Data has weight but only on SSDs
cubiclenate.com
cubiclenate.com
Therefore the mass (weight is a different thing, through it is proportional to mass at a given constant gravity potential) of the data on a SSD isn't fundamentally different from a HDD - they both are caused by a change of internal energy without any change in the number of fermions. I'd expect data on SSD to have larger mass change because a charged capacitor always store more energy than a discharged one, while energy of magnetic domains is less directional and depends mostly on the state of neighbor domains - but I'm not sure about this part.
[1] Thanks stackghost.
I believe TFA reads 2.43×10^-15 kg, not electrons. Unless SSDs are creating new and exciting physics, one can't have less than one electron, as it's an elementary particle.
That is definitely wrong! No way the source material has more electrons. The only way it could do that is by being charged.
Richard Feynman, The Feynman Lectures: "If you were standing at arm's length from someone and each of you had one percent more electrons than protons, the repelling force would be incredible. How great? Enough to lift the Empire State Building? No! To lift Mount Everest? No! The repulsion would be enough to lift a "weight" equal to that of the entire earth!"
From: https://tycho.parkland.edu/cc/parkland/phy142/summer/lecture...
Yes, but it's the same thing. The flux changes on the drive define the bits. It's probably true that a drive storing all 1's or all 0's would be quantitatively (but surely immesurably) lighter. But in practice a drive storing properly compressed high-entropy data is going to see a flux change every other bit on average. And all of those are regions of high magnetic field with calculable energy density. Same deal as charge in a capacitor, which also stores energy in the field.
TLDR: a given region of space can’t have more entropy than a black hole of the same volume. Rearranging terms, you find that N bits of information (for large N) has an equivalent black hole size, which in turn has a mass…
What happens when you XOR a 0 with a byte in a one time pad? You get the corresponding byte from the pad, right? And the block full of 1s or 0s is exactly a case the scrambler needs to be used.
It's not for compression or anything like that, it is used because the cells are packed so tightly that sharp changes in energy gradients over the space will even themselves out by donating electrons to their neighbors (flipping bits). It's kind of like rowhammer.
Instead of just a ont-time-pad, a more complicated and often proprietary scrambler is used. But it's towards the same end. It's a deterministic, pseudo-random bit sequence.
If you can bypass the controller (read directly from NAND) you will see the true values that it returns, and it will likely be scrambled even if the flash controller reports otherwise. This also allows you to recover the scrambler, since you know that the other side of the XOR operation is 0.
long since fixed but a common problem.
Interestingly, I heard that entrenched telco people were pushing for a much more complicated, SONET-ish approach. But classic Ethernet simplicity carried the day, and it's really nice...
That said, digital storage media has been somewhat pattern-sensitive for a century or more: https://en.wikipedia.org/wiki/Lace_card
I imagine MFM drives from 1985 might be a bit different from drives that are billions of times more data dense today. Back then, the drive didn't even control track width, the controller card did. And it was exposed to the OS.
I remember turning my "20MB", yes MB drive into a 30MB drive by messing with the track width. Of course, this was the time when people had Commodore 64 300baud modems, and would overclock them and get 450baud out of them.
In my computer club, we wrote a little piece of software to see which of us could get the highest bandwidth on a modem, one was even capable of just over 500baud!
After ranking, we all agreed to "trade down", so the guy with the fastest modem swapped his with the owner of the local Punter BBS. Everyone else traded so we still had the same ranking. That way, the BBS would always be able to support everyone at max speed, and everyone would still be "lucky" in terms of "next fastest modem".
I can't imagine that happening today.
EG, I literally thought of the controller and encoding as differing things, both separately called MFM. Ah well, it only took 40 years to discover differently.
Thanks for the link.
However, weight relative to what? All 0s on a chip will be heavier (the heaviest). All 1s would be the lightest. 50/50 1s and 0s would be the middle, which is where I’d expect generic “data” to fall.
In SLC flash, 10+ years is normal. Modern QLC is far more volatile: https://news.ycombinator.com/item?id=43739028
https://www.eejournal.com/fresh_bytes/how-do-you-weigh-a-pro...
Or the opposite, magnetic aligned fields for all 1’s or all 0’s?
Negligible now, but critically important effects to understand before we build a planet sized drive and wipe it!
Also, a planet sized drive will need to explicitly maintain large reserves of electrons. In theory, enough for an all ones (or zeros) state.
But that could be handled by tiling areas of one’s=high and zero’s=high. With tile charge flipping to maintain a balance in electron needs, locally and globally.
Time to replace "I'm zero surprised" with "That's a zero Shannon event"
Massless propulsion??
c is a really big number. c2 is a really really big number. E is small.
m is really really small.
Also would trimming cause a different value even though the data size remains the same ? I would think so, assuming I understand trim.
Every data storage media requires some work be done to it.
E=m^2
All data storage media has mass.
QED
As a counter-example, consider etching data in some form into another material, say stone or metal. You do work to remove the material you etch away, but you are removing material, so the final mass is actually less than you started.
That said, I believe most digital storage uses a high energy and low energy state to store 0 and 1, and in that case the high energy state will have (very, very slightly) more mass than the low energy state. But even then, having all bits in the high energy state would be the "heaviest", but would effectively have no data.
The first line
The data doesn't actually have weight because they aren't going to store a 1 or a 0, but rather do something like store 01 vs 10.
Perhaps I should have this carved on my tomb stone...
Assuming the platter is 100g, 42mm, spinning at 7200RPM, there is about 25J of rotational kinetic energy, whose mass equivalent is 2.8x10^-13g (0.28 femtograms).
Assuming 200 electrons per NAND floating gate with 3bits/cell TLC on a 2TB SSD, there would be 5.3x10^14 electrons, weighing about 0.5 femtograms.