# Regarding a 1 g rocket that "the amount you accelerate would be less due to relativity". Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference? Could a ship be accelerated at say, 10 g from our frame of reference on Earth, while the crew of the ship only experiences 1 g of acceleration in theirs? If this were possible, how far can we take this, and how quickly?

Regarding a 1 g rocket that "the amount you accelerate would be less due to relativity". Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference? Could a ship be accelerated at say, 10 g from our frame of reference on Earth, while the crew of the ship only experiences 1 g of acceleration in theirs? If this were possible, how far can we take this, and how quickly?
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Erika Gomez
No, the crew’s proper acceleration (the acceleration they feel) is greater than the coordinate acceleration (the derivative of their velocity in our frame) that we observe. They can continue at 1 g proper acceleration indefinitely, whereas we see their coordinate acceleration approach 0 g as they approach c.
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dansleiksj
There is no frame of reference in SR in which an acceleration can stay constant at a non zero value for an infinite amount of time. This would inevitably lead to a velocity that is greater than the speed of light.
It is well-known that when two frames $S$ and ${S}^{\prime }$ move relative to each other with velocity $v$ and a velocity is $w$ in $S$ then in ${S}^{\prime }$ it is
${w}^{\prime }=\frac{w-v}{1-vw/{c}^{2}}\phantom{\rule{thinmathspace}{0ex}}.$
To calculate the relations for the accelerations is straightforward:
${\stackrel{˙}{w}}^{\prime }=\stackrel{˙}{w}\frac{1-{v}^{2}/{c}^{2}}{\left(1-vw/{c}^{2}{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}.$
Note that ${\stackrel{˙}{w}}^{\prime }=d{w}^{\prime }/dt$ and therefore, $d{w}^{\prime }/d{t}^{\prime }=\gamma \phantom{\rule{thinmathspace}{0ex}}d{w}^{\prime }/dt=\frac{d{w}^{\prime }/dt}{\sqrt{1-{v}^{2}/{c}^{2}}}\phantom{\rule{thinmathspace}{0ex}}.$ It follows that
$\frac{d{w}^{\prime }}{d{t}^{\prime }}=\frac{dw}{dt}\frac{\sqrt{1-{v}^{2}/{c}^{2}}}{\left(1-vw/{c}^{2}{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}.$
It is easy to see that for very small $w$ the acceleration in ${S}^{\prime }$ is less than the one in $S$ which is probably what your citation meant.