Janus

No, it isn't, it is a time. All the values in this equation are times. There was never ever summation of speed in these equations, only summations of time. There is nothing wrong with the example. THis has been explained to you over and over again. The ship travels at a fixed speed back to the Observer, light travels at a fixed speed of twice the value. If the ship takes 50 days reach the observer from the turn around point, the light will take 25 days. All this means is that the ship is following the light back to the observer and arrives 25 days after the image does. This in no way means that the observer will conclude that the ship only took 25 days to make the trip. As an aside, you seem to be confusing "frame of reference" with "point of view". A frame of reference is not fixed to a particular observer or his position. A frame of reference would include any observer at rest with that frame, no matter where he is located. In other words, If that other planet you mentioned were at rest with respect to the Earth, it would be in the same frame of reference, and any spacetime diagram drawn from it would look exactly the same as the one drawn from the Earth.

What does that have to do with anything?

The diagram does not represent what is "observed" from the stationary observer's frame of reference, but what "happens" according to this frame of reference. What he "observes" would be determined by light propagation delay and Relativistic Doppler shift, both of which play no role in this diagram (Just because the observer "sees" the ship make the return trip in 25 days doesn't mean that the ship actually made the trip in 25 days according to the stationary observer). IOW, this diagram represents how events transpire according to anyone who is at rest with respect to the frame of reference, regardless of where they are spatially located with respect to the "stationary" observer, not what one observer "sees"..

The advantage of Loedel diagram is that the scales in both frames are the same. However, this only works if there are only two frames. If you add a third frame, it has to converted in a way so that its scales are not the same as the other two, thus removing the advantage. IOW, there is no good reason to do a three frame Loedel diagram, because a three frame Minkowski diagram is simpler and easier to work with.

The spacetime diagram shown is one drawn from the rest frame of the stay at home twin. There are two other frames that should have their S-T diagrams shown also: The rest frame of the traveling twin on his outbound leg, and his rest frame durng the return leg. It is this transistion between these two rest frames at point B by the traveling twin which is responsible for the breaking of symmetry between Stay at Home twin and Traveling twin. For example, here is set of S-T diagrams for this scenario in which I have added a couple of things: For one, I added events to each world line marking equal time intervals (they could be seconds, days, years, but they are the same from world line to world line.) For another, I added a light signal sent from the traveling twin to the stay at home twin at the moment he turns around. The first diagram is from the rest frame of the stay at home twin. Note that the traveling twin's time (tB), always lags behind the stay at home twins time (tA). When they meet up again, tA=8 and tB=7.36 also note that the traveling Twin turns around at tA=4 according to this frame and that the light signal from the Traveling twin is sent a little before Bt = 4 and arrives about halfway between tA=5 and tA=6. Same setup, but drawn from the rest frame of the traveling twin on his Outbound leg: This time we note that is tA that lags behind tB during the outbound leg of the Traveling twin. In fact, when the traveling twin reaches the turn around point, it is sometime between tA=3 and tA=4, not tA=4 like it was in the previous diagram. The traveling twin still reaches the turn around point a little before tB=4 and the light signal still would reach the stay at home twin halfway between tA=5 and tA=6. The traveling twin does his u-turn and starts his return leg, as shown in this diagram: Again, note that is still a little before tB=4 when he completes the turn and sends the light signal back to the stay at home twin and that signal still arrives between tA=5 and tA=6. However, whereas is was between tA=3 and tA=4 before he turned around, the instant after he turned around it is now between tA=4 and tA=5 (assuming negligible turn around time) . Time does run slower for the stay at home twin than for the traveling twin in this frame, but it never makes up for that "forward jump" and when the traveling twin meets up again with the stay at home twin, he will see his clock read 7.36 to his twin's clock reading of 8, exactly the same as his twin does in the first diagram.

DM doe not just effect the "outer reaches". It also effects stars closer in. The region where that effect of DM is lessened is close in towards the central bulge of the Galaxy where the high density of visble matter cause the gravitational effect of said matter to predominate. Also it is not just that the outer part move fast, but that the rotation curve is flat (Instead of stars orbiting at lower and lower speeds as you move outward from the core of the galaxy, they orbit at the same speed. Yes, DM clumps to a certain degree, it is just that it doesn't clump to the same degree or manner which baryonic matter does. This is because it only reacts gravitationally and not electromagnetically. If you take this property into account you would expect DM to form large clouds of an almost uniform density surrounding and permeating galaxies. If you then extrapolate how the mass of such a cloud would effect the rotation curves of the visible matter(stars and such) of the galaxy, you end up with a prediction that tallies up with what we observe. It seem to me that you need to spend more time studying what the DM model actually predicts and how it stacks up against observation rather than jumping to conclusions based on your own faulty knowledge of the subject.

When analyzing satellites, you should never be using the Ground as a reference frame but should be using the ECI frame or Earth center inertial frame, which does not rotate with the Earth. . It is also the best frame for considering what happens with your tower. You simply consider the effect of gravity on the tower at a given height vs. the effect of traveling around the Earth center in a circle. The Ground frame is not an inertial one, so analyzing things from this frame adds all kinds of unnecessary complications.

