Do origin and onboard rocket observers agree on velocity?

[BitWiz]

Given an ideal rocket in a friction-free, gravity-free environment:

A precise amount of energy E is applied to accelerating a rocket. An external observer at the origin (rocket velocity=0) watches the rocket accelerate to a final velocity V. As expected, the origin observer finds that V precisely matches the value required for E to equal the kinetic energy (KE) of the rocket.

An observer aboard the rocket only has an accelerometer and a proper clock. If the rocket observer integrates his accelerometer readings using the proper clock during the acceleration event to find his final velocity V', will V = V'?

Thanks!

 

[PeterDonis]

As expected, the origin observer finds that V precisely matches the value required for E to equal the kinetic energy (KE) of the rocket.

Yes, he does, but it's worth noting that the equation that relates the two is not the Newtonian $\frac{1}{2} m V^2$. The relativistic kinetic energy is $m \left( \gamma - 1 \right)$, where $\gamma = 1 / \sqrt{ 1 - \frac{V^2}{c^2} }$.

An observer aboard the rocket only has an accelerometer and a proper clock. If the rocket observer integrates his accelerometer readings using the proper clock during the acceleration event to find his final velocity V', will V = V'?

No. He will find $V = c \tanh \frac{V'}{c} = c \tanh \left( \frac{a \tau}{c} \right)$, where $a$ is the accelerometer reading and $\tau$ is the time elapsed on the proper clock.

(Btw, you can tell that the answer must be "no" even without knowing the formula: if the answer were "yes", then $V$ could become greater than $c$, since $V' = a \tau$ can certainly become greater than $c$ -- it can increase without limit since $a$ is constant and $\tau$ can increase without limit.)

 

[DrGreg]

BitWiz, PeterDonis has assumed constant a. But even if a isn't constant, what he says is still correct if you replace $a\tau$ with $$\int a \, d\tau.$$

 

[BitWiz]

BitWiz, PeterDonis has assumed constant a. But even if a isn't constant, what he says is still correct if you replace $a\tau$ with $$\int a \, d\tau.$$

Thank you, DrGreg. Any real-world attempt would have to do it this way.

Bit

 

[Meir Achuz]

.

An observer aboard the rocket only has an accelerometer and a proper clock. If the rocket observer integrates his accelerometer readings using the proper clock during the acceleration event to find his final velocity V', will V = V'?

No. If the accelerometer on the rocket reads a' for the acceleration, the rocket will have an acceleration a=a'/\gamma^3. Also the two times will be different.
Only if the rocketeer takes this into account, would she get the same velocity.

 

[BitWiz]

Thank you, Peter! You made my week. ;-)

Yes, he does, but it's worth noting that the equation that relates the two is not the Newtonian $\frac{1}{2} m V^2$. The relativistic kinetic energy is $m \left( \gamma - 1 \right)$, where $\gamma = 1 / \sqrt{ 1 - \frac{V^2}{c^2} }$.

I loved everything you said. I need to break up my response, though ...

To accelerate a 1Kg rocket to 10% c, using onboard propulsion and energy, and ignoring fuel mass losses (Tsiolkovsky):

By Newtonian standards, the energy required = Newtonian KE = 450 TJ with final momentum $p=V$?

By relativistic standards, KE is about 0.005 ... what units? ... and momentum $p = \gamma V$?

That's quite a difference and probably the basis of some spirited disagreements with my Newtonian friends regarding the energy required to accelerate mass -- from the relativistic rocket observer's point of view.

Do you know how I can determine the energy required to accelerate 1 kg to X% c from the POV of an onboard observer?

If appropriate, can you tell me how to respond to my Newtonian friends who insist that a rocket's Newtonian KE determines the minimum energy that must be expended to achieve both external and onboard (proper) velocity?

Thanks!

Happy Easter!

 

[BitWiz]

No. If the accelerometer on the rocket reads a' for the acceleration, the rocket will have an acceleration a=a'/\gamma^3. Also the two times will be different.
Only if the rocketeer takes this into account, would she get the same velocity.

Let me fix that: $a=a'/\gamma^3$ Correct? If so, that is a very, very handy equivalence. Thank you, sir!

Bit

 

[PeterDonis]

To accelerate a 1Kg rocket to 10% c, using onboard propulsion and energy, and ignoring fuel mass losses (Tsiolkovsky):

By Newtonian standards, the energy required = Newtonian KE = 450 TJ with final momentum $p=V$?

Assuming $m = 1$, yes.

By relativistic standards, KE is about 0.005 ... what units?

Units of $m c^2$, so KE is about $0.005 m c^2 = 0.005 \times 1 \times 9 \times 10^{16}$, i.e., about 450 TJ, just as you calculated. In other words, 10% c isn't fast enough for relativistic corrections to make much difference. (You would see the difference if you did the calculation with more precision.)

... and momentum $p = \gamma V$?

Assuming $m = 1$, yes. But since $\gamma \approx 1.005$, this isn't much of a difference, as above.

That's quite a difference

For 10% c, not really; see above. Try it with V = 90% c and see what you get.

Do you know how I can determine the energy required to accelerate 1 kg to X% c from the POV of an onboard observer?

My favorite resource for this topic is the Usenet Physics FAQ article on the relativistic rocket equation:

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

It covers everything, including what you're asking here; it's basically the correct relativistic version of the Newtonian rocket calculations that your friends are referring to.

can you tell me how to respond to my Newtonian friends who insist that a rocket's Newtonian KE determines the minimum energy that must be expended to achieve both external and onboard (proper) velocity?

Do they just not know about relativity, or do they think relativity somehow doesn't apply? If it's the former, you can just point them at correct relativistic calculations, like the article I linked to above. If it's the latter, any advice I can give would include terms like "crackpot", which you might not want to use, depending on how good friends these are.

 

[PeterDonis]

BitWiz, PeterDonis has assumed constant a. But even if a isn't constant, what he says is still correct if you replace $a\tau$ with $$\int a \, d\tau.$$

Yes, good point, I was assuming constant $a$ (as does the Usenet Physics FAQ article I linked to). Making $a$ variable doesn't really change anything conceptually, it just makes the math more complicated; but you're right, it's important to be aware of the effects of such complications.