Janus

Well Nimoy does have a significant role to play in this movie. (And Majel Barrett-Roddenberry reprises her role as the ship's computer's voice)

If you are thinking about electrons "jumping" to different orbitals of an atom, there is no actual physical movement involved. The orbitals are "probability clouds" They are the regions where you are most likely to, but not necessarily will, find an electron of a given energy level. So, when an electron "jumps" to a new orbital, what happens is that the probability that you would find the electron at a given position changes, not its actual position.

I hate to break it to you guys, but this is not exactly a new idea. Besides, a major flaw is that the Uncertainty Principle would not allow you to gather accurate enough information to make a exact copy at the end. You would be, in effect, killing someone and replacing them with an imperfect copy.

Incorrect. Using are using two different meanings of "divided". The first is "common usage", where "divided" can result in unequal shares, and the second is the mathematical definition, where it cannot. You can't just simply intermix the two. This is merely word play and not a logical argument. Merged post follows: Consecutive posts merged E=mc² says nothing about the acceleration of light. The c² is a conversion factor between mass and energy. Besides c² isn't even an acceleration as it is distance²/time² and acceleration is distance/time² If a car passes me standing on the road and one minute later is one mile from me, I can't say that the car accelerated at 2 miles/min², because the car could have moving at 1 mile/min when it passed me and been traveling at 1mile/min the whole time and not accelerated at all.

Yes there is a method to detect tachyons if there were to exist: Cerenkov radiation. When a charged particle travels through a medium at speeds greater than that of light in the medium, it produces a distinctive visible light. It is like a photonic shock wave.(This likely to be what Klaynos was referring to). One could detect a tachyon by noting Cerenkov radiation originating in a vacuum.

Actually, the correct expression is [math]\frac{U_1+ U_2}{1+\frac{U_1U_2}{c^2}} =V[/math] Where c is the speed of light in a vacuum. When U1 and U2 are small when compared to c then the answer does come out to be close to being equal to U1+U2. However, if either equals c, you get: [math]\frac{U_1+ c}{1+\frac{U_1c}{c^2}} =V[/math] [math]\frac{U_1+ c}{1+\frac{U_1}{c}} =V[/math] Which gives an answer of c for any value of U1

It comes down to how you view "time". Let's try an analogy. Start with two men standing at the same point on a featureless plane. They start walking in the same direction at the same speed. Now let's say that the direction they are walking is in the direction of "time". IOW, since they are walking in the same direction at the same speed, they are progressing through time at the same rate and aging at the same rate. Now assume that Man B turns and starts walking in a slightly different direction than Man A, but still at the same speed. What happens to Man B's progress through time? Well according to Man A, who judges progress through time as being in the direction that he is walking, Man B is falling behind and is progressing more slowly through time (aging slower). But, according to Man B, he judges progress through time as being in the direction that he is walking, and it is Man A, that is falling behind and aging more slowly. This is one of the standards of Relativity, there is no absolute time, but each frame judges time independently. Now assume that Man B changes direction again so that his new path intersects that of Man A. Again, according to Man A, Man B still progresses through time more slowly than he does and according to Man B, it is Man the progresses though time more slowly. When Man B crosses the path of A, he turns and walks in the same direction as A, but is behind him( younger then him). But if according to him, A aged more slowly than he did while they were walking, then he should be ahead of A and older than him. This is the Crux of the "Twin Paradox". What this ignores is what happens to Man A according to Man B when he makes that second turn. When he turns, Man A's position in time shifts. According to Man B, A starts out behind him, but ends up ahead of him after the turn. (imagine you are standing with an object to your left and slightly behind you. Slowly turn to your left. The object, from your perspective, will move to being in front of you). A then continues to progress more slowly through time until the paths intersect, but will have gained so much time during the turn, that when B reaches A's path, B will still be behind A. Thus both A and B agree that during the total elapsed period, B ages less than A, They will not however, agree on who was aging faster at any point in the trip, only that the combined effects led to the same result. According to each man, nothing happened to alter his own progress through time and all alterations in time rate happened to the other guy.

Actually, a straight from Earth surface to Moon shot is more complicated then the two step method mentioned. The launch window would have to be really precise. Not to mention having to account for effects such as wind aloft throwing your launch trajectory off.

