Naked Science Forum
On the Lighter Side => New Theories => Topic started by: butchmurray on 24/02/2014 05:58:30

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

http://en.wikipedia.org/wiki/Velocityaddition_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).

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

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

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.

Gravity is not a consideration in special relativity.
Thank you for your interest though.
Butch

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.

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 everchanging rate.

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 everchanging 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.

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 nonaccelerating.
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

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.

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

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.

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.

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.

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# (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*t²v²*t²
Simplify right side
C²*t’²= t²*(C² v²)
Divide both sides by C²
C²/C²*t’²= t²*(C²/C²v²/C²)
Simplify
t’²= t²*(1v²/C²)
Square root all
t’= t*sqrt(1v²/C²)
Divide both sides by sqrt(1v²/C²)
t’*1/sqrt(1v²/C²)=t
Substitute gamma for 1/sqrt(1v²/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 nonzero value of speed v.
Therefore, the equation from which gamma is derived is invalid.
Thorntone E Murray, Houston, Texas USA April 22, 2014

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 nonzero 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.

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.

Are the equations
t’= t*1/sqrt(1v²/C²) also stated as t’=t*gamma
and
t’= t*sqrt(1v²/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(1v²/C²) is greater than one and the value of 1/gamma, sqrt(1v²/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(1v²/C²) also stated as t’=t*gamma
and
t’= t*sqrt(1v²/C²) also stated as t’=t*1/gamma
both valid?
YES!
June 20, 2014
Butch Murray

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]

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.
Hi Butch,
I read your opening post and found it to be in error. This post explains the nature all those errors and gives the corrections to them.
First off, 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’. Let there be a frame of reference S which is moving relative to S’ parallel the x’axis. Label the events in frame S’ as (t’, x’) and in S as (t, x). Now consider two events A and B where
Event A: The clock ticks at time t’_{a}.
Event B: The clock ticks again at t’_{b}. This is next tick of the clock after t’_{a}.
The coordinates of these events are defined to be
Event A: In S’ the coordinates are (t’_{a}, x’_{a}). In S the coordinates are (t_{a}, x_{a}).
Event A: In S’ the coordinates are (t’_{b}, x’_{b}). In S the coordinates are (t_{b}, x_{b}).
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
Time Dilation Expression: dt = dt’/sqrt[1 – v^2/c^2]
QED: This gives yet another proof of the time dilation expression.
Since you’re talking about a particular derivation of the time dilation expression it’s best to have the derivation handy for those who aren’t familiar with it and as a reference for the various quantities. The derivation at hand is on my website at
http://home.comcast.net/~peter.m.brown/sr/time_dilation.htm
The right triangle that Butch is referring to is in Figure 1.
The hypotenuse is NOT the longest side of the right triangle.
What? Since when? It’s not only a universally known fact by every mathematician and physicist alive but very easy to prove. So why would you claim differently? The relationship between
a = adjacent
b = opposite
c = hypotenuse
is given by the Pythagorean theorem. See the proof, which shows in Eq. (3) at
http://home.comcast.net/~peter.m.brown/math_phy/pythagorean_theorem.htm
that
c^2 = a^2 + b^2
which clearly shows that c is larger than either a or b. If you claim otherwise then either prove the formula is wrong by proving the derivation of the theorem I created is wrong or find an example for which it fails.
It is a rule of physics that the hypotenuse is the longest side of a right triangle.
That’s incorrect. First off it’s a theorem in mathematics, not a “rule” in physics (other than geometry being a branch of physics what I just said is true). Physics doesn’t use “rules”. Please look up the term rule in the dictionary at
http://www.merriamwebster.com/dictionary/rule
Therefore, this right triangle is invalid.
That’s incorrect. See the proof I gave about and which can be found in math texts.
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.
First off you haven’t defined your terms making this post difficult to understand. What do these terms mean? I.e. what does t represent? What does t’ represent? Typically the three sides of the triangle are as defined in my web page that proves that your derivation and conclusions are wrong, i.e. at http://home.comcast.net/~peter.m.brown/sr/time_dilation.htm
L = Vertical leg = adjacent of the right triangle = ct’/2
vt/2 = Horizontal leg = opposite of right triangle
ct/2 = hypotenuse of right triangle
L = ct’/2
The derivation of the Pythagorean theorem is also on my web site at
http://home.comcast.net/~peter.m.brown/math_phy/pythagorean_theorem.htm
Previously stated, the lengths of the vertical leg and the hypotenuse are constant.
By this I assume that you mean that the distance traveled by the pulse of light during the relevant time period for each run of the experiment/light pulse remains the same for each run.
The length of the horizontal leg varies with speed v. The length of the horizontal leg is NOT constant.
That is incorrect. Since the rod is moving with constant speed the time for each run, and hence the length of the horizontal leg, is constant in time. All you can say here is that the horizontal leg is a function of speed. However that’s also true of the hypotenuse. So your conclusions are wrong again.
Therefore, the equation is invalid.
That’s not true at all. First off, if a derivation is faulty the result may or may not be correct. Just because a mistake was made during a derivation it doesn’t mean that the result is wrong. That is a flaw in your logic. However it really doesn’t matter since the derivation is not wrong. The equation is correct too as I demonstrated using only the Lorentz transformation at the beginning of this post.
1. INTRODUCTION: Is gamma, the time dilation factor, valid?
Yes. I proved that in two different ways.
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.
That is incorrect. The origin of the gamma factor is the Lorentz transformation, not this derivation.
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.
That’s an error in logic. You’re confusing “constant in time” with “independent of speed.” The two are not the same. In fact the length of the hypotenuse is a function of speed as well. For proof of that see
http://home.comcast.net/~peter.m.brown/sr/time_dilation.htm
See Eq. (4). It gives t as a function of v. I.e. (note that what you call t’ I call tau)
t = t’/sqrt(1 – v^2/c^2)
In fact that’s what we’re trying to do in this derivation, t as a function of v. It’s easy if you think about it. Look at it like this; the larger the value of v the longer the distance that the hypotenuse is because as the mirror is moving because of the speed, the faster it moves the longer the hypotenuse. But the length the stage where you said it was “constant” didn’t have a explicit expression for it being a function of v. That came in the end The point you chose to stop before getting to.
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.
Sorry, Butch. But you made a lot of serious mistakes in this post. I corrected as many as I had the patience for.
5. REFERENCES: All facts, stated or inferred, are well understood by those in the arena.
Not quite. Almost everything about this post was wrong and you certainly can’t be said to understand anything about the math of physics here.
Sorry, but dems the facts.

