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Why is the ratio of velocity to the speed of light squared in the Lorentz factor? $${\left( {{v \over c}} \right)^2}$$ My only guess is the value must be positive.

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Because time is orthogonal to every x component. –  DWin May 29 at 23:49
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5 Answers 5

It's due to the Lorentz invariance of the spacetime interval: $$\Delta s^2 = (c\Delta t)^2 - (\Delta x)^2 = (c\Delta t')^2 - (\Delta x')^2$$

Assume that, for example, $\Delta x = 0$ such that $\Delta t$ is the elapsed time according to a clock stationary in the unprimed frame of reference.

Thus

$$(c\Delta t)^2 = (c\Delta t')^2 - (\Delta x')^2$$

With a little algebra, we have

$$ \frac{(c\Delta t)^2}{(c\Delta t')^2} = 1 - \frac{(\Delta x')^2}{(c\Delta t')^2} = 1 - \frac{v^2}{c^2}$$

where

$$v = \frac{\Delta x'}{\Delta t'}$$

is the velocity of the clock in the primed frame of reference.

A little more algebra yields

$$\Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma \Delta t$$

which is the familiar time dilation formula

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"Assume that, for example, Δx=0." In that case, by definition, Δx'=0, which means $(cΔt)^2=(cΔt′)^2$. So, obviously, if there is no movement (Δx=0), there is also no time dilatation. –  bright magus May 29 at 21:34
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Same reason the speed of light is squared in the "Energy–momentum relation" equation: $$ E^2 = m_0^2 \cdot c^4 + p^2 \cdot c^2 $$ Same as $$ { E \over c } = \sqrt{ m_0^2 \cdot c^2 + p^2 } $$ It is a leg of a right triangle we cannot directly observe.

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Consider a particle of light moving along a sick of length $c\Delta t$ from the frame of reference of a stationary observer. i.e. The stick itself has velocity zero relative to the stationary observer.

Then consider another observer moving with a velocity perpendicular to the sick with speed $v$. The stationary observer sees the light move straight up. While the moving observer sees the light moving along a diagonal path.

enter image description here

Since Maxwell made it clear that the speed of light remains the same in all reverence frames we need to make a correction to the classical transformations.

The height of the sick also remains the same in both frames of references, because the non-stationary observer is moving with velocity perpendicular to the stick.

This means the distance along the stick in the stationary frame divided by the time in the stationary frame must equal to the same length of the stick in the moving frame considering the speed of light remains the same.

Using the right triangle shown we get:

$$ (c\Delta t_p)^2 = (v\Delta t_p)^2 + (c\Delta t)^2 $$

$$ \Delta t^2_p(c^2-v^2)=c^2\Delta t^2 $$

$$ \Delta t^2_p = \frac{c^2\Delta t^2}{c^2-v^2} = \frac{\Delta t^2}{1-\frac{v^2}{c^2}} $$

And hence the $\frac{v^2}{c^2}$ factor, Lorentz transformations, and the non-euclidean geometry of flat spacetime.

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It derives from the special relativistic version of the Pythagorean theorem.

The hypotenuse of a Euclidean triangle is given by

$$h^2 = a^2 + b^2$$

In Minkowski space (special relativity) you get a minus sign instead of a plus sign, but you still have to square everything:

$$\Delta s^2 = \Delta t^2 - \Delta x^2$$

(and then you work onwards as described in Alfred's answer.)

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I'll chime in with the hyperbolic geometry take:

The $\dfrac{v^2}{c^2}$ term in the Lorentz factor can be better understood if we look at the entire Lorentz factor:

$\gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Now we've got the term you asked about back into context. If we take a factor of $\frac{1}{c^2}$ out of the square root in the denominator we get

$\gamma = \dfrac{1}{\dfrac{1}{c}\sqrt{c^2-v^2}} = \dfrac{c}{\sqrt{c^2-v^2}}$

At this point, you might notice that we have something that looks an awful lot like a sine or a cosine definition. If only the two 'sides', v, and c in the square root were a sum instead of a difference, we'd be there. We're in luck though, because the definition of the hyperbolic cosine is exactly what we have: the adjacent side divided by the hyperbolic hypotenuse which is the square root of the difference, (not the sum) of the squares of the sides.

What's this tell us. Well first of all, $\gamma$ is known as the time dilation factor because it's the ratio of time experienced in the lab frame to time experienced by the particle, $\dfrac{dt}{d\tau}$. We can think of it as the speed of the moving particle through the time of the laboratory frame. If we write the sine function for the same triangle, we get

$\dfrac{v}{\sqrt{c^2 - v^2}}$, which is what we call the 'proper velocity', and denote by $\dfrac{dx}{d\tau}$. It indicates how quickly we move through the space of the laboratory frame with respect to the time of the moving frame.

So, to answer more concisely, the term you mentioned is squared because it's really the cloaked version of the square of a side of a velocity triangle in hyperbolic Minkowski space. Once we realize that, then we find out that the projection of our lab velocity $v$ onto the time velocity axis is our time dilation factor, and our projection onto the space velocity axis is our proper velocity. As a last note, proper velocity is often referred to as speedometer velocity, since it involves the moving frame measuring distance in the frame it started from, (think mile marker signs), and dividing by time in the moving frame, (the clock in the car). Proper velocity can exceed c. This is in with keeping wit special relativity, you just have to remember that your time's dilated, so you didn't exceed c in the laboratory frame.

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