[butchmurray (OP)]

The interim replacement for “The Time Dilation Factor – gamma Oversight (Draft)“ is reply number 27 of this thread. This version is only here for the record.

The Time Dilation Factor – gamma Oversight (Draft)
Thorntone Errick Murray   February 24, 2014

ABSTRACT: The origin and foundation of the time dilation factor, gamma, is a particular equation based on a particular right triangle and the Pythagorean theorem. The hypotenuse of the right triangle is in the inertial frame in which the observer is at rest. For that reason its length is constant. The vertical leg is perpendicular to the direction of motion. For that reason its length is constant. When the relative speed (speed v) is zero the vertical leg is in the frame in which the observer is at rest. In that frame the length of the hypotenuse and the length of the vertical leg are equal. As both lengths are constant, they are equal for any relative speed less than C. The hypotenuse is NOT the longest side of the right triangle. It is a rule of physics that the hypotenuse is the longest side of a right triangle. Therefore, this right triangle is invalid. In the equation the terms Ct, Ct’, and vt are the three sides of the right triangle, the hypotenuse, vertical leg and horizontal leg, respectively. Previously stated, the lengths of the vertical leg and the hypotenuse are constant. The length of the horizontal leg varies with speed v. The length of the horizontal leg is NOT constant. It is a rule of physics that in an equation of only three terms when two of the terms are constant the third term is also constant. Therefore, the equation is invalid.

1. INTRODUCTION: Is gamma, the time dilation factor, valid?

2. THE EQUATION: The equation (vt)²+(Ct’)²=(Ct)² is based on the right triangle in which Ct’ is the vertical leg, Ct is the hypotenuse and vt is the horizontal leg. Solved for t, t=t’*(1/sqrt(1-(v²/C²))) or t=t’*gamma, it is the origin of gamma, the time dilation factor. The length of the vertical leg is the length of Ct’. Since the vertical leg is perpendicular to the direction of motion the length of Ct’ is constant. The length of the hypotenuse is the length of Ct. Since the hypotenuse is in the frame in which the observer is at rest the length of Ct is constant. The horizontal leg is in the direction of motion. The length of the horizontal leg is the length of vt, which varies with speed v. The length of vt is not constant. It is a rule of physics that in an equation of only three terms when two of the terms are constant (Ct’ and Ct) the third term is also constant. Therefore, the equation is invalid.

3. THE RIGHT TRIANGLE: In the frame in which the observer is at rest t’=t. Then, in the frame in which the observer is at rest the length of Ct’ is equal to the length of Ct. The length of the vertical leg is the length of Ct’. When speed v=0 the vertical leg is in the frame in which the observer is at rest and the length of the vertical leg, the length of Ct’, is equal to the length of Ct. Since the vertical leg is perpendicular to the direction of motion its length is constant. Then, the length of the vertical leg is equal to the length of Ct, which is constant. The length of the hypotenuse is also the length of Ct, which is also constant. The hypotenuse is not the longest side of the right triangle. It is a rule of physics that the hypotenuse is the longest side of a right triangle. Therefore, the right triangle is invalid.

4. RESULTS: The right triangle and the equation, which are the basis of gamma, are proved invalid. Therefore, the time dilation factor and gamma as derived are invalid.

5. REFERENCES: All facts, stated or inferred, are well understood by those in the arena.

Butchmurray – Thorntone Errick Murray

[jeffreyH]

http://en.wikipedia.org/wiki/Velocity-addition_formula

The form of the relativistic composition law can be understood as an effect of the failure of simultaneity at a distance. For the parallel component, the time dilation decreases the speed, the length contraction increases it, and the two effects cancel out. The failure of simultaneity means that the fly is changing simultaneity slices as the projection of u onto v. Since this effect is entirely due to the time slicing, the same factor multiplies the perpendicular component, but for the perpendicular component there is no length contraction, so the time dilation multiplies by a factor of 1/V0 = √(1 − v2).

[butchmurray (OP)]

The reference cited explains addition of relativistic velocities. That is not questioned.

Of concern is the validity of the right triangle and the equation from which the time dilation factor, gamma, is produced.

The Lorentz transformation stipulates y’=y. That is, a given length that is perpendicular to the direction of motion, the length of y’, is equal to its “at rest” length, the length of y. Since y is in the frame in which the observer is at rest, the length of y is the same for all relative speeds. Again, y’=y. Therefore, the length of y’ is the same for all relative speeds, its “at rest” length, the length of y.

