Janus

No, they don't. There is no hierarchy that places laws above theories. Laws are just formal statements of observed or inferred relationships. An example would be Kepler's Laws of planetary motion. These were purely based on observation. It wasn't until Newton's theory of Universal gravity came around that we understood why these relationships held. The Laws merely stated the observed properties, the theory explained them.

according to the rocket, A and C are 1.732 light days from the rocket at the moment of emission, due to length contraction. Planet A is receding at 0.5c and planet C is approaching at 0.5c Given the speed of light is c, the time I gave are how log it takes for each planet to meet up with its respective signal. There is no such thing as the "real" distance between the planet's, only the distance as measured by each frame ( or conversely, the distance as measured in each frame is equally "real".) And no, for the astronau,t the signals do not travel at anything but c. the closing speed between A and the signal is 0.5c and between C and the signal is 1.5 c, but closing speed does not reflect any real change in c The difference is that "v" is not an invariant so the car will not measure the speeds of the motorcycles as being a constant with respect to himself. However, that does not mean that according to him, that the motorcycle heading for C will not get there first. In this situation you have to use the addition of velocities theorem: [math]w = \frac{u+v}{1+\frac{uv}{c^2}}[/math] To get the velocity of the motorcycles with respect to the car as measured from the car frame. Given that 0.5v is the relative velocity of the car to the road and v is the relative speed of the motorcycles, then: The speed of the motorcycle heading toward A relative to the Car is: [math]w = \frac{0.5v+v}{1+\frac{0.5v^2}{c^2}}[/math] which is less than 1.5v and the speed of the motorcycle heading towards C relative to the car is: [math]w = \frac{v-0.5v}{1-\frac{0.5v^2}{c^2}}[/math] which is more than 0.5v If you take these speeds and the speed of the car relative to the roadway( 0.5v), you get the closing speeds between the motorcycles and the towns according to the car. The closing speed between motorcycle and A will be lower than that of between motorcycle and C. Given that A and C were equal distances from the car when the motorcycles left, a motorcycle will reach C before it reaches A, according to the Car. Now it would be almost impossible to measure the difference in timing at the speeds at which cars and motorcycles really travel, so we would not tend to notice it. For instance if v =200 kph then [math]\frac{200 kph+100 kph}{1+ \frac{200kph(100kph)}{c^2}}[/math] = 299.999999999995 kph. So close to 300 kph, that you would never have a chance of noticing it in everyday life. But a v and u get closer to c the difference becomes more and more measurable.

It's because the speed of light is constant that the astronaut will determine that the signals reach the planets at different times. The signals are emitted halfway between A and C. The signals travel away from the rocket at c in both directions. (after 1 sec, both the signals sent towards A and C will be 299,792,458 meters from the rocket.) Planets A and C do not maintain a constant distance from the Rocket. After 1 sec planet A is 149,896,299 meters further away, and planet C is 149,896,229 meters closer. IOW, according to the astronaut's frame, planet A is running away from the signal and Planet C is rushing to meet it. By the time the signal and planet C meet, Planet C will be much closer to the rocket than it was when the signal was emitted, and by the time it reaches planet A, planet A will be much further away. Thus from the rocket's frame, the signal traveling to A travels further than the signal to A, and since both signals travel at the same speed, the signal reaches A after it reaches C.

According to which frame? According to the rest frame of planets A and C, the signals from both the Rocket and Planet B all arrive simultaneously at A and C. According to the rest frame of the rocket, the signals sent from the both rocket and B arrive at planet C in 1.15 days and at planet A in 3.46 days. IOW, while the signals from both the rocket and planet B will arrive at each planet at the same time, the signals will not arrive at planet A at the same time as they do at planet C.

