Is velocity an angle?

Question

I am not a physicist. I have only rudimentary notions about the following.

I looked for similar questions on SE but I did not find any. I also tried search engines but results relate to angular velocity which has nothing to do with my question.

As far as I know, the energy of a particle at rest $E = mc^2$. When the particle moves, there is an additional factor, i.e. Lorentz factor, $\gamma = 1 / \sqrt{ 1 - \frac{v^2}{c^2}}$, so the energy varies with the particle's velocity, $E = \gamma m c^2$.

I do not know Lorentz transformation well enough to understand how the Lorentz factor emerges from it.

What I know though, is that this Lorentz factor looks like the equation of a circle, and could be written in the form: $$\gamma = 1 / \cos( \phi ) = 1 / \sqrt{ 1 - \sin^2 \phi }$$ with $0 <= \phi <= \pi/2$. The relationship between $\frac{v^2}{c^2}$ and $\phi$ is relatively straightforward.

So, does that mean $\phi$ is a more fundamental dimension than velocity?

Edit: found this: http://reciprocalsystem.org/PDFa/RS2-108%20The%20Lorentz%20Factor%20(Peret,%20Bruce).pdf It says Lorentz factor is a unit circle.

Edit2: I found this picture

Answer 1

You are groping towards something interesting - representing the Lorentz factor as $\sec \phi$, where $\sin \phi = v/c$. Note that $\phi$ here is dimensionless and varies between 0 and $\pi/2$. In some senses this is more fundamental than $v$, since the absolute value of the speed light is just an artefact of the system of units we use and often (in theoretical work), one opts to let $c=1$ in any case and then $v = \sin \phi$. This then lets you represent the Lorentz transform of distance/time, which is written $$ \begin{bmatrix} t' \\ x' \end{bmatrix} = \begin{bmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{bmatrix} \begin{bmatrix} t \\ x \\ \end{bmatrix} $$ where $\beta = v$, as $$ \begin{bmatrix} t' \\ x' \end{bmatrix} = \begin{bmatrix} \sec \phi & -\tan \phi \\ -\tan \phi & \sec \phi \end{bmatrix} \begin{bmatrix} t \\ x \\ \end{bmatrix} $$ for which I don't see any obvious geometric interpretation.

You say that "the Lorentz factor looks like the equation of a circle". I don't think so. $$ \gamma^2\left( 1- v^2\right) =1$$ is not the equation of a circle; it is the equation of a hyperbola of the general form $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ and parametric form $x=a\cosh \phi$, $y=b\sinh \phi$, where here $\gamma = \cosh \phi$. This turns out to be a much neater representation, and with a more fundamental topological/geometric interpretation. If $\gamma = \cosh \phi$, then $\beta = \tanh \phi$, where $\phi$ is known as the rapidity. This then lets you write the Lorentz transformation as $$ \begin{bmatrix} t' \\ x' \end{bmatrix} = \begin{bmatrix} \cosh \phi & -\sinh \phi \\ -\sinh \phi & \cosh \phi \end{bmatrix} \begin{bmatrix} t \\ x \\ \end{bmatrix} $$ which is a hyperbolic rotation.

This definition has lots of useful products, including that adding velocities in relativity means that $$\tanh \phi_{\rm sum} = \tanh(\phi_1 + \phi_2)$$ $$ \phi_{\rm sum} = \phi_1 + \phi_2\ .$$ i.e. You can just add rapidities, just like you can add rotation angles to get the total rotation angle.

Other useful and elegant results are that the Doppler factor due to a rapidity $\phi$ is just $\exp (\phi)$ and that the proper acceleration is just $d\phi /d\tau$, where $\tau$ is the proper time.

Answer 2

So, does that mean $\phi$ is a more fundamental dimension than velocity?

I wouldn't say that $\phi$ is more fundamental than velocity, but it is certainly a useful way to represent the quantity of motion.

