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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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    〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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    Noether's Theorem is perhaps the most beautiful mathematical theorem in physics •It states that: Every continuous symmetry (T) of the Lagrangian has a corresponding conservation lawpic.twitter.com/RrsS1QKm4A

    12:32 PM - 15 Jan 2020
    • 304 Retweets
    • 1,435 Likes
    • Iñigo Robredo Eoin Griffin Lorenzo Graziotto Sam Benami Donald Mitchell Miguel Paps Ot Garcés
    19 replies 304 retweets 1,435 likes
      1. New conversation
      2. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        Caveats (1/2) The astute reader will notice that technically Noether's theorem applies to invariance of the action, which is the spacetime integral over the Lagrangian density Hence, you will also need to make sure that the measure is invariant under the transformation as well

        1 reply 6 retweets 98 likes
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      3. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        Caveat (2) Since surface terms do not affect the Euler Lagrange equations, the equations of motion are always trivially invariant upon adding a 4-divergence to the Lagrangian density

        2 replies 4 retweets 60 likes
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      4. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        also the math just looks cool

        2 replies 2 retweets 51 likes
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      5. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        In Quantum Field Theories, however, Noether's theorem can be violated due to quantum effects! A symmetry at the classical level that is broken by quantum corrections is referred to as 'Anomalous' Anomalous symmetries are one of the many great predictions of the Standard Model!pic.twitter.com/Xku5q3MERl

        1 reply 16 retweets 100 likes
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      6. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        The reason for this is because Noether's theorem relies on the Euler-Lagrange equations, which need not be satisfied at the quantum level!

        6 replies 4 retweets 73 likes
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      7. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 16
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        If Noether’s theorem gets 1k likes I will lose my marbles (in a good way)

        4 replies 1 retweet 51 likes
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      8. End of conversation
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      2. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        yes! the classical electrodynamics lagrangian has a U(1) symmetry which corresponds to the conservation of charge!

        0 replies 3 retweets 21 likes
      3. 4 more replies
      1. New conversation
      2. Scott Little‏ @JScottLittle Jan 15
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        Replying to @InertialObservr

        I could be wrong; however, I believe this form of the current assumes the Lagrangian density is only a functional of the field and the derivatives of the field. Again, if I’m not mistaken, the form can be generalized if the Lagrangian density depends on higher derivatives.

        2 replies 0 retweets 0 likes
      3. 〈 Berger | Dillon 〉‏ @InertialObservr Jan 15
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        Replying to @JScottLittle

        if there are higher derivatives of the field then they will show up when you take the partial with respect to the field derivatives .. in QFT this isn't an issue if we only consider renormalizable operators

        2 replies 0 retweets 4 likes
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