Janus

The distance between c and E and The distance between c and X are the same. So are you claiming that an observer at c will say an object going from E to X ( along the red arrow) traveled zero distance? Of course not. The same holds for an observer at d, or e, f and g. They all would say that an object moving from E to B traveled the same distance in the same time. It doesn't matter how far you put your observer away, the traveled distance doesn't converge to 0, but remains the same.

No, the geometry does not evidently show the relative velocity as be almost null. As the observer gets further away, the angular velocity he would measure would decrease, But since the actual relative velocity is w x d*, Where w is the observed angular velocity and d is the distance to b, the increase in d cancels out the decrease in w and gives a constant answer for the relative velocity no matter how far the distance d is. Also, the distance between E and X doesn't change with an increase of d, and the fact that the E- observer distance and the X-observer distance is the same, just makes its easier for the observer to determine that B traveled at 0.8c . For example, lets assume that the observer is 1000 ly away from both. B leaves E, and the light of this event heads off towards our observer, it takes 1000 yrs to reach him. It takes 1 1/4 hours for B to reach X, so 1 1/4 hr after the light of B leaving E is emitted, the light of B arriving at X is emitted. It also takes 1000 yrs to reach the observer. This means that the light telling the observer that B arrived at X arrives 1 1/4 hrs after the light telling the observer that B arrived at X. Since the observer knows that he is an equal distance from both, he know that the light from these events took equal times to reach him, so he knows that the 1 1/4 difference between his seeing these event is exactly the same as the time it actually took B to go from E to X. The distance between E and X is 1 light hour, so he knows be traveled at 0.8 relative to them, and since he is at rest with respect to them, 0.8c relative to himself. This is all pretty straight forward, and in no way supports your conclusion. * technically, this equation assumes that B is traveling along an arc of a circle with our observer at the center. B is actually traveling a straight line. This would have the effect of causing the measured angular velocity to change as B went from E to X. Starting at a one value, increasing to a maximum at the midpoint, then decreasing again. This variance in angular speed would be largest with small values of D and decrease ( but never totally vanish) at larger values of D. If we assume D is very large, we can safely ignore this without too much loss of accuracy.) While complicates things a bit, it does not change the fact that our observer will always get a value of 0.8c for the velocity of B

Again, You are not getting the concept of inertial frames of reference. Planet X and Earth ( and their associated clocks) are at rest with respect to a single inertial reference frame. According to that reference frame the Earth clock and Planet X clock are in sync. B is also "in" that inertial frame, but not at rest with respect to it. Likewise, there is a reference frame where B is at rest, while Earth and planet X are in motion ( in fact, there are an infinite number of inertial frame two choose from, It is just that these two are the most convenient to work with.) So in the second animation, we are making our determination from the later of these two frames. In that frame, the distance between Earth and Planet X is 0.6 light hr. And is as measured from that frame that planet X's clock is 48 min ahead of Earth's clock. In the "rest frame" of Planet X and the Earth, the distance is 1 light hr, and the clocks are in sync ( planet X's clock reads 12:00 when B passes Earth) What you are trying to do is called "mixing frames" ( trying to use measurements from two different frames at the same time.). This only leads to confusion.

No, not really. In most discussions dealing with Relativity, we factor out the light signal delay effects. Thus if the Earth "sees" 11:00 on Planet X's clock when the Earth clock read 12:00, The Earth knows that the light carrying that image left planet X 1 hr ago when the Earth clock read 11:00 and thus both clocks read 11:00 at that moment. And assuming that Planet X remained at rest with respect to the Earth, that in the time it took the light to reach the Earth, the Planet x clock ticked off 1 hr and thus reads 12:00 when that light arrives at Earth and the Earth clock reads 12:00. Thus the Earth would conclude that the the two clocks are in sync. Planet X would conclude the same from its observations of Earth's clock from a distance of 1 light hr. Relativity is what is "left over" after you account for light signal delay.

I don't think you understand what "frame of reference" means in Relativity. It is not the same as a "point" of reference. A frame of reference includes all points regardless of their relative position that are at rest with respect to each other. So as long as Earth and X are motionless with respect to each other, they share the same frame of reference, and there is not an "Earth frame of reference" and a "Planet X frame of reference", but a single reference frame which Earth and Planet X are at rest with respect to. And in that reference frame, the Earth and Planet X clocks are synchronized. The fact that different observers placed at different positions in that reference frame would visually see different times on those clocks does not change the fact that all these observers would agree that these clocks are "in fact" synchronized with each other. Differences in position has no effect on the determination of simultaneity, only differences in velocity.