The right triangle is one in which Ct’ is the vertical leg, Ct is the hypotenuse and vt is the horizontal leg. Since Ct’ is perpendicular to the direction of motion all that apply to y’ also apply to Ct’. Then, the length of Ct’ is equal to its “at rest” length, the length of Ct, which is the same at all relative speeds. Ct’=Ct.

The length of the hypotenuse is Ct. The length of the vertical leg is Ct or Ct’ (Ct=Ct’). The lengths are the same. That is in violation of rules of physics that define right triangles. Consequently, the right triangle that gamma, the time dilation factor, is based on is invalid.

Further, the equation based on that right triangle, (vt)²+(Ct’)²=(Ct)², from which gamma, the time dilation factor is derived is also invalid. When Ct’ is replaced in the equation with its equal, Ct, the equation is (vt)²+(Ct)²=(Ct)². The equation is unbalanced and, as such, invalid.

The right triangle and the equation, which are the basis of gamma, are proved invalid. Therefore, the time dilation factor and gamma as derived are invalid.

butchmurray

[butchmurray (OP)]

Given for this purpose:
     Inertial frame K’ is in motion relative to inertial frame K at speed v.
     Frame K is the frame in which the observer is at rest.
     Lengths in frame K are proper lengths.
     Lx and Ly are lengths in frame K.
     Lx’ is a length parallel to the direction of motion in frame K’.
     Ly’ is a length perpendicular to the direction of motion in frame K’.
     Time in frame K is t.
     Time in frame K’ is t’.
     Lx/t=C and Ly/t=C
     Constant – in this context: The same at any relative speed of two inertial frames.

Time Dilation Directly Conflicts With Constant Length Perpendicular to the Direction of Motion

Thorntone Murray
March 16, 2014

Lx’ is a length parallel to the direction of motion in inertial frame K’. Seen by the observer at rest in inertial frame K the length of Lx’ is its proper length Lx factored by 1/gamma. Lx’=Lx*1/gamma. Time in frame K’ is t’. Relative to time t in frame K time in K’ is factored by 1/gamma. t’=t*1/gamma. It is given that in frame K light travels the length of Lx in the time t. C=Lx/t. Relative to frame K, in frame K’:
      Lx’/t’=Lx*1/gamma/t*1/gamma=Lx/t=C.
Light travels the length of Lx’ in the time t’. The speed of light parallel to the direction of motion in frame K’ is the same as the speed of light in frame K.

Ly’ is a length perpendicular to the direction of motion in inertial frame K’. It is a stipulation of the Lorentz transformation that y’=y. Then, seen by the observer at rest in inertial frame K the length of Ly’ is its proper length, Ly. Time in frame K’ is t’. Relative to time t in frame K time in K’ is factored by 1/gamma. t’=t*1/gamma. It is given that in frame K light travels the length of Ly in the time t. C=Ly/t. Relative to frame K, in frame K’:
       Ly’/t’= Ly/t*1/gamma - NOT equal to C.
Light does not travel the length of Ly’ in the time t’. The speed of light perpendicular to the direction of motion in frame K’ is NOT the same as the speed of light in frame K.

The speed of light parallel to the direction of motion in frame K’ is the same as the speed of light in frame K. The speed of light perpendicular to the direction of motion in frame K’ is NOT the same as the speed of light in frame K. Time dilation directly conflicts with constant length perpendicular to the direction of motion.

It is astonishing that until now the physics community was unaware of what I have discovered.

Thorntone Errick Murray
butchmurray

[jeffreyH]

What you are forgetting is it is easier to travel perpendicular to a gravitational field than away from it. Therefore light will find it easier to travel perpendicular to a direction of motion. You are also forgetting the warping of space with is a smooth gradient. All squares would become rectangles but still measured as squares in the moving frame as the observer is distorted in the same manner. Therefore the photon is also distorted.

[butchmurray (OP)]

Gravity is not a consideration in special relativity.

Thank you for your interest though.

Butch

[jeffreyH]

You can't ignore gravity. For the purposes of your theory it is pervasive. Just saying that special relativity doesn't include it is not sufficient to exclude it. After all in theory it is the cause of the very length contraction and time dilation you are discussing. We are all affected by it. Your head experiences an infinitesimal difference in length contraction to your big toe. Just because your head is further from the centre of gravity of the earth.