As I said in a comment, velocity is the spacetime slope of a worldline and at relativistic speeds it is better to work with the angle than the slope. However, there's a reason that we generally prefer to use the hyperbolic angle (which as Rob Jeffries mentions is termed rapidity) rather than your $\phi$.

The circular functions are fundamentally connected with the notion of distance in the Euclidean plane, (and by extension, to distance in Euclidean space of any number of dimensions). The equation of the circle comes from Pythagoras' theorem. The point $$(x=r\cos\phi,y=r\sin\phi)$$ is obviously at a distance $r$ from the origin. If we use a rotated coordinate system (with the same origin) we get coordinates

$$(x'=r\cos\phi',y'=r\sin\phi')$$ where $\phi'-\phi$ is the angle between the old axes and the new ones, but clearly the distance to the origin will remain $r$.

Now let's see how this connects to SR (Special Relativity).

Let's say that we are two inertial observers moving relative to one another. That is, we're not experiencing any acceleration, but you're moving with a speed of $v$ relative to my frame, and conversely I'm moving at $-v$ relative to your frame. We can each choose the direction of motion to be our X axis (and to keep things simple we can ignore the other 2 space directions).

Let A and B be two events (eg, two flashes of light). In my frame, the spatial distance between A & B is $\Delta x_0$, and the time interval between them is $\Delta t_0$. In your frame, you'll measure a spatial distance of $\Delta x_1$ between A & B, and a time interval of $\Delta t_1$. In traditional Galilean / Newtonian physics, we'd expect $\Delta t_0 = \Delta t_1$, but in relativity that's not the case (unless $v=0$).

I won't derive it here, but it can be shown that:

$$\begin{align}(\Delta s)^2&=(c\Delta t_0)^2-(\Delta x_0)^2\\&=(c\Delta t_1)^2-(\Delta x_1)^2\end{align}$$ Any other inertial observer who witnesses A & B and makes measurements $(\Delta t_2,\Delta x_2)$ will get the same value

$$(\Delta s)^2=(c\Delta t_2)^2-(\Delta x_2)^2$$

that is, $(\Delta s)^2$ is the same in all frames, so it's a fundamental measure of the spacetime geometry of A & B. We call it the spacetime interval between A & B. The formula for the spacetime interval is almost the standard Pythagorean formula for distance squared in Euclidean space, apart from that minus sign. We can eliminate that minus sign by using complex numbers:

$$\begin{align}(\Delta s)^2&=(c\Delta t_0)^2-(\Delta x_0)^2\\&=(c\Delta t_1)^2+(i\Delta x_1)^2\end{align}$$

With this setup, $\beta=\frac{v}{c}=\Delta x/\Delta t$ of a particle travelling (in uniform motion) from A to B is essentially the slope (tangent) of the worldine from A to B (apart from that factor of $i$). In Einstein's classic The Meaning of Relativity you'll find numerous mentions of these imaginary tangents.

That's ok in simple scenarios where we only need 1 space dimension (like the above scenario), but it gets messy when we need to work with all 3 space dimensions. (Also, it's nice to avoid complex numbers if we can). Fortunately, we can invoke the hyperbolic functions, which are analogous to the circular functions, except they have the minus sign that we need:

$$\begin{align} 1 & = \cos^2(\theta)+\sin^2(\theta)\\ 1 & = \cosh^2(\phi)-\sinh^2(\phi)\end{align}$$

And now we can use $\beta=\frac{v}{c}=tanh(\phi)$ which has useful mathematical properties. At low speeds, $\beta\approx\phi$, and we can combine speeds by simple addition. At relativistic speeds, just adding slopes is no longer an adequate approximation, we need to add the (hyperbolic) angles.

Let's say there's a body A moving at $\beta_A=\tanh(\phi_A)$ in the lab frame, and body B moving at $\beta_B=\tanh(\phi_B)$ in A's frame. Then the speed of B in the lab frame is

$$\tanh(\phi_A+\phi_B) = \frac{\beta_A+\beta_B}{1+\beta_A\beta_B}$$ that formula is exactly analogous to

$$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$$

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However, there's nothing wrong with using the circular functions to do simple relativistic calculations involving $\beta$ and $\gamma$. It's just the standard these days to use the hyperbolic functions.

