Timeline for Is mass still a scalar in special relativity?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 6, 2021 at 15:22 | answer | added | Laff70 | timeline score: 0 | |
| May 6, 2021 at 13:06 | comment | added | Prahar | @Frobenius - I suppose it depends on how one defines the 4-vector $p^\mu$. The definition that one is presented with when starting to learn special relativity does involve $m_0$ in which case my argument would be circular and since this is the context that is relevant to OP, I'll have to agree with you. | |
| May 6, 2021 at 13:02 | comment | added | Voulkos | @Prahar Mitra : But the 4-momentum $\,p_{\mu}\,$ is defined in dependence to the a priori considered invariant rest mass $\,m_{\rm o}$, see equation \eqref{08a} in my answer. So I don't think that this is a proof, unless there exists a different definition of $\,p_{\mu}\,$ independent of the rest mass $\,m_{\rm o},$ which I miss. But then from this different definition must be proved first that $\,p_{\mu}\,$ is a Lorentz 4-vector in order for your proof to be valid. | |
| May 6, 2021 at 11:37 | vote | accept | Himanshu | ||
| May 6, 2021 at 11:37 | |||||
| May 6, 2021 at 10:01 | comment | added | Prahar | The mass of a particle is related to the square of the four momentum $m^2 = p^\mu p_\mu$ which makes it clear that $m$ is Lorentz invariant. | |
| May 6, 2021 at 7:26 | history | edited | Qmechanic♦ |
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| May 6, 2021 at 7:17 | answer | added | Voulkos | timeline score: 8 | |
| May 5, 2021 at 14:44 | comment | added | nwolijin | Mass is one of the Casimirs of ISO(d-1,1) | |
| May 5, 2021 at 14:23 | history | asked | Himanshu | CC BY-SA 4.0 |