Register to reply

What happens in the restframe with lightsource?

by faen
Tags: lightsource, restframe
Share this thread:
John232
#73
Mar10-12, 01:35 AM
P: 249
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
#74
Mar10-12, 02:07 AM
P: 249
Come to think of it I can beleive Einstein had this same discussion.....

Maybe it could mean something...
John232
#75
Mar10-12, 02:11 AM
P: 249
Maybe they had to have this same discussion because he didn't get the same value for velocity!
John232
#76
Mar10-12, 09:05 AM
P: 249
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
#77
Mar10-12, 11:37 AM
PF Gold
P: 4,781
Quote Quote by John232 View Post
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.
Quote Quote by John232 View Post
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:
Quote Quote by ghwellsjr View Post
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
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?
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.
Quote Quote by John232 View Post
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.
Quote Quote by John232 View Post
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
#78
Mar10-12, 06:59 PM
P: 249
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 dialation 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.
DaleSpam
#79
Mar10-12, 07:16 PM
Mentor
P: 17,526
Quote Quote by John232 View Post
the effects of gravity are negligible so it is not included in my theory yet
Personal theories are not permitted topics on Physics Forums.

Quote Quote by John232 View Post
So if the photons frame of reference
Photons do not have a frame of reference:
http://www.physicsforums.com/showthread.php?t=511170
DaleSpam
#80
Mar10-12, 07:19 PM
Mentor
P: 17,526
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.
...
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
#81
Mar10-12, 08:54 PM
P: 249
Quote Quote by DaleSpam View Post
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
#82
Mar10-12, 09:31 PM
P: 249
How does a theory become accepted by physics forums?
DaleSpam
#83
Mar10-12, 09:50 PM
Mentor
P: 17,526
Quote Quote by John232 View Post
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.
DaleSpam
#84
Mar10-12, 09:52 PM
Mentor
P: 17,526
Quote Quote by John232 View Post
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
#85
Mar10-12, 11:05 PM
P: 249
Quote Quote by DaleSpam View Post
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.
DaleSpam
#86
Mar11-12, 06:13 AM
Mentor
P: 17,526
Quote Quote by John232 View Post
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
#87
Mar11-12, 06:29 AM
P: 249
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?
DaleSpam
#88
Mar11-12, 06:54 AM
Mentor
P: 17,526
Quote Quote by John232 View Post
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
#89
Mar11-12, 09:34 PM
P: 249
Quote Quote by DaleSpam View Post
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
#90
Mar12-12, 05:53 AM
P: 3,188
Quote Quote by John232 View Post
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...aham_Michelson
Apparently you are making statements about his papers of 1881 and 1887.


Register to reply

Related Discussions
As you move away from a lightsource... Introductory Physics Homework 4