Just some more detailed points in regards to the flaws in the linked paper. 1. In the formulas for clocks 3 and 4, the author simply replaces c with Cs (assumably the speed of light in the glass). This is incorrect as the formulas using c are based on the idea the c is invariant, and Cs is not. The proper formulas would have to reflect this and incorporate the velocity addition formula. 2. Simply applying the Doppler shift formula and plugging in the velocity of the rocket does not give you the time rate difference between the clocks. Instead, what you need to do is to calculate the total accumulated time for both clocks over the course of a round trip of the rocket. (and if you do, you will find it in complete agreement with time dilation. ) 3, The formula given is fine for the time as measured in the rocket, but fails to take into account any relativistic effects. A proper analysis of the period of the pendulum would have to take into account the effect of the velocity addition theorem on the pendulum bob. In addition, when the author "corrects" for length contraction, all they did was shorten the length of the pendulum arm and reapply the period equation. Since the arm of the pendulum would not always be parallel to the motion of the rocket, length contraction on it would not constantly act along its length. (in other words as the pendulum swings from side to side, the length of the arm would vary. The author oversimplifies a complex problem and as a result come to the wrong conclusion.

Whether or not such an test has actually been done on the ISS is moot, since the same experiment is essentially done every day with GPS satellites. The clocks on these satellite have to be adjusted to compensate for time dilation in order for the system to remain accurate.

The "experiment" is nonsense. It talks about "measurements", but no measurements were made because the experiment was never actually performed. He ignores relativistic effects for some of the clocks in the calculations of their time rate and claims because these results don't match up with the Relativistic prediction, time dilation does not exist. In other words, he misapplies Relativity to a thought experiment and then thinks that the since the results he thinks he would get don't agree with Relativity that he's proved SR wrong, when all he's proven is that he doesn't understand Relativity.

[math]\frac{Gm}{d^2}[/math] would give you the acceleration of an object at a distance of d from the center of mass 'm'. For instance, if d is the radius of the Earth, and m the mass of the Earth, you would get ~9.8 meters/sec called 1g. It is generally what is used when you what to compare the gravity of two planets, Etc. because it doesn't depend on any other mass than that of the planet. F=ma tells you how much force it takes to accelerate a given mass by a given acceleration So for instance if we make m=m1, we can say that [math]m_1a = \frac{Gm_2m_1}{d^2}[/math] Since m1 appears on both sides of the equation, it cancels out leaving. [math]a = \frac{Gm_2}{d^2}[/math] Which is the formula for acceleration I gave above. No. The electromagnetic force also follows the inverse square law.

You are mistaken. For one, if the sun just went out, it would not have a change on the Earth gravitationally. For the the other, even if the Sun "went away", the Earth would still feel the effects of its gravity for ~8 min after it vanished. Gravity waves (which are what carry the information of changes in gravity) travel at the speed of light.

Based on what? Anything that come within the vicinity of the moon is subject to the same orbital mechanic as the moon. So you have four possibilities: Objects that pretty much have the same orbit as the moon which will have very little velocity with respect to the moon and just will fall in on the moon due to its gravity. The moon will pull this stuff in equally from all directions and clear it out pretty quickly. Objects in orbit around the Planet which cross the moon's orbit but have an average distance closer to the planet. These objects will be moving slower than the moon and will more likely collide with its leading edge. Objects in orbit around the planet which cross the moon's orbit but have an average distance further from the planet. These objects will be moving slower that the moon and are more likely to hit from behind. Of the last two, whether or not there are more of one that than the other depends on the Moon's own position in orbit. If it is near the planet, there will be more of the later and it will get hit from behind more often, pushing to a higher orbit. If it in a high orbit, there will be more of the former which will push it into a lower orbit. Generally the action of these two will tend to move the planet towards a medium orbit. But as in the first case, these types of objects get cleared out fairly early on. Last we have objects that come from outside the planetary system. All of these will have fallen through the planet's gravity and picked up quite a bit of speed. They also fall into the first three categories as above with their orbits being with respect to the Sun. They will have various speeds and trajectories with respect to the planet and can hi the moon at any point of its orbit. A faster moving body coming from one direction is just as likely to hit the moon when it is on any side of the planet and add or subtract from the moon's velocity, the same can be said for a slower moving moving object. There is no strong statistical preference for one over the other. You have to look at things from a orbital mechanical view. Look at any moon for which we have a picture that shows cratering. None of them show preferential cratering on their leading edge. This would have nothing to do the likelyhood of direction of impact. Objects coming in from behind or from ahead of the Planet are just as like to pass inward of the planet as outward, and the moon is just as likely to be inward or outward of the planet during a collision. What the orbital velocity of the Moon does determine is how much of a velocity change it would take to significantly change the orbit. For Io it would take a 7 km/sec change to kick it out of orbit and a 9 km/sec change to get it to fall in to Jupiter.

