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3 votes

About Lorentz Invariant

It means $\phi(x^\mu) = \tilde{\phi}(\tilde{x}^\mu) = \tilde{\phi}((\Lambda^{-1})_\nu^\mu x^\nu)=\phi(x^\mu)\neq \tilde{\phi}(x^\mu)$
Wolphram jonny's user avatar
2 votes

Second postulate of Special Relativity - Finite Invariante Speed?

It possible that you can't define the 2nd postulate, since according to https://math.stackexchange.com/questions/650718/does-infinite-equal-infinite $$ \infty \ne \infty $$ Regarding the 1st postulate,...
JEB's user avatar
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1 vote

When an electron is moving at relativistic speeds, is the electric field length-contracted as well?

For an electron that is moving at relativistic speeds, ${\bf E}=\frac{q\gamma{\bf r}}{({\bf r_\perp}^2 +\gamma^2{\bf r_\parallel}^2)^{\frac{3}{2}}} = \frac{q{\bf r}} {\gamma^2[r^2-({\bf v\times r})^2]^...
Jerrold Franklin's user avatar
1 vote
Accepted

Estimating particle mass from bubble chamber

You certainly have a good point, and I apologize for my confusing/wrong leading comment, a knee-jerk reaction to the mention of the deprecated/taboo term "relativistic mass" in lieu of $E/c^...
Cosmas Zachos's user avatar
1 vote
Accepted

When an electron is moving at relativistic speeds, is the electric field length-contracted as well?

Yes, the electric field is length-contracted because it lives in physical space just like the charge itself. In other words, each point in space-time in which the electric field is present is a space-...
Brian Bi's user avatar
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1 vote

${}$Doppler shift

The form of the Doppler shift tells you a lot about your system. If the equations are symmetric in transmit (Tx) and receive (Rx), then all motion is relative and there is no preferred rest frame. If ...
JEB's user avatar
  • 34.8k
1 vote
Accepted

Lorentz Boosts and Hyperbolic Quaternions

I'd never heard of hyperbolic quaternions, and they seem very strange – not even an alternative algebra, which is worse than the octonions. The subspace $a+br$ is isomorphic to the split-complex ...
benrg's user avatar
  • 27.3k
1 vote
Accepted

Long-range approximations of the Uehling interaction

I found out the answer; it might be useful to someone in the future. The two approximate interactions, $$ U_\delta=-\frac{4}{15}\frac{\alpha(Z\alpha)}{m^2}\delta(\vec{r}) \ \ \ , \ \ \ U_\text{h}=-m\...
dennismoore94's user avatar

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