Relativity of simultaneity. Since the clocks are synchronized in the Rest frame of Earth and X, they can't be in the rest frame of B. Below we have three clocks all a t rest in the same frame. The clocks are synchronized in this frame. The central clock emits a pulse of light which expands outward at c, and when it reaches the other clocks, they emit each emit a light pulse. These pulses meet at the center clock. Note that int this frame, all three clocks read the same when the pulses are emitted and when they meet at the central clock. Now we consider the exact same scenario, except now we consider events according to a frame of reference that is moving relative to the clocks. Unlike the first animation, the clocks do not remain at the center of the light pulses they emit. This is because light must move at c relative to this frame as measured from this frame. Thus each pulse expands at c from a point that remains at rest with respect to this frame of reference, and the clocks move away from that emission point after emitting the pulse. In this frame, the clocks run slower, they and the distance between them is length contracted. What does remain consistent is what time each clock reads when a light pulse reaches it or it emits a pulse. Now, when the center clock emits its pulse, the clock at moving relative to the center emission point from which the pulse is expanding at c. as a result, the leftmost clock runs into the first pulse and emits its pulse before the rightmost clock does. Each of these pulses expand out at c from their emission points and meet up at the center of the central clock. In order for the pulses to arrive at each clock when that clock is reading the same time as it was in the first animation, the clocks cannot be in sync in this frame. This the reality of Relativity. You can complain all you want about how "this doesn't make sense" or that you can't accept this as being true, but that will not change the reality of it.

You have to apply them properly, and you need to allow for the relativity of simultaneity while also taking into account what frame you are measuring from. This is how events unfold according to Earth and Planet x. B and it clock are length contracted and B's clock is time dilated. The length contraction doesn't really play a role in the out come here. B leaves the Earth and take's 1 hr 15 min as measured by Earth's and planet X's clock to cross 1 Lh at 0.8c. B's clock is time dilated and only ticks of 45 min during the trip. Note that the animation pausing at the end is not meant to mean B stopping at X, only that the animation "freezes" at moment so that we can compare clock readings for that moment. If we now consider things from B's inertial frame of reference, you get this. B is at rest while the Earth and planet X move to the left at 0.8c. the Earth, planet X, their clocks, and the distance between them is length contracted. B is not. It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c Earth and planet X's clocks are time dilated and only accumulate 27 min during this time. However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B. The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation.

It's the answer I get.

The formula N =N0 e-ct Gives you the number of remaining atoms(N) starting from an Original number of atoms(N0) when you know the time of decay(t) and the decay constant (c) You are given t and N/N0. From this, you should be able to solve for c Further, cT = ln(2), where T is the half-life. This should be enough to solve the problem (though as you suspected, 173,000 yrs is a "rounded out" answer.)

The speed of light in a vacuum is invariant as it is c. This not only means that the speed you measure for it doesn't depend of the motion of the source, or your motion. Light in a medium doesn't travel at c, but as Swansont said, is c/n. Any speed less than c is not invariant. In this case, the measured speed will depend on whether it is the observer or source that it moving relative to the medium. But also as already mentioned, you need to use Relativistic velocity addition to get the correct answer. Relativity doesn't "just go away" because light is now longer moving at c. It is also important to understand that is the speed "c" that is important and not the speed of light itself. When someone says that you cannot exceed the "speed of light", they mean "c", which is the speed light travels in a vacuum. It is possible to exceed the speed of light for a given medium, even in that medium. For instance, there is a type of nuclear reactor which is surrounded by water. It produces subatomic particles which travel through the water at speeds greater than the speed of light through the water. This produces a type of electromagnetic "shock wave", which is seen as a faint blue glow known as Cerenkov radiation.

ZENO: The chicken couldn't have crossed the road. It would have first had to cross halfway, then halfway the remaining distance...

This reminds me of a quote by Robert Heinlein: “Expertise in one field does not carry over into other fields. But experts often think so. The narrower their field of knowledge the more likely they are to think so.” van Flandern got his degree in Astronomy, specializing in celestial mechanics. Yet he thought this expertise carried over to understanding the nature of gravity itself.

The perspective effect is due to the fact that objects of the same size subtend a smaller angle of your visual field as the distance to them increases. If the observer is moving relative to the object, a second visual effect also has to be taken into account: aberration. Aberration affects the angle at which you will see the light coming from. In the case where you are looking at an object coming towards you, the angle tightens inward making the object "look" smaller. So let's go back to my last example: A and B both see light coming from planet X as they pass each other. They see the same light. The difference is that due to aberration, A sees the light come from a smaller angle, and visually "sees" a smaller image. Now this is not inconsistent with what I said about what he concludes, which is that the light he is now seeing left planet X when it was 3 light hrs from him, thus he would expect to "see" a smaller image. The point is that there is nothing about what A "sees" that runs contrary to, or is inconsistent with, what Relativity says is happening. Relativity usually ignores these secondary "visual" effects, not because it can't deal or account for them, but because at best, They don't tell us anything important, and at worst, they add unnecessary "clutter" to the scenario.

