Existence

Hi Son, I have traveled for 365.2425 days in which I have traveled 2,934,796,536,000 miles.You are saying the light beam will catch up to me in 182.62125 days.A light beam at the speed of c can only travel 2,934,796,536,000 in which time I have traveled a further 1,467,398,268,000 so no the light beam has not caught up to me yet.Now if you demand that it catches up to me in 182.62125 days that means for the light beam to cover the 2,934,796,536,000 miles + the 1,467,398,268,000 miles I have traveled since the light beam left the original location the light beam has to travel 4,402,194,804,000 miles to reach my location. That means the light beam will have to be traveling at 279,000 miles per second. But the speed of light is 186,000 mps.Your math is wrong as light can not travel at 279,000 mps. The reality is that I have traveled an additional 1,467,398,268,000 miles since the light beam left the original position.The reality is that the light beam can not travel at 279,000 mps.Therefore it can not reach my location of 4,402,194,804,000 miles in
182.62125 days.The reality is that the light beam can only travel 2,934,796,536,000 miles in 182.62125 days.The reality is that I will travel 5,869,593,072,000 miles in 730,485 days.The reality is that it will take the light beam 365.2425 days to travel
5,869,593,072,000 miles to the point I have reached in 730,485 days since I have left the original starting position.So if you add the 365.2425 days head start I had to the 365.2425 days it takes the light beam to travel 5,869,593,072,000 miles you will get the light beam becoming visible to me after 730.485 days of my journey.Where is my math wrong.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Your math is wrong in that it doesn't take into account the fact that the speed of light is constant in EVERY frame of reference (even if this frame of reference is moving). The light will have moved 2,934,796,536,000 miles in half a year in YOUR frame of reference, and since it started 2,934,796,536,000 miles away from you, it will have caught up in half a year. The reality is that GPS use this and works whereas your physics don't (actually my numbers are a bit off since I didn't take into account length contraction but since you don't seem to understand basics, I didn't feel the need to go into details).

You perform the experiment. You choose a distance, d, and measure how long it takes light approaching from behind you to travel that distance, t. You divide the distance by the time and it comes out as c. We get the same measurement for the speed of light relative to us when we travel towards it. This is unlike with cars. You are perfectly justified in having difficulty with this, but that's what we observe and what is meant by the constancy of the speed of light.There is a solution, and it is not intuitive.

The speed is the same. The time it took to get there is what is different.

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keep your mind from this way of enquiry, for never will you show that not-being is
~parmenides

Hi Mod, I did and you did not refute the math.But I will elaborate for you, using 186,000 mps as c for ease of figuring the math.I leave my wife on earth, and head out on my journey.Distance = I have traveled for 365.2425 days in which I have traveled 2,934,796,536,000 miles.At this point a beam of light leaves a source beside my wife.If I stop and sit still in space it will take the light beam 182.62125 days to travel the distance 2,934,796,536,000 miles, to my location.
The speed of the light beam is 186,000 mps.That is where the problem comes in as I am on a journey and have not stopped so when the light beam reaches where I was located 182.62125 days ago. I have traveled an additional 1,467,398,268,000 miles.The light beam continues it's course for 1,467,398,268,000 miles which takes 91.310625 days.
The speed of the light beam is 186,000 mpsBut in that 91.310625 days I have traveled a further 733,699,134,000 miles, which takes the light beam 45.6553125 days to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 366,849,567,000 miles, which will take the light beam 22.82765625 days to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 183,424,783,500 miles which will take the light beam 11.413828125 to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 91,712,391,750 miles, which will take the light beam 5.7069140625 days to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 45,856,195,875 miles, which will take the light beam 2.85345703125 days to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 22,928,097,937.5 miles which will take the beam of light 1.426728515625 days to reach that location.
The speed of the light beam is 186,000 mps.At which time I will have traveled 11464048968.75 miles, which will take the light beam 0.7133642578125 of a day to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 2,866,012,242.1875 miles, which will take the light beam 0.35668212890625 of a day to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 1,433,006,121.09375 miles, which will take the light beam 0.178341064453125 of a day to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 716,503,060.546875 miles, which will take the light beam 0.0891705322265625 of a day to reach that location.
The speed of the light beam is 186,000 mpsAt which time I will have traveled 358,251,530.2734375 miles, which will take the light beam 0.0445852661132813 of a day to reach that location.
The speed of the light beam is 186,000 mpsI travel (distance) 5,869,593,072,000 miles in (time) 730,485 days, the speed I have traveled is 93,000 mps.The light beam travels the distance of 5,869,593,072,000 miles, in the time of 365.2425 days the speed of the light beam is 186,000 mpsThe light beam reaches the halfway point of my journey at the same time I do.So yes there is a solution. It just is not the one you would prefer.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Hi Son, Your assertion is just that an assertion.The first mention of my journey was in message 307 and you have had from then until now to refute my math. You have asserted it was wrong several times.But you have not proved the math wrong. The only way the beam of light can reach my location in 182.62125 days is if I don't move. But I am moving at .5c.Your assumption is what is wrong.If you disagree then prove the math wrong.God Bless,

