# Why must $v$ be $< c$ in the Lorentz transformations? Do these equations not apply to light? [duplicate]

I was trying to understand how things look from the perspective of light. Looking at the Lorentz transformations, it seems that the universe would contract along the direction of movement into a plane, and time would stop. But I have heard that these equations cannot be applied to light speed, when $v=c$.

Why don't the Lorentz transformations apply when $v=c$?

• Take a look at the formula for gamma in the Lorentz transformation and tell me what happens when $v=c$. It should be clear then. – Asher Jun 23 '16 at 20:01
• – Winther Jun 23 '16 at 20:02

To see what's going on, it's enough to do this in two dimensions, with the Lorentz form $\pmatrix{-1&0\cr 0&1\cr}$. (I've set $c=1$.)

The Lorentz group is the group that preserves this form. A typical element is $$\pmatrix{\pm\sec\theta&\tan\theta\cr \tan\theta&\pm\sec\theta\cr}$$ where $\theta$ runs through the open interval from $-\pi/2$ to $\pi/2$.

The subgroup that preserves the direction of time is the connected component of the identity, where the $\pm$ sign is positive. This subgroup is also sometimes called the Lorentz group.

Now given an element of the Lorentz group, we can define the corresponding velocity to be $v=\sin(\theta)$, so that $v$ is automatically (strictly) bounded between $-1$ and $1$.

• Why in my 52 years have I never thought of writing $\sec\theta = \cosh \eta$ here? A neat trick! Is that taken from a text? Actually I have seen it before in another context: $\theta = \operatorname{gd}(\eta)$, where $\operatorname{gd}$ is the Gudermannian function. Apparently your $\theta$ is called the "oblique angle" or "velocity angle" and was introduced by Karapetof. Apparently also useful if you want to fire a relativistic projectile for maximum range in a (huge) region of uniform gravity: see .... – Selene Routley Jun 24 '16 at 0:35
• ...see here. I don't know about you, I'm off to get my empty coke bottle and air compressor to test it out! There is also a $\operatorname{gd}$ scale on a couple of very clever slide rules (a couple of my quirks are that I collect slide rules and matrioshka dolls). Together with a pythagorean sum (implemented by two sliding scales), $\operatorname{gd}$, $\sin$ and $\cos$ let you calculate all the trig and hyperbolic functions with about the same effort as referring to twelve different scales for each. – Selene Routley Jun 24 '16 at 0:38
• @WetSavannaAnimalakaRodVance: I don't normally write $\sec\theta$, but I don't normally write $\cosh\eta$ either. On the rare occasions when I need to write anything like this at all, I usually just write something like $\beta$ or $1/\sqrt{1-v^2}$, which is what I did in the first draft of this post. Looking over that draft, it just kind of jumped out at me that $v/\sqrt{1-v^2}$ looks a lot like a tangent and $1/\sqrt{1-v^2}$ is the corresponding secant, and I figured this was probably one of the many things that everybody in the world except me already knew, so I edited accordingly. – WillO Jun 24 '16 at 1:07
• Awesome! Nice trick. Now you know how to launch a relativistic coke bottle for maximum range! – Selene Routley Jun 24 '16 at 1:31

Aside from simply looking at the Lorentz transformation, seeing a divergence and concluding "meh, it doesn't work", another way to gain insight into the divergence is through the statement:

no finite sequence of finite boosts will get you to a speed $c$ relative to your beginning inertial frame.

Imagine yourself in a spaceship with orientation controls and a booster such that you can accelerate yourself to any speed in some finite interval (let's say $[0,\,\Delta v]$ with $\Delta v\ll c$) in any direction relative to your present momentarily co-moving inertial reference frame in a unit time as measured by your on-board spaceship clock.

Group theoretically, this is equivalent to the assertion that after unit time, there is some neighborhood $\mathcal{N}_\mathrm{id}$ of the identity in $SO(1,\,3)$ such that I can impart any Lorentz transformation in that neighborhood to present my reference frame. As time (as measured by my trusty on board clock) goes by, I can impose any sequence of these neighborhood members; the overall transformation relative to my beginning frame is their product and so my overall transformation follows some continuous path through $SO(1,\,3)$. Our initial statement is equivalent to:

No member of the identity-connected component of $SO(1,\,3)$ corresponds to a relative speed of $c$

(indeed this is of course true of any member of $SO(1,\,3)$, but the identity component is the transformations we can physically reach with out controls, given enough fuel).

In particular, imagine heading off in a steady direction; and each unit time you are going to impose the same boost. As in WillO's answer, we concentrate on one spatial dimension, so our unit boost is:

$$\Delta\Lambda = \exp\left(\delta \eta\left(\begin{array}{cc}0&+1\\+1&0\end{array}\right)\right)=\left(\begin{array}{cc}\cosh\delta\eta&\sinh\delta\eta\\\sinh\delta\eta&\cosh\delta\eta\end{array}\right);\quad \delta\eta = \operatorname{artanh}\frac{\Delta v}{c}\approx \frac{\Delta v}{c}\tag{1}$$

The same, finite, $\Delta v$ relative to our present frame imparted $n$ times over is $\exp\left(n\,\delta \eta\left(\begin{array}{cc}0&+1\\+1&0\end{array}\right)\right)$. So, when we accelerate at a uniform velocity $\Delta v$ per unit time by to our clock, so we feel a constant accelerating force from our seat, an observer in our initial frame sees as accelerate so that our rapidity $\eta = n\,\delta\eta$ changes by an amount $\Delta v/c$ in a time interval $\cosh\eta$ relative to their frame. So the change in overall velocity between the two frames is

$$c\,\left(\tanh\left(\eta+\operatorname{artanh}\frac{\Delta v}{c}\right) -\tanh\eta\right) \approx \Delta v\,\operatorname{sech}^2\eta\tag{2}$$

and our apparent acceleration from our beginning frame is $\Delta v \,\operatorname{sech}^3\eta$; We seem to be accelerating more and more slowly, both because it is our rapidity, not speed, that is changing uniformly with boost interval number and also because these boost intervals are getting longer and longer relative to the beginning frame.

In neither frame will the overall speed difference ever reach $c$.

Note that the above arguments apply even if $\Delta v$ is a large fraction of $c$. When $\Delta v\ll c$ the approximation in (2) holds.

Lorentz transformations apply to objects with nonzero mass. For an object with mass, it would require an infinite amount of energy to reach light speed.

• The Lorentz transformation describes a change of coordinates when you switch from one coordinate system to another. It doesn't make sense to say it applies to "objects with nonzero mass". It's like saying that the formula for converting from centigrade to degrees doesn't apply to heavy objects. And it doesn't require any energy to change coordinate system. – Dan Piponi Jun 24 '16 at 0:05
• @DanPiponi you are of course correct. Rather than give the mathematical physics point of view I was trying to give more of an "in practice" and intuitive point of view: a physicist would not normally try to apply the Lorentz transformation to a massless object, knowing what would happen. – RiskyScientist Jun 24 '16 at 10:17