Teleporting an object 1 meter up gives it more energy the closer it is to the planet, because gravity gets weaker the further away it is. If you’re at infinity, it adds 0 energy to move further away.
I think your error is in not putting real axes on your phase space diagram. If going to the right increases your potential energy, and the center has 0 potential energy, then being to the left of the origin means you have negative potential energy? This is not how orbits work; a real orbit would never leave the top right quadrant of the phase space since neither quantity can be negative.
You also simply assume that arrows of the same length are imparting the same amount of energy, but don’t check; in reality, if you want the constant-energy contours to be a circle, the axes can’t be linear. (Since if they were linear, an object that has half its energy as potential and half as kinetic would be at [0.5, 0.5], which is inside the unit circle.)
(I’m assuming that when you say “momentum” you mean kinetic energy, but those are different things. You claim that any point on the Y axis has equal momentum and energy, but setting aside the fact that these quantities use different units, momentum is proportional to speed, while kinetic energy scales quadratically.)
You have misunderstood me in a couple of places. I think think maybe the diagram is confusing you, or maybe some of the (very weird) simplifying assumptions I made, but I am not sure entirely.
First, when I say “momentum” I mean actual momentum (mass times velocity). I don’t mean kinetic energy.
To highlight the relationship between the two, the total energy of a mass on a spring can be written as: E=(1/2)p2/m+(1/2)kx2 where p is the momentum, m the mass, k the spring strength and x the position (in units where the lowest potential point is at x=0). The first of the two terms in that expression is the kinetic energy (related to the square of the momentum). The second term is the potential energy, related to the square of the position.
I am not treating gravity remotely accurately in my answer, as I am not trying to be exact but illustrative. So, I am pretending that gravity is just a spring. The force on a spring increases with distance, gravity decreases. That is obviously very important for lots of things in real life! But I will continue to ignore it here because it makes the diagrams here simpler, and its best to understand the simple ones first before adding the complexity.
If going to the right increases your potential energy, and the center has 0 potential energy, then being to the left of the origin means you have negative potential energy?
Here, because we are pretending gravity is a spring, potential energy is related to the square of the potion. (x2). The potential energy is zero when x=0. But it increases in either direction from the middle. Similarly, in the diagram, the kinetic energy is related to the square of the momentum, so we have zero kinetic energy in the vertical middle, but going either upwards or downwards would increase the kinetic energy. As I said, the circles are the energy contours, any two points on the same circle have the same total energy. Over time, our oscillator will just go around and around in its circle, never going up or down in total energy.
If we made gravity more realistic then potential energy would still increase in either direction from the middle (minimum as x=0, increasing in either direction), instead of being x^2 it would be some other equation.
The x-direction is position (x). The y-direction is momentum (p). The energy isn’t shown, but you can implicitly imagine that it is plotted “coming out the page” towards you and that is why their are the circular contour lines.
I am making a number of simplifying assumptions above, for example I am treating the system as one dimensional (where an orbit actually happens in 2d). Similarly, I am approximating the gravitational field as a spring. Probably much of the confusion comes from me getting a lot of (admittedly important things!) and throwing them out the window to try and focus on other things.
Well that explains why you got the wrong answer! Springs, as you now point out, work opposite the way gravity does, in that the longer a spring is, the more energy it take to continue to deform it. (Assuming we mean an ideal spring, not one that’s going to switch to plastic deformation at some point.) So if we were talking about springs, you would be correct that the most efficient time to teleport the spring longer would be when it’s already as long as possible.
But we are not talking about springs, we are talking about gravity, which works differently. (Not only is the function going in a different direction, but also at a different rate. Gravity decreases as the inverse square of the distance, whereas spring force increases linearly with distance.) So your “simplification” is just wrong. You stated:
A weird consequence. Say our spaceship didn’t have a rocket, but instead it had a machine that teleported the ship a fixed distance (say 100m). (A fixed change in position, instead of a fixed change in momentum). In this diagram that is just rotating the arrows 90 degrees. This implies the most efficient time to use the teleporting machine is when you are at the maximum distance from the planet (minimum kinetic energy, maximum potential). Mathematically this is because the potential energy has the same quadratic scaling as the kinetic. Visually, its because its where you are adding the new vector to your existing vector most efficiently.
This is false. It takes more energy to move an object up by 1 meter on the surface of Earth than it does a million km away, because gravity gets weaker as you go further away. So if you want to maximize the gain in potential energy you get from your teleportation machine, you want to use it as close to the planet as possible.