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"John 5:39 (KJS) Search the scriptures; for in them ye think ye have eternal life: and they are they which testify of me."

Your answer is exactly correct!! You are the first to give the correct answer. But I'd be completely remiss if I didn't note that you claimed that other, quite different answers were correct and that you have used those answers to refute relativity. Am I correct in assuming that you collected a hint from my comment to tesla? Can you now acknowledge that your other answers were incorrect?I'll also point out that a little algebra would have avoided all of those iterative steps.What is not correct is that your answer is not what is predicted by SR. I'll indicate why your prediction that I don't prefer this answer is wrong. Not exactly. Again, light year is a unit of distance. Light can travel 1 light year in one year. I presume without checking that 5,869,593,072,000 miles is the length a light year, so light travels that distance in one year. Similar comment. How can I get you to stop making this mistake? For the purpose of this problem when you refer to a light year, I'm going to assume you mean a solar year, and I'll refer to time in years. But please make the correction for future posts.Given that correction, you have correctly calculated the answer in the frame of reference of the wife. I'm going to assume again that the time and distances given are all in the wife's frame of reference and that she is in an inertial frame of reference.Due to the high velocity, 0.5c of ICANT's ship relative to the wife, when we compare coordinates we use relativistic transforms. Let x and t represent distance and time in the wife's frame, and x' and t' are the coordinates in the wife's time frame, and v is the relative velocity.I'll also assume that wife's clock and the space cycle clock both indicated 0 at the time ICANT began the trip at 0.5c. We'll also put the origin of our trip at coordinate x'=x=0 with the x axis directed along the direction of travel. Finally, I'll assume that the light beam is traveling along x axis as is ICANT's cycle.It turns out that the easiest way to answer ICANT's question is to convert to the space cycle coordinate system to determine the time and distance when the light beam reaches the space cycle, and then to convert back to wife coordinates so that we have a complete answer in each frame.We can then check our answer to see if the speed of light really is the same as measured in both frames. ICANT seems to agree that velocities greater than c are impossible, so we sure don't want to find any of those.The relativistic coordinate transformation equations are as follows: (1) (2)It can be shown that these equations follow directly from the assumptions I discussed in Message 418 and Message 425. For anyone interested, the following links are videos that cover the derivation in detail.https://www.youtube.com/watch?v=qaT6YKrKQEg
https://www.youtube.com/watch?v=X_3lTXMg6ZwBefore proceeding, it might be worthwhile to compare those equations with the transforms one might use if all of the velocities in the problem are small compared to c. In that case we can approximate the transforms in equations 1 and 2 as follows:t' = t
x' = x - vtThe equations numbered 1 and 2 above work for any velocity, while the simpler, but approximate equations work just fine for problems where everything moves slow with respect to the speed of light. For example in those time and distance problems encountered in elementary school, the approximations work just fine.It would be an error (label the error as you will) to select the approximate equations (either explicitly or implicitly) for use in a problem where the approximation does not hold, such as when the velocities in question are 1c and 0.5c.ICANT implicitly makes this mistake when he determines relative velocities by adding or subtracting v from the