Hi, Peter.
Rather than address the issues you raise individually, I will prove one of my points with the information you provide on your Time Dilation web page http://home.comcast.net/~peter.m.brown/sr/time_dilation.htm.
Let “T” represent proper time.
Let “L(o)” represent proper length.
L=L(o) [ref you] “…distance L remains unchanged due to the motion.”
L=ct; The BIMP length formula
Inertial frame S’ is in motion relative to inertial frame S at speed v.
In the light clock which is perpendicular to the direction of motion and stationary in frame S, a photon at speed c travels the length of the clock, L(o), in the time T/2.
L(o)=cT/2 or T/2=L(o)/c In reference frame S
As “distance L remains unchanged…” [ref PMB] the length, L(o), of the light clock which is perpendicular to the direction of motion in frame S and the length, L, which is perpendicular to the direction of motion in frame S’ are equal.
L(o)=L at speed v>0
At speed v=0 time T in the reference frame S and time t in frame S’ are of the same duration relative to each other.
T=t when speed v=0
At speed v=0 a photon at speed c travels the length of L, which is perpendicular to the direction of motion in frame S’ in the time t/2.
L=ct/2 or t/2=L/c at speed v=0
Speed v is of no consequence to L or c then,
L=ct/2 or t/2=L/c at speed v>0
Previously shown:
T/2=L(o)/c
Speed v is of no consequence to L(o) or c then,
T/2=L(o)/c at speed v>0
Substitute L(o) with its equivalent L
T/2=L/c at speed v>0
Previously shown:
t/2=L/c at speed v>0
As t/2=L/c and T/2=L/c when speed v>0, then:
t/2=T/2 at speed v>0
Simplify
T=t at speed v>0
If the length is not relatively contracted, the time is not relatively dilated.
The speed of light is the same in both frames. Relative to each other, the lengths that are perpendicular to the direction of motion are equal in both frames (neither is contracted relative to the other). Therefore, the times for light to travel the equal lengths are equal relative to each other. The speed of the frames relative to each other is of no consequence.
The time between ticks is not dilated in either frame relative to the other. Therefore, time dilation as specified by the Lorentz transformation is invalid.
.