(An easy way to see why this must be true is that an object’s potential energy at infinity is finite, so each additional interval of distance must decrease in energy in order for the sum of all of them to stay finite.)
Teleporting an object 1 meter up gives it more energy the closer it is to the planet, because gravity gets weaker the further away it is. If you’re at infinity, it adds 0 energy to move further away.
I think your error is in not putting real axes on your phase space diagram. If going to the right increases your potential energy, and the center has 0 potential energy, then being to the left of the origin means you have negative potential energy? This is not how orbits work; a real orbit would never leave the top right quadrant of the phase space since neither quantity can be negative.
You also simply assume that arrows of the same length are imparting the same amount of energy, but don’t check; in reality, if you want the constant-energy contours to be a circle, the axes can’t be linear. (Since if they were linear, an object that has half its energy as potential and half as kinetic would be at [0.5, 0.5], which is inside the unit circle.)
(I’m assuming that when you say “momentum” you mean kinetic energy, but those are different things. You claim that any point on the Y axis has equal momentum and energy, but setting aside the fact that these quantities use different units, momentum is proportional to speed, while kinetic energy scales quadratically.)
You have misunderstood me in a couple of places. I think think maybe the diagram is confusing you, or maybe some of the (very weird) simplifying assumptions I made, but I am not sure entirely.
First, when I say “momentum” I mean actual momentum (mass times velocity). I don’t mean kinetic energy.
To highlight the relationship between the two, the total energy of a mass on a spring can be written as: E=(1/2)p2/m+(1/2)kx2 where p is the momentum, m the mass, k the spring strength and x the position (in units where the lowest potential point is at x=0). The first of the two terms in that expression is the kinetic energy (related to the square of the momentum). The second term is the potential energy, related to the square of the position.
I am not treating gravity remotely accurately in my answer, as I am not trying to be exact but illustrative. So, I am pretending that gravity is just a spring. The force on a spring increases with distance, gravity decreases. That is obviously very important for lots of things in real life! But I will continue to ignore it here because it makes the diagrams here simpler, and its best to understand the simple ones first before adding the complexity.
Here, because we are pretending gravity is a spring, potential energy is related to the square of the potion. (x2). The potential energy is zero when x=0. But it increases in either direction from the middle. Similarly, in the diagram, the kinetic energy is related to the square of the momentum, so we have zero kinetic energy in the vertical middle, but going either upwards or downwards would increase the kinetic energy. As I said, the circles are the energy contours, any two points on the same circle have the same total energy. Over time, our oscillator will just go around and around in its circle, never going up or down in total energy.
If we made gravity more realistic then potential energy would still increase in either direction from the middle (minimum as x=0, increasing in either direction), instead of being x^2 it would be some other equation.
The x-direction is position (x). The y-direction is momentum (p). The energy isn’t shown, but you can implicitly imagine that it is plotted “coming out the page” towards you and that is why their are the circular contour lines.
If you haven’t seen phase space diagrams much before this webpage seems good like a good intro: http://www.acs.psu.edu/drussell/Demos/phase-diagram/phase-nodamp.gif.
I am making a number of simplifying assumptions above, for example I am treating the system as one dimensional (where an orbit actually happens in 2d). Similarly, I am approximating the gravitational field as a spring. Probably much of the confusion comes from me getting a lot of (admittedly important things!) and throwing them out the window to try and focus on other things.
Well that explains why you got the wrong answer! Springs, as you now point out, work opposite the way gravity does, in that the longer a spring is, the more energy it take to continue to deform it. (Assuming we mean an ideal spring, not one that’s going to switch to plastic deformation at some point.) So if we were talking about springs, you would be correct that the most efficient time to teleport the spring longer would be when it’s already as long as possible.
But we are not talking about springs, we are talking about gravity, which works differently. (Not only is the function going in a different direction, but also at a different rate. Gravity decreases as the inverse square of the distance, whereas spring force increases linearly with distance.) So your “simplification” is just wrong. You stated:
This is false. It takes more energy to move an object up by 1 meter on the surface of Earth than it does a million km away, because gravity gets weaker as you go further away. So if you want to maximize the gain in potential energy you get from your teleportation machine, you want to use it as close to the planet as possible.
(An easy way to see why this must be true is that an object’s potential energy at infinity is finite, so each additional interval of distance must decrease in energy in order for the sum of all of them to stay finite.)