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Zo the Relativist
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Force is not invariant, it is covariant. Force is a vector, and it should be treated as such. therefore, Newton's second law generalizes to:

$F^{a} = \frac{d }{d\tau}\left(mv^{a}\right)$

Where $\tau$ is the affine parameter of the curve, and $v^{a}$ statisfies $v^{a}v_{a} = -1$. Even in special relativity, this has a bunch of consquences, including that you can no longer use that quadratic form for the position, even for constant force.

But, for your case, the most important thing is that if $F^{a}$ is ONLY in the $x$ direction in an unboosted frame, it will pick up a $t$ component in the boosted frame, whihc means that you can't just simply do the type of multiplication you do in the OP.

Also, for accelerated observers in special relativity, experiments like the one you describe can run into problems from this mixing of space and time, including the case that some reference frames actually CANNOT see some accelerated observers, thanks to Rinder horizons

Force is not invariant, it is covariant. Force is a vector, and it should be treated as such. therefore, Newton's second law generalizes to:

$F^{a} = \frac{d }{d\tau}\left(mv^{a}\right)$

Where $\tau$ is the affine parameter of the curve, and $v^{a}$ statisfies $v^{a}v_{a} = -1$. Even in special relativity, this has a bunch of consquences, including that you can no longer use that quadratic form for the position, even for constant force.

But, for your case, the most important thing is that if $F^{a}$ is ONLY in the $x$ direction in an unboosted frame, it will pick up a $t$ component in the boosted frame, whihc means that you can't just simply do the type of multiplication you do in the OP.

Force is not invariant, it is covariant. Force is a vector, and it should be treated as such. therefore, Newton's second law generalizes to:

$F^{a} = \frac{d }{d\tau}\left(mv^{a}\right)$

Where $\tau$ is the affine parameter of the curve, and $v^{a}$ statisfies $v^{a}v_{a} = -1$. Even in special relativity, this has a bunch of consquences, including that you can no longer use that quadratic form for the position, even for constant force.

But, for your case, the most important thing is that if $F^{a}$ is ONLY in the $x$ direction in an unboosted frame, it will pick up a $t$ component in the boosted frame, whihc means that you can't just simply do the type of multiplication you do in the OP.

Also, for accelerated observers in special relativity, experiments like the one you describe can run into problems from this mixing of space and time, including the case that some reference frames actually CANNOT see some accelerated observers, thanks to Rinder horizons

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Zo the Relativist
  • 41.9k
  • 2
  • 79
  • 146

Force is not invariant, it is covariant. Force is a vector, and it should be treated as such. therefore, Newton's second law generalizes to:

$F^{a} = \frac{d }{d\tau}\left(mv^{a}\right)$

Where $\tau$ is the affine parameter of the curve, and $v^{a}$ statisfies $v^{a}v_{a} = -1$. Even in special relativity, this has a bunch of consquences, including that you can no longer use that quadratic form for the position, even for constant force.

But, for your case, the most important thing is that if $F^{a}$ is ONLY in the $x$ direction in an unboosted frame, it will pick up a $t$ component in the boosted frame, whihc means that you can't just simply do the type of multiplication you do in the OP.