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## What happens in the restframe with lightsource?

Are you looking at the current English version of the wikipedia.org article on Time Dilation? I cannot see much of what you say you are seeing.
 Quote by John232 Okay, it says the observer in motion observers the photon to travel a distance cΔt'.
I don't see any mention of the word "photon" in the article and I don't see any mention of the expression cΔt'. Tell me where you are getting this from.
 Quote by John232 I am saying that it doesn't. The observer at rest measures the hypotenus to be cΔt.
The observer at rest does not see a hypotenuse. Where are you getting this from?
 Quote by John232 The length of they hypotenus divided by the change in time would give the speed of light for the observer at rest and not the observer in motion.
The section starts off by saying that the speed of light is constant in all reference frames. It doesn't say anything about calculating the speed of light. Where are you getting this from?
 Quote by John232 The observer in motion would measure the vertical distance divided by time prime to be the speed of light, not the other way around. I think the error comes in when you say that the observer at rest see's the photon to travel a vertical distance, the observer in motion also see's this as well, but is being compared to what an observer at rest see's(the hypotenus).
You have to be looking at something different than what I'm seeing. None of what you are talking about is in the article that I'm looking at.
 Quote by John232 I am saying that the light clock doesn't measure ticks, but in fact that time is actually warped by the same amount as the relation between the sides of two light triangles. Where each side represents how the speed of light is measured from each frame of reference. So by putting cΔt' as the vertical distance it explains how time has to be adjusted to allow for the observer in motion to measure c even though he see's the photon travel a shorter distance. So then he has to experience less time to fill in this shorter side of the triangle. The observer at rest has to measure more time in order to account for the greater distance the photon is seen to travel. That is how it is possible to get the proper time out of the light clock example simply by assigning the time variables differently. Also, Δt√(1-v^2/c^2) ≠ √(1-v^2/c^2)/Δt v ≠ ΔL√(1-v^2/c^2)/(Δt/√(1-v^2/c^2) v ≠ ΔL√(1-v^2/c^2)/(√(1-v^2/c^2)/Δt) v = ΔL/τ the gamma cancels, (v ≠ ΔL/Δt) So since this does not equal the velocity put into the equation, if you found a velocity with dialated spacetime using ΔL and Δt, the theory breaks down. You could then take that new velocity and then find new values for ΔL and Δt, and then find another new velocity and so on etc. This is why I think the time dialation and the length contraction equation should both be directly porportional to gamma. Δt'=Δt√(1-v^2/c^2) solving for the time variables reversed ΔL'=cΔt' I assume the distance traveled by the photon is a different value since the speed of light is constant ΔL'=cΔt√(1-v^2/c^2) substitute Δt' from the first equation ΔL'=ΔL√(1-v^2/c^2) ΔL=cΔt, this also works for the direction of motion ΔL'=vΔt' if you assume v=v' Solving for length in the vertical dimension doesn't assume space contraction in that direction since both observers would agree that the photon reached the same position after being measured. Neither one says the photon traveled a greater vertical distance and is a requirement of this calculation to form the right triangle. I then found an equation that describes time dialation for an object under constant acceleration. It assumes that because of the Michelson-Morely experiment that an object under acceleration still measures the photon to travel in a straight line since the Earth was accelerating and the effects of gravity where balanced out. Δt'=Δt√(1-(vi+vo)^2/4c^2) I still need to look into how this differs from constant motion
You're going to have to clue me in to where you are getting all this from. I have no idea.

 Quote by ghwellsjr So I guess when I say "time spent at a higher speed" and when you include three different speeds, that entitles you to pick which two speeds to use, ignoring the effect from the third speed. [..] Notice how I included all three speeds and got the correct answer. I didn't ever say or imply that you could ignore the effect of one of those speeds or make some erroneous attempt to "average" two of those speeds like you did. If you follow the precise procedure I outlined earlier in post #16, there won't be any errors.[..]
Just a little support from me: I find it unfair to hang someone up on a sound bite that refers to a certain problem and is meant to clarify the essence of a mathematical analysis. For example "the speed of light is constant" is of course wrong as general statement but I bet that we all use that sound bite now and then correctly with a certain meaning in a certain context.

 Quote by ghwellsjr Are you looking at the current English version of the wikipedia.org article on Time Dilation? I cannot see much of what you say you are seeing. I don't see any mention of the word "photon" in the article and I don't see any mention of the expression cΔt'. Tell me where you are getting this from.
Yes, but a lot of it comes from my own simple proof of relativity that is just similiar to the page described. In my own proof I say that the distance traveled by the photon is cΔt. Then the length in a direction that I just used as L not considering direction, so that L=cΔt. This equation suggest that you can take L/Δt and then calculate the speed of light.