What makes you think that all those collisions are head-on? Just look at our own Moon, are all the craters on its leading side?. Some collisions come from bodies traveling in the same direction as the moon and actually add orbital energy. During the early stages of formation there were likely a great many more collisions and some of those might have resulted in a moon getting too close to the planet or being ejected. But you have to remember, in order to be ejected from orbit the moon would have to increase its velocity by over 40%. ( For our own Moon that would work out to be around 400 m/sec) Conversely, for our Moon to lose enough orbital speed to crash into the Earth it would have to lose over 80% of its velocity. In addition, over time the moons and planets sweep their orbits clean of most of the debris. So any moons left over after the initial dust up are pretty stable. For example: Every day the Earth collides with some 100,000 kg of meteorites. Even if all of these robbed the Earth of orbital speed, is in significant to Earth mass. It would take trillions of years before this could reduce Earth's orbital velocity by even 1%

You have to take into account the effect of the material moving outward or inward with respect to the planet's gravitational field and the effect that this has on the material's orbital velocity as it approaches the moon's orbit. As the moon accumulates mass via gravitational collapse, it begins to draw in mass from higher and lower orbits. But when it draws in an object from a higher orbit, that object has to lose potential energy with respect to the planet, and has to increase its kinetic energy and thus its speed, to compensate. By the time it reaches the orbit of the Moon is will be moving faster than the orbital velocity of the moon. Going from the other direction, pulling the object from lower orbit causes it to gain potential energy and lose kinetic energy speed, and it will be moving slower than the moon. Thus material coming in from the planet side is moving slower and material coming in from the other side is moving faster. This tends to give the forming moon a pro-grade rotation.

As Swansont pointed out, Mary and Sam do not have a zero relative velocity according to everyone. If they were in an inertial frame, (traveling in a straight line at constant speed), then all observers that are themselves in inertial frames would also agree that they had zero velocity with respect to each other. As it is, they wouldn't. Here's another example of such a case. Mary and Sam are in a rocket that is accelerating (according to them) at a constant rate. Mary is in the nose and Sam is near the tail. According to them, they maintain a constant distance from each other and are at rest with respect to each other. However, according to someone in an inertial frame, the rocket is constantly changing velocity. As a result, the length of the rocket is always changing due to length contraction. This means that Mary and Sam do not maintain a constant distance from each other and are not at rest with respect to each other according to this frame. Another thing to note is that, as measured by anyone in the rocket frame, Mary's clock runs faster, despite the fact that Mary and Sam are undergoing the same acceleration and the only difference between them is their spatial separation.

Again, you must be careful here. Acceleration, in of itself, does not cause time dilation. Example, if you put a clock on a centrifuge, and compare it rate to one at the center, it will run more slowly, Now let's say that you change the parameters of the centrifuge so that it has a different length of arm and spins at a different angular velocity in such a way that the clock experiences the same acceleration force as it did before. You will find that the clock will run at a different rate than it did before. Conversely, You could change things so that the clock travels at the same speed on the end of the arm but experiences a different acceleration, and it will run at the same rate as it did before.