butchmurray  It's common in this forum for people who create threads claiming that all the relativists on earth have been wrong for the last 100 years and that the one person posting the thread is the sole person to know what's going on as if the rest of us are some sort of dummies. I'll spend one post correcting the mistakes made by such people like I already explained your mistake to you. However nearly all the people who create threads in this forum don't accept the corrections made to their claims or they're unable to grasp it. I don't know which it is with you but I won't bother trying to keep correcting all of the mistakes you've already made and will keep making. I corrected them once already and I won't do it again. In any case the correction was not meant for you but for the people who might be mislead by your bogus derivation. Long story short  I won't dignify your new error with another correction. It's the experience of every physicist who has ever posted in a forum that once someone makes a mistake like this it's next to impossible for them to get them to either see or admit their mistake. I see no reason why I should expect anything different given the way you responded, i.e. with new errors.

The last reply to “The Time Dilation Factor Oversight” thread by PmbPhy was more than insulting. He clearly attacked and intentionally offended me. He made it obvious that he decided for every one else that I have zero credibility. If people are attacked for thinking outside the box in “New Theories” what is the point?
Thank you, Butch

If you keep on posting errorridden stuff that takes ages for anyone to work through, they will inevitably get fed up of it and your credibility will be shot. That's a shame if you then go on to get something right and to come up with something profound, but that's how the world works. You now need to stop relying on other people to put in an enormous amount of effort to follow your arguments and do your utmost to make it easy for them to find out what you're saying. Start out by spelling everything out in ordinary langauge with good diagrams and then go into your impenetrable jargon at the end. As it stands, what you're doing is like throwing faulty program code at people without putting any comments in it and expecting them to fix it for you. It's ruddy bad manners. If you wrote your arguments out clearly in normal language, you'd probably also find the mistakes all by yourself and not need to bother anyone else with them at all.

The last reply to “The Time Dilation Factor Oversight” thread by PmbPhy was more than insulting.
You're wrong. Please stop trying to put words into my mouth. I made it very clear that I have years of experience with people who claim that everyone in the physics community is wrong and that they're right. I've tried a great deal in the past to help these people see the errors that they're making and in all cases they were unable to grasp the explanation. It's easy for any physicist to understand it but impossible for such people to. Not everyone has what it takes to be a physicist. That's not an insult. It's just a fact of life.
He clearly attacked and intentionally offended me.
Clearly that's nonsense and a perfect example of why I don't talk to people like you. When your mistakes are pointed out to you and we show you that you're not the incredible genius that you thought you were you claim that we're insulting you. There's no way I'd continue to feed into this kind of nonsense and rude accusations.
He made it obvious that he decided for every one else that I have zero credibility.
Only in your own paranoid mind.
If people are attacked for thinking outside the box in “New Theories” what is the point?
That was hardly outside the box thinking. It was full of huge holes and had absolutely no validity to it. Please stop insulting the people who point out your mistakes or nobody will help you and we'll let you continue to think whatever you want to and live with the fantasy that you're right.
It's I who am insulted. I did you a major favor digging through your socalled "proof" and took a great deal of time reading it and paying close attention to it. That's how I found all the errors in it. Did you even thank me for taking the time to find the problems with it? Nope. All you did was to misread my last response about why I'll only point out mistakes in this subforum once because of my experience with other members of this and other forums. You didn't read it carefully that I was talking about other people and jumped to the false conclusion that I was talking about you. Then you insulted me for it. Shame on you. If you're going to act like that I'll be ignoring all of your posts from now on!
Pay close attention to what David Cooper said above. They are wise words which you should pay attention to!!

