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A New Theory of Motion and the Speed of Light


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Posted

To Studiot

" I am still waiting for a mathematical description of the light front surface for observer 2 "

 

I will explain the light wave front equation for moving Observer 2 below.

Let a light source and Observer 1 be at the origin O of coordinate system ( x, y, z) . Observer 2 is located at the origin O’ of coordinate system ( x’, y’, z’).  +x and +x’ are aligned and parallel.

We assume that Observer 1 is at absolute rest and Observer 2 is moving with (absolute) velocity V in the +x direction in the coordinate system (x, y, z).  Any velocity relative to the ( x, y, z ) frame is absolute velocity.

As the second system ( x’, y’ , z’ ) origin O' passes through and is coincident with the first system ( x, y, z ) origin O, it happens that observer 1 emits a pulse of light, at time t = 0.

The problem is to find the equation of the light front as seen by Observer 2.

Let us first consider the problem only in one dimension, (x, y, z ) = (x, 0, 0).

Observer 1 , who is at absolute rest, sees a light front travelling out from his source in the +x and –x directions and obeying the equation:

                                    x2 = c2 t2                   ( y = z= 0 )

To find the equation of the light front as seen by Observer 2 , we consider two observers co-moving with observer 2 , one at x’ = +D1 and the other at x’ = -D2 . That is, the observer at x’ = -D2 is behind Observer 2 ( and behind the source ) and the observer at x’ = D1 is in front of Observer 1, with respect to the direction of absolute motion.

We determine the time of detection of light by these observers, from which we get the light front equations as ‘seen’ by Observer 2.

For the observer at x’ = +D1 :

This observer has the same absolute velocity (V) as Observer 2 because both are co-moving. The light source is behind him, with respect to the direction of absolute velocity.

At the instant of light emission, the actual/physical distance of the source from this observer is D1.

Apparent Source Theory states that the effect of absolute motion of an inertial observer is to create an apparent change in point of light emission relative to that observer. The group and phase velocity of light relative to that point and relative to the observer is always equal to c. This means that once the apparent point of light emission relative to the observer is determined, the experiment is analyzed by assuming that the speed of light is constant relative to that point. Note that we are only considering light coming directly from source to observer.

Therefore, since this observer is in absolute motion and since the source is behind him, the apparent point of light emission will be at a distance of D1’ behind him where,

                           D1’ = D1 c / ( c – V )

Note that the point of light emission apparently shifts away from the observer because D1’ > D1 .

Although light was physically emitted from point x’ = 0 , for this observer light behaves as if it was emitted from  x’ = - Δ = D1 – D1’

It can be shown that:

                            Δ  =   D1 V / ( c –V )

The time elapsed for the observer to detect the light pulse is:

                             t  = D1’ / c  =  D1 / ( c – V )

To get the time elapsed, we divided the apparent distance by the speed of light c. We are applying the AST postulate that the velocity of light relative to the observer is always constant c , irrespective of absolute or relative motion of the observer. Note that, crucially, observer in AST is the detector of the light. The observer/detector is the person, device, particle or atom directly detecting the light. This is unlike SRT in which the ‘observer’ is not necessarily the detector of light. In SRT the ‘observer ‘ is an inertial reference frame. This distinction is crucial.

Therefore, the equation of the light front as ‘seen’ by Observer 1  ,for any point x’ on the positive x’ axis , is obtained as follows: 

                            t = D1 / ( c – V )     t = x ’ / ( c –V )     x ’ = ( c –V ) t

Note that  I have quoted ‘seen’ above ; I will explain the reason at the end.

For the observer at x’ = -D2

This observer also has the same absolute velocity V as Observer 2 . For this observer the light source is in front of him. Even though light was emitted from physical distance D2 in front of him  ( from x’ = 0 ), light behaves as if it was emitted from distance D2’ in front of him, for this observer, where

                                    D2 ‘  = D2 c / ( c + V )

In this case the point of light emission apparently shifts towards the observer, since D2’ <  D2 .

Although light was emitted from x’ = 0 , it appears for this observer that light was emitted from x’ = - Δ , where

                           Δ = D2 – D2’ = D2  V / ( c + V )

Therefore, the time taken by light to be detected by this observer is:

                              t  = D2’ / c = D2 / ( c + V )

From the above equation, the equation of the light front as ‘seen’ by Observer 2 , for any point x’ on the negative x’ axis , will be obtained as follows:

t  = D2 / ( c + V )       t  = - x’ / ( c + V )       x’ = - ( c+V ) t

Summary :

Therefore, the equation of the light front as seen by Observer 2 will be:

    x ’  =  ( c –V ) t         (  for x’ > 0 )

     x’  =  - ( c + V ) t         ( for x’ < 0 )

This means that, in the absolutely moving (x’, y’ , z’ ) frame,  an observer behind the light source will detect the light before an observer in front of the source , even if their physical distances from the light source are equal . As ‘seen’ by Observer 2 , light moves faster in the backward direction relative to the source than in the forward direction, with respect to the direction of absolute motion. 

