Timeline for Heisenberg uncertainty and Lorentz contraction
Current License: CC BY-SA 4.0
16 events
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| Aug 26, 2019 at 15:53 | comment | added | Sidharth Ghoshal | seeing that there is no accepted answer has anyone successfully managed to conduct the OP's thought experiment in QFT or some version of RQM? | |
| Jun 25, 2019 at 10:24 | comment | added | Paradoxy | Wave function in QFT is invariant under such a transformation, which means there shouldn't be a problem like that. In non relativistic QM theory you will get a wrong answer. | |
| Jun 25, 2019 at 9:10 | comment | added | user87745 | +1: Interesting question! I presume that a satisfactory answer cannot come from simple QM and I am not exactly clear on how this question gets translated into QFT (and back). | |
| Jun 25, 2019 at 8:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 8, 2019 at 12:00 | history | tweeted | twitter.com/StackPhysics/status/1103988913042010116 | ||
| Feb 21, 2019 at 19:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 16, 2019 at 0:14 | history | edited | Ján Lalinský | CC BY-SA 4.0 |
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| Jan 15, 2019 at 19:34 | history | edited | Qmechanic♦ |
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| Jan 15, 2019 at 19:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 16, 2016 at 21:14 | comment | added | Bill N | No. When you increase v you also increase $\gamma$, and that's a multiplicative change to p. | |
| Sep 16, 2016 at 4:41 | history | edited | knzhou | CC BY-SA 3.0 |
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| Sep 16, 2016 at 4:34 | answer | added | knzhou | timeline score: 2 | |
| Sep 15, 2016 at 20:47 | answer | added | WillO | timeline score: 0 | |
| Sep 15, 2016 at 18:42 | comment | added | Oti | But we are adding a constant value to p, not multiplying every value by a constant. 1,2,3,4,5 vs 2,3,4,5,6 have the same standard deviation! Let's say that in the particle's co-moving frame the momentum is distributed symmetrically around p=0. The values of momentum in the lab frame should be distributed also symmetrically but around $p=\gamma mv$, where v is the relative speed of the frames. So I don't understand your reasoning why $\Delta p$ should increase. | |
| Sep 15, 2016 at 17:08 | comment | added | Bill N | $p=\gamma m v$. That's very basic special relativity. If you make all the momenta bigger, you make the standard deviation bigger. Try it: find the standard deviation of 1, 2, 3, 4, 5 vs. 2, 4, 6, 8, 10. | |
| Sep 15, 2016 at 16:45 | history | asked | Oti | CC BY-SA 3.0 |