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.htmThe 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.htmthat

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.merriam-webster.com/dictionary/ruleTherefore, 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.htmL = 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.htmPreviously 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.htmSee 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.