Existence

True. So Let's examine the photon.A photon enters an electron it gains energy, when a photon is expelled, it looses energy. So photons are the engines of electrons.and then there is radiation:( I know it's WIKI, please forgive)"In physics, radiation is a process in which energetic particles or energy or waves travel through a medium or space. There are two distinct types of radiation; ionizing and non-ionizing. The word radiation is commonly used in reference to ionizing radiation only (i.e., having sufficient energy to ionize an atom), but it may also refer to non-ionizing radiation (e.g., radio waves or visible light). The energy radiates (i.e., travels outward in straight lines in all directions) from its source. This geometry naturally leads to a system of measurements and physical units that are equally applicable to all types of radiation. Both ionizing and non-ionizing radiation can be harmful to organisms and can result in changes to the natural environment"and:
"Beta-minus (−) radiation consists of an energetic electron. It is more ionizing than alpha radiation, but less than gamma. The electrons can often be stopped with a few centimeters of metal. It occurs when a neutron decays into a proton in a nucleus, releasing the beta particle and an antineutrino." Yes but Time is relative to distance. I'm not attacking the math. I'm reviewing assumptions for the behavior.

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

ICANT,Now that we've answered your question, I am curious about how you would answer a similar question. What if the photon was emitted at the three year point in the four year journey so that the cycle was moving towards the light source. You've previously indicated that velocities greater than the speed of light are impossible, so how would you determine the time for the photon to reach the space cycle in this instance?I'd be happy to provide the SR answer, although I haven't done the work yet. Edited by NoNukes, : photo ==> photon.

Yes, that's what you were doing, it doesn't bode well that you didn't notice. I suggest that if you move towards light at 0.5c you will measure oncoming light as moving at c relative to you and not 1.5c as we might expect as in the example of cars. If you are now conceding that we will observe the light passing us at c then, I challenge you to make the maths work so that c is the same for all observors without changing the rate of time passage for someone. I agree that if you measure the time it takes for light to travel any given distance you will always get c, regardless of your direction and magnitude of your speed, unlike with any other thing in the universe.What happens if I set up a bi-directional laser on a one light year long train travelling at 0.5c, and put light sensors at either end of the train and then turn the laser on?If I am on the train, both sensors come on at the same time, since I measure the laser travels 0.5ly forwards and backwards taking half a year to do so: both light sensors will come on in half a year and I will notice it in one year. The speed I calculate light to be travelling relative to me is c.If I am standing still relative to the train, I will see the rear sensor light come on first since the distance between the sensor and the light decreases as the rear of the train 'catches up' to the light. Likewise the front sensors are moving away from the place where the light was emitted and so it takes longer than half a year for the light to get there. The speed I calculate light to be travelling relative to me is is c.This is the constancy of the speed of light and the consequences thereof. If you think otherwise, you are indeed 'trying to refute the constancy of the speed of light '

Hi NoNukes, I should be able to do that. Still using 186,000 as c for ease of math.Things that are known from discussion of light beam coming from behind the cycle.Speed of the light beam = 186,000 mps (miles per second)
Speed of cycle = 93,000 mps.Distance light travels in 365.2425 days as figured in first example equals 5,869,593,072,000 miles.Distance light beam can travel in 1 day = 16,070,400,000.Distance cycle can travel at 93,000 mps in 365.2425 days = 2,934,796,536,000 miles.Distance cycle can travel in 1 day = 8,035,200,000.So if the light beam leaves the source beside my wife when I am 365.2425 days from the end of my journey you want me to solve for when they will meet.I am traveling at .5 c and the light beam is traveling at 1 c. Thus the light beam is traveling twice as fast as I am and will travel 2 miles for every 1 mile I travel.So if I divide the distance to be traveled by 3 I will get 978,265,512,000 miles that I will travel before I meet the light beam which will have traveled 1,956,531,024,000 miles.If I divide my traveled distance by 8,035,200,000 = 121.7475 days.So I add 3 years which = 1,095.7275 days + 121.7475 = 1,217.475.That is how I would come to the conclusion that on day 1,217.475 of my trip I would meet the light beam.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."

As I've already said, for every results you put, you must tell which frame of reference you use. Are your results for the cycle's frame of reference, your wife's or both? It's imporant when you describe when you meet the light. Do you mean to say on the day 1,217.475 in your cycle's frame of reference, your wife's or both?