 Quote by ghwellsjr The observer at rest does not see a hypotenuse. Where are you getting this from?.
If a photon is sent from a moving object it does in the light clock example, since the object in motion observers the photon to travel straight out perpendicular to its direction of motion. So then a observer at rest relative to this object sending the photon see's it to travel straight out along with the object in motion.
 Quote by ghwellsjr The section starts off by saying that the speed of light is constant in all reference frames. It doesn't say anything about calculating the speed of light. Where are you getting this from?
That is a problem I saw with the proof on the page. In making my own proof I found that the distances of the sides of the triangle on the page where not the correct size. Then I realized that it doesn't account for light being a constant speed in each frame of reference, because the variables where not correctly assigined to do so. So what I did was instead of counting the number of times the light clock ticks, I found the relation between two distances that actually measure the value c to come out correctly.

My simple proof.

An observer in motion observers a photon to travel straight out in a line a distance cΔt'. An observer at rest measures the photon to travel out at an angle along with the objects direction of motion cΔt. The observer at rest also measures the object to travel a distance vΔt. This gives three sides of a right triangle and then the relation to how they measure Δt can be found.

(ct')^2+(vΔt)^2=(cΔt)^2 a^2+b^2=c^2

c^2Δt'^2=c^2Δt^2-v^2Δt^2 distribute the square and subtract both sides by (v^2Δt^2)

Δt'^2=(c^2Δt^2-v^2Δt^2)/c^2 divide both sides by c^2

Δt'^2=c^2Δt^2(1-v^2/c^2)/c^2 factor out a (c^2Δt^2) from (c^2Δt^2-v^2t^2)

Δt'=Δt√(1-v^2/c^2) the c^2 cancels and take the square root of both sides

It then follows that the Michealson-Morley experiment also measured the photon to travel out in a straight line even though the experiment was under constant acceleration due to the Earths movement. So then, the distance traveled by the object in constant acceleration can be replaced with Δt(vi+vo)/2.

s=Δtvo+(aΔt^2)/2

s=Δtvo+(Δt^2(vi-vo)/t)/2 substitute from a=(vi-vo)/Δt

s=Δtvo+(Δtvi-Δtvo)/2 cancel a Δt from the substitution and distribute the Δt

s=2Δtvo/2+Δtvi/2-Δtvo/2 multiply Δtvo by 2/2 and seperate the factor

s=Δt(vi+vo)/2 add like terms and factor out Δt

(cΔt')^2+(Δt(vi+vo)/2)^2 = (cΔt)^2 Pythagorean Theorem

c^2Δt'^2 = c^2Δt^2 - Δt^2(vi+vo)^2/4 distribute the square

c^2Δt'^2 = c^2Δt^2(1-(vi+vo)^2/4c^2) factor out c^2Δt^2 from the right side

Δt'^2 = Δt^2(1-(vi+vo)^2/4c^2) divide both sides by c^2

Δt'=Δt√(1-(vi+vo)^2/4c^2) take the square root

Then you have an equation for the relation between time dialation and constant acceleration. I noticed that by adding the initial velocity and the final velocity that you can get double that amount of velocity than just the time dialation equation alone. Then when it is squared it comes to be counteracted by the 4 in the denominator. This is because 2 in the acceleration equation is squared to give 4. So you have 4c^2 instead of just c^2, that seems to work out. But my proof seems to imply that as a velocity becomes c then the amount of time that is dialated comes to zero instead of being undefined.

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 Quote by John232 Yes, but a lot of it comes from my own simple proof of relativity that is just similiar to the page described. In my own proof I say that the distance traveled by the photon is cΔt. Then the length in a direction that I just used as L not considering direction, so that L=cΔt. This equation suggest that you can take L/Δt and then calculate the speed of light. If a photon is sent from a moving object it does in the light clock example, since the object in motion observers the photon to travel straight out perpendicular to its direction of motion. So then a observer at rest relative to this object sending the photon see's it to travel straight out along with the object in motion. That is a problem I saw with the proof on the page. In making my own proof I found that the distances of the sides of the triangle on the page where not the correct size. Then I realized that it doesn't account for light being a constant speed in each frame of reference, because the variables where not correctly assigined to do so. So what I did was instead of counting the number of times the light clock ticks, I found the relation between two distances that actually measure the value c to come out correctly.
The wiki article is not presenting a proof. You seem to be taking exception with it because it is an inadequate proof but you have a fundamental misunderstanding of what it is doing. There can be no proof of the speed of a photon or of the propagation of light as is used in the article. So rather than try to prove how the light moves in the light clock under different circumstances, they use Einstein's (unprovable) postulate that light propagates at c in any Frame of Reference. That's why they talk about viewing the light clock from two different Frames of Reference, first from a Frame in which the light clock is stationary and then from a Frame in which the light clock is moving. This enables them, without any proof, to assert that the light is propagating at c in both cases.

