Timeline for Construction of the supersymmetric Faraday tensor
Current License: CC BY-SA 3.0
11 events
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| Jan 13, 2018 at 5:48 | comment | added | WunderNatur | @QuantumDot The commutator is due to expressing the curvature in terms of connection, where the gauge field strength is the curvature of the gauge connection on the corresponding fiber bundle. Can the super field strength be expressed as the curvature of something? | |
| Feb 17, 2013 at 22:49 | comment | added | QuantumDot | @LubošMotl After weeks of thinking about your response, I finally understand it. Basically you're saying there's no canonical anything: what's simplest is what's usually relevant, and what's more complicated is (hopefully) less relevant. | |
| Feb 14, 2013 at 15:21 | vote | accept | QuantumDot | ||
| Jan 21, 2013 at 2:48 | comment | added | QuantumDot | @LubošMotl Wait a sec... the commutator trick works in GR too, doesn't it? Don't I get the Riemann tensor that way? I guess SUSY is just special...? | |
| Jan 20, 2013 at 21:01 | comment | added | QuantumDot | @LubošMotl Very clear explanation; especially useful to think of the $D_\alpha$ as the 'square-root' of the ordinary derivative. Thanks! | |
| Jan 20, 2013 at 10:25 | comment | added | Luboš Motl | Ordinary covariant derivatives are bad in SUSY because the $\partial_\mu$ derivative is no longer the fundamental "minimal" derivative. Instead, one may find a square root of it - and the superderivatives $D_\alpha$ etc. are square roots of the ordinary derivatives, and they're therefore more fundamental. So you haven't really started to think in the SUSY way if you still want to place ordinary derivatives everywhere. | |
| Jan 20, 2013 at 10:24 | comment | added | Luboš Motl | The universal conditions we expect from field strength is not that it is a commutator - this is just a solution in the case of non-SUSY gauge theories or theories in non-SUSY formalism. If we want field strength in general, it is a field that transforms covariantly - for U(1), it is gauge-invariant - and that is minimum possible so that one may conveniently construct free Lagrangians from it etc. For non-SUSY, it's solved via your commutator route, in N=1 SUSY, Frederic wrote you the next step in the solution. | |
| Jan 20, 2013 at 10:22 | comment | added | Luboš Motl | QuantumDot, it's more appropriate to say that the field strength may be required to be a chiral superfield, so if the constraint is possible, one should clearly impose it to work with the minimal possible - maximally constrained - representations. Quite generally, your attempt to write the field strength as a commutator of covariant derivatives is misguided. That's how the gauge-covariant objects are constructed in non-SUSY theory but there's no reason that this is the right universal template for all theories, e.g. for SUSY. | |
| Jan 17, 2013 at 11:03 | comment | added | Yuji | @QuantumDot Read Wess-Bagger. The answer is there. | |
| Jan 14, 2013 at 17:18 | comment | added | QuantumDot | Ok! is there a reason why the supersymmetric field strength tensor must be a chiral superfield? Is there a geometric way of constructing the supersymmetric field strength tensor? Should I post a new question? | |
| Jan 14, 2013 at 11:31 | history | answered | Frederic Brünner | CC BY-SA 3.0 |