Hi Mod, And if you account for the distance the sensor has traveled since the ligtht was emitted from the rear of the train you will understand why it takes longer than half a year for the beam to reach the front of the train.The distance the light has to travel has increased, the speed of the beam remains the same.You will also understand why it takes less time for the beam from the front to reach the sensor in the rear of the train when you account for the distance the sensor in the back of the train has traveled since the beam was emitted from the front of the train.The distance the beam has to travel has been reduced, the speed of the beam remains the same.If you do not account for the distance the sensors have traveled since the beams was emitted you will come to the wrong conclusion.So what you have done is intentionally defined a false perception to replace reality.And as far as the man standing at the train station when you went by at .5 c he did not even see you go by must less see the beams of light or observe their actions.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 ICANT,Your answer is correct assuming that it represents events as viewed in the wife's frame of reference. Of course I have some questions for you regarding how things are viewed from hubby's inertial frame. As I stated when posing the question, your solution requires that the cycle and photon close the distance between them at 279,000 miles per second. In fact your 2:1 partitioning method is equivalent to dividing the initial distance between the husband and wife at the time the photon is released by 1.5c (279,000/186,000 = 1.5c).t = (0.5 light year/1.5c) = 1/3 year => 121.747 solar days.Are you not bothered by the consequence that the relative speed between the photon and the space cycle is greater that the speed of light. In other words, in the cyclist's frame of reference, husband stands still, while the photon approaches him at 279,000 miles per second. After all, you did say in Message 481And you were exactly right. If instead of merely striking the cycle, the photon had entered a light speed measuring apparatus mounted on and moving along with the cycle, in your version of events, the equipment would measure 279,000 miles per second. Experimental results show that this does not occur and that the speed of light in a vacuum is the same for all observers. I can provide some pointers to some results if this issue is in dispute.Let me present the SR version of the analysis. For convenience, I have chosen the origin of the wife's coordinate system to coincide with the space ship which is 0.5 light years away. So the wife is actually located at x=-0.5 light years. t = 0 corresponds to the time, in the wife's frame, when the photon is released.In the wife's frame of reference, the photon is released from wife's coordinates attime = 0, x = -0.5 light years.Converting these coordinates to hubby's frame reference using the SR equations presented previously tells us that the coordinates of the photon at time of release, expressed in hubby's inertial frame are as follows:t' = -.288675 years, x' = -0.57735 light years.The photon will strike the ship after it has covered those 0.57735 light years in 0.57735 years. At which time the coordinates of the photon and of the ship will be the same and are given as follows.t' = +.288675 years, x' = 0. (The x coordinate of the ship is always zero in the ship's inertial frame).Returning the coordinates to the wife's frame produces.
t = 0.333333 years x = -.166667 light years.A bit more work than is required using a non SR method, but more information is produced. We can see that in the wife's inertial frame, the photon hits the ship in 1/3 of a year (121.747 days) at a point 1/6 light year ahead of the cycle's original position. Those numbers agree with your calculated values.In the husband's inertial frame, the space cycle does not move and the photon does all of the traveling. However, due to time dilation effects and length contraction, the husband sees things differently. Most importantly, though, the speed of light in the husband's frame is only 1c.

These kinds of answers are not productive. Surely we can imagine electronic equipment in either frame to allow detecting the sequence and timing of events.Edited by NoNukes, : No reason given.

Hi Son, Since the only frame we have experimental evidence for the speed of light being c is the earth, then since I am talking about the speed of the light beam being c I must be talking about the beams speed relative to the earth. As well as my speed relative to the earth, as I am traveling at .5 c.Just as the chosen frame of reference is the earth clock, as the frame of reference in the GPS system.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."

