What happens in the restframe with lightsource?

[John232]

Just to make sure we're on the same page here, you are referring to the wikipedia article on Time Dilation in the section called "Simple inference of time dilation due to relative velocity", correct?.

Correct.

OK, well in the picture of the stationary light clock, they say that the period of the clock is Δt and they calculate it based on the speed of light and the distance between the mirrors. So I agree that t is the amount of time it takes for light, traveling at c, to travel a certain distance. But I don't see why you bring up solving for c--isn't c a known constant and you're solving for Δt, the time between two ticks? So I don't know why you think this leads to a wrong value for c. I have no idea what your concern is.

Okay, it says the observer in motion observers the photon to travel a distance cΔt'. I am saying that it doesn't. The observer at rest measures the hypotenus to be cΔt. 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 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).

In any case, the only way you would get a larger value for Δt (why are you dropping the delta?) is if the distance between the mirrors is greater. And if that is the case, the light clock is taking longer between ticks and therefore ticking at a slower rate.

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 dilation 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 dilation 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

 

[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.

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.

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?

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?

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.

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 dilation 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 dilation 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 into where you are getting all this from. I have no idea.

 

[harrylin]

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.

 

[John232]

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 similar 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.

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.

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 separate 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 dilation 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 dilation 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.

 

[ghwellsjr]

Yes, but a lot of it comes from my own simple proof of relativity that is just similar 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.

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.

(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 separate 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 dilation 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 dilation 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]

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]

[..] 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]

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]

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]

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]

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]

I fear this is just going to be way over your head...

It is, but not the way you envision it. :)

 

[ghwellsjr]

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]

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]

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]

http://en.wikipedia.org/wiki/Particle_acceleration

Here is another link, do these equations look familiar?

 

[ghwellsjr]

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.

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.

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.

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]

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?

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]

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 dilation 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]

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?

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?

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 dilation 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]

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]

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.

 

[John232]

Finally, you state that you actually know something about physics. So why ask me, if you already know? Is there a point to all this? If you can prove that you can't measure the distance a photon has traveled, then you would have proved that relativity itself is wrong because it does the same thing. So then what makes you think you can disprove 100 years of accepted physics? I find it upsetting to work hard in thinking about how to solve many of the problems faced in physics and then find an answer, just to have someone insult me the whole time about it. I tried as hard as I could to explain it well enough to make someone else understand it, but apperently it takes two geniuses to create new science, one to figure it out and another to say yes that is right.

http://en.wikipedia.org/wiki/Einstein_synchronization

He finds that you can add time 1 and time two and multiply it by one half. This only takes the average of the two times. Like you would find the avearge velocity in Newtons equations, for this to be true the velocity would have had to have been the same both ways. It is just a lot ado about nothing. It would be like telling Isaac Newton that his theory's of motion didn't mean anything because he can't prove that it works the same both ways...
According to Albert Einstein's prescription from 1905, a light signal is sent at time from clock 1 to clock 2 and immediately back, e.g. by means of a mirror. Its arrival time back at clock 1 is . This synchronisation convention sets clock 2 so that the time of signal reflection is .[1]

 

[John232]

Come to think of it I can believe Einstein had this same discussion...

Maybe it could mean something...

 

[John232]

Maybe they had to have this same discussion because he didn't get the same value for velocity!

 

[John232]

I think the only way someone could depict a more accurate theory using this method of algebra, would be to consider the coordinate system of a photon traveling at the speed of light. But, in the equations length and time would be contracted to zero. In another coordinate frame their would exist real distance. The problem is that these two coordinate systems do not agree with each other. In one the triangle would have a side with the length of zero (it would no longer form a triangle), and in the other it would have some other real value. Then you would have to find the relation between these two system so that in some way they agree on the speed of light, even though one of those planes was fully contracted. I don't think there is a way it can be done mathematically as we know it, but it may be possible to describe the length along with quantum uncertainty in some other way.

 

[ghwellsjr]

Finally, you state that you actually know something about physics. So why ask me, if you already know? Is there a point to all this?

The point of these forums is to teach people who want to learn physics. I didn't make that rule but I follow it.

If you can prove that you can't measure the distance a photon has traveled, then you would have proved that relativity itself is wrong because it does the same thing.

I never said you couldn't measure the distance a photon travels, where'd you get that idea? I just said in post #67:

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.