There is nothing wrong with the article but, as I said before, you have to understand what their variables stand for. If you think there is something wrong with the article, please don't mix up your complaint with your own version, just point out where you think it is in error and we can deal with that separately from your own version.
 Quote by John232 My simple proof. An observer in motion observers a photon to travel straight out in a line a distance cΔt'. An observer at rest measures the photon to travel out at an angle along with the objects direction of motion cΔt. The observer at rest also measures the object to travel a distance vΔt. This gives three sides of a right triangle and then the relation to how they measure Δt can be found.
Your simple proof starts out confusing me and I have several problems with it.

First off, I know that you cannot observe a photon or measure its progress so you're going to have to show me how you plan on doing this before I can get motivated to try to understand the rest of your proof.

Secondly, it seems like you have it backwards. Why do you say an observer in motion observes the photon traveling straight out but the observer at rest measures it at an angle? I'm trying to associate your scenarios with the diagrams in the wiki article and maybe that's a mistake. Maybe you are presenting something totally different. So I cannot make sense of your equations that follow. You need to provide your own pictures and relate your equations to them so that I can follow your proof. But remember, you cannot observe the motion of a photon so deal with that before continuing on.
 Quote by John232 (ct')^2+(vΔt)^2=(cΔt)^2 a^2+b^2=c^2 c^2Δt'^2=c^2Δt^2-v^2Δt^2 distribute the square and subtract both sides by (v^2Δt^2) Δt'^2=(c^2Δt^2-v^2Δt^2)/c^2 divide both sides by c^2 Δt'^2=c^2Δt^2(1-v^2/c^2)/c^2 factor out a (c^2Δt^2) from (c^2Δt^2-v^2t^2) Δt'=Δt√(1-v^2/c^2) the c^2 cancels and take the square root of both sides It then follows that the Michealson-Morley experiment also measured the photon to travel out in a straight line even though the experiment was under constant acceleration due to the Earths movement. So then, the distance traveled by the object in constant acceleration can be replaced with Δt(vi+vo)/2. s=Δtvo+(aΔt^2)/2 s=Δtvo+(Δt^2(vi-vo)/t)/2 substitute from a=(vi-vo)/Δt s=Δtvo+(Δtvi-Δtvo)/2 cancel a Δt from the substitution and distribute the Δt s=2Δtvo/2+Δtvi/2-Δtvo/2 multiply Δtvo by 2/2 and seperate the factor s=Δt(vi+vo)/2 add like terms and factor out Δt (cΔt')^2+(Δt(vi+vo)/2)^2 = (cΔt)^2 Pythagorean Theorem c^2Δt'^2 = c^2Δt^2 - Δt^2(vi+vo)^2/4 distribute the square c^2Δt'^2 = c^2Δt^2(1-(vi+vo)^2/4c^2) factor out c^2Δt^2 from the right side Δt'^2 = Δt^2(1-(vi+vo)^2/4c^2) divide both sides by c^2 Δt'=Δt√(1-(vi+vo)^2/4c^2) take the square root Then you have an equation for the relation between time dialation and constant acceleration. I noticed that by adding the initial velocity and the final velocity that you can get double that amount of velocity than just the time dialation equation alone. Then when it is squared it comes to be counteracted by the 4 in the denominator. This is because 2 in the acceleration equation is squared to give 4. So you have 4c^2 instead of just c^2, that seems to work out. But my proof seems to imply that as a velocity becomes c then the amount of time that is dialated comes to zero instead of being undefined.
I really have a problem taking seriously anyone who thinks after more than a century of thousands of the best minds in science coming to the same conclusions about MMX and SR, that he has discovered a fundamental mistake and wants to publish his own theory of Special Relativity. I can help you understand Special Relativity but I have no interest in trying to understand your alternative theory, only in discovering where you are making your own errors, and the first one is thinking that you know how a photon propagates.