[David Cooper]

Butch is right on this point - gravity has no role in it. This confusion comes out of Einstein's claim that acceleration and gravity are equivalent, but they are radically different. If a clock sits in a gravitational field where it feels the same "acceleration" force on it all the time, it will tick at a constant rate, whereas if a clock is actually being accelerated it will tick at an ever-changing rate.

[jeffreyH]

Butch is right on this point - gravity has no role in it. This confusion comes out of Einstein's claim that acceleration and gravity are equivalent, but they are radically different. If a clock sits in a gravitational field where it feels the same "acceleration" force on it all the time, it will tick at a constant rate, whereas if a clock is actually being accelerated it will tick at an ever-changing rate.

That all depends upon the relationship between gravity and momentum. As that is an unanswered question it cannot be determined either way. Unless you can state categorically that a gravitational field is unaffected by acceleration then you cannot make that claim. The mechanism of gravitational feedback has to be affected in some way as you are moving a mass through space and the mass is the source of gravitation. This is a complex issue that cannot have a simple answer. This is why the equations of GR are so difficult to solve. In my view that makes them too complex but that is another matter.

[butchmurray (OP)]

David,
Thanks, Butch

JefferyH,

That all depends upon the relationship between gravity and momentum.

There is no mass, so there can be no momentum.

Unless you can state categorically that a gravitational field is unaffected by acceleration then you cannot make that claim.

All frames are inertial which means non-accelerating.

has to be affected in some way as you are moving a mass through space and the mass is the source of gravitation.

Again, there is no mass.

Although your position is nullified by the fact there is no mass, thus no momentum, it does suggest clarification of my finding may be in order.

Thank you,
Butch

[jeffreyH]

You don't have either time dilation or length contraction without a gravitational field PERIOD. That requires mass. You know that stuff that is all around you, pervading the universe.

[butchmurray (OP)]

In the discussion, there are two inertial frames which have no mass, light which has no mass, time which has no mass, speed which has no mass, length (distance) which has no mass and a light path which consists of two points (coordinates) which have no mass. Nothing being discussed here has mass.

Thank you,
Butch

[jeffreyH]

In the discussion, there are two inertial frames which have no mass, light which has no mass, time which has no mass, speed which has no mass, length (distance) which has no mass and a light path which consists of two points (coordinates) which have no mass. Nothing being discussed here has mass.

Thank you,
Butch

Well then you don't have time dilation which I thought was the point of the whole thread.

[alan hess]

0×0 always equals 0 change in the names, doesn't change the facts. But I have to take a slight disagreement. Mass has gravity, gravity has no mass.

[MathLessWorld]

Murray, it is difficult to follow your argument because they are all in words.
A math equation becomes much clearer if you show it also in  graph.

Make a JPG or PNG drawing using MS paint or something and attach it with your argument.

[butchmurray (OP)]

The formulation of gamma.

This formulation is from the critically acclaimed Cal Tech lecture series “The Mechanical Universe and Beyond” Lecture No. 42 “The Lorentz Transformation” starting about 12 minutes into the video. Dr. Feynman and Dr. Einstein held professorships at Cal Tech University. Here is a link to Lecture No. 42.http://www.learner.org/resources/series42.html#


One of the methods to formulate the Lorentz factor gamma uses a right triangle and the Pythagorean theorem.

Inertial frame K’ is in motion relative to inertial frame K at speed v. The observer is at rest in frame K.
   In the right triangle the vertical leg, Ct’, is the length of a light path perpendicular to the direction of motion in frame K’.
   The length of horizontal leg, vt, is the distance frame K’ advanced relative to frame K at speed v in the time t.
   The hypotenuse, Ct, is the length of the path of the light in light path Ct’ seen in frame K by the observer.

The seminal equation:
       (Ct’)²+(vt)²=(Ct)²
Subtract (vt)² from both sides
       (Ct’)² =(Ct)²-(vt)²
Remove parentheses
       C²*t’²= C²*t²-v²*t²
Simplify right side
       C²*t’²= t²*(C²- v²)
Divide both sides by C²
        C²/C²*t’²= t²*(C²/C²-v²/C²)
Simplify
        t’²= t²*(1-v²/C²)
Square root all
        t’= t*sqrt(1-v²/C²)
Divide both sides by sqrt(1-v²/C²)
        t’*1/sqrt(1-v²/C²)=t
Substitute gamma for 1/sqrt(1-v²/C²)
        t’*gamma=t
Divide both sides by gamma
        t’=t*1/gamma