So then what makes you think you can disprove 100 years of accepted physics? I find it upsetting to work hard in thinking about how to solve many of the problems faced in physics and then find an answer, just to have someone insult me the whole time about it. I tried as hard as I could to explain it well enough to make someone else understand it, but apperently it takes two geniuses to create new science, one to figure it out and another to say yes that is right.

Huh? Why do you think I'm trying to disprove 100 years of accepted physics? I'm trying to help you learn it. You're the one that wants to publish your own theory of Special Relativity, not me. I'm trying to dissuade you from that endeavor. Einstein is pretty much accepted as a genius by a great many other people that I would also consider to geniuses. What other geniuses are you talking about here? Do you see yourself in the role of a genius creating new science? Is that why you want to publish your own theory of Special Relativity?

http://en.wikipedia.org/wiki/Einstein_synchronization

Yes, the wiki article does not say that Einstein is measuring the time it takes for the light to go from clock 1 to clock 2 as you have been claiming. Rather it says τ₁ is the time on clock 1 at the start of the light pulse and τ₂ is the time on the same clock 1, not clock 2, after it has been reflected back, in other words, he is measuring the round-trip time of the light pulse. He also records the time on clock 2 when the light was reflected. Then he calculates the average of those two times measured on clock 1 and sees how far off the recorded time on clock 2 was from the average. He makes an adjustment to clock 2 (this is where he sets the time on clock 2). Now clock 2 should be synchronized to clock 1. If he repeats the experiment and if he did everything right the first time, then the second time, clock 2 should display the average of the two times on clock1 and he can now say, by definition--not by measurement, that the time it takes for light to go from clock 1 to clock 2 equals the time it takes for the light to go from clock 2 to clock 1. You should not think of the process of synchronization as a way to discover the truth about the speed of light but rather as a way of creating truth about the speed of light.

He finds that you can add time 1 and time two and multiply it by one half. This only takes the average of the two times. Like you would find the avearge velocity in Newtons equations, for this to be true the velocity would have had to have been the same both ways. It is just a lot ado about nothing.

It's not much ado about nothing. It's the foundation of Special Relativity and without Einstein's insight into the fact that until and unless you create the meaning of the time on clock 2, it can have no meaning. After he makes that definition, then you can conclude that the light takes the same amount of time to go both directions but only in that one frame of reference for which the definition holds true. In another frame of reference with its own application of the definition, the time it takes for light to get from clock 1 to clock 2 is not the same as it is for the light to get back from clock 2 to clock 1.

It would be like telling Isaac Newton that his theory's of motion didn't mean anything because he can't prove that it works the same both ways...
According to Albert Einstein's prescription from 1905, a light signal is sent at time from clock 1 to clock 2 and immediately back, e.g. by means of a mirror. Its arrival time back at clock 1 is . This synchronisation convention sets clock 2 so that the time of signal reflection is .[1]

This is incomplete so I don't know what you are saying here.

But the bottom line is that anytime you want to measure how long it takes for light to go from point A to point B with two different clocks, you have to first synchronize those two clocks via round-trip light signals that are assumed to travel at the same speed in both directions and therefore take the same time in both directions, then, of course, you will "measure" the speed of light to be the same in both directions, how could it be otherwise?

This whole discussion is a result of your rejection of the wikipedia article on time dilation in its explanation of a light clock based on Einstein's definition of remote time in a Frame of Reference and the constant speed of light, and your insistence that there was a better way in which you could measure the one-way speed of light apart from previously defining it.

 

[John232]

I read a book a long time ago, don't recal what one it was, but it said that the writer new about the instance where Einstein's theory was rejected by a particle physisist and the theory didn't work out with what they found in the experiment. They then had an argument about it because there was no clear way to define how someone could know that Newtonian physics still applied to quantum mechanics. It really started to make me wonder if you where that same guy because of the insidious questions about Newtonian physics. If so I apoligize if you ended up getting in a argument with both of us. But, i think he may have passed away, don't remember exactly who that was.

I guess the wiki claims that any type of lorentz transform theory would not follow the two way speed of light, but I think mine can because I derived gamma differently. The equations for velocity would not change if the value's canceled so in effect the equations that deal with velocity could stay the same, but then someone could calculate how long a particle lived by finding the amount of time dilation it experienced while under acceleration. Also the effects of gravity are negligible so it is not included in my theory yet, also it would work accurately for sure for any experiment done on Earth since the conditions of the observer would be guarnteed to be the same as the Michealson-Morley experiment with the case that an observer traveling relative to the MMX would detect the outcome to come out differently. I also think that the relation itself just does not exist anymore for an object traveling at the speed of light, since the triangle itself no longer exist. So if the photons frame of reference is in no way related to an observer at rest, then any value we find in our frame would not affect any value in the photons frame.