Here's a cute way (using standard Pythagoras' theorem) to avoid square roots when working with $\beta$ and $\gamma$ for bodies at relativistic speeds. For all $k$,

$$(k^2+1)^2=(k^2-1)^2+(2k)^2$$

Let $$\beta=\frac{k^2-1}{k^2+1}$$ then $$\gamma=\frac{k^2+1}{2k}$$

For large $k, \gamma\approx k/2$. Eg, let $k=10$. Then

$$\beta=\frac{99}{101}$$ and $$\gamma=\frac{101}{20}=5\frac1{20}$$

To combine two speeds using this $k$ parameter, we multiply the parameters. Eg, if body A has

$$\beta_A=(a-1)/(a+1)$$ in the lab frame, and body B has $$\beta_B=(b-1)/(b+1)$$ in A's frame, then the $\beta$ of B in the lab frame is $$(ab-1)/(ab+1)$$

As robphy mentions in the comments, this $k$ is used in Bondi's $k$-calculus. $k$ turns out to be the radial Doppler factor, and it's related to the rapidity via

$$k=e^\phi$$

Note that the reciprocal of $k$ can be used define a negative velocity of equal magnitude but opposite sign to the velocity defined by $k$.

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FWIW, there's a closely related trick for accurately calculating $\gamma$ at low speeds, please see my answer here for details.

Answer 3

The following may be useful. If you consider the Lorentz transformation as a matrix operation, you obtain the following form (considering only time and one space dimension):

$$ \begin{bmatrix} ct' \\ x' \end{bmatrix} = \begin{bmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{bmatrix} \begin{bmatrix} ct \\ x \\ \end{bmatrix} $$

where $\beta=\frac{v}{c}$. If you plot up the transformation applied to a grid of $\left(ct,x\right)$ points, you obtain a remapping as shown below. Note however, that the diagonal lines which represent the constant velocity of light only compress or expand the points. This figure was calculated for a $\beta=0.3$.

I hope this helps.

Answer 4

I have a rather crude but geometric interpretation as to why the velocity and the angle seem to be connected:

In special relativity, we 'map out' all the paths and interactions that happen in the universe in a spacetime diagram. This is essentially a graph, but with time running vertically, and the space directions perpendicular to it.

Here the vertical axis represents time, and the horizontal one is one dimension of space. The dashed diagonal you see is the path traced out by a ray of light. This is called its worldline. You can see that it makes an angle of 45 degrees (which I will now refer to as $\pi / 4$). Now, assuming that $c = 1$, this can interpreted by saying that light travels 1 unit distance in 1 unit time (in natural units)

The interesting point is that angle. Light makes an angle of $\pi / 4$ on the space time diagram. Any particle moving slower than $c$ will cover lesser units of distance in more units of time, thus giving a steeper slope.

Here the blue particles goes slower than light, so it makes a steeper slope in the diagram. Which is analogous to saying that the angle with respect to the time axis has decreased. So, if you define $\phi$ to be the angle between the time axis and the worldline, then $$v \propto \phi$$

Now, of course you can choose that $v = \sin \phi$. It matches the "condition" that when $\phi = 0$ then $v = 0$. But, it does not match up properly. We defined the angle light makes to be $\pi / 4$. Try plugging it in the Lorentz transformation:

$$ 1 / \gamma = \sqrt{1 - \sin^2 \phi} = \sqrt{ 1 - 0.5} = \sqrt{0.5} = 1 / \sqrt{2}$$

which is a positive definite result. In fact the actual Lorentz transformation for light speed gives you $1 / 0$, which is not defined.

A rather more appropriate term would be $$v = \tan \phi$$. This matches the Lorentz transformation a well. Of course, this is still far away from the formal Lorentz transforms using $\cosh$, $\sinh$, and $\tanh$.

These were my two cents on why the angle pops up.