We had bad air quality for ~ a week, staying in the "Very unhealthy" to "Hazardous" ranges. We tried to seal things up as best we could, but the smell of smoke still found its way in. We were stuck in a thermal inversion that was holding the smoke near the ground. We finally broke out of it Friday, when a system finally blew in. Had good air quality for a few days, then it jumped back up to over 130 today again, before dropping back to under 100 this evening.

Again A and B would visually "see" the same thing. They would conclude a different state of things. B is at rest with, and 1 light hour from planet X as measured by either. A is moving at 0.8c relative to both, and is passing B on his way to planet X. As A and B pass, they see the same light that came from Planet X. B concludes that the light left 1 hr ago and took one hour to reach him across the 1 light hr distance separating them. A however, has to reason like this: "I am now seeing light from Planet X, while I am next to B, This light was traveling at c relative to me, and had to have left planet X before I was next to B, and when the distance between Planet X and myself was much greater than it is at this moment now, when I am next to B Planet X was 3 light hrs away from me when the light left, to be exact. It took 3 hrs for that light to reach me, during which time, The distance between planet X and myself decreased at 0.8c to 3lh -(3h*0.8c) = 0.6 light hrs. Thus as I am passing B, planet X is 0.6 light hrs from me ( even though the light I am now seeing left it when it was 3 light hrs from me) " A and B see exactly the same light from Planet X as they pass each other, but reach different conclusions as to how far away planet X is at that moment. A can reaffirm his conclusion by waiting until he and planet X meet up, which will occur in 45 min by A's clock. 45 min * .8c = 36 light min = 0.6 light hrs.

Yes, you are wrong. You are again confusing what someone visually "sees", and what they would determine as being true at that moment. He would "see" exactly the same light as he saw before he made the jump* However, the conclusions he would make based on what he sees and his relative velocity to planet X would be that Planet X is closer and length contracted at that moment. It is important to separate the "optical" effects due to Relativistic velocity differences from the actual Relativistic effects. For example while according to you, an object flying by you at .8c would be length contracted, visually you would see it rotating as it passed you. This "Terrell rotation" is an optical effect, which doesn't represent an actual rotation of the object according to you, but the length contraction is measured as a physical change. This distinction between what one "sees" and what they "determine" seems to be something you struggle with. * There would be an aberration effect which would distort the image of planet X for him, But that would occur with or without Relativity( just to a different degree). And this aberration would vanish just as quickly if he were to come to a rest again with respect to Planet X.

As Bufofrog has already pointed out, this would mean that two people, at the same spot and at rest with respect to each other would see two different things happening to the Earth for an hour. This is impossible, as the same light from the Earth is reaching both of them simultaneously at any given moment of that hr. Your claim leads to a physical contradiction.

Or maybe it has something to do with the fact that even without taking relativity into account, even if we had rocket engines 100 time more efficient than the most efficient one presently available, such a rocket would have to mass over 19 times the mass of the Earth, just to get a 1 kg payload up to the speed of light. However, we do have countless practical observations and experiments that do give results that lend support to Einstein's Theory. This is why it is accepted, for the very reason that it has passed every test thrown at it. It would have never gained the level of acceptance it has without such experimental evidence. And science continues to throw tests at it.

No he doesn't. Because, since the traveler has to undergo a change of velocity at the end of his outbound leg in order to return to the Earth, he will see a change in the frequency of light he receives from the Earth when he makes that velocity change. There is no delay. He sees the Earth clock tick at 1/3 speed for 45 min by his clock and thus accumulating 15 min, and then sees it ticking 3 times as fast for 45 min, accumulating 2 hrs, 15 min. Thus he sees the Earth clock accumulate 2 1/2 hrs while his own clock ticked off 1 1/2 hrs. The Earth observer has to wait to see the result of the velocity change because it takes place 1 light hr from him. The traveler doesn't have to wait, because he is the one making the velocity change, so it is happening where he is. I never addressed what the traveler would have seen before this. You jumped to an erroneous conclusion regarding what the traveler would see.

