High precision calculation in Mathematica

by brian0918
Tags: calculation, mathematica, precision
 P: 44 I'm trying to calculate the velocity of 10^20 eV cosmic rays, but the precision in Mathematica and Google is not high enough: both give velocities of exactly 1*c. I've tried putting SetAccuracy[...,1000] around all the constants and the final result, and it still doesnt work. It now gives 1.0000000000000...000*c I know Mathematica can do this, so what is the proper command? Thanks.
 HW Helper Sci Advisor P: 3,149 Why don't you try subtracting c from the answer to find what Mathematica thinks the difference is?
 P: 44 Well, I've tried N, SetAccuracy, and SetPrecision. For a precision of 1000 decimals, I subtracted c (which also has a precision of 1000 decimals) and it got zero out to all the digits. I'm pretty sure I'm just not using the correct command, since I've had this problem before and figured out what to do.
P: 44

High precision calculation in Mathematica

Nevermind, I figured it out. You have to use SetPrecision on basically every number.

Oh, and for the record, 10^20 eV corresponds to 0.999999999999999999999955982283131960440050993898862440697486711315663 19262949728462461319064015929690c
 P: 59 hmm, aren't those rays photons? aren't they supposed to move at the speed of light? or are they propagating through some medium or otherwise interacting in a way that would slow them down?
 P: 44 No, they're ultra-high energy cosmic rays (most likely protons). I'm using the relativistic kinetic energy equation to get the velocity: Attached Thumbnails
 P: 44 I wonder if this necessary high precision is going to slow down numerical analysis, since everything will have to be out to ~30 decimals.
 Sci Advisor P: 902 It's not numerically efficient to deal directly with v/c when energy is very high. You should get into the habit of using the relevant expansions. So for example, $$\gamma={1\over\sqrt{1-\beta^2}}$$ where $\gamma=\mbox{Energy}/(mc^2)$, and $\beta=v/c$. Then $$\beta=\sqrt{1-1/\gamma^2}\approx 1-{1\over 2\gamma^2}$$ This gives you easily sufficient accuracy, and is very fast.

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