Also, “erasure signals” aren’t a thing in my model, see below.
You need an erasure signal to erase a bit.
Let’s step back for a second and more formally define “erasure” and “bit”.
I see several meanings of “bit”:
A True Bit—A bit in computer RAM/registers etc—read/write controlled by arbitrary signal bits on signal lines.
A signal line bit propagating on a wire
A temporary ‘bit’ which is not necessary for 1 or 2 above, and thus isn’t really an irreducible bit
I believe to the extent your spinning wheel thing works, it is only ‘erasing’ type 3 non-bits, which is irrelevant. You need to show erasure of at least type-2, and really type-1 bits.
A reversible computer can already (in simplified theory at least) avoid all type 2 and 3 erasures completely, so it’s irrelevant how much energy they use. You need to show reliable type 1 bit erasure, or you should change your post to add disclaimers that your claimed ‘erasure’ doesn’t actually work for bits in RAM, and can’t actually reduce energy use for a reversible computer.
The entire wheel apparatus in your scheme is just a complex way of reversible propagation of bits on signal lines. We could just simplify that down to a single particle propagating on the signal line.
Yes, I also agree that moving bits around is not erasing them.
Right, and your flywheel thing so far is just moving an (irreducible) bit.
Generally, I don’t see a problem with using my procedure to erase a register in a computer. Seems like it would work fine for that, just like Landauer’s double well procedure would. It would also work fine to erase data directly in RAM, though I personally don’t think anyone’s going to ever build a reversible computer that works that way. Anyways, what I’m saying is my procedure would suffice to erase a Cannellian Type-1 bit.
The point of the wheel example is that you claimed that: “manipulating the wells is just equivalent to moving particles out of wells”. The wheel device provides a counterexample, since it changes the energy levels of the various states just by spinning. (It’s also an example of a situation where an “erasure signal” is not necessary to erase a bit, since it erases a bit on every clock tick.)
But I’m in general totally fine with having an erasure signal. When I say it’s not in my model, I mean that the basic erasure procedure I describe is independent of that kind of external detail.
Right, and your flywheel thing so far is just moving an (irreducible) bit.
Where is it moving to in your opinion? I provided you an argument that the answer isn’t the wheel. What’s left other than the environment?
Just also replying to a chunk of your earlier comment:
To actually reliably erase all the bits, the wheel needs to eject them from their low energy wells, which thus requires ΔE≈40KBT per bit, because each bit had to be in a well with energy that much lower than the neighborhood. So to the extent your wheel works it’s just equivalent to the domino erasure mechanism (which could be having all the dominoes linked to axle rods and levers to a motor etc).
You could try to use the wheel to implement a temporary well, but that doesn’t work because it is not sufficient to simply lower the well as that doesn’t give the particle enough energy to move out of its location reliably.
After lowering the wall between the two states, I do then actually raise up one of the states high enough to kick the particle out of it with high reliability. The reason this doesn’t cost the full 40kT is that as the energy of the state gets higher, the probability of the particle occupying that state is already decreasing. I only pay any energy cost if the particle happens to be occupying the state I’m currently lifting. So you do an integral to get the expected energy cost of kicking this particle all the way out, and it comes out to way less than 40kT. In fact, it comes out to exactly kTlog2 if we’re moving so slowly that the particle is always distributed over states according to the Boltzmann distribution. But if you want to finish quickly, it comes out to a little bit more.
Generally, I don’t see a problem with using my procedure to erase a register in a computer. Seems like it would work fine for that, just like Landauer’s double well procedure would. It would also work fine to erase data directly in RAM,
You haven’t described any mechanism to do this yet.
Your flywheel thing (to the extent I now understand it), just seems to be equivalent to representing a signal bit moving down a wire, which could just be simplified to a single minimal particle. You haven’t described anything that is capable of true bit erasure.
True bit erasure would require having a control signal which then controls the wheel, or something like that.
You haven’t described any mechanism to do this yet.
However you’re imagining Landauer’s double well procedure raises the energy level of one of the wells conditional on a control signal, you should be able to imagine my procedure being implemented in a similar way. What obstacle do you run into when you try to do this?
Just to give an explicit example, though, you can imagine the control signal being used to determine the position of the particle that encodes the bit. You have some states close to the wheel, and some far away. Far away states aren’t affected by the spinning of the wheel and if the particle is in the far states rather than the near states, then the wheel isn’t slowed down and the bit isn’t erased.
Choosing whether the particle goes into the near states or the far states conditional on a control bit is a reversible process, of course, so that completes the implementation.
However you’re imagining Landauer’s double well procedure raises the energy level of one of the wells conditional on a control signal, you should be able to imagine my procedure being implemented in a similar way.