THE OVERSIGHT:
The seminal equation:
       (Ct’)²+(vt)²=(Ct)²
At speed v=0, (vt)²=0  substitute 0 for (vt)²:
       (Ct’)²+0=(Ct)²
Simplify:
       (Ct’)²=(Ct)²
Square root all.
       Ct’=Ct  At speed v=0

    C is the constant speed of light.
    t is time in frame K. Time t in frame K is constant with respect to speed v.
    The length of Ct is then constant with respect to speed v.
    The length of Ct’ is perpendicular to the direction of motion. In accordance with the y’=y equation of the Galilean transformation and the Lorentz transformation its length is constant with respect to speed v.

Calculated above:
        Ct’=Ct at speed v=0.
The lengths of Ct’ and Ct are constant with respect to speed v for the reasons stated above. Then:
        Ct’=Ct at speed v>0.
Therefore:
        Ct’=Ct.

The seminal equation:
       (Ct’)²+(vt)²=(Ct)²
Substitute Ct’ with its equivalent, Ct, in the equation:
       (Ct)²+(vt)²=(Ct)²  The equation is invalid for any non-zero value of speed v.

Therefore, the equation from which gamma is derived is invalid.

Thorntone E Murray, Houston, Texas USA  April 22, 2014

[butchmurray (OP)]

 
An exact copy of the quantitative description you sent to me in or about August last year, 2013, follows a brief analysis of it.
 
AN ANALYSIS OF THE QUANTITATIVE DESCRIPTION:
1. The final equation in the description, square root[(Cy)² + (v)²] t = Ct, can be stated as square root[(Cyt)² + (vt)²]= Ct.
 
2. In the quantitative description, stated or implicit, the lengths of d’, d and Cyt are equal.
       d’=d
       d=Cyt
 
3. For the relative speed v=0, square root[(Cyt)² + (vt)²]= Ct, can be stated as square root[(Cyt)² + 0]= Ct. Then:
       Cyt=Ct when v=0
 
As the lengths of d’, d and Cyt are equal:
       Cyt=Ct when v=0
       d=Ct when v=0
       d’=Ct when v=0
 
4. As d’ is perpendicular to the direction of motion in frame K’, the length of d’ is constant with respect to relative speed v. Then as d’=Ct when v=0, d’=Ct when v=0 and when v>0. Then:
       d’=Ct when v=0 and when v>0
 
As the lengths of d’, d and Cyt are equal:
       d’=Ct when v=0 and when v>0
       d=Ct when v=0 and when v>0
       Cyt=Ct when v=0 and when v>0
 
5. The final equation in the description is square root[(Cy)² + (v)²] t = Ct. It can be stated as square root[(Cyt)² + (vt)²]= Ct. When v=0 and when v>0, Cyt=Ct. With the final equation stated as:
       square root[(Cyt)² + (vt)²]= Ct
Substitute Cyt with its equivalent Ct
       square root[(Ct)² + (vt)²]= Ct
 
6. For any non-zero value of speed v, the equation square root[(Ct)² + (vt)²]= Ct is invalid. Therefore, the final equation in the quantitative description, square root[(Cy)² + (v)²] t = Ct is invalid.
 
7. Additionally, in the quantitative description it is stated “the time of travel is t = d/Cy”. With respect to relative speed v, t and d are constant. Then, with respect to relative speed v, Cy is constant. With respect to relative speed v, square root[C² -v²] is not constant. Therefore, the statement “Cy can be obtained from the relation [Cx]² + [Cy]² = [C]². This gives us Cy = square root[C² -v²]” is incorrect.
 
Is this analysis correct?
 
THE QUANTITATIVE DESCRIPTION YOU SENT TO ME:
Frames K and K' have their origins O and O' coincident at time t=t'=0. Let x the direction of motion of K' with respect to K being v the speed.

A ray of light leaves the origin and propagates in the vertical direction y with speed C, as seen by an observer at rest in the K' frame. We have C = d'/t', where d' is the distance travelled in the time t'. The distance d' could be represented by a vertical rod of length d' = O'A'.

In order to describe light propagation in frame K we need the components of the velocity C. Let us call them Cx (component in the x direction) and Cy (component in the vertical or y direction). Obviously, we have Cx = v. Cy can be obtained from the relation [Cx]² + [Cy]² = [C]². This gives us Cy = square root[C² -v²].