The path along which the shortest distance between to points lies is a geodesic. The path that constitutes a geodesic depends on the geometry of the space involved. All particles without forces acting on them follow a geodesic. Of course when particles interact, then there are forces on them, so it is no surprise that light which encounters matter may deviate from a straight line.Matter/energy distorts space-time (despite ICANTs protests) changing the 'shape' of the geodesic in the surrounding space. For the SR problems involving space cycles, these effects have been neglected and the problems have been worked under the assumption that space-time is flat. I think the assumption is appropriate for these problems.What I don't understand is where you are going with this. For example in some of these discussions, we are talking about time dilation effects which are on the order of 15 per cent. Are you suggesting that there is some kind of light curvy thing that produces deviations on the order of 15% and more, yet which is visibly undetectable? Color me skeptical.For another thing, although the most recent discussion involves light beam travel, relativity works for all high speed objects, so it isn't about light not traveling in a straight line. It would have to be any particle and every object. How would curvy light affect that?Also recall that special relativity is used with quantum mechanics to explain sub atomic phenomena. It isn't just about how photons travel.

The question is about an experiment with your wife and the cycle, there's only 2 frame of references there(the cycle and your wife). And if you're only discussing about a single frame of reference, what does it have to do with relativity?I also don't understand why you're saying we only know of the Earth reference frame? There's also the solar one, your car's, a train, an aircraft's, etc.... There are lots of referential frame from which we can prove that the speed of light is c.As for GPS, we already showed you that it takes into account time dilatation which would make no sense if it only used a single frame of reference.I would also like to know what would be your answer in the cycle's frame of reference. When will the cycle and the beam meet in this frame of reference?Edited by Son, : No reason given.

Hi NoNukes, I have no problem with what you would preceive sitting on my cycle. Especially when it is refuted by reality.I know I am moving at .5 c and the light beam is traveling at 186,000 mps in relation to its frame. We have assumed at the beginning that this trip is in a vaccum. Light has been experimentally shown to be c in relative to the earth.So since the distance is being reduced at 1.5 c 279,000 miles per second, and I am traveling at 93,000 mps the light beam is traveling at 186,000 mps.Whats the problem? But that would not be measuring the speed of light. It would be measuring the rate at which the distance between the light beam and the cycle was decreasing.If that is not the case, how does my math work? This is a false assumption as I am traveling at 93,000 mps, regardless of your assumption. I am not stationary.If I was stationary as you claim it would take the light beam:Distance = 2,934,796,536,000
Speed = 186,000 mps.
Days to cover distance = 182.62125.The light beam would reach my location at day 1095.7275 of my journey.But that can't be right because 182.62125 days would have passed making it 1278.34875 days into my journey.That can't be right either, because I am still 2,934,796,536,000 miles
from my destination which will take 365.2425 days to travel.But that is really getting confusing as I am traveling at .5 c which will take me 1460.97 to travel the 11,739,186,144,000 miles of my trip. But now with you having me stopped it will take 1460.97 days plus the 182.62125 days for the light beam to get to me while I was stopped which = 1643.59125 days for my trip.But that can't be right as I would not have traveled my entire trip at 93,000 mps.Thus you have proved what you observed is indeed the imaginations of your mind. As I stated above I have no problems with the light beam traveling at me at 186,000 mps while I am traveling toward the light beam at 93,000 mps and the distance decreasing at 279,000 mps.The problem only exists in the imagination of your mind. Time dilation does not occur.If it did your math would disagree with my math.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, Are you forgetting the source of the light beam that is sitting beside my wife? Give me a physical experiment that is not earth bound that instruments are used to determine the speed of light rather than a thought experiment. It is duly noted that you have asserted that on several occasions.When I have time to lookup the web site I will give you the link that refutes your assertion.Got to take care of some business now.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."

I 've already shown where GPS use time dilatation in message 395 from the very document you brought up. I even reminded you of it in message 478. Panda had first brought it up in message 373. You didn't respond to this point in all three of those messages. To remind you, the quote used was:

quote:

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( 2 ) Second-older Doppler shift. A clock moving with respect to an ECIF runs slower
relative to coordinate time in that ECIF than if it were at, rest in the ECIF. This is the
time dilation effect due to the magnitude of the relative velocity, sometimes called the
second-order Doppler effect.

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As for the frame of reference, you still didn't answer my question (you can even consider the reference frame of the light source, if you want).For the speed of light, do you then disagree that the speed of light is constant? How do you then explain the accuracy of the GPS (that is not earthbound) if the speed of light is not constant in a vacuum?Edited by Son, : No reason given.

Now I see what was the problem, you don't actually know what a reference frame is. How can any object be moving in it's own frame of reference?I guess you never had a scientific education beyond middle school then?