No. The gravity the object feels has nothing to do with time dilation. While a clock sitting on the surface of the Moon feels less gravity than one on the surface of the Earth, and runs faster than one on the Earth, a clock sitting on the surface of Uranus would also feel less gravity than one sitting on the surface of the Earth, yet a clock on the surface of Uranus would run slower than one on the Earth. Time dilation isn't about something physically altering how the clock operates, it has to do with the very nature of "Time". No, what happens is that if you accelerate up to the near the speed of light (relative to your starting point) and then return to your starting point, you will find that your clock has accumulated less total time then one that stayed at your starting point. Why this happened would be a matter of dispute between you and someone who stayed behind. They would say that time ran slow for you while you were going away and coming back. You on the other hand would say that time ran slow for them while you were going away and coming back, but their time "jumped forward" during the turn around period when you went from going away to coming back, and that this jump forward resulted in your clock reading less time than theirs when you meet up again. Again, it has nothing to do with anything physically acting on the operation of your or his clock, but with the way we compare "time" between different reference frames.

Also, wait, making it go "faster"? I thought the closer you approached the speed of light, the slower the rate of your time flowed to other people. If you are moving with respect to someone else (or they are moving with respect to you, there is no difference) you both will determine that the other person's clock is running slower than your own. It isn't until one of you undergoes an acceleration that matches their speed to the other that you will end up with a situation where both of you will agree that his clock has accumulated less time.

The thing is that they aren't in the same inertial frame. The satellite is in a forced orbit, which puts it in a rotating frame and under constant acceleration (as evidenced by the fact that the ship has to be constantly firing its engines is order to maintain its orbit.)

An mass of 100° steam does contain more heat than an equal mass of 100° water. This is due to the heat of vaporization needed to convert water to steam. This works out to being ~540 kilo-calories per kilogram. Put another way, it takes over 5 times as much heat to convert water to steam than it does to initially raise the water's temperature from 0 to 100°. That heat is released when the steam condenses.

It has to do with the very "nature" of time itself. Here's an analogy: You have two men standing on a featureless plane. They start walking in slightly different directions at the same rate. After having walked a certain distance they check up on each others progress. Man 1 looks around to find that Man 2 is behind him and to his left. Man 2 looks around to find that Man 1 is behind him and to his right. Each man judges forward progress as progress in the direction that he, himself is walking. By this judgement,each man considers the other man as having made less forward progress than himself and while walking is make progress more slowly than he is. Since they are on a featureless plane, there is no method of determining which is correct in any absolute manner and each man's definition of forward progress is as good as the others. Now replace forward progress with time passage and walking in different directions with having relative velocities with respect to each other, and you have an idea of how time dilation works. Nothing physically interacts with a clock to slow it down (just like nothing interacts with either of the men to slow his pace), but because of the nature of space-time, each clock measures its own time passage as being faster than time passage for a clock moving relative to it. Then why is it the if we send a clock away at high speed and then return it, it comes back showing less time? If both clock see each other as running slow, how can one end up with less time at the end? To answer that, consider whats happens if one of the men in our example turns so that his new path takes him back to intersect the other man's path. Keeping in mind, that he judges forward progress as being in the direction he is facing at the time, what happens to the other man's "forward progress" when the man turns toward him? (If an object is behind you and to your right, what happens if you turn to your right?)

Be careful here. A difference in potential does not require a difference in local gravity. As an extreme example, you can imagine a uniform gravitational field (one that does not change strength with altitude). Two objects placed at two different heights in this field will be at different potentials, and two clocks place at different heights in this field will run at different speeds even though they feel exactly the same gravitational force. It doesn't. Dilation from motion is just due to the fact that clocks in inertial frames with relative motion measure time and space differently from each other.

That 0.1 calorie per leg stroke is just the energy you are expending to overcome friction, etc, and isn't any true measure of how fast you are moving. Increase or decrease the friction and you will use up more or less calories per leg stroke. In fact, put the bike on a treadmill and you can expend exactly the 0.1 calorie without moving with respect to the road at all.

The difference is that the structures in these pictures were made by a a civilization that was technologically advance enough to be able to afford to divert resources for such luxuries as large structures which are just "art for art's sake." A pre-industrial civilization wouldn't have that kind of freedom with it resources.