Since the vertical distance travelled is d = d', equal in both K and K', the time of travel is t = d/Cy, when light arrives at A=A' at the top of the rod d.
The length OA = square root[d² + (vt)²] = square root[(Cyt)² + (vt)²] = square root[(Cy)² + (v)²] t = C t, as expected.

You may wish to calculate the time and speed for the return trip. You will find again C = length AO/t.

End of quantitative description.

[jeffreyH]

What you are forgetting is the sides of your triangle needs to be 299792458 m long in order for it to make sense. In which case the length contraction would flatten the path to be vertical as light speed would have been reached. Your frame of reference really needs to be that big to make sense. In one second a photon has moved 299792458 m so any hypotenuse is going to be about vertical instantly. Any other way of looking at this is wrong.

This also means that length contraction is insignificant and can be ignored until you reach high relativistic speeds.

[butchmurray (OP)]

Are the equations
         t’= t*1/sqrt(1-v²/C²) also stated as t’=t*gamma
and
         t’= t*sqrt(1-v²/C²) also stated as t’=t*1/gamma
both valid?

For frame K’ in motion relative to frame K at a speed v greater than zero but less than C, the value of gamma, 1/sqrt(1-v²/C²) is greater than one and the value of 1/gamma, sqrt(1-v²/C²) is less than one.

The equation t’=t*gamma refers to the duration of a unit of time t’ (second, year etc.) in frame K’ relative to the duration of a like unit of time t in frame K. The duration of a unit of time in frame K’ is relatively dilated (greater) by the factor gamma. For example, for speed v equal to .866C the value of gamma is 2. One second in frame K’ is t’. One second in frame K is t. Then for:
          t’=t*gamma
Substitute 2 for gamma
          t’=t*2 or t’=2t
Substitute 1 second in K’ for t’ and 1 second in K for t
          1 second in K’=2*1 second in K
Simplify
          1 second in K’=2 seconds in K
Then, the duration of one second in frame K’ and the duration of 2 seconds in frame K are of the same duration relative to each other.

The equation t’=t*1/gamma refers to the rate time advances in frame K’ relative to the rate time advances in frame K. The rate time advances is relatively slower (less) in frame K’ by the factor 1/gamma. For example, for speed v equal to .866C the value of 1/gamma is .5. The rate time advances in frame K’ is t’. The rate time advances in frame K is t. Then for:
          t’=t*1/gamma
Substitute .5 for 1/gamma
          t’=t*.5 or t’=.5t
Substitute the rate time advances in K’ for t’ and the rate time advances in K for t
          the rate time advances in K’=.5*the rate time advances in K
Then, the rate time advances in K’ is slower relative to the rate time advances in K by the factor .5.

Then, are the equations
         t’= t*1/sqrt(1-v²/C²) also stated as t’=t*gamma
and
         t’= t*sqrt(1-v²/C²) also stated as t’=t*1/gamma
both valid?

YES!

June 20, 2014
Butch Murray

[PmbPhy]

The Time Dilation Factor – gamma Oversight (Draft)
Thorntone Errick Murray   February 24, 2014

ABSTRACT: The origin and foundation of the time dilation factor, gamma, is a particular equation based on a particular right triangle and the Pythagorean theorem.

The time dilation formula wasn’t derived like that by Einstein in his paper on relativity. He used the Lorentz transformation. Here’s one of the ways to use the Lorentz transformation to derive the time dilation expression: Let b = v/c, g = 1/sqrt(1 – b^2). Let there be a clock at rest in the inertial frame S’. Now consider two events A and B where event A is “clock ticks at time t’_a” and Event B is “clock ticks at t_b = t_a + dt’ where dt’ = t’_b – t’_a is called the proper time between these two events. Since the clock is at rest in S’ it follows that the location of the clock is the same at both events, i.e. x’ = x’_a = x’_b.

The Lorentz transformation for time is

t = g(t’ – vx’/c^2)

then

t_a = g(t’_a – vx/c^2)
t_b = g(t’_b – vx/c^2)

Subtracting the second from the first gives

t_b – t_a = g(t’_b – vx/c^2) - g(t’_a – vx/c^2) = g(t’_b – t’_a) = gdt’

The time increment in frame S between Event A and Event B is then found to be

t_b – t_a = dt = gdt’

Therefore

dt= gdt’

or

dt = dt/sqrt[1 – v^2/c^2]