speed of light. For example, he makes this mistake when he adds 186,000 to 93,000 to conclude that his answers refute the constancy of the speed of light because the velocity sum is greater than 186,000. If you want to prove that relativity doesn't work, you must show that equations 1 and 2 fail.In short, ICANT's math error is his assumption that he learned everything you need to know about math and physics in the sixth grade. That mistake will not be made here.On the other hand, equations 1 and 2 always work. You could use them on sixth grade test problems if you wanted to. But you might not finish your test in the allotted time.Interestingly enough, I did not see ICANTs final answer when I was working out the math, so I was pleasantly surprised to find the message to Modulus where he obtains the correct answer. Unfortunately for ICANT, SR gives the same answer AND by using equations 1 and 2, we can show that the constancy of the speed of light is not violated.Now to answer the question the SR way.In the wife's frame, the coordinates of the photon at the time it is emitted aretime t = 1 year, and
position x = 0 light yearsIn wife's frame the space cycle is at coordinatest = 1 year,
x = +0.5 light years.Converting the photon coordinates to the ICANT's frame using equations 1 and 2, we see that the coordinates for the photon at emission x' and t' in ICANT's frame can be found as follows. t' = 1.1547 *(1 year - (0.5c*0.5 light years)/c^2) = 1.1547 years.
x' = 1.1547 * (0 - 0.5c*1 year)/c^2 = - 0.57735 light years.Similarly the coordinates for the space cycle in the space cycle frame are at the time the photon was released are.t' = 1.1547 *(1 year - (0.5c*0.5 light years)/c^2) = 0.8660 years.
x' = 1.1547 * (0.5 light years - 0.5c*1 year)/c^2 = 0 light years.We don't really need the coordinates of the space cycle, but they do provide a sanity check. As expected, the space cycle is at x' = 0 in the space cycle frame.Now consider the flight of the light beam as seen in the space cycle frame. In ICANTs frame, the space cycle's velocity is zero. The photon approaches the cycle at 1c, and accordingly will traverse 0.57735 light years from the emission point in exactly 0.57735 years in the space cycle coordinate system. The x coordinate of the space cycle in the space cycle is of course always zero. So when the photon hits the space cycle, the photon's lateral coordinate in the space cycle frame is also zero.Accordingly, in space cycle coordinates the coordinates for the event of the photon hitting the space cycle arex' = 0 and
t' = (1.1547+.57735) years = 1.73205 years.There is no need to check that the velocity of the light beam is 1 c in the cycle frame because we imposed that criteria explicitly. But for completeness the speed of the light beam relative to the space cycle is(0 + 0.57735 light years)/(1.73205 years - 1.1547 years) = 1c.But what is the speed of light as measured in the wife's frame. Let's convert our coordinates back to the wife's frame.To perform the reverse transform, we use equation 1 and 2 again, but we note that from the relative velocity between the frames from the perspective of the cycle is -0.5c.The coordinates of the light hitting space cycle event in the wife's frame.t = 1.1547(1.73205 years + (0.5c *0 light years)/c^2) = 2 years
x = 1.1547(0 light years + 0.5c * 1.73205 years) = 1 light year.So the light beam reaches the space cycle exactly two years into the journey, and exactly 1 year after the photon was emitted. The velocity of the photon in the wife's frame is)(1 light year - 0 light years)/(2 years - 1 year) = 1 c.I do have to acknowledge that my earlier, off the cuff attempt to give the time and distance was total nonsense.So to answer the time when the photon hits the ship. The answer according to SR is 730.4844 days into the trip.(2 years)*365.2422 days/solar year = 730.4844 days.Props to ICANT for posting the first correct answer.