I do not imagine it this way, as trying to manipulate the barrier just punts the issue. It’s simpler to imagine static wells which the particle/bit moves between, as in dominoes.
Just to give an explicit example, though, you can imagine the control signal being used to determine the position of the particle that encodes the bit.
This is not a true memory bit storage mechanism unless the particle is in some sort of loop in a low energy potential well about 1eV lower than the background. The control signal could come at any time or never and the bit must be stable for very long periods of time.
“Determining the position of the particle” requires deflecting it into 1 of 2 paths with high reliability, which requires energy on order 1eV. If you can reliably alter the particle’s path—you no longer need the wheel, as one path can just exit the device or thermalize to erase. Any attempt to recover the energy of the erased particle is a dead end which just punts the problem to a new mechanism for erasure.
The gate which is formed from this process has 2 unique bit inputs (memory bit, control wire bit) and only 1 unique bit output (the control wire bit), as the other is erased. It is fundamentally irreversible—it doesn’t matter how many complex additional mechanisms (flywheels) you tack on, they are just distractions.
I do not imagine it this way, as trying to manipulate the barrier just punts the issue. It’s simpler to imagine static wells which the particle/bit moves between, as in dominoes.
A machine which turns a domino right side up is in fact raising the energy of the “domino tipped over” state. If the domino stayed tipped over, the mechanical arm pushing the domino up would at some point be intersecting with the domino, a state of very high energy indeed. Unless you’re imagining some other way for the domino resetter to function?
This is not a true memory bit storage mechanism unless the particle is in some sort of loop in a low energy potential well about 1eV lower than the background. [...] “Determining the position of the particle” requires deflecting it into 1 of 2 paths with high reliability, which requires energy on order 1eV.
Yes, you need to be 1eV below extraneous nearby states to be doing anything sensible at all. This is an energy difference and not an energy cost. You’re not dissipating any energy because the particle is not occupying those states. Determining the position of the particle based on a control bit is a reversible operation. If you want to argue that it must cost 1eV, you need to explain either why the operation is not actually reversible, or why it must have a thermodynamically unavoidable energy cost anyways, despite being reversible.
The control signal could come at any time or never and the bit must be stable for very long periods of time.
So the particle is in the far away states by default and the control signal kicks it into the nearby states when the bit has to be erased.
If you can reliably alter the particle’s path—you no longer need the wheel, as one path can just exit the device or thermalize to erase. Any attempt to recover the energy of the erased particle is a dead end which just punts the problem to a new mechanism for erasure.
I’m certainly not saying the wheel part is necessary, it’s just a particular example I gave. I do think it’s allowed by thermodynamics to reliably and costlessly alter the particle’s path in ways that are reversible. I’m not trying to recover the energy from the erased bit (I assume you meant “bit” here, as particles aren’t erased). I’m spending the kTlog2 (on average), and then that’s the end of it.
The gate which is formed from this process has 2 unique bit inputs (memory bit, control wire bit) and only 1 unique bit output (the control wire bit), as the other is erased. It is fundamentally irreversible—it doesn’t matter how many complex additional mechanisms (flywheels) you tack on, they are just distractions.
No, the whole point of controllable erasures is that sometimes the bit isn’t erased. So we need 2 outputs, one for the control wire as you say, and one for the original bit since sometimes the erasure signal is 0 and so the bit isn’t erased. The gate is still irreversible, of course, but it seems like your mental model of the situation here is quite different than mine, so I’m pointing out the difference in the hopes of some clarification.
You need an erasure signal to erase a bit.
Let’s step back for a second and more formally define “erasure” and “bit”.
I see several meanings of “bit”:
A True Bit—A bit in computer RAM/registers etc—read/write controlled by arbitrary signal bits on signal lines.
A signal line bit propagating on a wire
A temporary ‘bit’ which is not necessary for 1 or 2 above, and thus isn’t really an irreducible bit
I believe to the extent your spinning wheel thing works, it is only ‘erasing’ type 3 non-bits, which is irrelevant. You need to show erasure of at least type-2, and really type-1 bits.
A reversible computer can already (in simplified theory at least) avoid all type 2 and 3 erasures completely, so it’s irrelevant how much energy they use. You need to show reliable type 1 bit erasure, or you should change your post to add disclaimers that your claimed ‘erasure’ doesn’t actually work for bits in RAM, and can’t actually reduce energy use for a reversible computer.
The entire wheel apparatus in your scheme is just a complex way of reversible propagation of bits on signal lines. We could just simplify that down to a single particle propagating on the signal line.
Right, and your flywheel thing so far is just moving an (irreducible) bit.
Generally, I don’t see a problem with using my procedure to erase a register in a computer. Seems like it would work fine for that, just like Landauer’s double well procedure would. It would also work fine to erase data directly in RAM, though I personally don’t think anyone’s going to ever build a reversible computer that works that way. Anyways, what I’m saying is my procedure would suffice to erase a Cannellian Type-1 bit.
