Elixir
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Somebody’s coffee is going to get cold.
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What would it take to sustain a 1g acceleration over a 75 year period?
- Thread starter SLD
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Somebody’s coffee is going to get cold.
lpetrich
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I checked Barbos's calculations, and I found a rapidity of only 7.22 c.
That means v = (1 - 1.07*10-6) c and γ = 683.
Rapidity = (acceleration) * (time)
That means v = (1 - 1.07*10-6) c and γ = 683.
Rapidity = (acceleration) * (time)
lpetrich
Contributor
I will now calculate the energy efficiency of a rocket. It is (kinetic energy of the rocket when finished) / (kinetic energy of its exhaust relative to the rocket). Working it out, and using c = 1, it is
\( \displaystyle{ eff = \frac{ (\gamma - 1) m_f }{ (1 - \gamma_e^{-1} ) (m_i - m_f) } } \)
where the velocity v is given by
\( \displaystyle{ v = \tanh w ;\ \gamma = \cosh w ;\ w = v_e \ln \frac{m_i}{m_f} } \)
where ve is the effective exhaust velocity: (momentum of exhaust) / (total mass-energy of exhaust)
and where mi and mf are the rocket's initial and final masses. Using the mass ratio \( r = m_i/m_f \)
\( \displaystyle { eff = \frac{ \gamma - 1 }{ (1 - \gamma_e^{-1} ) (r - 1) } ;\ w = v_e \ln r } \)
\( \displaystyle{ eff = \frac{ (\gamma - 1) m_f }{ (1 - \gamma_e^{-1} ) (m_i - m_f) } } \)
where the velocity v is given by
\( \displaystyle{ v = \tanh w ;\ \gamma = \cosh w ;\ w = v_e \ln \frac{m_i}{m_f} } \)
where ve is the effective exhaust velocity: (momentum of exhaust) / (total mass-energy of exhaust)
and where mi and mf are the rocket's initial and final masses. Using the mass ratio \( r = m_i/m_f \)
\( \displaystyle { eff = \frac{ \gamma - 1 }{ (1 - \gamma_e^{-1} ) (r - 1) } ;\ w = v_e \ln r } \)
barbos
Contributor
Well, to have your starship moving with almost speed of light you would have to have your ship made of almost exclusively antimater fuel. No calculations is necessary.
So the only practical way to reach such speed is to not carry any actual fuel.
So the only practical way to reach such speed is to not carry any actual fuel.
lpetrich
Contributor
Making the efficiency a function of rapidity,
\( \displaystyle{ eff = \frac{ \cosh w - 1 }{ (1 - \sqrt{1 - v_e^2} ) (e^{w/v_e} - 1) } } \)
In the Newtonian limit, v = w,
\( \displaystyle{ eff = \frac{ (v/v_e)^2 }{ e^{v/v_e} - 1 } } \)
The maximum value is 0.64761, for v/ve = 1.59362 and mass ratio r = mi/mf = ev/ve = 4.92155
In the opposite limit, for ve = 1 (c = 1 units), we find
\( \displaystyle{ eff = \frac12 \left( 1 - e^{-w} \right) = \frac12 \left( 1 - \frac1r \right) } \)
For ve close to 1, I find maximum efficiency close to 1/2, for rapidity - ln(1 - ve), velocity 1 - 2(1 - ve)2, and mass ratio 1/(1 - ve)
\( \displaystyle{ eff = \frac{ \cosh w - 1 }{ (1 - \sqrt{1 - v_e^2} ) (e^{w/v_e} - 1) } } \)
In the Newtonian limit, v = w,
\( \displaystyle{ eff = \frac{ (v/v_e)^2 }{ e^{v/v_e} - 1 } } \)
The maximum value is 0.64761, for v/ve = 1.59362 and mass ratio r = mi/mf = ev/ve = 4.92155
In the opposite limit, for ve = 1 (c = 1 units), we find
\( \displaystyle{ eff = \frac12 \left( 1 - e^{-w} \right) = \frac12 \left( 1 - \frac1r \right) } \)
For ve close to 1, I find maximum efficiency close to 1/2, for rapidity - ln(1 - ve), velocity 1 - 2(1 - ve)2, and mass ratio 1/(1 - ve)
Jimmy Higgins
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I was pondering the causality issue someone, I think bilby raised. And I wondered if that was a legitimate argument. We know the speed limit for electromagnetism is a fundamental property of the fabric of spacetime. Therefore, that indicates things can't physically go faster than the speed of light, period. But is an argument via causality really an issue? Can things not mathematically go backwards in time?