Let us consider the problem again in one dimension, this time for ( x, y, z ) = ( 0, y, 0 ) , i.e.  x = z = 0.

Let an observer be at y’ = +D .This means that the line connecting this observer with the source is orthogonal to the direction of absolute velocity.  This observer also has the same absolute velocity ( V ) as Observer  2 .

Again, although the physical distance of the source from this observer is equal to D , light behaves as if it is at a distance of D‘ relative to this observer, where :

                                                    D‘ = D c / sqrt ( c2 – V2)

In this case, the point of light emission will apparently shift backwards in the –x’ direction.  The line connecting the observer and the apparent point of light emission is no more orthogonal to the direction of absolute velocity. That is, although light was physically emitted from point ( x’,y’ ) = ( 0,0), the apparent point of light emission for this observer will be (x’, y’) = (-Δ , 0 ) , where

                           Δ = sqrt ( D’2 – D2 ) = D V/ sqrt ( c2 – V2)

 The   time taken by light to be detected by the observer will be:

                             t  = D’/c  =  D / sqrt ( c2 – V2)

From the above equation, the equation of the light front as ‘seen’ by Observer 2  is obtained as follows:

                            t  = D / sqrt ( c2 – V2)      t  = y’ / sqrt ( c2 – V2)     y’ = (  sqrt ( c2 – V2) )  t

 Due to symmetry, the result is the same for an observer at y‘ = -D. The light front equation for an observer on the negative y’ axis will be:

                                y’ = - (  sqrt ( c2 – V2) )  t

According to AST the light wave front is no more spherical as ‘seen’ by absolutely moving Observer 2. Spherical wave front can be ‘seen’ only by an observer at absolute rest.

From Studiot’s  question, I think SRT requires that the wave front ‘seen’ by Observer 2 is also spherical , as for Observer 1 , because SRT postulates that the velocity of light in all inertial frames is constant c.

We have considered one dimensional problem only, to explain the distinction between AST and SRT.      The general equation for the light wave front in two or three dimensions is a bit involved but can be derived by using the AST procedure.

 Crucial difference between AST and SRT.

According to SRT, the velocity of light is the same constant c in all inertial reference frames.

According to AST, the (apparent) velocity of light is constant relative to all inertial observers.

In AST , unlike SRT , an observer is the person, device, particle, atom, e.t.c  directly detecting the light ! 

In SRT , the observer is the reference frame. See on Wikipedia , ' Observer ( special relativity) '

"  In special relativity, an observer is a frame of reference from which a set of objects or events are being measured. Usually this is an inertial reference frame or "inertial observer".  "

In AST , the velocity ( both phase and group ) of light is always constant relative to all inertial observers.

According to AST , the velocity of light in an absolutely moving reference frame is not constant ! But the (apparent) velocity of light relative to all inertial observers ( light detectors) is always constant , whether they are in absolute or relative motion or not.

Therefore, since the reference frame ( x’, y’ , z’ ) above is in absolute motion, only light coming to Observer 2 (to the origin O’ ) has constant (apparent) velocity. 

The reference frame concept is deeply flawed and is the source of all confusions regarding the problem of motion and the speed of light in physics during the past century. It is the reference frame approach that predicted a large fringe shift for the Michelson-Morley experiment, leading to length contraction and time dilation theory. Reference frames can be used accurately enough as an approximation. For example, they are used in classical physics, such as in Newton’s laws of motion and gravitation. The reference frame paradigm ( including the absolute reference frame) is fundamentally wrong and cannot be used to formulate the most fundamental laws of nature, such as in light and electromagnetism and gravity. Light, electromagnetism and gravitation phenomena are so subtle that their real nature simply eludes the ‘third’ observer, that is the reference frame. 

The reference frame concept should be replaced by Apparent Source Theory.

Please read my paper on Vixra:

The irrelevance of abstract reference frames in physics

Posted
52 minutes ago, lidal said:

I will explain the light wave front equation for moving Observer 2 below.

 

Thank you for a response deserving serious consideration.

It will certainly receive that.

Meanwhile a couple of quick first impressions.

