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What happens in the restframe with lightsource?

by faen
Tags: lightsource, restframe
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ghwellsjr
#55
Mar8-12, 10:03 AM
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Quote Quote by John232 View Post
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 Quote by John232 View Post
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 Quote by John232 View Post
(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.
John232
#56
Mar8-12, 04:09 PM
P: 249
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.
harrylin
#57
Mar8-12, 04:27 PM
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Quote Quote by John232 View Post
[..] 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?
ghwellsjr
#58
Mar8-12, 04:43 PM
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Quote Quote by John232 View Post
All he would have to do is find the amount of time it took to travel across the experiment...
How does he do that?
John232
#59
Mar8-12, 07:47 PM
P: 249
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...
PAllen
#60
Mar8-12, 07:54 PM
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Quote Quote by John232 View Post
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.
John232
#61
Mar8-12, 09:54 PM
P: 249
Quote Quote by PAllen View Post
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...
Passionflower
#62
Mar8-12, 10:01 PM
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Quote Quote by John232 View Post
I fear this is just going to be way over your head...
It is, but not the way you envision it. :)
ghwellsjr
#63
Mar8-12, 10:19 PM
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Quote Quote by John232 View Post
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.
John232
#64
Mar9-12, 01:04 AM
P: 249
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.
John232
#65
Mar9-12, 01:22 AM
P: 249
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
John232
#66
Mar9-12, 01:26 AM
P: 249
http://en.wikipedia.org/wiki/Particle_acceleration

Here is another link, do these equations look familiar?
ghwellsjr
#67
Mar9-12, 03:08 AM
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Quote Quote by John232 View Post
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 Quote by John232 View Post
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 Quote by John232 View Post
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 Quote by John232 View Post
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.
ghwellsjr
#68
Mar9-12, 03:30 AM
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Quote Quote by John232 View Post
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 Quote by John232 View Post
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".
John232
#69
Mar9-12, 04:34 PM
P: 249
Einstein found the two way speed of light just by adding the times it reached both clocks and dividing it by two, in no way does this imply that the speed of light in two directions is different. The Michealson-Morley experiment proves that the two way speed of light is the same as the one way even in different directions. In no way does the Einstein synhronisation imply that velocity is different. I still don't get why you have a problem with this. Pretend your Isaac Newton, you measure a ball to be shot across the room. You find that it always travels a distance vt. Now you check every other distance across the room that it could travel and you find that the distance equals vt in every case. So then you know that d=vt. Now say the photon is the same ball Newton was measuring. You measure the photon to travel the same distance according to d=vt, and then do this at different distances. You find it always travels at a constant speed no matter what distance it traveled, so then you know that d=vt is true. How could you have some other equation where light comes to the same speed and the distance it travles is not inversely related to that velocity? If it traveled at different speeds in different directions then the speed of light wouldn't be constant and Michealson-Morley would have had a whole different story. I think you should look into the experiment into further detail and not mind any mention of aether...

It also says on the two way speed of light that an equation that describes this is the lorentz transformations, well what I did was in effect solving for my own lorentz transformation. But the main difference being that my velocity is different so that the two way speed of light is equal to the one way speed of light. That was actually found by the experiment it comes from. Like I said before v doesn't equal the length contraction equation divided by the time dialation equation you can find this by putting a velcoity in both equations and then solving for length and time then useing that to find the velocity, it doesn't come out to be the same, but in my theory it does. So then my theory it says the two way speed of light is the same.
ghwellsjr
#70
Mar9-12, 04:46 PM
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Quote Quote by John232 View Post
Einstein found the two way speed of light just by adding the times it reached both clocks and dividing it by two, in no way does this imply that the speed of light in two directions is different.
Can you show me a reputable reference for this?
Quote Quote by John232 View Post
The Michealson-Morley experiment proves that the two way speed of light is the same as the one way even in different directions.
Can you show me a reputable reference for this?
Quote Quote by John232 View Post
In no way does the Einstein synhronisation imply that velocity is different. I still don't get why you have a problem with this. Pretend your Isaac Newton, you measure a ball to be shot across the room. You find that it always travels a distance v/t. Now you check every other distance across the room that it could travel and you find that the distance equals v/t in every case. So then you know that d=v/t. Now say the photon is the same ball Newton was measuring. You measure the photon to travel the same distance according to d=vt, and then do this at different distances. You find it always travels at a constant speed no matter what distance it traveled, so then you know that d=vt is true. How could you have some other equation where light comes to the same speed and the distance it travles is not inversely related to that velocity? If it traveled at different speeds in different directions then the speed of light wouldn't be constant and Michealson-Morley would have had a whole different story. I think you should look into the experiment into further detail and not mind any mention of aether...

It also says on the two way speed of light that an equation that describes this is the lorentz transformations, well what I did was in effect solving for my own lorentz transformation. But the main difference being that my velocity is different so that the two way speed of light is equal to the one way speed of light. That was actually found by the experiment it comes from. Like I said before v doesn't equal the length contraction equation divided by the time dialation equation you can find this by putting a velcoity in both equations and then solving for length and time then useing that to find the velocity, it doesn't come out to be the same, but in my theory it does. So then my theory it says the two way speed of light is the same.
Until you show me the two references I asked for, nothing else in your post matters.
John232
#71
Mar9-12, 07:45 PM
P: 249
I think even if I did give you references to that it wouldn't even matter, I am done talking to you. Google it and find out for yourself. How you question the michealson-morley experiment as not proving that light travels the same speed when sent into different directions is just mind boggling.
ghwellsjr
#72
Mar9-12, 08:19 PM
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Quote Quote by John232 View Post
I think even if I did give you references to that it wouldn't even matter, I am done talking to you. Google it and find out for yourself. How you question the michealson-morley experiment as not proving that light travels the same speed when sent into different directions is just mind boggling.
MMX concluded that the two-way speed of light is the same in all directions. It said nothing about the one-way speed of light. The scientists of the day, especially Lorentz concluded that the length of the arm traveling along the direction of the ether was contracted which gave the illusion that the two-way speed of light was the same as in the direction perpendicular to the direction of motion through the ether.

It was Einstein who came along later that postulated that in any state of inertial motion, you could define the one-way speed of light to be equal to the measured two-way speed of light, in other words, the light is defined to take the same amount of time to propagate from the single clock to the reflector as it takes to come back from the reflector to the clock. Einstein never started with two clocks located at either end of the experiment and measured the two times for the light to travel and then add them together to get the two-way speed of light. You have it backwards.

You have a lot to learn and it's a shame you want to leave in this state of ignorance. I hope you will reconsider. I've invested many hours of my time to help you and I'd hate to see it go to waste.


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