The problem is that velocities don't add up like that. For instance, assuming that the velocity of the second gun was "u" and the velocity that it fires the third gun out is "v", then the velocity that the third gun relative to the first gun won't be: u+v but (u+v)/(1+uv/c²) so if u is 0.5c and v is 0.5c then they add up to 0.8c not 1c 0.8c and 0.8c add up to 0.9756c 0.9756c and 0.9756c add up to 0.9997 c etc. The sum will never add up to c

Difference in gravitational potential can be characterized as the amount of work needed to lift an object from one point to another. To demonstrate how this applies to here, first imagine lifting an object from the surface of the Earth to the distance of the Moon. As you move further from Earth, the force of gravity declines until, at the distance of the Moon it has dropped to 1/3609 of that at the surface of the Earth. IOW, it takes less work to lift the object from the surface to 1km altitude than it does to lift it from 1 km to 2 km, etc, with the amount of energy needed to lift it each additional Km decreasing. The total energy needed to lfit the object would be the sum of all these intermediate energies. Now imagine you are lifting the object against a 1g field for the entire distance, The froce of gravity is 1g at the surface, 1g at the distance of Moon and 1g at every point in between. It would take the same amount of work to lift the object through each successive km. Again the total energy needed would be the sum of all the smaller steps As you can see, it would take more work to lift the object agains 1g the whole way than it would to lift it against Earth's decreasing energy. And even though thare is no difference in g-force in the uniform field between the surface and the distance of the Moon, there is a large differnce in potential, which is what effects gravitational time dilation. With the spaceship accelerating at 1g, you have to assume a uniform gravitational field that extends for inifinity, and that the Milky Way is falling through that field if you want to use the GR approach to the problem. The standard Gravitational time dilation equation is for dealing with a typical gravity field that decreases in strength over distance.

Length contraction is not a "visual distortion", according to the crew, the Milky way really shrinks in the direction of travel. You can't mix measurements from different frames (distance from the frame of the Galaxy and Time from the frame of the ship) and get a result that has a physical meaning. How did you arrive at this answer? Did you use the standard gravitational time dilation formula? If so, it is invalid for this situation. It is for use for gravitational fields that fall off by the square of the distance. Remember Gravitational time dilation is dependent on difference in gravitational potential, not differences in local gravitational strength or local acceleration due to gravity. In this situation you can either calculate strictly using speed and SR or using GR (ship standing still in 1g uniform gravity field while the galaxy falls past you), but you do not combine them. This is a result of the clock postulate. IOW, acceleration does not cause any additional time dilation beyond that due to the change in relative. This has been demonstrated by using high speed centrifuges (up to 10^18g). It has been shown that the time dilation only depends on the speed at which the object is moving and not the g's it experiences. With different radii of centrifuges you can have the same speed but experience different g forces, but the sample on the centrifuge will show the same time dilation. As for the differences you note, they are just due to rounding errors. For instance, with the formula given in the web page quote, for a 2 year trip, v actually works out to 0.9680303033155948312177397420156c which he rounds up to 0.97c. But the difference in gamma between these two velocities is not insignificant. Using the more accurate value for v gives you a gamma of 3.986711890106634759197778382839 when using the Lorentz transformation, and calculating using the acceleration formula gives you 3.986711890106634759197778382839, the exact same answer.

The bending of light's path when passing a massive object is due to the curvature of space-time. The light follows a geodesic, which is the shortest path it can take. When seen from outside of the curved space-time region, the path appears curved. You see the same thing when you look at maps of airline flight paths. Especially over long flights, jets fly along a geodesic which is the shortest route. If you plot these paths on a map, the lines looked curved: http://www.aaronkoblin.com/work/flightpatterns/FlightMapColor1.jpg So, for instance, if you want to travel to a point due West of your present position, in the Northern hemisphere you would fly a course that starts out at a little North of West, and then gently curves to just South of West on arrival. ( On a globe, geodesics follow "Great Circles" which are circles that bisect the globe. For instance, the Equator and all the lines of longitude follow great circles, the Northern and Southern lines of latitude do not.)

Sorry, no. This can be shown by considering two ships flying to Alpha Centauri. 1 at .866c and the other at .1c The first takes 4.96 yrs Earth time to make the trip and records the trip as taking 2.48 years. The second takes 43 years Earth time and records the trip as taking 42.78 years. So when the first ship arrives it reads 2.48 yrs and the second ship has traveled 4.96 yrs of its 43 year long trip. Thus the first ship's clock will record an addtional 43-4.96 = 38.04 years while waiting for the second ship to arrive and will read a total of 38.04+2.48 = 40.52 yrs when the second ship arrives. As noted above, the second ship's clock records 42.78 years upon arrival, so there will be a 2.26 year difference between the clocks. The same would be true of clocks in planes flying around the world at different speeds; the clocks would differ in the end.