54 minutes ago, lidal said:

Therefore, the equation of the light front as seen by Observer 2 will be:

 

    x ’  =  ( c –V ) t         (  for x’ > 0 )

 

     x’  =  - ( c + V ) t         ( for x’ < 0 )

This implies a discontinuity at x' = 0.

unless we add these equations (equivalent to taking the average)

2x' = 2ct

x' = ct

This is, of course identical to the equation for observer1 viz  x = ct

58 minutes ago, lidal said:

Apparent Source Theory states that the effect of absolute motion of an inertial observer is to create an apparent change in point of light emission relative to that observer. The group and phase velocity of light relative to that point and relative to the observer is always equal to c. This means that once the apparent point of light emission relative to the observer is determined, the experiment is analyzed by assuming that the speed of light is constant relative to that point. Note that we are only considering light coming directly from source to observer

You seem to have finally correctly identified the necessary second postulate of SRT.

If this is your replacement postulate it doe not provide the means to numerically evaluate the apparent change of position.

Where do your formulae come from?

Posted
1 hour ago, lidal said:

According to AST, the (apparent) velocity of light is constant relative to all inertial observers.

 

...

 

In AST , the velocity ( both phase and group ) of light is always constant relative to all inertial observers.

So the apparent velocity and absolute velocity are invariant.

 

1 hour ago, lidal said:

According to AST , the velocity of light in an absolutely moving reference frame is not constant ! But the (apparent) velocity of light relative to all inertial observers ( light detectors) is always constant , whether they are in absolute or relative motion or not.

...except that it's not.

 

Which is it?

 

1 hour ago, lidal said:

 Please read my paper on Vixra:

Please stop doing this. The rules state that the discussion takes place here, and that you must provide enough information to have the discussion without people clicking on any links.

Posted
4 hours ago, lidal said:

According to AST , the velocity of light in an absolutely moving reference frame is not constant ! But the (apparent) velocity of light relative to all inertial observers ( light detectors) is always constant , whether they are in absolute or relative motion or not.

This sounds a lot like you are saying there is an absolute frame and that the speed of light is not constant in that frame. But it always appears constant. Does this mean that, in your model, it is actually impossible to detect this absolute frame reference?

(And, again, I assume when you say "constant" you actually mean invariant. These are different concepts. It worries me that you don't know the difference.)

Posted (edited)

[math]\acute{D1}=\frac{D1 c}{(c-V)}[/math] do you think this statement is correct ?

14 hours ago, lidal said:

 

 

Apparent Source Theory states that the effect of absolute motion of an inertial observer is to create an apparent change in point of light emission relative to that observer. The group and phase velocity of light relative to that point and relative to the observer is always equal to c. This means that once the apparent point of light emission relative to the observer is determined, the experiment is analyzed by assuming that the speed of light is constant relative to that point. Note that we are only considering light coming directly from source to observer.

 

Therefore, since this observer is in absolute motion and since the source is behind him, the apparent point of light emission will be at a distance of D1’ behind him where,

 

                           D1’ = D1 c / ( c – V )

Note that the point of light emission apparently shifts away from the observer because D1’ > D1 .

 

Sorry but this just doesn't make any sense. Why didn't you just use Galilean relativity  to start with ?

[math]\acute{x}=x-vt [/math]

[math]\acute{t}=t [/math]

[math]\acute{y}=y [/math]

[math]\acute{z}=z [/math]

 

Edited by Mordred
Posted
20 hours ago, lidal said:

Therefore, the equation of the light front as seen by Observer 2 will be:

 

    x ’  =  ( c –V ) t         (  for x’ > 0 )

 

     x’  =  - ( c + V ) t         ( for x’ < 0 )

these equations are very concerning since they imply that space is neither isotropic nor homogeneous.

The situation is even worse when we extend it to 3D since in your model when we are aligning the x and x' axes and considering V directed along these, we have

y= ct

z=ct

 

  • 1 year later...
Posted

 

I have prepared a simulation of Apparent Source Theory (AST) and posted it on Youtube. Kindly take your time to see the simulation.

Comments are welcome. I would like to thank you all for the comments given so far.

                                    

Link deleted

 

Posted
3 hours ago, lidal said:

 

I have prepared a simulation of Apparent Source Theory (AST) and posted it on Youtube. Kindly take your time to see the simulation.

Comments are welcome. 

!

Moderator Note

My comment: rule 2.7 says “Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos. Videos and pictures should be accompanied by enough text to set the tone for the discussion, and should not be posted alone.”

So just asking us to watch the video does not comply with the rules.

 
Posted (edited)

 

Apparent Source Theory (AST) is a new theory proposed as an alternative explanation to the Michelson-Morley experiment (MMX). I will formulate AST (w.r.t  MMX ) once again as follows:

The effect of absolute motion of the MM interferometer is to create an APPARENT change in position of the light source as seen by the detector.

