Timeline for What is the meaning of the negative sign in $\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 - (c\Delta t)^2$?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2019 at 2:42 | comment | added | Chiral Anomaly | @Allure Yes, that is correct. | |
| Jan 4, 2019 at 22:45 | comment | added | Allure | Looking at the proof, it seems the difference arises as well from the fact that there's only one time dimension. If there's more than one time dimension, then any individual dimension can change sign. | |
| Nov 24, 2018 at 13:33 | vote | accept | Khaled Oqab | ||
| Nov 24, 2018 at 12:54 | vote | accept | Khaled Oqab | ||
| Nov 24, 2018 at 12:54 | |||||
| Nov 24, 2018 at 8:50 | comment | added | Philip Wood | Excellent! The result is such an important one that it's great to see how it is required by your equation (1). | |
| Nov 24, 2018 at 1:42 | comment | added | Chiral Anomaly | @PhilipWood That's correct. Since a worldline is a one-dimensional trace through spacetime, it can always be parameterized by $u$ so that different values of $u$ give different points along the worldline. Now that you mention it, I see that my wording was too careless, because it doesn't rule out an isolated point where all of the derivatives are zero. I should have said that any worldline (for any sign of the RHS of eqn (3)) can always be parameterized such that the derivatives $dt/du$, $dx/du$, etc, are not all zero anywhere. Then (4) implies that $dt/du$ can never be zero. | |
| Nov 24, 2018 at 0:09 | comment | added | Philip Wood | Thanks again. The key step seems to be, "If the x,y,z derivatives are all zero, then this requires dt/du≠0." Are you saying that unless this were the case, then, locally, we wouldn't have different space-time points corresponding to different values of $u$? | |
| Nov 23, 2018 at 23:55 | history | edited | Chiral Anomaly | CC BY-SA 4.0 |
Fixed a typo
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| Nov 23, 2018 at 23:53 | comment | added | Chiral Anomaly | @PhilipWood Thanks for the gracious feedback. Actually, I wasn't quite happy with the "proof" in my comment, because it doesn't explain how those abstract transformations relate to the motion of physical objects, and it's unnecessarily specific to flat spacetime. That was bugging my conscience, so I added an appendix to the main answer that addresses your question more directly and that can be generalized to curved spacetime (though I'm still only showing the flat case). | |
| Nov 23, 2018 at 23:46 | history | edited | Chiral Anomaly | CC BY-SA 4.0 |
Added appendix with a proof
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| Nov 23, 2018 at 22:54 | comment | added | Chiral Anomaly | @PhilipWood The plus sign in $x^2+y^2$ implies that this quantity is invariant under ordinary rotations: $x\mapsto x\cos\theta-y\sin\theta$, $y\mapsto x\sin\theta+y\cos\theta$. We can go from $\theta=0$ to $\theta=\pi$ to "turn around", reversing the signs of $x$ and $y$. But the minus sign in $(ct^2)-x^2$ means that this quantity is invariant under hyperbolic rotations (Lorentz transf's): $ct\mapsto ct\cosh\theta+x\sinh\theta$, $x\mapsto ct\sinh\theta+x\cosh\theta$, with $\cosh^2\theta-\sinh^2\theta=1$. No value of $\theta$ can change the sign of $t$, so we cannot "turn around" in time. | |
| Nov 23, 2018 at 22:51 | comment | added | niels nielsen | Dan, this is the clearest exposition of this I have seen to date, and reading it reminds me of why I started hanging around this site in the first place. Thanks for taking the time to post this answer and have a happy holiday. | |
| Nov 23, 2018 at 22:44 | comment | added | Philip Wood | "This minus sign is the reason we can't "turn around and go the opposite direction" in time, like we can in space." Would you be kind enough to explain how this follows? | |
| Nov 23, 2018 at 21:53 | history | answered | Chiral Anomaly | CC BY-SA 4.0 |