GETTING RID OF EINSTEIN RELATIVITY
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From: Jeckyl on 23 Sep 2007 19:07 "sean" <jaymoseley(a)hotmail.com> wrote in message news:1190560590.631815.78940(a)d55g2000hsg.googlegroups.com... >> Sagnac falsifies emission theory. >> No question about it. > So far you have supplied no evidence to back up this claim. > Wheras Ive supplied mathematical proof that emmision theory CAN > explain sagnac. Really .. please post the link or repost the calcs (I don't want ot have to go hunting through brain-dead google groups t ofind it) All the explanations of Sagnac the include emission theory that I have seen clearly show that it is refuted by Sagnac, as emission theory predicts no interference, no phase shift, no different in times.
From: sean on 25 Sep 2007 06:26 On 23 Sep, 20:06, Dono <sa...(a)comcast.net> wrote: > On Sep 23, 8:16 am, sean <jaymose...(a)hotmail.com> wrote: > > > So far you have supplied no evidence to back up this claim. > > Wheras Ive supplied mathematical proof that emmision theory CAN > > explain sagnac. Simply .. a correct simulation > > of sagnac shows that light at c in the source frame travels > > farther on one path then the other. Which gives a path > > difference.. > > Ummm,no. It is really simple , Shawn : > The forward light beam "chases" the Sagnac disc at the "ballistic" > speed c+v, while the disc moves forward at speed v, so: > > tf*(c+v) =2*pi*r + tf*v > > meaning that : > > tf = 2*pi*r/c > > Likewise, the beam that moves in the opposite direction with c-v, > satisfies the motion equation: > > tb*(c-v) = 2*pi*r - tb*v > so: > tb = 2*pi*r/c > > delta_t = tf - tb = 0 Your calculations are wrong simply because you have incorrectly assumed that in the lab frame light in emmision theory is either c+ or c-v.But it isnt. If light is at c in the source frame then in the *lab frame* for emmision theory the light travels away from the rotating source (in sagnac) at a variable speed , not a constant speed of c+ or c-v. Why do you make this mistake? Because you forget that in the lab frame the source *rotates*. Therefore any path calculated has to take account of the fact that in any one direction the source is actually slowing down in the lab frame because its direction of travel is constantly changing. Do your maths properly for a change. Try including a correct vector calculation for the lightspeed in the lab frame. Your above calculations would only work if the source in sagnac were travelling in a straight line. But it isnt. Sean see my correct sagnac simulations at http://www.youtube.com/profile?user=jaymoseleygrb Notice these calculations correctly include the rotational or vector speed of the source in sagnac. Notice how I do not incorrectly assume the sagnac source travels in a straight line when calculating the speed of the light for emmision theory in the lab frame.
From: George Dishman on 25 Sep 2007 07:44 On 25 Sep, 11:26, sean <jaymose...(a)hotmail.com> wrote: > On 23 Sep, 20:06, Dono <sa...(a)comcast.net> wrote: > > > > > On Sep 23, 8:16 am, sean <jaymose...(a)hotmail.com> wrote: > > > > So far you have supplied no evidence to back up this claim. > > > Wheras Ive supplied mathematical proof that emmision theory CAN > > > explain sagnac. Simply .. a correct simulation > > > of sagnac shows that light at c in the source frame travels > > > farther on one path then the other. Which gives a path > > > difference.. > > > Ummm,no. It is really simple , Shawn : > > The forward light beam "chases" the Sagnac disc at the "ballistic" > > speed c+v, while the disc moves forward at speed v, so: > > > tf*(c+v) =2*pi*r + tf*v > > > meaning that : > > > tf = 2*pi*r/c > > > Likewise, the beam that moves in the opposite direction with c-v, > > satisfies the motion equation: > > > tb*(c-v) = 2*pi*r - tb*v > > so: > > tb = 2*pi*r/c > > > delta_t = tf - tb = 0 > > Your calculations are wrong simply because you have incorrectly > assumed that in the lab frame light in emmision theory is either c+ or > c-v.But it isnt. If light is at c in the source frame then in the *lab > frame* for emmision theory the light travels away from the rotating > source (in sagnac) at a variable speed , not a constant speed of c+ or > c-v. Sean, you still haven't explained how the source can influence the light after it has been switched off. Suppose we pulse the source to give individual flashes of light. For each one the source is off, then on creating the flash, then off. The light moves away in a straight line in the lab frame. Switching the source on to create the next flash cannot affect the previous flash that is already some distance away. > Do your maths properly for a change. Try > including a correct vector calculation for the lightspeed in the lab > frame. Your above calculations would only work if the source in sagnac > were travelling in a straight line. But it isnt. No, the maths is correct if the _light_ travels in a straight line _after_ it has been emitted regardless of what happens to the source. That is of course what happens, unles you can explain how the source causes the distant light to move sideways. George
