Why is the energy of the OMG particle so low compared to its high velocity?

[Reallyfat]

Hi everybody, I'm relatively new to particle physics. So I was reading up some stuff on high-velocity particles, and I found something on Fourmilab. The report is of a so-called OMG particle, traveling at some 0.9999999999999999999999951c. That's a ridiculously high speed, as far as I can tell. But anyways. I read that this particle was a proton, and it had an energy of around 50 joules. So I decided to use the classical physics equation:
E_Kinetic = m*v²*0.5
Just to test it. Needless to say, it did not work. I looked it up, and high-velocity particles have a relativistic formula where:

However, when I substitued my values for v and m, my energy came out as value which was to the power -10. Is there a reason that my energy value is some 10 orders of magnitude lower than it should be? And if so, could someone please guide me through the equation step-by-step so that I can work out where I went wrong?
I appreciate the responses, thanks.

 

[adjacent]

Do you think protons have a big mass?

 

[Reallyfat]

I used the standard proton mass, 1.67e-27 kg. Whether or not you'd call it big is relative.

 

[jtbell]

0.9999999999999999999999951c

That's 25 significant figures. Squaring it accurately and then subtracting from 1 requires at least 50 significant figures. Can your calculator handle that many?

 

[Einj]

You can obtain a more or less accurate estimate writing: \beta=1-\epsilon with \epsilon=49\times 10^{-25}.
Then:
$$
E=\frac{m}{\sqrt{1-(1-\epsilon)^2}}=\frac{m}{\sqrt{2\epsilon-\epsilon^2}} \simeq \frac{m}{\sqrt{2\epsilon}}.
$$
In this way you only need to compute a number with 12 significant digits. Using the mass of the proton to be roughly 1GeV I obtained E\simeq 3.2\times 10^{10} GeV, which is more or less 5 Joule. I'm still out by an order of magnitude.

 

[jtbell]

Doing the calculation in SI units directly:
$$E \simeq \frac{mc^2}{\sqrt{2\epsilon}}$$
I get 48 J.

 

[Einj]

I probably screwed some calculation :D

 

[Reallyfat]

I used an online calculator with max precision.
Anyway, I see what you've done, but could you explain why you did it? What exactly do Beta and Epsilon represent? And also, why does the formula simply not work with the
E=E0/sqrt(1-v^2/c^2)
formula I used?
Thanks in advance.

 

[dipole]

You are assuming your calculator has unlimited precision - it doesn't. Your calculator can only store about 16-17 digits reliably.

When you take a number like 1 - (0.99999999...9)^2, you're going to end up with a number which looks like 0.000000000..., however if your numerical accuracy can't store more than the first 16-17 digits, then the result becomes identically zero.

This is why you can't just blindly plug numbers in a computer - you have to actually think about the calculation a bit, which is what Einj was demonstrating.

 

[Einj]

In special relativity one defines \beta=v/c. I just defined \epsilon=49\times 10^{-25} as a small number, such that \beta=1-\epsilon.

 

[Bill_K]

The report is of a so-called OMG particle, traveling at some 0.9999999999999999999999951c. That's a ridiculously high speed, as far as I can tell.

No, it's not even as much as the speed of light. But it's a good illustration of why people in high energy physics don't concern themselves with velocity! It's impossible to measure v to 20 decimal places or whatever this is, and equally impossible to do anything useful with it.

But anyways. I read that this particle was a proton, and it had an energy of around 50 joules... I used the standard proton mass, 1.67e-27 kg.

We also don't use SI units, which are suitable for everyday measurements but bring in very large exponents when we try to apply them to particle physics.

The mass of a proton is about 1 GeV/c². The energy of this OMG particle, according to the article, was 3 x 10⁸ TeV. Since E = γmc², the gamma factor is the ratio of these two numbers, 3 x 10¹¹.

Now γ = (1 - v²/c²)^-1/2 so (1 - v²/c²)^1/2 = 3 x 10^-12, and (1 - v²/c²) = 10^-23. Thus v²/c² = one part in 10²³ less than 1, and v/c is half a part in 10²³ less.

 

[Reallyfat]

Ah, I see where my mistake lies. I did my calculation in electron volts and it works now. Thanks so much everybody!