Hallelujah!!!Edited by NoNukes, : Change tone

Well, you are wrong in that in every results you posted, you never pointed out the frame of reference you used. In that, I disagree when Nonukes says you posted the correct answer, your answer can only be at best incomplete if you don't write which frame of reference you use.
I don't know how it is in your country, but in France, I would have been lucky to even have been given half of the points in physics if I didn't give the frame of reference I use when writing my answer.It seems that you believe there is an absolute speed but even in your car example, it's not true. When you give the speed of the car, you do it in the frame of reference of the Earth. It would be incorrect to say that the speed of your car is 60mph if you took a solar frame of reference. That's why that for every results you give, you must also give the frame of reference you use.As for your math, if it is meant to model reality, it is wrong, you prove it wrong yourself every time you use a GPS. What you need is to prove that your math accurately model reality, not that it is correct by itself. For example: 3+2=5 but 3 oranges + 2 apples =/= 5 oranges.

So your refutation of the constancy of the speed of light is to measure the speed of light at several points during the experiment and calculating that the speed of light is the same at each point?

ICANT, this blank screen stuff is nonsense. At the turnaround point you will be two light years away from your wife and light will take an additional 1 year to get to her, thus arriving at 3 years from your departure. So that much is correct, at least from the wife's perspective.But what about the images from immediately before your turn around point. When do they reach wife's screen? Let's look at things in your wife's frame of reference before and after the turn around point. Since your wife is stationary in this frame, the calculations are quite easy.Ten minutes before turnaround, in your wife's frame you are five light minutes closer to your wife than to the turn around point. So light from that point would reach earth in 5 minutes less than one year. But you have also been traveling 10 minutes less than two years. So light from that point will reach earth in 15 minutes short of three years from journey start. No black screen at 15 minutes before day 1095.72+We agree that the screen is not blank at turn around. No blank screen at day 1095.72+Finally at 10 minutes after turn around, the wife places you again at 5 light minutes short of one light year from earth. But you have been traveling for 2 years and 10 minutes. Accordingly, light from this point of your journey reaches you in 3 years and five minutes after your journey starts. So your screen is not blank at day 5 minutes after day 1095.72+You can continue in this vein to show that the screen is never blank during the return journey. At all times during your journey, the light from some other portion of your journey can reach your wife. Yes, some of the images will be compressed in time. But images prior to turn around time are stretched in time.

I suppose this is correct. ICANT might well think that his answer is the same in the husband's frame of reference. It will be interesting to see his answer to what the speed of light appears to be in the space cycle frame. I plan to ask him that if he doesn't pick up on it first.If that value is less than c, then seemingly, he would have to agree that his answer is wrong. True, but let's look at it this way. There are only two obvious frames of reference to use for this problem. ICANT was the first person to post and answer that was correct in one of the two frames even if he did not specify which one.