From: sean on 26 Sep 2007 09:38 On 25 Sep, 12:44, George Dishman <geo...(a)briar.demon.co.uk> wrote: > On 25 Sep, 11:26, sean <jaymose...(a)hotmail.com> wrote: > > > > > > > On 23 Sep, 20:06, Dono <sa...(a)comcast.net> wrote: > > > > On Sep 23, 8:16 am, sean <jaymose...(a)hotmail.com> wrote: > > > > > So far you have supplied no evidence to back up this claim. > > > > Wheras Ive supplied mathematical proof that emmision theory CAN > > > > explain sagnac. Simply .. a correct simulation > > > > of sagnac shows that light at c in the source frame travels > > > > farther on one path then the other. Which gives a path > > > > difference.. > > > > Ummm,no. It is really simple , Shawn : > > > The forward light beam "chases" the Sagnac disc at the "ballistic" > > > speed c+v, while the disc moves forward at speed v, so: > > > > tf*(c+v) =2*pi*r + tf*v > > > > meaning that : > > > > tf = 2*pi*r/c > > > > Likewise, the beam that moves in the opposite direction with c-v, > > > satisfies the motion equation: > > > > tb*(c-v) = 2*pi*r - tb*v > > > so: > > > tb = 2*pi*r/c > > > > delta_t = tf - tb = 0 > > > Your calculations are wrong simply because you have incorrectly > > assumed that in the lab frame light in emmision theory is either c+ or > > c-v.But it isnt. If light is at c in the source frame then in the *lab > > frame* for emmision theory the light travels away from the rotating > > source (in sagnac) at a variable speed , not a constant speed of c+ or > > c-v. > > Sean, you still haven't explained how the source > can influence the light after it has been switched > off. Suppose we pulse the source to give individual > flashes of light. For each one the source is off, > then on creating the flash, then off. The light > moves away in a straight line in the lab frame. > Switching the source on to create the next flash > cannot affect the previous flash that is already > some distance away. > > > Do your maths properly for a change. Try > > including a correct vector calculation for the lightspeed in the lab > > frame. Your above calculations would only work if the source in sagnac > > were travelling in a straight line. But it isnt. > > No, the maths is correct if the _light_ travels > in a straight line _after_ it has been emitted > regardless of what happens to the source. That > is of course what happens, unles you can explain > how the source causes the distant light to move > sideways. Ill reply to Bz relevent posts soon but a quick response here. I dont have to show how the light follows the source after its been emitted. In the source frame,..the source never moves. Therefore the emitted light doesnt have to `drag`. Its only if one looks at teh path from another frame does there apppear this `drag`. But those other frames never effect the light path in the source frame. In the same way that if I throw a ball to you and a 3rd person watches as he rotates in a circle. He sees the ball curve and change speed in his frame just as you see my light curve and wobble in your sirius frame. But does the ball I throw sudenly go straight in the 3rd observors frame just because he looked at it? No. In the sma eway as if you were on the moon and looked at a MMx experiment on earth. Youd see the light travelling to and from the mirrors in the earth MMx in curved lines.. Does the light suddenly go straight just because you looked at it from the moon ? No Sean
From: George Dishman on 26 Sep 2007 14:03
"sean" <jaymoseley(a)hotmail.com> wrote in message news:1190813901.668792.46230(a)w3g2000hsg.googlegroups.com... > On 25 Sep, 12:44, George Dishman <geo...(a)briar.demon.co.uk> wrote: ... >> Sean, you still haven't explained how the source >> can influence the light after it has been switched >> off. Suppose we pulse the source to give individual >> flashes of light. For each one the source is off, >> then on creating the flash, then off. The light >> moves away in a straight line in the lab frame. >> Switching the source on to create the next flash >> cannot affect the previous flash that is already >> some distance away. .... > I dont have to show how the light follows the source after its been > emitted. In the source frame,..the source never moves. Therefore the > emitted light doesnt have to `drag`. Its only if one looks at teh path > from another frame does there apppear this `drag`. But those other > frames never effect the light path in the source frame. In the same > way that if I throw a ball to you and a 3rd person watches as he > rotates in a circle. He sees the ball curve and change speed in his > frame just as you see my light curve and wobble in your sirius frame. > But does the ball I throw sudenly go straight in the 3rd observors > frame just because he looked at it? > No. In the sma eway as if you were on the moon and looked at a MMx > experiment on earth. Youd see the light travelling to and from the > mirrors in the earth MMx in curved lines.. Does the light suddenly go > straight just because you looked at it from the moon ? > No OK, maybe we can use the thrown ball to break through this. Suppose you decide you don't like me and throw a ball to hit me. To do that you face me. I am 50m away and you throw it at 10m/s. "S" is Sean, "G" is George and "B" is the ball: After 1s it looks like this: S B . . . G The dots are just to mark off units of 10m. After 2s: S . B . . G After 3s you change your mind and decide I'm OK after all so you turn 90 degrees to your left. You are saying the picture is now like this: . . B . . S . . . . G and after 4s like this . B . . . S . . . . G and after 5s, when the ball should have hit me, it is like this: B . . . . S . . . . G Do you see how silly that sounds? Now instead of a ball, imagine a flash of light leaving a laser. The flash lasts just 0.1ns so is just 30mm long. You fire the laser at me and after about 33ns it has moved 10m: S - . . . G After 67ns you are still facing me and it has moved this far: S . - . . G After 100ns you decide you wish you hadn't fired it at me so you turn through 90 degrees in a fraction of a nanosecond. Does the laser flash swing round like the ball above so it is now moving up the screen and I am safe like this: . . | . . S . . . . G Then this: . | . . . S . . . . G and so on, or is it going to carry on towards me and put a hole through me like this: S . . - . G My opinion is that it keeps going regardless of what you do and I should duck or else. George |