 It is Newton's equation for distance... I don't really know what more there is to say about it. d=vt. I am saying that you can replace v with the speed of light in order to find the distance the photon has traveled. Both observers would then use their own time to measure this distance. So then if the photon traveled for one secound then you would multiply that times the velocity and get about 300,000 km. So then say Michealson stays with his experiment and then measures the speed of light traveling in it. All he would have to do is find the amount of time it took to travel across the experiment and the distance across the experiment to find the speed of light. So then c=d/t. Now say Morley pass's the experiment in his car and watches it through the window. He notices that the experiment still measured the photons to travel in a straight line in the experiment to reach the end at the same time. Traveling at a relative velocity doesn't effect the outcome of the experiment. He then draws the path of the photon on his car window as it passes by with a magic marker. The line he just drew was moveing at an angle along with the motion of the car. Morley then concludes that the photon traveled a larger distance in his frame of reference so then he solves for the speed of light. So then he takes this longer distance and the time he measured it to take for it to travel and he gets the same answer for the speed of light. So then light has been seen to travel two different distances and comes out to be the same speed. They then compare their watches to find that the amount of time it took to reach across the experiment was different when Morley was in the car. So then they could both set up a right triangle to give their relation to the time that is experienced in the car and just sitting next to the experiment. But, this could get too confusing because I switched the observer at rest and the observer in motion, so I will stop here. But the point is mainly that the observer in motion wouldn't observe the results of this experiement to come out differently. In Einsteins world the experiment would have been seen to shoot the photon at an arc, but that wasn't the case. An observe in motion can't change the outcome of a Michealson-Morley experiment just by accelerating and looking at it from afar or by accelerating the experiment further. That is one of the main differences of my theory.

 Quote by John232 [..] That is one of the main differences of my theory.
Sorry if I'm a game breaker here, but that sounds as if you are not playing by the Physicsforums rules. Which theory that is presented in the scientific literature do you want to discuss or ask questions about?

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 Quote by John232 All he would have to do is find the amount of time it took to travel across the experiment...
How does he do that?

 You guys are really no fun at all, I wasn't aware that physics forums was not to include anything about the scientific process. You need to go back to high school and relearn the first property of physics, I fear this is just going to be way over your head...

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 Quote by John232 You guys are really no fun at all, I wasn't aware that physics forums was not to include anything about the scientific process. You need to go back to high school and relearn the first property of physics, I fear this is just going to be way over your head...
Redirect this comment at yourself, where it belongs. ghwellsjr simply asked you how something is measured, and you suggest that is a bad question. It is the most important question in physics - you can' measure it, it isn't physics.

 Quote by PAllen Redirect this comment at yourself, where it belongs. ghwellsjr simply asked you how something is measured, and you suggest that is a bad question. It is the most important question in physics - you can' measure it, it isn't physics.
The speed of light hasn't been measured now? Do I really need to get into all that? It seems like it would be something for another forum. Why don't you actually do the math of what I am saying and not just assume that it is wrong without even checking into it? I see why now how this has come to be over a hundred years this has been overlooked.

Okay, a pitcher throws a ball it takes 2s to reach the catcher. He threw the ball at 10m/s. How far is the catcher from the pitcher? (2s)(10m/s) = 20 meters The value for secounds cancels and then you are left with only the unit meters...

 Quote by John232 I fear this is just going to be way over your head...
It is, but not the way you envision it. :)

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 Quote by John232 The speed of light hasn't been measured now? Do I really need to get into all that? It seems like it would be something for another forum. Why don't you actually do the math of what I am saying and not just assume that it is wrong without even checking into it? I see why now how this has come to be over a hundred years this has been overlooked. Okay, a pitcher throws a ball it takes 2s to reach the catcher. He threw the ball at 10m/s. How far is the catcher from the pitcher? (2s)(10m/s) = 20 meters The value for secounds cancels and then you are left with only the unit meters...
When a pitcher throws a ball, you can watch it, you can see when he throws it and you can see when the pitcher catches it. But you are using light which is traveling millions of times faster than the ball to see those two events and so your error in timing is negligible and you don't have to factor out the time it takes for the light to travel from those two events to your eye.

But when Michelson's experiment emits a flash of light, how does he see it to know when that happened? When the light pulse hits the other end of the experiment, how does he see it to know when it arrived? He can only watch it with light, correct? So I need for you to tell me how he can factor out the light travel time from those two events so that he can get a meaningful measurement.