No, the path doesn't have to go through the center of the Earth, just on a straight line from town to town.(see attachment).

The North magnetic pole isn't at true North even now, nor is it fixed, as it wanders. The attached image shows its relative positions in 2001 and 2005. This polar drift is not due to an impending pole shift.

None of this can be used to carry information faster than c. For example, a person at the bottom of the incline can not use this to send information to a person at the top of the incline. He has no control over over the information in the light beam. That is controlled at the source. He is merely an inactive observer.

I think what they meant was that the center of the Earth was close to the temperature of the surface of the Sun,

He's probably thinking about the gravitational time dilation equation: [math]T = \frac{t_0} {\sqrt{1- \frac{2Gm}{rc^2}}}[/math] which is what you get when you substitute the equation for escape velocity: [math] \sqrt{\frac{2GM}{r}}[/math] for "v" in the SR time dilation equation,

If it were gravity affecting the mechanism then the time difference would vary according to the local strength of the gravitational field. Instead, Relativity predicts, and we find, that the time difference varies according to the difference in gravitational potential. IOW, it is quite possible to have two clocks experiencing exactly the same gravitational force, yet be at different gravitational potentials and to run at different rates.

http://theory.caltech.edu/people/patricia/lclens.html

The amount of energy needed to accelerate a mass up to a given velocity can be found by: [math]E = mc^2 \left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right )[/math] Notice what happens as v nears c. The bottom half of that fraction approaches zero, which means the energy approaches infinity.

The key here is the word "tangentially". A photon moving at a "tangent" inside the photon sphere follows a geodesic which intersects the event horizon. But a photon traveling at some other angle relative to the tangent may not. An extreme example would be a photon traveling directly away from the BH. Its geodesic carries it out and away from the BH. It is only within the event horizon itself that all geodesics lead back to the BH.

It does come down to a matter of economics, just not in the production of the electricity. It works like this, the lower the voltage, the higher the current needed to run a comparable appliance (see my last post). The thinner the wire, the higher the resistance. I^2R determines how much wattage is used up by the wire. Thus at a higher voltage, you can use a thinner wire to deliver the same power. Now for the economics: When electricity use began to come into general usage, copper and cotton were inexpensive in the US, So it was cheaper to use cotton in the insulator, which was a good insulator up to 120v, and use a thicker copper wire. In the UK, copper was expensive, so it made more economic sense to use a higher voltage which allowed them to use thinner wires, The trade off was that they needed to use a better insulator (rubber for instance), but it was over-all still cheaper than using thicker wire.

If you want to, you can blame Ben Franklin for giving the Electron a negative charge. During his early experiments with electricity he assumed it was a flow from and excess(+) to a deficit(-). However, he had no way of determining which was the actual direction of the flow. He took a guess and labled one terminal (+) and the other (-). His convention for labeling the terminals was carried on by others. By the time it was determined that the actual current flow ran in the opposite direction, the practice was too deeply entrenched to change easily. The particle found to be the carrier of electricity was of course called the "electron". It has sometimes been called a negatron. If Franklin had guessed the other way, Electrons would have a positive charge and we would likely have negatrons in the nucleus.

No, there is more copper to conduct electrons. Think of it in terms of a multi-lane freeway vs. a single lane road. The wider freeway can carry more traffic easier. Wound up, the extension cord acts as an inductor. The electric field around the cord cuts through adjacent coils of the cord. Since we are dealing with AC, the polarity of these fields are always changing. Changing electric fields induce a current in conductors. This current will be in opposition to the regular current. This in effect causes the wound up cord to act as if it has a higher resistance, which reduces its loading ability (to be correct, the opposition to current due to the coil is called "reactance" and the combination of the reactance and normal resistance is called impedance. The current drawn by the appliance depends on its resistance, which is fixed and the voltage applied. Thus if you take a US appliance to the UK, it will try to draw more current. For instance a 1200 watt US hair dryer will normally draw about 11 A, meaning it has a resistance of 10 ohms. Hook this up to a European 240 V outlet, and it will try to draw 24 A, You will most likely blow a fuse, and damage your dryer. On the other hand, a 1200 W European dryer normally draws 5 A and has a resistance of 48 ohms. Plug this into an US outlet and it will draw only 2.3 A, and only use 252 W. It would perform weakly or not at all. Luckily, it is not easy to accidentally plug a US appliance into a UK outlet or vice versa, as they also use different styles of plugs and outlets.