This propulsion system won’t work like traditional systems that rely on a reaction mass and, therefore, conservation of momentum. Instead it will work more like reverse osmosis.
Reverse osmosis devices are used to make fresh water. They also conserve momentum.
Conservation of momentum isn’t just how conventional rockets work. It’s a law that we suspect applies universally and without exception.
In the reverse osmosis analogy, the “solution” is space-time and the “solute” is the vacuum fluctuations. In osmosis, the solution will try to equalize the concentration of solute.
Wordy analogy based reasoning of this kind does not reliably produce correct answers.
At best, reasoning like this can be used to generate a suggestion for what equations to consider. Because if the upside is a nobel prize, and the downside is wasting a few hours, it’s worth a go even if it’s probably wrong.
This relies on a person that understands the maths of quantum field theory.
In practice, there are a lot of people going “I have the ideas, I just need someone to add the maths”, and not that many people who understand the maths.
Looking at your equations, I think I can spot at least 1 mistake.
You say F=TdSdx from “Verlinde’s entropic gravity”. Verlinde’s work isn’t something I am familiar with, so I can’t say whether this is correct or not.
But, if it is, this is the force at 1 point. To calculate the overall force, we must take the integral.
If S tends to a constant (the background entropy of empty space) sufficiently fast as the distance from your spacecraft increases, then we can use the gradient theorem https://en.wikipedia.org/wiki/Gradient_theorem to show that all the forces must inevitably cancel out.
Consider this diagram.
The entropy needs to be the same at the far left and far right of the graph, because empty spacetime, far from any influence, has a fixed entropy. Your spacecraft sits in the middle. A small amount of your spacecraft on the far left experiences a steep gradient, and so a strong rightwards force. A larger amount of your spacecraft in the middle and right experiences a weaker leftward force.
And so it all adds up to 0 force in total.
This sort of everything cancelling out behavior is an inevitability assuming the equation F=TdSdx and
Flat spacetime
T and S are converge to a constant (and do so at least cubically fast, 2x the distance means at most 1⁄8 th as much variation in T and S.)
Reverse osmosis devices are used to make fresh water. They also conserve momentum.
Conservation of momentum isn’t just how conventional rockets work. It’s a law that we suspect applies universally and without exception.
Wordy analogy based reasoning of this kind does not reliably produce correct answers.
At best, reasoning like this can be used to generate a suggestion for what equations to consider. Because if the upside is a nobel prize, and the downside is wasting a few hours, it’s worth a go even if it’s probably wrong.
This relies on a person that understands the maths of quantum field theory.
In practice, there are a lot of people going “I have the ideas, I just need someone to add the maths”, and not that many people who understand the maths.
Looking at your equations, I think I can spot at least 1 mistake.
You say F=TdSdx from “Verlinde’s entropic gravity”. Verlinde’s work isn’t something I am familiar with, so I can’t say whether this is correct or not.
But, if it is, this is the force at 1 point. To calculate the overall force, we must take the integral.
If S tends to a constant (the background entropy of empty space) sufficiently fast as the distance from your spacecraft increases, then we can use the gradient theorem https://en.wikipedia.org/wiki/Gradient_theorem to show that all the forces must inevitably cancel out.
Consider this diagram.
The entropy needs to be the same at the far left and far right of the graph, because empty spacetime, far from any influence, has a fixed entropy. Your spacecraft sits in the middle. A small amount of your spacecraft on the far left experiences a steep gradient, and so a strong rightwards force. A larger amount of your spacecraft in the middle and right experiences a weaker leftward force.
And so it all adds up to 0 force in total.
This sort of everything cancelling out behavior is an inevitability assuming the equation F=TdSdx and
Flat spacetime
T and S are converge to a constant (and do so at least cubically fast, 2x the distance means at most 1⁄8 th as much variation in T and S.)
Donald, thanks for taking the time to explain this to me. I can see now why this won’t work.
I can’t thank you enough for patiently explaining the errors in my reasoning. Your thoughtful analysis is exactly what I was hoping for.
I had suspected that if it were this easy someone else would have surely thought of it by now and pursued this angle. And now I know why they didn’t.
I’ll leave this stuff to the physicists and mathematicians and get back to renovating our old house—I know that would certainly make my wife happier!