# Is there an approximation for the Lorentz factor for very large velocities?

I am aware of the approximation generally used for low speeds to calculate the Lorentz factor, that being,

$$\gamma \approx 1 + \frac{1}{2} \left(\frac{v}{c} \right)^2$$

But I need the exact opposite thing -- is there any suitable approximation for when v is extremely close to c?

This is roughly the simplest you can get it: $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-v/c} \sqrt{1+v/c}} \approx \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-v/c}}.$$ In other words, if $$\Delta v$$ is how far the speed is below $$c$$, then $$\gamma \approx \sqrt{\frac{c}{2 \Delta v}}.$$

• The case when the approximation is not really simpler than the original expression... The problem is basically that at $v\approx c$, $\gamma$ has an algebraic branch point, so any asymptotic expansion at that point involves a root. – Ruslan Feb 23 at 9:23
• @Ruslan It is simpler, in the sense that it's much easier for a computer to calculate without error, which is what the OP is after. – knzhou Feb 23 at 22:26
• I wouldn't be so sure that the OP is looking to avoid the loss of precision due to addition of numbers of different magnitudes. The question is asking for approximation, not for a way to preserve precision. – Ruslan Feb 23 at 22:40
• @Ruslan Following their comment, "I was trying to simulate [...] and it seemed impossible to calculate the Lorentz factor for it always shot up to infinity because of precision errors." It is very common in computational physics to use approximations to enhance numeric stability. – knzhou Feb 23 at 22:41
• Ah indeed, hadn't noticed that comment. – Ruslan Feb 23 at 22:42

For ultra-relativistic particles, $$c-v$$ basically stops being an experimentally accessible observable. Unless you are extremely careful about timing, you assume that the beam is traveling at $$c$$ and measure the Lorenz factor by comparing the kinetic energy per particle to the particle mass, $$\gamma = E/mc^2$$. (If you care about the difference between total energy $$E=\gamma m c^2$$ and the kinetic energy $$K=(\gamma-1)mc^2$$, you're not ultra-relativistic yet.)

I first understood this when I got to Jefferson Lab, which has two antiparallel 1 GeV electron linacs connected like a racetrack. The electrons are injected at 50 MeV. After a lap they're at 2 GeV. After five laps, they're at 10 GeV. The accelerator feeds beam to four different halls at once, each of which can request a different energy by accepting the beam after a different number of passes around the track. So at any given instant while the beam is on, the linac might have beam bunches with five or six different energies interleaved. And --- here's what's interesting to you --- the fast ones don't have different timing than the slow ones. Even the electrons from the 50 MeV injector already have $$\gamma = 100$$, and there are a bunch of 1% issues that make it hard to distinguish between their speed and exactly $$c$$.

Search term: "XY problem."

• Thanks! I understand that it's not experimentally verifiable, but say I was trying to simulate electrons in a magnetic field at 200 MeV (so the Lorentz force stuff but now relativistic) and it seemed impossible to calculate the Lorentz factor for it always shot up to infinity because of precision errors. – kitizl Feb 23 at 11:21
• Consider using the product $\gamma\beta$ as an asymptote-free dynamical variable in your simulation, and extracting $\gamma$ or $\beta$ from it when you need them. I remember that working out kind of nicely because $F=\dot p$ is still relativistically correct and $p=\gamma\beta m c$. A follow-up question about such conceptual issues in your simulation would be interesting to me. – rob Feb 23 at 13:32
• @rob the so-called "XY problem" is a bane on SE. The OP has been clear and concise about what they want, and instead of giving an answer you took it as an opportunity to waffle about something that interests you. – Myridium Feb 23 at 21:51
• @Myridium In 1000+ posts and 4400+ comments this is my first use of the term on the site. Furthermore the asker responded with gratitude by elaborating on the hidden motivation for their question, and I was able to offer them further, more useful advice and a way that a productive follow-up question could usefully be framed. Your point about dismissive uses of the term elsewhere is well-taken, but I invite you to consider that my choice here was judicious and well-motivated. If you'd like to talk more, let's do so on Physics Meta. – rob Feb 23 at 22:09