No, because by his measurement, the distance has contracted to 50 ly.(this not exactly right because the time dilation ratio in this example is not correct. For example, at 99% of c the factor would be 7.) Space and time are relative. Because of this there is no "who's perception is 'real'". If you and are are passing it other at 87% of the speed of light we each will see the other's clock as running half as slow as ours, and see each others length contracted in half, and each of our perceptions is equally 'real'. With the example of the space traveler, the only way to compare who aged less is to bring them back together, and to do this the space traveler has to turn around and return to Earth, and this turn around is what decides who's clock accumulated less time during the trip.(basically due to something called the "Relativity of Simultaneity")

If I get what you are saying, you are asking what would happen if you didn't disconnect the vloage source from the capacitor, and the voltasge source never exceeds the capacitor rating. In that case we are dealing with the RC time constant. Given a circuit with resistance R and and capacitance C, the time it would take for the capacitor to charge to ~63.2% of the source voltage is equal to R x C. This is called the time constant. It will then take an addtional R x C seconds for the capacitor to charge an addtional ~63.2% of the difference between the present charge and the source voltage. In other words if the time constant were 1 second, and the source voltage was 10v, it would take 1 sec to charge to 6.32V, another second to charge to 8.65 V, and another to charge to 9.5v. etc, etc. Theorectically, the capacitor would continue to charge forever, getting ever closer to the source voltage, but never quite reaching it. In practice, a capacitor is considered fully charged after a certain number of time constant periods.

Try reducing [math]\frac{3^3x^9y^{12}}{(2x)^3y^6}[/math] Further. Hint: what does [math]\frac{y^{12}}{y^6}[/math] reduce to?

This does not change the fact that you are confused. The fact that, in the EO frame, A transmits information from its starting point to some point at c and that B travels from its starting point to that same point in the same time, is not the same as A transfering information to B at greater than c. But it doesn't. Scaling up adds nothing significant to the discussion.

This still doesn't work. For one, no collection of intersteller particles will absorb/scatter all frequencies of electromagnetic radiation equally. Secondly, these particles would tend to be denser in the area immediately surrounding galaxies, plus they would be receiving a stronger intensity of light. As a result, you would see a brighter "glow" surrounding galaxies.

No, they will also disagree as to how much time elasped for the EO , and What time the EO's watch read when they crossed the beams. I went over this in a earlier post. For instance, if Both A and B start their watches from 0 the moment they cross the reference beam and the EO sets his watch to zero the instant A and B cross the reference beams according to him( so that according to the EO, all three watches start from zero at the same instant), then according to A or B the EO starts his watch before the respective ship crosses its reference beam. Not only that, but according to A, B crosses its reference beam and starts its watch before A or the EO start their watches, and according to B, A crosses its reference beam and starts its watch before B or the EO start their watches.

Okay, here goes. Say you are in a space ship out in darkest space. You fire a laser from the back of the ship to the front, and time how long it takes and measure its speed to be c relative to the ship. You drop off a buoy and then fire your engines and accelerate in the same direction as you fired the laser. You turn off your engines and coast. You now have a different velocity than you did before ( as evidenced by the relative velocity difference between you and the buoy) You fire the laser again and measure its speed relative to the ship. You again get an answer of c. Now you signal the buoy and have it fire a laser in your direction. You measure the laser light's speed as it passes the ship, you again get an answer of c relative to the ship (If you fired the laser at the tail of your ship at the same instant the laser from the buoy past it, both lasers would reach the nose of the ship at the same instant. ) Thus no matter what you do, you cannot measure your own motion using light, because you always get the same answer. Now, what exactly about this do you think contradicts my animation, keeping mind that it was made from the inertial frame of the Earth observer.

As far as TON 202 goes, the calculated proper velocity comes from its proper motion( its side to side motion against the background of the sky), and assuming that its redshift is indicative of it being at a great distance. That is, if TON202 is as far away as its redshift seems to predict, and the measured proper motion is accurate, then it would be moving at 1100c. However, I don't think that there is any serious consideration that this is the case. In fact, this calculated proper velocity is taken to mean that one or both of these assumptions are flawed.

My animation adheres to this. If I made an animation showing the exact same situation according to the left ship, it would show the left ship stationary with the light expanding outward from it at c (the ship always remaining at the center of the circle) and the Right ship traveling to the Left at 0.94c to meet the light. An animation for this situation according to the Right ship would show the light expanding in a circle from the point where the left ship emittied it and the left ship moving to the right at 0.94, following closely behind the right edge of the circle of light. In all three cases the light travels at c relative to the observer and the light intercepts the Right ship before the ships collide. Closing speed is the difference in speed between two objects as measured by someone who is not at rest with repect to either object. Realtive speed is the difference in speed between the objects as measured by either of the objects. These two values will not be the same. If they were, then the whole concept of closing speed would be superfluous, and it would be relative speed in both cases.