 

[Dale]

the effects of gravity are negligible so it is not included in my theory yet

Personal theories are not permitted topics on Physics Forums.

So if the photons frame of reference

Photons do not have a frame of reference:
https://www.physicsforums.com/showthread.php?t=511170

 

[Dale]

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.
...
So then, the distance traveled by the object in constant acceleration can be replaced with Δt(vi+vo)/2.
...
(cΔt')^2+(Δt(vi+vo)/2)^2 = (cΔt)^2 Pythagorean Theorem

This assumes that the postulates of relativity apply to non-inertial frames, which is wrong.

 

[John232]

This assumes that the postulates of relativity apply to non-inertial frames, which is wrong.

I don't think it does. A non-inertial frame would not have a length (vt), so then instead of being able to insert c as the velocity, you would have to say that the triangle does not exist so that you can't obtain an answer as zero for the formula to work. So then going back to the inital question of does the distance light travel equal its velocity times time would be yes, because of the nature of the algebra itself. So then Δt=t√(1-(vi+vo)/4c^2) would only be true if vi≠c and vo≠c. That agrees with the intial assumption that the distance the photon travled is the same of the Newtonian equation d=vt.

I will stop here, I guess i would have to ask where I could be redirected to where these topics could be discussed?

 

[John232]

How does a theory become accepted by physics forums?

 

[Dale]

you would have to say that the triangle does not exist

I wouldn't say that it doesn't exist, just that light doesn't travel in a straight line in terms of non inertial coordinates.

 

[Dale]

How does a theory become accepted by physics forums?

Read the rules link at the top of each page. It has to be published in the mainstream scientific literature.

 

[John232]

I wouldn't say that it doesn't exist, just that light doesn't travel in a straight line in terms of non inertial coordinates.

So then wouldn't that mean that the MMX wasn't in a non inertial frame, but was actually in an inertial frame because the outcome of the experiment showed that light beams sent in two different directions ended up haveing their wavelengths match up as though one beam had not traveled a longer distance than the other when they split up?

The mainstream literature I read on it suggest that the experiment actually was in a non-inertial frame and that the beam did travel in a straight line, and that Einstein himself didn't base his theory on that experiment, but the experiment itself is always mentioned in physics literature.

 

[Dale]

So then wouldn't that mean that the MMX wasn't in a non inertial frame, but was actually in an inertial frame because the outcome of the experiment showed that light beams sent in two different directions ended up haveing their wavelengths match up as though one beam had not traveled a longer distance than the other when they split up?

The experiment was not in an inertial frame according to general relativity, but the apparatus was not sensitive to the gravitational effects. You can calculate the general relativistic corrections if desired, but you will see that they are far smaller than the noise.

 

[John232]

How could MMX not be in a inertial frame according to general relativity if there is acceleration of the rotation/revolutions of Earth? Putting gravity aside, you would think that since the theory predicts that a photon would propogate at a curve if the experiment was only accelerating at a different velocity. So how then can MMX get the result it did and still be in accordence with the general theory?

 

[Dale]

So how then can MMX get the result it did and still be in accordence with the general theory?

As I said before, it is simply not sensitive to general relativistic effects.

 

[John232]

As I said before, it is simply not sensitive to general relativistic effects.

Then why did MMX make the claim that they could calculate the acceleration relative to the aether when it did not accelerate enough to find the difference of acceleration according to GR? Seems to me that they would have had to have found an arc in the beam that was the same as GR predicted in order for the theory to be correct. Then they would have had to say that light bends when Earth accelerates through the aether. But, this is not what happened. I think this is why GR is not compatable with quantum mechanics, because in this sense it is just wrong. The beam of light isn't effected the same way from acceleration as it is affected by gravity.

 

[harrylin]

Then why did MMX make the claim that they could calculate the acceleration relative to the aether [..]

That's not exact, they hoped to be able to detect the speed of the Earth. You can read it here:
http://en.wikisource.org/wiki/Author:Albert_Abraham_Michelson
Apparently you are making statements about his papers of 1881 and 1887.

 

[Dale]

Then why did MMX make the claim that they could calculate the acceleration relative to the aether when it did not accelerate enough to find the difference of acceleration according to GR?

The apparatus was designed to measure the velocity relative to the aether, not the acceleration relative to the aether. Furthermore, just because something is not sensitive to GR effects does not imply that it is not sensitive to SR effects.