The point of the wheel example is that you claimed that: “manipulating the wells is just equivalent to moving particles out of wells”. The wheel device provides a counterexample, since it changes the energy levels of the various states just by spinning. (It’s also an example of a situation where an “erasure signal” is not necessary to erase a bit, since it erases a bit on every clock tick.)
But I’m in general totally fine with having an erasure signal. When I say it’s not in my model, I mean that the basic erasure procedure I describe is independent of that kind of external detail.
Where is it moving to in your opinion? I provided you an argument that the answer isn’t the wheel. What’s left other than the environment?
Just also replying to a chunk of your earlier comment:
After lowering the wall between the two states, I do then actually raise up one of the states high enough to kick the particle out of it with high reliability. The reason this doesn’t cost the full 40kT is that as the energy of the state gets higher, the probability of the particle occupying that state is already decreasing. I only pay any energy cost if the particle happens to be occupying the state I’m currently lifting. So you do an integral to get the expected energy cost of kicking this particle all the way out, and it comes out to way less than 40kT. In fact, it comes out to exactly kTlog2 if we’re moving so slowly that the particle is always distributed over states according to the Boltzmann distribution. But if you want to finish quickly, it comes out to a little bit more.
You haven’t described any mechanism to do this yet.
Your flywheel thing (to the extent I now understand it), just seems to be equivalent to representing a signal bit moving down a wire, which could just be simplified to a single minimal particle. You haven’t described anything that is capable of true bit erasure.
True bit erasure would require having a control signal which then controls the wheel, or something like that.
However you’re imagining Landauer’s double well procedure raises the energy level of one of the wells conditional on a control signal, you should be able to imagine my procedure being implemented in a similar way. What obstacle do you run into when you try to do this?
Just to give an explicit example, though, you can imagine the control signal being used to determine the position of the particle that encodes the bit. You have some states close to the wheel, and some far away. Far away states aren’t affected by the spinning of the wheel and if the particle is in the far states rather than the near states, then the wheel isn’t slowed down and the bit isn’t erased.
Choosing whether the particle goes into the near states or the far states conditional on a control bit is a reversible process, of course, so that completes the implementation.
I do not imagine it this way, as trying to manipulate the barrier just punts the issue. It’s simpler to imagine static wells which the particle/bit moves between, as in dominoes.
This is not a true memory bit storage mechanism unless the particle is in some sort of loop in a low energy potential well about 1eV lower than the background. The control signal could come at any time or never and the bit must be stable for very long periods of time.
“Determining the position of the particle” requires deflecting it into 1 of 2 paths with high reliability, which requires energy on order 1eV. If you can reliably alter the particle’s path—you no longer need the wheel, as one path can just exit the device or thermalize to erase. Any attempt to recover the energy of the erased particle is a dead end which just punts the problem to a new mechanism for erasure.
The gate which is formed from this process has 2 unique bit inputs (memory bit, control wire bit) and only 1 unique bit output (the control wire bit), as the other is erased. It is fundamentally irreversible—it doesn’t matter how many complex additional mechanisms (flywheels) you tack on, they are just distractions.
A machine which turns a domino right side up is in fact raising the energy of the “domino tipped over” state. If the domino stayed tipped over, the mechanical arm pushing the domino up would at some point be intersecting with the domino, a state of very high energy indeed. Unless you’re imagining some other way for the domino resetter to function?
Yes, you need to be 1eV below extraneous nearby states to be doing anything sensible at all. This is an energy difference and not an energy cost. You’re not dissipating any energy because the particle is not occupying those states. Determining the position of the particle based on a control bit is a reversible operation. If you want to argue that it must cost 1eV, you need to explain either why the operation is not actually reversible, or why it must have a thermodynamically unavoidable energy cost anyways, despite being reversible.
So the particle is in the far away states by default and the control signal kicks it into the nearby states when the bit has to be erased.
I’m certainly not saying the wheel part is necessary, it’s just a particular example I gave. I do think it’s allowed by thermodynamics to reliably and costlessly alter the particle’s path in ways that are reversible. I’m not trying to recover the energy from the erased bit (I assume you meant “bit” here, as particles aren’t erased). I’m spending the kTlog2 (on average), and then that’s the end of it.
No, the whole point of controllable erasures is that sometimes the bit isn’t erased. So we need 2 outputs, one for the control wire as you say, and one for the original bit since sometimes the erasure signal is 0 and so the bit isn’t erased. The gate is still irreversible, of course, but it seems like your mental model of the situation here is quite different than mine, so I’m pointing out the difference in the hopes of some clarification.