Because If A sees B's clock tick slow while he travels out to a distance of 1 light hour, he will see it ticking slow for 2 and 1/4 hrs and accumulate 45 min of time. He will then see B's clock tick 3 times as fast for 15 min. End result, he sees B's clock accumulate 1 1/2 hrs in the time it takes for his own clock to tick off 2 1/2 hrs. In order to get a result where B's clock ticks off as much as A's, you would need to use the Newtonian equation for Doppler shift, or f0 = fs (c/(1±v/c) Where v is positive if the source is receding. In this case, A would see B's clock tick 5/9 as fast on the outbound trip, accumulating 1 1/4 hrs in 2 1/4 hrs by A's clock, and then tick 5 times faster for 15 min, accumulating another 1 1/4 hrs, for a total of 2 1/2 hrs, the same that accumulates for A. However, this is not what we measure in real life. We measure a Doppler shift that matches the Relativistic version, which ends up giving an answer of less accumulated time for B.

To understand time, you have to understand Relativity, because, at its heart, Relativity is all about the very nature of Time and Space.

And, in the rest frame of clock 3, it is clocks 1 and 2 that are moving at 0.8 c, and thus are length contracted, and this includes the distance between them. Thus by the 3rd clock's measure, the distance between clocks 1 and 2 is only 0.6 lh. So, as it passes clock 1, clock 2 is 0.6 lh away, is coming towards him at 0.8c, and takes 0.6lh/0.8c = 0.75 hr = 45 min.

1. No, the 3rd clock traveled 1 light hr in 1hr 15 min as measured by you, it just also ran slow by a factor of 0.6 and thus only accumulated 45 min , as measured by you. Visually, you saw it take 2 hr and 15 min to reach that 1 light hr distance. 2. Again, it took the clock 1hr 15 min to travel the distance. The fact that it arrived only 15 min after you saw it start its trip is just because it was following closely behind it's own light. Two cars leave a point 100 miles from you, driving towards you. One car, traveling at 100 mph carries a message that says the other car is also just leaving. The second car travels at 80 mph. The faster car arrives at 1:00, while the second car arrives 15 min later at 1:15. Now just because the message saying that the second car was leaving arrived just 15 min before the second car arrives, this does not mean that you will conclude that the second car only took 15 min to cross the distance and had to be traveling at 400 mph. The second car took 1 hr 15 min to make the trip, it was just following not too far behind the faster car.( By the time the first car arrives, the second car is only 20 miles away) Again, you are conflating "what someone visually sees", with what they say is actually happening. If I am 1 light hour from a clock stationary with respect to me, and see it reading 11:00, I don't think that the clock reads 11:00 at that moment, but that, assuming the clock continued to run, it ticked off an additional hr since the image I am now seeing left it, thus I will say that the clock now reads 12:00. Conversely, if your clock reads 12:00, you know that the light left when your clock read 11:00 and thus both clocks read 11:00 at that moment. This works if even if the other clock is moving relative to you. Light that left it when it was 1 light hr away will carry the image of what that clock read 1 hr ago by your clock. So if your clock reads 2:15 when you see the image of the 3rd clock carried by light that left it when it was 1hr away, You know that image left when your clock read 1:15. and if that image is of the 3rd clock reading 12:45, you know that the 3rd clock read 12:45 at the same moment your clock read 1:15.

The Doppler effect does, but the time dilation does not. The relativistic Doppler shift is due to two compounding effects: Changing light propagation delay due to changing distance, the effect of which is determined whether the propagation delay is increasing or decreasing. Time dilation, which only depends on relative speed and is independent of direction. So with the above example, Doppler shift gives a value of 1/3 on the for the receding leg and 3 for the return leg, but time dilation has a factor of 0.6 for both legs.

Again. Relativistic Doppler effect. (which is what you get by combining both light propagation and time dilation) If the clock is receding at 0.8 c, you see it ticking 1/3 as fast as your own. If it is approaching at 0.8 c, you see it running 3 times faster. One way to look at it is that by the time you visually see that clock begin its return trip, the clock has already been on its return leg for an hr, and only 15 min remain until it arrives. All the light emitted from it during the return leg is squeezed into that 15 min by your clock. If you see the returning clock tick 3 times faster, 3*15 min = 45 min is the time you see pass on that clock during its return leg. If you factor out the light propagation delay you will conclude that the clock left reading 12:45, when your clock read 1:15. It took 1 hr 15 min to cross the 1 light hr at 0.8c, and arrives when your clock reads 2:30. The clock read 12:45 when it left, and reads 1:30 when it arrives, thus ticked off 45 min in that 1 hr 15 min by your clock, or 6/10 of 1 hr 15 min(75 min) The time dilation factor for 0.8c is 0.6.