac{ \cos^2 v^2}{c^2} \right)Y^2\alpha [/math] re-arranging: [math]c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( \sin^2 \alpha + \cos^2 - \frac{ \cos^2\alpha v^2}{c^2} \right) [/math] [math]c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) [/math] Gathering the t1^2 terms together: [math]c^2t_1^2-v^2t_1^2 = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) [/math] [math]t_1^2 (c^2-v^2) = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) [/math] Moving all the terms to the left side: [math]t_1^2 (c^2-v^2) - 2 \gamma Y v t_1 \cos \alpha -Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) =0 [/math] We get an equation that can be solved for t1 using the quadriatic equation thusly: [math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4 \gamma^2 Y^2 v^2 \cos^2 \alpha + 4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math] [math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4\left( 1- \frac{ v^2}{c^2} \right)Y^2 v^2 \cos^2 \alpha -+4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math] [math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + (c^2-v^2) \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math] [math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \frac{(c^2-v^2)}{c^2} c^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math] [math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \left( 1- \frac{ v^2}{c^2} \right) \left( c^2- \cos^2\alpha v^2 \right) } }{2(c^2-v^2) }[/math] [math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + \left( c^2- v^2 \cos^2\alpha \right) } }{(c^2-v^2) }[/math] [math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + c^2- v^2 \cos^2\alpha } }{(c^2-v^2) }[/math] [math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }} {(c^2-v^2) }[/math] [math]t_1 = \frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } [/math] Now consider image 2 which shows the path the light takes on its return to the center of the sphere: Here we can see the light travels a distance of ct2 in frame S, while the sphere moves a distance of vt2. As a result we can see that: [math]c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2 [/math] and following the same steps as above we find that solving for t2 we get: [math]t_2 = \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }[/math] The total time T it takes for the light to leave the emitter and return is [math]T = t_1 + t_2[/math] Thus [math]T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } + \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math] [math]T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}} - \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math] [math]T=\frac{ \pm cY \sqrt {1- \frac{ v^2}{c^2}} \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math] atr this point we have a choice to make. There is only one possible answer for T, but there are four ways to solve this equation due to the "plus or minus" signs. Luckily we can automatically elliminate 3. If both are minus we end up with a negative time for an answer, so that doesn't work. And if one is minus and the other plus we end up with a answer of 0 for T and that makes no sense. this leaves us with both being plus giving us: [math]T=\frac{ cY \sqrt {1- \frac{ v^2}{c^2}} + cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math] [math]T=\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } [/math] Now let's relate T to T' the total round trip time in S' In S' the light travels a distance of Y to the sphere and then travels a distance of Y back again. this gives us: [math]T' =\frac{ 2Y}{c}[/math] and [math]\frac{T'}{T}= \frac{\frac{ 2Y}{c}}{\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } }[/math] [math]\frac{T'}{T}= \frac{ 2Y}{c}} \frac {c^2-v^2 }{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } [/math] [math]\frac{T'}{T}= \frac {c^2-v^2 }{ c^2 \sqrt {1- \frac{ v^2}{c^2}} } [/math] divide top and bottom by c²: [math]\frac{T'}{T}= \frac {1- \frac{v^2}{c^2} }{ \sqrt {1- \frac{ v^2}{c^2}} } [/math] simplify and re-arrange: [math]T = \frac{T'}{\sqrt{1- \frac{v^2}{c^2}}}[/math] And we see that no matter what angle alpha the light is aimed, the stardard time dilation holds, thus falsifying the claim.

[math]c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2 [/math] [math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }}}{(c^2-v^2)}[/math] [math]t_1 = \frac{\gamma v Y \cos \alpha \pm cY\sqrt{1-\frac{v^2}{c^2}}}{(c^2-v^2)}[/math]

Definitely read the book, if for no other reason than a sense of history. Written in 1898 by H.G. Wells, a man ahead of his time who more or less invented many of the "themes" of modern SF. (Alien Invasion, time travel, antigravity, invisibility)