Not entirely. Your root assumptions are incorrect. Your math assumes that the motion and physics of light behave in the same way that the physics and motion of a baseball behave. If you stand on a moving train and throw a ball in the direction of motion, the velocity of the ball relative to the ground will be the sum total of the velocity at which you throw baseballs plus the velocity of the train. It's simple addition, the like of which you've posted throughout this thread. Balls and trains moving at normal Earth-bound speeds easily work with Newtonian mechanics.But light does not at all behave this way. I know it sounds intuitively bogus. I know it doesn't seem to make sense at first. But the simple, directly observed fact is that light maintains its speed regardless of the frame of reference of the observer.If you're on a train moving at .5c towards me, and I'm on stationary platform,l and you shine a laser at me, the light from the laser will appear to you to move at c. Intuitively that should make the laser beam move at 1.5c relative to me on my stationary platform - the speed of light plus the speed of your superfast .5c train. But that's not what happens. Instead, I detect the light moving at exactly c. If you were one light-year away when you flashed the laser, regardless of how fast you were moving towards me, I would see that laser light exactly one year later. But you would still see the laser moving at c from your perspective, as well.The only way to make that observation mathematically consistent is to apply time and space dilation according to inertial frames. The amount of time you experience has to be different on a .5c train than for me on my stationary platform in order for both of us to measure the light from the laser as moving at exactly c relative to both of us.We've directly observed and measured that effect, multiple times from multiple different experiments, satellites, etc.That means that one of the basic axioms of your calculations is wrong, and all reasoning stemming from that false axiom is invalid, regardless of whether your arithmetic is actually correct or not. The math becomes irrelevant - your assumptions need to be consistent with observed reality before mathematics can make any predictions. You're adding up apples when you actually have oranges. Your math shows nothign regarding the constancy of c. Instead, it operates from the axiomatic assumption that c is different in different reference frames; your math assumes that the velocity of the light leaving a spaceship that is traveling at .5c will be c+-.5c depending on whether the ship is approaching or leaving. This is a false assumption, and so your math is irrelevant. We know it's a false assumption by directly observing the behavior of light.The acceleration factor of Earth's standard gravity is 9.8m/s^2. You can use that factor to easily calculate how fasy a dropped object will be moving from second to second.But what happens if you used the wrong factor, say 5m/s^2? All of the math you do, even though your arithmetic would be correct, would give the wrong answer, because one of your root assumptions was incorrect.So too with this example. Your math makes certain assumptions, and one of them is completely wrong. Therefore your mathematics are irrelevant - they stem from a false premise, and are no more valid than the premise itself. Your disagreement is irrelevant because that fact has been confirmed by direct observation and measurement in multiple independent experiments. You can argue that the moon is made of cheese all you want, but you'd be just as wrong. When Nature tells us something via direct observation, there's simply no use arguing.Rather than insisting we prove your math wrong, you should be insisting that we demonstrate by providing peer-reviewed material that the speed of light is constant in all reference frames. If we can do so, then your base assumption is wrong. If we fail to do so, you're justified in tentatively keeping your hypothesis until an experiment can be devised and run to specifically test that assumption.That's just a little debating hint, ICANT. Debate what's actually being argued and demand evidence for a claim. Technically the burden of proof is on you because you're the one claiming the speed of light is not constant in all reference frames, but we can give you this one for free, just this once. If you ask, that is. Which, again, is based on Newtonian mechanics. They don;t teach relativity in 6th grade at many schools, largely because Newton works just fine for the average word problem in a 6th grade textbook dealing with trains that move at 60mph rather than significant fractions of c.And Newton was wrong.And yet an observer on that moving object will still observe the light approaching and retreating at c relative to himself, just as an observer in a stationary reference frame would observe the same light moving at c relative to themselves.That's just observational fact, ICANT. Your root assumption, that the speed of light relative to an observer depends partially on the speed of that observer, is incorrect.Light is the constant. We and our reference frames are what change. c is measured at exactly c regardless of whether your moving towards or away from the light source, or whether you're stationary. The implication of this, as verified directly in numerous experiments and used every day in the GPS system, is that time moves at different rates in different inertial frames. Yes. Whether your spaceship is moving towards me or away from me at .5c or just stationary, both you and I will detect that the light emitted from your ship is moving at exactly c, not c-.5c when you're moving away, not c+.5c when you're moving towards me. We will both measure the same speed of the light even though you're moving, because the speed of light is constant in all reference frames.

Let’s say an object is defined by:_____________x______________X 1 X2The X's are a point of interest. Triangulations can derive the path of object X_____________(x)_____x(new position)_____X1 (relative point) x2 (relative point)This is using three straight lines, and the movement of the object is a straight line, under the assumption light is in a vacuum and is CNow using a curved line increases the distance which would increase the time of travel, even if it does not affect speed. (This curve would probably be miniscule for short distances, but wouldn’t it be important for long distances?)How is this worked out?

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keep your mind from this way of enquiry, for never will you show that not-being is
~parmenides

That makes no sense at all to me. Please clarify.

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"I hate to advocate the use of drugs, alcohol, violence, or insanity to anyone, but they always worked for me." - Hunter S. ThompsonAd astra per asperaNihil curo de ista tua stulta superstitione.