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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. 〈 Berger | Dillon 〉‏ @InertialObservr 12 Mar 2019
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      A Nasty Integral & An Interesting Observation It's interesting that equation (2) is independent of the choice of alpha.. I can't immediately see why this should be true.. I tried proving scale-invariance, but to no avail. Can any of you see why this should be true?pic.twitter.com/yircjqHmG0

      5 replies 24 retweets 113 likes
    2. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 12 Mar 2019
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      Replying to @InertialObservr

      if by scale-invariance you mean f(a*x) = a*f(x) then the reason you can't prove it is that it's not true (i made plots)

      1 reply 0 retweets 0 likes
    3. 〈 Berger | Dillon 〉‏ @InertialObservr 12 Mar 2019
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      Replying to @Corey_Yanofsky

      No, that’s linearity and is not true for either sine or exponents. Scale invariance is f(ax)=f(x)

      1 reply 0 retweets 2 likes
    4. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 12 Mar 2019
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      Replying to @InertialObservr

      that's not true either it's a wiggler

      1 reply 0 retweets 0 likes
    5. 〈 Berger | Dillon 〉‏ @InertialObservr 12 Mar 2019
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      Replying to @Corey_Yanofsky

      The integrand is a wiggler..Did you plot the integrand or the integral as a function of alpha? Because the latter is the thing that should be scale invariant

      1 reply 0 retweets 0 likes
    6. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 12 Mar 2019
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      Replying to @InertialObservr

      i didn't exactly plot it as a function of alpha, but it's everywhere positive (in the domain of integration) for alpha = 1 and dips below zero for alpha = 2

      1 reply 0 retweets 0 likes
    7. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 12 Mar 2019
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      Replying to @Corey_Yanofsky @InertialObservr

      see? wiggler

      1 reply 0 retweets 0 likes
    8. 〈 Berger | Dillon 〉‏ @InertialObservr 12 Mar 2019
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      Replying to @Corey_Yanofsky

      Right, but that doesn’t say anything about scale invariance

      1 reply 0 retweets 0 likes
    9. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 12 Mar 2019
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      Replying to @InertialObservr

      oh, the integral is definitely scale invariant, sorry ...but then isn't scale invariance exactly what your differentiation under the integral sign proves? /me is confused

      3 replies 0 retweets 1 like
      〈 Berger | Dillon 〉‏ @InertialObservr 12 Mar 2019
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      Replying to @Corey_Yanofsky

      In a way, yes.. but it’s not exactly obvious why it ought to be

      5:25 PM - 12 Mar 2019
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      • Corey Yanofsky truly and sincerely
      1 reply 0 retweets 1 like
        1. Corey Yanofsky truly and sincerely‏ @Corey_Yanofsky 12 Mar 2019
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          Replying to @InertialObservr

          i thought you meant something else by "prove scale invariance" because it seemed clear to me that you *had* proved scale invariance of the integral

          0 replies 0 retweets 1 like
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