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guslCompare "accelerating under constant power"
with "accelerating under constant force" (constant acceleration)
My intuition tells me that they should be the same, but kinetic energy considerations show that the acceleration is decreasing on the first one (it takes 4 times the energy to get 2 times as fast).
This would seem to contradict "velocity is relative": if velocity were relative, then the energy needed to get faster by 1m/s would be the same whether you are stationary or already at 1 m/s.
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My intuition also tells me that I should be able to come up with a similar paradox about predicting the outcome of a 1-dimensional elastic collision. If you do it with energy vs momentum.
Conservation of energy:
v1_before^2 + v2_before^2 = v1_after^2 + v2_after^2 (if we fix one side of the equation, then the point (v1,v2) falls in a circle)
Conservation of momentum:
v1_before + v2_before = v1_after + v2_after (if we fix one side of the equation, then (v1,v2) falls in a straight line)
The solutions are where circle and line intersect. I guess there's no paradox afterall.
I would like to do a transform to a moving reference frame, to make sure that everything is still alright. Transforming to a fast-moving reference frame will just make the circle bigger. Basically, the point and the line all get transposed diagonally up and to the right. The distance between the intersections still remains the same.
Oh I see, physics is fine. Nothing to worry about.
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The concept of kinetic energy has always been problematic for me. Given the choice, I'll integrate over force instead.
with "accelerating under constant force" (constant acceleration)
My intuition tells me that they should be the same, but kinetic energy considerations show that the acceleration is decreasing on the first one (it takes 4 times the energy to get 2 times as fast).
This would seem to contradict "velocity is relative": if velocity were relative, then the energy needed to get faster by 1m/s would be the same whether you are stationary or already at 1 m/s.
---
My intuition also tells me that I should be able to come up with a similar paradox about predicting the outcome of a 1-dimensional elastic collision. If you do it with energy vs momentum.
Conservation of energy:
v1_before^2 + v2_before^2 = v1_after^2 + v2_after^2 (if we fix one side of the equation, then the point (v1,v2) falls in a circle)
Conservation of momentum:
v1_before + v2_before = v1_after + v2_after (if we fix one side of the equation, then (v1,v2) falls in a straight line)
The solutions are where circle and line intersect. I guess there's no paradox afterall.
I would like to do a transform to a moving reference frame, to make sure that everything is still alright. Transforming to a fast-moving reference frame will just make the circle bigger. Basically, the point and the line all get transposed diagonally up and to the right. The distance between the intersections still remains the same.
Oh I see, physics is fine. Nothing to worry about.
---
The concept of kinetic energy has always been problematic for me. Given the choice, I'll integrate over force instead.
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Energy
Date: 2005-09-27 02:50 pm (UTC)Re: Energy
Date: 2005-09-27 02:56 pm (UTC)Re: Energy
Date: 2005-09-27 04:26 pm (UTC)Re: Energy
Date: 2005-09-27 04:33 pm (UTC)Re: Energy
Date: 2005-09-27 04:38 pm (UTC)Re: Energy
Date: 2005-09-27 04:51 pm (UTC)(keep the yeses coming...)
My argument is that the stationary car is identical to the moving car, modulo a transformation of reference frames.
Since we have no friction, we can imagine that the two cars move by shooting identical cannonballs at the same relative speed (e.g. 100 m/s away from the car). Does this shooting do the same amount of work in both cases?
Re: Energy
Date: 2005-09-27 04:54 pm (UTC)Re: Energy
Date: 2005-09-27 04:59 pm (UTC)But what happens if the cars are pushing the Earth? Is the difference in your answer microscopic, due to the movement of the Earth?
Re: Energy
Date: 2005-09-27 05:05 pm (UTC)Oh, if we are firing rockets then the fuel needed will be the same.
If the cars are pushing the Earth, the difference is huge, because in one reference frame the entire earth is moving at 10 m/s. That’s 5.9742 × 1026 joules (http://www.google.ca/search?q=%2810+m%2Fs%29%5E2+*+mass++of+earth+)!