Wrong. For one, DM does form macro-structures, in the from of galactic halos and around galaxy clusters. Secondly, it does interact with itself gravitationally. The reason that it does not form smaller structures is due to the very fact that it does not interact electromagnetically and thus cannot not avail itself of that mechanism to shed energy. When baryonic particles make a near approach to each other, their electromagnetic fields interact causing a deflection of the particles. This deflection results in the particles also emitting electromagnetic radiation, which comes at the expense of the kinetic energy of the particles and they leave each other with less speed then they met. If the the new velocity is low enough the two particles may become gravitationally bound. If the approach is close enough or the relative velocity small enough, the particles "stick" together electromagnetically and start to shed their energy as electromagnetic radiation. Constant interactions of this type lead to compact bodies such as planets (part of the internal heat of out own planet is due to it still shedding the energy of its formation) When DM particles make a close approach, they only interact gravitationally, this can still cause a deflection, but this deflection doe not result in the release of EMR, and subsequent loss of energy/velocity. There will be a emission of gravitational radiation, but since this is ~1039 times weaker than electromagnetic radiation, its effect is very small. For the most part, DM particles will just approach each other pass, and then leave at the same speed at which they approached. They shed some energy via gravitational radiation, and gravitational interaction between a group of particles can lead to some losing velocity others gain it, leaving the slower particles in a loose gravitationally bound group. It these types of interaction that lead to the macroscopic structures that DM does form. But such weak interactions take a long time to achieve any type of noticeable result, which is why DM has so far only formed the type of loose structures we see.

The classic zombie of walking dead flesh is just not possible. Once you die and respiration and circulation ceases, the cells begin to die. A part of this process is rigor mortis. In order for muscles to contract and relax the cells have to be alive and chemically active. When they die they go into a state of locked contraction. It isn't until the cells decay that they release again, but after that they can no longer contract. The chemical cycle that allows for continued contraction/relaxation that causes movement can't occur anymore. So, the only type of "zombie" that could even theoretically be possible would be the "still living" type. Higher brain function gone in a living breathing person. This type of zombie would be just as easy to kill as a normal human would be.

First off, I like to apologize for an error in my numbers. There are two formulas that can be used to determine the Doppler shift in this situation. One uses the actual angle between source and observer and the other uses the angle as seen by the observer, due to aberration. I accidentally used the angle for one in the formula for the other. The correct values hover much closer to 2 throughout the whole trip. That being said, My original numbers still gave a a final count value for the pulses counted on the trip that equaled the count made at home. the numbers I gave were for 1 leg of the trip. Thus while traveling from Earth to midpoint (midpoint being where the perpendicular line form the pulsar crosses the path of the ship) she would see the flash rate that was as high as 3.73, it wasn't until she actually reached Alpha C that the rate drops to 0.28. So for half the trip shes sees the flashes at a high rate and half at a low rate. When you summed up the flashes seen in both halves you got a number that matches the number counted on Earth. Since this final result was what I expected, I didn't bother to recheck my steps, which is why I didn't realize my mistake until later when something began to gnaw at the back of my head and I went back to double check. So to recap, she will see the pulse rate at just a hair over 2 for half a leg and just a hair under 2 fro half a leg. IOW, If Earth counts 1 flash per sec for 10 yrs, she will count ~2 flashes per sec for 5 yrs and end up counting the same number of flashes. Again, I apologize for any confusion my error may have caused. I'm not quite sure what your point is here. Over a small enough region, a segment of a circle looks like a straight line. But if you extrapolate out that section treating it as a straight line, you will deviate from the actual curve of the circle. This is the equivalent of Physics in the 19th century. At the speeds and energies they were able to study, time and space appeared constant. But then some experiments began to give results that were not as predicted. What Einstein showed was that the results of these experiments made sense if you abandoned the idea that time and space were constant. ( in essence, he was saying that the line that appeared straight at normal energies and speeds was in fact a segment of a curve, and when extrapolated to high energies and speeds lead to quite different consequences than expected if you considered it straight. But you have to accept the possibility that the "not making sense to you" may be your fault and not the fault of the concepts. At the very start of this thread I told you that in order for Relativity to make sense for you, you would have to abandon some notions you likely had about time and space. So far, you have been reluctant to do so.