Why is the ratio of velocity to the speed of light squared in the Lorentz factor?

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

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.

Answer 4

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.)

Answer 5

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.

Answer 6

"The value must be positive" is a nonsensical explanation, velocity is a vector, the speed is always positive.

These sorts of explanations (something is so because "so" satisfies such properties, even though we haven't proved that other stuff doesn't satisfy such properties) are universally bad, and so are the questions that lead to them. Did you actually derive one of the formulae that contains the Lorentz factor and get $1/\sqrt{1-v/c}$? Didn't think so.

You can define anything to mean anything, but the quantity $1/\sqrt{1-v/c}$ simply isn't particularly useful, except as the denominator to the eigenvalue of the Lorentz transformation. You can derive $\gamma=\frac1{\sqrt{1-v^2}}$ fairly easily (see e.g. Section 4 of my introduction to special relativity), but if you want to intuit it through, read on.

The Lorentz factor is the scale factor on the transformed axes. What does this mean? Consider a rotation. You can think of it as turning the axes around, or as pushes the x-axis in the positive y-direction and the y-axis in the negative x-direction by the same amount, then scaling each axis down. How much are you scaling the axes down? Well, by $\cos\theta$. Conversely, the scale put on the orthogonal axis that gets added to this axis is $\sin\theta$.

$$\begin{array}{l}x' = x\cos \theta - y\sin \theta \\y' = x\sin \theta + y\cos \theta \end{array}$$

A Lorentz transformation is actually quite similar -- instead of pushing the axes in the same direction, you push them in the opposite directions -- i.e. towards each other or away from each other. This kind of a transformation is known as a skew -- and although skews and rotations somewhat feel like the exact opposite of each other, they're actually pretty related/complementary, much like real and imaginary numbers are. Indeed there are plenty of analogs between the two -- while the Euclidean/Pythagorean norm $\Delta x^2 + \Delta y^2$ (and corresponding dot product) is preserved by rotations, you have the Minkowski norm $\Delta t^2 - \Delta x^2$ preserved under skews. Both link to different, complementary conic sections as their invariant shapes. Rotations preserve angular differences $\Delta\theta$, skews preserve differences in a quantity called "rapidity" or hyperbolic angle, $\Delta\xi$.

Most importantly in this context, however, the hyperbolic transformations can be written as:

$$\begin{array}{l}x' = x\cosh \xi + t\sinh \xi \\t' = x\sinh \xi + t\cosh \xi \end{array}$$

Then velocity $v=\tanh\xi$ (instead of any $\tan\theta$ nonsense), the scale on any axis is

$$\cosh\xi=\frac1{\mathrm{sech}\xi}=\frac1{\sqrt{1-\tanh^2\xi}}$$

Which gives us our desired result, so that:

$$\begin{array}{l}\cosh \xi = \gamma = \frac{1}{{\sqrt {1 - {v^2}} }}\\\sinh \xi = v\gamma = \frac{v}{{\sqrt {1 - {v^2}} }}\end{array}$$

This is where the significance of $\gamma$ and $v\gamma$ comes from, and they also show up in dynamics, because energy and momentum can also be written as a time-space like pair. In none of this would you see $1\sqrt{1-v/c}$ show up as the scale on anything.