 

AST is a very elusive theory, so I will use an analogy and a simulation to explain it further. ( I hope I have complied with the rules this time)

The simplest way to explain AST is as follows: 

Suppose that, instead of setting the MMX apparatus in absolute motion, the position of the light source is slightly changed about its original position, say, by 1mm backward, forward, upwards, downwards, etc.  What will be the effect of this ACTUAL/PHYSICAL change in source position on the position of the interference fringes?

Intuitively, we can see that changing the position of the light source along the longitudinal axis (forward and backward) will have no effect on the fringe positions because both light beams are affected (advanced and delayed, respectively) identically. In the case of upward or downward change of source position (i.e. along the transverse direction) , from geometrical optics, we can intuitively see that a small fringe shift will occur. This is because the two light beams will be affected slightly differently. ( see the video simulation in the link at the end of this post )

Apparent Source Theory (AST) states that the APPARENT change of source position ( caused by absolute motion) has the same effect as a corresponding ACTUAL/PHYSICAL change of source position.

This theory explains and predicts not only the null fringe shift of the MM experiments but also the small fringe shifts observed in the Miller experiments. AST turns out to be a fusion of ether theory and emission theory.

Let us use an analogy.

Two persons S and O are standing on a moving cart. We will consider three cases:

CASE 1. O is in front of S 

CASE 2. S is in front of O 

CASE 3. The line connecting O and S is perpendicular to the direction of motion of the cart.

 

Person S acts as a 'source' throwing balls towards person O who acts as an 'observer'. First consider CASE 1. Assume that S always throws balls with constant velocity c relative to himself/herself. Two synchronized clocks, one at S another at O, are used to measure the time delay of a ball going from S to O. Now we want the ball to behave both according to emission theory and according to ether theory, at the same time.

 

At first assume that the cart is at rest. Let the distance between S and the O be D. When the cart is at rest, the time taken by the ball to move from S to O is, t = D/c. Now suppose that the cart starts moving forward with velocity V. Since, according to emission theory, the velocity of the ball relative to the source S is always constant c , then the time t will still be equal to D/c, regardless of the velocity of the cart. But we want the time delay t to change due to the motion of the cart, to make the ball appear to behave according to ether theory also. How can this be done? To make the ball behave according to ether ( absolute motion) theory, the ball should take more time to catch up with observer O when the cart is in motion (because O and the ball are moving in the same direction). Since the source S always throws the ball with constant velocity c relative to himself (relative to the cart), the only way to make the time t longer is for the source S to move back away from observer O, to a point a distance D’ away.

 

 

In this case, the time taken by the ball will be:

 

                                                       t = D' / c

Therefore, the velocity of the ball relative to the (apparent) source S' is still equal to c , but the point of ball ‘emission’ has changed from S to S’. Thus, the effect of ‘absolute’ motion of the cart is to change the point of ‘emission’ of the ball. The velocity of the ball relative to the observer O is always constant c, regardless of the ‘absolute’ velocity of the cart.

 

 

Now, for the ball to exactly simulate its ‘wave’ nature, i.e. to behave according to ether theory, the time delay should be as predicted by ether theory. According to ether theory, the time delay is equal to the actual distance D divided by the velocity of the ‘ball wave’ relative to the observer O, which is equal to c – V in this case. Therefore:

 

                                                      t =  D / ( c - V)

From the above two equations,

                                                 D' /c  =  D / ( c - V)

From which,

 

                                               D'  =  D c / ( c - V )

 

Note that the velocity of the ball as ‘seen’ by an ‘observer’ at rest on the ground is equal to  c + V .

 

 

Next we consider CASE 2 . For this we assume the same arrangement as above except that the cart moves backwards.

In this case, motion of the cart will make the time delay t shorter because O is moving towards the ball. By the same argument as above, S needs to change his/her position to a distance D’, where:

                                                D'  =  D c / ( c + V )

The profound result we found is that the speed of the ball is always constant relative to the observer O, regardless of the velocity of the cart. Light behaves in the same way as the ball in the above analogy.

 

Now consider CASE 3. With the same arguments as above, it can be shown that:

                                               D'  =  D c / ( c2 - V2 ) 1/2

This theoretical model reveals the mystery of the speed of light and why the Michelson-Morley experiment gave a null result and failed to detect absolute motion. One can imagine doing a ‘Michelson-Morley’ experiment by using the above ball analogy and can see why it gives ‘null’ results.

 

 I have made a simulation of how (changes) in absolute velocity of the MM apparatus changes the apparent position of the light source and hence affect the light paths. The simulation has been uploaded on Youtube.

 “ A New Theory of Motion and the Speed of Light – Special and General Relativity Disproved”

 

 

              https://youtu.be/W0r-UHAk_us        

 

Edited by lidal

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