Re: Energy
Date: 2005-09-27 05:23 pm (UTC)I'm not sure what you were doing, but I think you forgot the "1/2" factor.
But the point is that you don't need to make the Earth move at 10m/s in order for you to move, since its momentum (or more precisely, its moment of inertia) is huge. Cars obviously have no trouble accelerating to 20m/s, don't spend anywhere near that much energy.
Re: Energy
Date: 2005-09-27 05:38 pm (UTC)Right, I forgot the 1/2 factor.
Okay, consider a car at rest on the earth at rest. The car pushes off the earth and moves at 10 − &epsilon m/s, and the earth moves at ε m/s. (&epsilon is determined by conservation of momentum). Before the was no kinetic engery, now there is some. The engery was gained from burning the fuel. I believe the car has most of the kinetic engery, and the earth has almost none. This may be worth checking.
Anyhow, now consider a car at rest, on an earth moving at 10 m/s. The car pushes off the earth and moves 10 − &epsilon′ m/s. Now the earth is moving at 10 + &epsilon′ m/s. Now almost the same amount of engery is gained by the car in as before (maybe exactly the same if ε′=ε), but the energy gained by the earth is much more is this second case. That is where the extra fuel is needed.
Re: Energy
Date: 2005-09-27 05:56 pm (UTC)interesting. If you mean from the Earth-before's frame of reference, then I agree. This is due to energy being proportial to v^2 while momentum is proportional to v.
I see how this follows: for the same reason that 100^2 - 99^2 is ~200 times greater than 2^2 - 1^1. Likewise the ratio there will be 2*(10m/s) / epsilon.
But I'm not sure this would hold up in all reference frames (although it should if I trust physicists)
Ok. So I failed to make KE contradict itself.
But why should it make a difference whether you shoot rockets or push the Earth? Should this be our new energy efficient form of transportation? :-P
Re: Energy
Date: 2005-09-27 06:04 pm (UTC)The problem with rockets is that you need to carry your propellent with you. So the reality of rocket science is that you need exponentially more fuel the faster you want to go.
In our rocket example we were comparing two identical cars, one going from 0 m/s to 10 m/s, and the other 10 m/s to 20 m/s. But if you want to go from 0 m/s to 20 m/s, then you need to carry enough fuel so that when you reach 10 m/s, you have enough fuel to go to 20 m/s. That extra fuel is heavy, so it will cost more fuel to go from 0 m/s to 10 m/s then it will to go from 10 m/s to 20 m/s. This is the fundamental source of the exponential blow up.
Re: Energy
Date: 2005-09-27 06:12 pm (UTC)i.e. does it make a difference what order you throw things out?
constant-energy, or constant-speed?
I think it isn't... neither at constant-speed nor constant-work.
Re: Energy
Date: 2005-09-27 06:18 pm (UTC)Re: Energy
Date: 2005-09-27 06:20 pm (UTC)Re: Energy
Date: 2005-09-27 06:21 pm (UTC)Re: Energy
Date: 2005-09-27 06:22 pm (UTC)Re: Energy
Date: 2005-09-27 06:24 pm (UTC)Re: Energy
Date: 2005-09-27 08:05 pm (UTC)Re: Energy
Date: 2005-09-27 08:10 pm (UTC)Suppose you spend E1 when shooting the first cannonball, and E2 shooting the second.
Now go back. If you had shot them both at once, spending E1 + E2, then you should end up with the same speed.
Correct?
Re: Energy
Date: 2005-09-27 09:03 pm (UTC)Re: Energy
Date: 2005-09-29 08:36 pm (UTC)From your reasoning before, about pushing oneself off the Earth, it follows that it's more energy-efficient to shoot heavy things: you get more acceleration per rocket this way.
So if you had 2 identical cannonballs and two cannons, then you should shoot them simultaneously, rather than one after the other.
Oh, that makes sense, actually.
...maybe
Re: Energy
Date: 2005-09-27 07:16 pm (UTC)Re: Energy
Date: 2005-09-27 05:39 pm (UTC)