Anything that absorbs light (other than black holes) would have to radiate away the energy it absorbed, this would cause it to "glow" at some electromagnetic frequency. Since we explore the universe over a large range of the electromagnetic spectrum and not just the visible part, we would dectect this radiation.

With Relativity, it is always the measuring sticks that move with respect to you that contract, never your own. In Relativity, different frames will disagree as to times and distances, but they will always agree as to what happened. For example, let's replace your space ships with cars driving along a road. The cars carry clocks and the road is lined by km posts, each with its own clock. According to someone stationary with respect to the road, the km post clocks are synchronized. The km post where the cars collide is 0, and the rest are numbered outward from there. Thus each car is at a km post marked 210,000 when they emit a light towards the other. Both cars and the roadside observer will agree to that. They will also agree that the clocks on each car and the 210,000 km post it passes read zero at the moment the car emits the light. Everyone will also agree that the cars collide at km post zero and that the cars' clocks read 0.714 sec when this happens and that the clock at km post 0 reads 1 sec. Everyone will agree that the light emitted by the cars will reach km post 0 when the clock at km post 0 reads 0.7 secs. However, means that according to each car, the distance between km markers is only 0.714 km and the distance between them and km post 0 is 149970 km at the time they emit the light, due to length contraction. Also, according to each car, the clocks on the km posts run 0.714 as fast as their own due to time dilation. This means that only 0.51 sec pass on the 0 km clock while each car travels between it and the 210,000 km post, according to each car. Since the 0 Km clock reads 1 sec when the cars collide at it,this means that according to each car, when it passes the 210,000 km post and the clock this post reads 0, the clock on the 0 km post already reads 0.49 sec. This is called the Relativity of Simultaneity. According the the roadside observer all the clocks on the mile posts read the same time, but according to the cars, they do not. He will also conclude that the other car passed its 210,000 km post and emitted its light before they did. The upshot is that while everyone will agree what clocks at a given location read when events occur at that location, They will not agree as to what clocks at other locations, including their own, read when those events occur. Where in the world do you get the light traveling at 1.7c in this animation? Einstein's postulate says that light travels at c with respect to any inertial reference frame when measured from that frame. In this animation, the inertial reference frame is that of the Earth. Since the left ship is traveling at 0.7 c to the right in this frame and the light is moving at c to the right in this frame, the light will travel ahead of the ship. There is no traveling at 1.7c involved.

No, because in that same second, B will have traveled to the midpoint. Since the laser was intially fired when A was 210,000 km from the midpoint, and travels 300,000 km in one sec, it will reach the midpoint in less than 1 sec and before B does. This can only mean that the laser will intercept B at some point between where B started and the midpoint (the point of collision). The laser will reach B before it collides with A. Here's quick animation showing how events transpire according to the Earth observer. The white circle represents a flash of light that radiates in all directions.

So what? Force is not energy. The potential energy of the ball was added to it when the plane lifted it to 30,000 ft. The plane had to expend energy to do so. The ice already had potential energy with respect to the surface of the Earth. For the hunk of ice it would be equal to: [math]E_p = \frac{GMm}{R_e} - \frac{GMm}{R_m}[/math] Where G is the gravitational constant, Re is the radius of the Earth, Rm is the distance of the ice from the center of the Earth, M is the mass of the Earth and m is the mass of the ice

It depends on the type of accelerator you are using and how it is oriented. If you had a linear accelerator which accelerated the muons in the direction the ship was traveling, then according to the ship, the muon lifetime would be 15.57 ns and according to the Earth, their lifetime would be 219 ns (as they would traveling at 0.99995c wrt Earth.) If you your accelerator was pointing in the opposite direction, the lifetime would still be 15.57 ns according to the ship, but would be 2.197 ns according to the Earth( as they would be at rest wrt Earth. If you had a circular accelerator and the axis around which the particles circled was oriented along the direction of the ships motion, then the lifetime onboard ship would be 15.57 ns and the lifetime according to Earth would be ~110 ns. If the axis was perpendicular to the line of motion, then the lifetime on ship would still be 15.57 ns, but the lifetime according to the Earth would vary depending upon which point the muon was in its circuit around the accelerator.

OK, but he's no Speedwalker

First off, it is important to note that no material object or observer can travel at exactly the speed of light relative to another. So let's say that they are moving at 0.999999c, meaning that according to the first observer they have a closing speed of 1.999998c with respect to each other. For this observer, the light from each ship will arrive just a little bit ahead of the ship. For an observer in either ship, the relative speed between ships will be 0.99999999999949999949999975c, and since the light from the other ship travels at c, it will arrive just the tiniest bit before the other ship does.

Once again you display a profound misconception of how science works. Even if the LHC did not find even one predicted particle, that in itself would be information worth knowing. An experiment that does not produce the expected result can push back the boundaries of knowledge just as much, if not more, than one that does.

Statements like these simply betray ignorance on your part and only serve to undermine your credibility.