As explained the ivaders were destroyed by microbes. This is explained in the voise over at the end of the movie, which, with your weak French, probably why you missed that part. The destruction of one of the machines by Cruise had nothing to do with the rest of the machines malfunctioning. It did not happen simultaneously. (You have no idea how much time passes between the destruction of the first machine and the scene where they show the rest inactive. I think the machine destroyed by cruise and the one done in later were nods to the novel again. In it, the Machines were not invunerable(they had no shields) and the humans were able to take a few out; it was just that it wasn't enough to stop or slow the advance of the invaders. Also, once we found an effective tactic against them, the Martians would come up with something to circumvent it. The shields were added in the first movie made in the 50's and retained in this movie to give the invaders the same degree of "edge" over modern weapons as the original novel gave the Martian's over late 1800's military technology. The birds were merely an indication that the shield was not functioning for that machine(most likely due to the ailing invaders inside), and thus it was vunerable to weapons now. It is also a nod to the original novel, where the protagonist sees birds circling the hood of a machine and picking at the remains of the dead Martians inside. In the original novel, the martians "fed" on the blood of others, so they used humans as a food source. In order to keep the aspect of the story where the invaders captured live humans, the director just changed this a little so that they used the blood as fertilizer instead. The Red weed is taken from the novel also, so he incorporated the blood into its cultivation.

Yes it can. According to brother B, it is only the relative velocity that causes A to age less.

But that reference point is can only be consider "stationary" for convenience's sake we can know nothing about our absolute motion through space A speed through that distance is absolute. Since there is no absolute reference to judge that speed by, speed can never be absolute.

Sorry, but you are mistaken. This is easily demonstrated every day with particle accelerators. The Earth itself accelerates as it orbits the Sun, thus the Earth's orbital velocity of 30km/sec would be added and substracted to the particle's final velocity as it leaves the accelerator during different points of its orbit. Today's accelerators can reach velocities close enough to c that this 60 km/sec variation would cause a noticeable difference in relativistic effects seen in those particles over the course of a year. No such variations are ever noticed.

The above is only true as measured by an observer to which you are moving at c/2. As measured by you, the light will have a relative speed of c with respect to yourself, per the second postulate of Relativity.

Yes it does reach the other particle in this frame, just as it does in every frame.

Saint also goes on to say "The backbone of SR is not simply the constancy of c, but also that it is the universal "speed limit". Everything was developed based on that idea. Since it was clearly developed with the assumption that FTL speeds are impossible to achieve, it cannot provide useable answers to inputs beyond c." Where he quite literally says that SR was based on the Speed of light being a universal speed limit.

Common misconception. Time would not go backwards at FTL speeds' date=' it would become imaginary (a product of the square root of -1) It would be undefined. Look at the time dilation equation: [math]T = \frac{T'}{\sqrt{1- \frac{v^2}{c^2}}}[/math] Note that as v equals c, the equation becomes [math]T = \frac{T'}{0}[/math] And division by zero is undefined. But since nothing with mass can travel at c, it'll never come up.

As close to infinite as you want to get. Temperature is related to the Average kinetic energy of the particles involved. Now while you can't get particles up to light speed you can get as close as you want. Kinetic energy, taking Relativity into account is [math]KE = mc^2 \left( \frac{1}{1- \frac{v^2}{c^2}}-1 \right)[/math] note that as v approaches c, the kinetic energy approaches infinity and so does the temp.

Your third POV would see things depending on its relative velocity to the ship and Earth. It is no different than either the POV of the Earth or the POV of the ship. There is no POV that can say that it sees what "actually" happens. All POV's are equally valid, and what each POV measures as happening is what "actually" happens in that frame of reference.

To add to what Swansont has already stated. The speed of light in a vacuum is always c, where c is 299,792,458 meters/sec as measured by any observer. Thus say you have three observes: A; standing next to the light source, B; moving away from the light source, and C; moving towards the light source. Each has a device capable of measuring the speed of the light from the source relative to themselves. They will each measure that the light has a velocity of c to relative to themselves.

It depends on your orientation of the ship's gravity to the impact. If the impact is at a right angle to the "up" of your ship's gravity and strong enough, then, yes, you will be thrown off your feet. While the gravity will pull you down to the deck plates, it has no sidewise component. If the ship is hit hard enough to shift it sideways, your feet, due to the friction between them and the deck plate, will go with it. your upper body, however will try and stay in place (law of inertia). Your feet come out from under your center of gravity and you lose balance. If you were hit from the "bottom", you would feel heavier for a moment. If the impact is large enough and you weren't prepared, you could also end up falling. If you were hit from the top, you would feel lighter, and again, depending on the severity of the impact and what you were doing at the time, it could throw off your balance.