# I get tweets: How close can you get to the speed of light?

As I hope you know by now, I’m on twitter, and you should totally follow me. While it’s not always a great idea to ask physics questions in tweets (as 140 character answers are unlikely to be informative), a follower named @MikeKayatta asked:

If math has proven that .999 (repeating) = 1, then why is a particle accelerating to .999c not actually just at c?

He also added “<3 AUGTTU!”, but really, why wouldn’t he?

The short answer to his question is that there is a difference between something going 0.999c and something going 0.999999999… c, and that difference tends to be important. To give you an idea why, I’m going to have to tell you a few things about the energy of relativistic particles. I should also warn you that this is going to be a wee bit technical, so the math-phobic may want to skip it.

In your high school or college physics course, you’re likely to have learned an equation for kinetic energy that looked something like this:

$K=frac{1}{2}mv^2$

Not to put to fine a point on it, but this equation is wrong. Einstein derived a couple of very important relationships for particles moving close to the speed of light. First, he showed that even particles at rest have energy:

$E=mc^2$

which I suspect you were familiar with already. What you may not have known is that for moving particles, the total energy is:

$E=mc^2gamma$

where the gamma-factor is:

$gammaequiv frac{1}{sqrt{1-v^2/c^2}}$

This is a crazy looking function. Plotted against velocity, it looks like:

(The y-axis is $gamma$, but just seen in the context of time-dilation, rather than energy).

At speeds much less than c, gamma is almost exactly 1, but at speeds close to c, it explodes. One of the cool things that you can show is that at small speeds (say, at less than 10% the speed of light), the energy is approximately:

$Esimeq mc^2+frac{1}{2}mv^2$

the rest energy plus kinetic, and the kinetic looks exactly like what you learned in school. It’s only at speeds close to the speed of light where this is completely useless.

The point of all of this is that if we do a fair amount of plug and chug, we’ll find:

 Speed Energy 0.01c=$3times 10^6m/s$ $1.000050004 mc^2$ 0.87c $2 mc^2$ 0.99c $7.1 mc^2$ 0.999c $22.3 mc^2$ 0.9999c $71 mc^2$

If you notice the pattern, every two additional 9’s at the end costs you a factor of 10 in energy! In other words, it’s not just a matter of not being able to go the speed of light, it takes an arbitrarily large amount of energy to get arbitrarily close.

As a real world example, protons in the LHC currently have an energy of about 3.5TeV, but their rest energy ($mc^2$) is only 935 MeV. Plugging that in, we get:

$gamma_{lhc}=3743$

which means that the protons in the LHC are currently flying around at:

$v_{protons,lhc}=0.9999999643c$

Every little decimal place counts!

-Dave

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### One Response to I get tweets: How close can you get to the speed of light?

1. Frank says:

Wasn’t too mathematical, even I understood 😉