 He could move an entanglement experiment next to the Michealson-Morley experiment and then put an atomic clock on the end of it. Then have a switch linked to both experiments so that it is the same distance to each. When the switch turns on it measures the entangled particle and turns on the Michealson-Morley experiment. The atomic clock would then have a detector to see when the photon reaches it, it then turns on when the other entangled particle changes it's spin from being measured on the other side. The atomic clock stops and read time when the photon reaches the detector. That would be a bunch of trouble just to find that a particle follows d=vt. I thought it should be a given. Or he could just shoot a photon at a piece of material that alters when hit and then time how long it takes it to do it, other way could be a lot of trouble. I thought they already have done this and should be a given. Like when they found the wave properties of light. Why does it take over a hundred years for everyone to know that it was measured to always travel at the same speed? Why would particles travel with anything other than their velocity? Is there some kind of particle velocity I have not heard about? Maybe you could try using my equation to find the velocity in respect to time and get the same answer as the velocity you originally put into it and find that the other equation doesn't. If v=v', then gamma will always cancel so then it doesn't matter if you used the equation for dialated time or not they will reduce to the same equations. Sounds like that should be something you should see wrong with accepted equation.
 I found one link to particle velocity but I don't think it applies here, because I said it was in a vacuum with no resistance. Laws of motion don't even take into account resistance anyways.... http://en.wikipedia.org/wiki/Particle_velocity
 http://en.wikipedia.org/wiki/Particle_acceleration Here is another link, do these equations look familiar?

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 Quote by John232 He could move an entanglement experiment next to the Michealson-Morley experiment and then put an atomic clock on the end of it. Then have a switch linked to both experiments so that it is the same distance to each. When the switch turns on it measures the entangled particle and turns on the Michealson-Morley experiment. The atomic clock would then have a detector to see when the photon reaches it, it then turns on when the other entangled particle changes it's spin from being measured on the other side. The atomic clock stops and read time when the photon reaches the detector. That would be a bunch of trouble just to find that a particle follows d=vt. I thought it should be a given.
We are not trying to measure how long it takes a massive particle to travel some distance, that would be like a baseball where you use light, which travels faster than the massive particle to identify when the particle left one end of the experiment and you use light at the other end to identify when the particle arrives there and so that both light signals can be used to start and stop a timer with only a minimal error caused by the ratio of the speed of light to the speed of the particle being less than infinity.
 Quote by John232 Or he could just shoot a photon at a piece of material that alters when hit and then time how long it takes it to do it, other way could be a lot of trouble.
How do you time how long it takes for the photon to leave the photon gun and arrive at the target? Let's assume that you have at the gun a very fast electronic circuit that produces a pulse precisely when the photon is fired and you have at the target another very fast electronic circuit that produces a pulse precisely when the photon hits the target. The problem is how do you use these two pulses separated in space to start and stop a timer to make your measurement? That's what I need for you to describe for me.
 Quote by John232 I thought they already have done this and should be a given. Like when they found the wave properties of light. Why does it take over a hundred years for everyone to know that it was measured to always travel at the same speed? Why would particles travel with anything other than their velocity? Is there some kind of particle velocity I have not heard about?
All measurements of the speed of light involve a round-trip for the light so that the two fast electronic circuits that I described earlier are located at the same place, at the photon gun. So the gun shoots a photon and starts the timer. The photon hits a reflector some measured distance away and a photon returns. This photon hits the second detector and stops the timer. Now we can calculate the "average" speed of light. It turns out that this value is a constant equal to c as long as the experiment is inertial. But we cannot know whether it took the same amount of time for the photon to travel from the gun to the reflector as it took for the photon to travel from the reflector back to the detector colocated with the gun.
 Quote by John232 Maybe you could try using my equation to find the velocity in respect to time and get the same answer as the velocity you originally put into it and find that the other equation doesn't. If v=v', then gamma will always cancel so then it doesn't matter if you used the equation for dialated time or not they will reduce to the same equations. Sounds like that should be something you should see wrong with accepted equation.
What is your equation that works and what is the other equation that doesn't work?

Gamma was proposed by scientists prior to Einstein as a way to explain how the measurement of light would always yield the same answer even though they thought light was only traveling at c in a fixed ether medium. So if the photon travels at a fixed speed with respect to the ether, that could cause the time for the photon to travel from the gun to the reflector to be shorter than the return trip or vice versa.

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 Quote by John232 I found one link to particle velocity but I don't think it applies here, because I said it was in a vacuum with no resistance. Laws of motion don't even take into account resistance anyways.... http://en.wikipedia.org/wiki/Particle_velocity
 Quote by John232 http://en.wikipedia.org/wiki/Particle_acceleration Here is another link, do these equations look familiar?
These two wiki articles are about massive paritlcles and don't apply to the one-way speed of light. You should instead look up and study the wiki article called "One way speed of light".