Hi David,
I appreciate your advice and encouragement.
Thank you,
Butch

Thorntone E. Murray
July 20, 2014
The Lorentz Transformation Contradiction
The Lorentz transformation equations:
1. x’=xvt/sqrt(1(v²/c²))
2. y’=y
3. z’=z
4. t’=t(vx/c²)/sqrt(1(v²/c²))
Let v represent the speed with which inertial frame K’ is in motion relative to inertial frame K.
In accordance with the Lorentz transformation, judged from frame K, a clock in frame K’ goes more slowly than a clock at rest in frame K. Also in accordance with the Lorentz transformation, a measuring rod perpendicular to the direction of motion in frame K’ and an identical measuring rod perpendicular to the direction of motion in frame K are of equal length relative to each other. As the speed of light is constant and the same for all observers, judged from frame K the time for light to travel the length of each rod is identical in K and K’. Therefore, judged from frame K, the clock in frame K’ goes at the same rate as a clock at rest in frame K, not more slowly.
Then, in accordance with the Lorentz transformation, judged from frame K, the clock in frame K’ goes more slowly than the clock at rest in frame K and, concurrently, that same clock in frame K’ goes at the same rate as the clock at rest in frame K. Following is mathematical confirmation of that contradiction.
Parallel to the Direction of Motion:
The following is a quote from Dr. Albert Einstein’s 1916 Relativity – The Special and General Theory, Section 12 – The Behavior of Measuring Rods and Clocks in Motion; translated by Robert W. Lawson.
“Let us now consider a secondsclock which is permanently situated at the origin (x’=0) of K’. t’=0 and t’=1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t=0
and
t=1/sqrt(1(v²/c²))
As judged from K, the clock is moving with the velocity v; as judged from this referencebody, the time which elapses between two strokes of the clock is not one second, but
1/sqrt(1(v²/c²))
seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest.” End quote.
Then, judged from K, one second in K’ equals 1/sqrt(1(v²/c²)) seconds in K. The operative equation is t’=t*1/sqrt(1(v²/c²)).
Perpendicular to the Direction of Motion:
Judged from K with speed v=0, in frame K light at speed c travels the length L of a measuring rod perpendicular to the direction of motion in the time t.
L=ct or t=L/c with speed v=0
Judged from K with speed v=0, light at speed c travels the length L’ of an identical measuring rod perpendicular to the direction of motion in frame K’ in the time t’.
L’=ct’ or t’=L’/c with speed v=0
Judged from K with speed v=0, length L’ of the measuring rod in frame K’ is equal to length L of the identical measuring rod in frame K.
L’=L with speed v=0
With speed v=0, frame K’ and frame K are at rest relative to each other. With speed v=0, time t’ in frame K’ is equal to time t in frame K.
t’=t with speed v=0
In accordance with the second (and third) equation(s) of the Lorentz transformation, with speed v=0 and speed v>0, the length of a measuring rod perpendicular to the direction of motion in frame K and the length of an identical measuring rod perpendicular to the direction of motion in frame K’ are of equal length relative to each other.
y’=y (z’=z) with speed v>0
Then, judged from K with speed v>0, length L’ of the measuring rod in frame K’ and length L of an identical measuring rod in frame K are equal.
L’=L with speed v>0
As the speed of light is constant and the same for all observers, then:
L’/c=L/c with speed v>0
Shown previously, t=L/c and t’=L’/c. As L/c=L’/c with speed v>0, then:
t’=t with speed v>0
Then, judged from K, one second in K’ equals one second in K. The operative equation is t’=t.
Shown previously:
Judged from K, one second in K’ equals 1/sqrt(1(v²/c²)) seconds in K. The operative equation is t’=t*1/sqrt(1(v²/c²)).
The Contradiction:
As a consequence of the Lorentz transformation, judged from frame K, t’=t*1/sqrt(1(v²/c²)) and t’=t concurrently.
Therefore, the Lorentz transformation is invalid.
Thorntone E. Murray