Can a person arrive somewhere "before" leaving, on paper? They actually move back in time as they travel, but don't actually experience the change in time as going backwards, they are just moving.
In practice, this isn't possible, spacetime has the physical speed limit, making the question worse than hypothetical. But if a ship went faster than c via a warp bubble, which I think is bubkis, is there really a causality rule?
Can a person arrive somewhere "before" leaving, on paper? They actually move back in time as they travel, but don't actually experience the change in time as going backwards, they are just moving.
In practice, this isn't possible, spacetime has the physical speed limit, making the question worse than hypothetical. But if a ship went faster than c via a warp bubble, which I think is bubkis, is there really a causality rule?
skepticalbip
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It's probably not possible to understand the effects of relativity without a scrapping of Newtonian 'intuition' and an acceptance of what the math tells us about reality.
This link does a fair job of describing relativistic reality. You may enjoy it:
WATCH: Why The Speed of Light Is NOT About Light
The Universe is pretty nuts when you think about it.
www.sciencealert.com
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
It is not just EMR. Hit a ling stiff metal rod on one end with a hammer and C is the limit at which efeect can reach the far end.
Regardless of C I'd argue things can not happen instantaneously. It would require infinite energy.
If you are at C what do you see? For light from behind you see a standing wave, but the models break diwn.
Regardless of C I'd argue things can not happen instantaneously. It would require infinite energy.
If you are at C what do you see? For light from behind you see a standing wave, but the models break diwn.
bilby
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In Newton's universe, if you are at c (relative to the univeral reference frame of space), light from behind appears stationary.
In Einstein's universe (which is demonstrably the one we inhabit), light from behind you appears to catch up to you at c regardless of what speed you move at relative to anything else, and it's completely meaningless to say "moving at x miles per hour" without saying what that speed is relative to. We get away with this in everyday life by always assuming the local patch of the surface of the Earth as our datum, but that's a purely local phenomenon.
In Einstein's universe, light always moves at c in a vacuum, as measured by any observer no matter how that observer is moving.
If your spaceship is flying away from Earth at 0.9c, and you turn on your headlights, you will see photons streaming away from you at c relative to you. An astronomer on Earth will see those same photons streaming away from him at c, and report that they are only going 0.1c faster than your spaceship.
Both observations are correct; The reason they seem to disagree is that speed is distance traveled in a given time, and time isn't the same for both of the observers in this situation.
No matter what accelerations you undergo, you will never observe light that you are trying to run away from failing to catch up to you at exactly c.
The speed of propagation of particles with zero rest mass in a vacuum is a constant for all observers. That was Einstein's big idea. And he was right.
What if we did not have to convert any of the spaceship's mass into energy to generate acceleration? What if "beamed energy" could be a solution (pushing a solar sail with a laser that is powered independently of the spacecraft, for example). Not that we can pull 1 g with that technology today, but what if the mass conversion was not an issue.... then does the feasibility rise significantly?
Swammerdami
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I think the answer to your reddened question is Yes. But the faster-than-light travel it implies may depend on entanglements that prevent any faster-than-light communication of useful information.
The action-at-a-distance interpretation of the EPR paradox or GHZ experiment is often thought to imply faster-than-light "travel" BUT that speed limit IS obeyed (in each zig or zag) if the causality chain somehow zigzags through entanglements, alternating forward and backward causality.
IIUC, many top physicists regard the arrow of time NOT as a fundamental law of physics but as the result of the strong "tailwind of entropy" we endure ("by happenstance") in our environment, due to the sun's evolution and traceable back to the Big Bang.