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The OMG particle's energy

  1. May 19, 2014 #1
    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, travelling 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:
    EKinetic = 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.
    Last edited: May 19, 2014
  2. jcsd
  3. May 19, 2014 #2


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    Gold Member

    Do you think protons have a big mass?
  4. May 19, 2014 #3
    I used the standard proton mass, 1.67e-27 kg. Whether or not you'd call it big is relative.
  5. May 19, 2014 #4


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    Staff: Mentor

    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?
  6. May 19, 2014 #5
    You can obtain a more or less accurate estimate writing: [itex]\beta=1-\epsilon[/itex] with [itex]\epsilon=49\times 10^{-25}[/itex].
    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 [itex]E\simeq 3.2\times 10^{10} GeV[/itex], which is more or less 5 Joule. I'm still out by an order of magnitude.
  7. May 19, 2014 #6


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    Staff: Mentor

    Doing the calculation in SI units directly:
    $$E \simeq \frac{mc^2}{\sqrt{2\epsilon}}$$
    I get 48 J.
  8. May 19, 2014 #7
    I probably screwed some calculation :D
  9. May 19, 2014 #8
    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
    formula I used?
    Thanks in advance.
  10. May 19, 2014 #9
    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 [itex] 1 - (0.99999999...9)^2 [/itex], you're going to end up with a number which looks like [itex] 0.000000000...[/itex], 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.
  11. May 19, 2014 #10
    In special relativity one defines [itex]\beta=v/c[/itex]. I just defined [itex]\epsilon=49\times 10^{-25}[/itex] as a small number, such that [itex]\beta=1-\epsilon[/itex].
  12. May 19, 2014 #11


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    No, it's not even as much as the speed of light. :smile: 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.

    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.
  13. May 20, 2014 #12
    Ah, I see where my mistake lies. I did my calculation in electron volts and it works now. Thanks so much everybody!
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