Okay; admittedly this guy has a PhD in Computer Science while I merely have a Bachelors in CS, but I’m sorry, this does not sound right to me.
Does the solid state memory in my Kindle really require a charge to maintain it’s state? It doesn’t use non-volatile memory? If the battery goes completely dead do I have to download all of my books again?
Even if an e-book doesn’t use non-volatile memory, and even if it uses dynamic ram, isn’t the charge of a given bit maintained whether or not the index claims it contains usable data? Is it really the case that a block of memory that isn’t indexed isn’t maintained?
Even if he’s correct about that. What about the charge in the battery? Doesn’t the charging and discharging of the battery cause greater swings in weight than the amounts he’s talking about?
And once again, someone proves that just because they have a doctorate in a subject doesn’t mean that they can’t be incredibly misinformed, or just plain stupid.
Well, a compressed spring is supposed to have a tiny bit more mass than an uncompressed one. So it is not beyond belief that a filled up flash memory would weigh a tiny but more (or less) than unfilled - since switching each bit state involves manipulation and trapping electrons which would change the total electric potential energy. If the resultant energy state when switching from 0 to 1 is higher, you’d get more mass when filling it up. If lower, less. Of course, that’s assuming that before stuff being stored there it is all zeroed-out (could be just left in indeterminate state at manufacture, I guess). Not sure exactly how to figure out just how much energy is involved though or where to find that out.
Oh and I would guess charging/discharging the battery would require higher energy changes than filling up memory.
As Terr said, the weight is from potential energy trapped in the device, which can exist without the device needing to be continuously recharged.
And once again, we’ve proved that its usually worthwhile to actually see what a person said rather then relying on the a fourth hand media report to relay relatively technical information before assuming that someone who presumably has spent a sizable portion of their life studying and thinking about a given topic is incorrect about that same topic.
I might have missed something in my last ten years on the board, but i don’t recall you having a PhD in physics or computer science. Do you have some level of expertise that qualifies you to evaluate the professor’s claim?
I’m not arguing that he’s right. He might, as the OP suggested, have missed something. I don’t know.
What i do know is that this sort of physics can be incredibly complicated, and that a guy who is a faculty member at one of the top universities in the country probably didn’t make the claim without at least some reasonable argument for its validity.
Do you have a particular reason for suggesting that he’s “misinformed” or “stupid,” or are you simply saying that because his ideas do not conform to your layperson’s understanding of physics?
And what are your credentials? I have a couple of engineering degrees and I fucking hate it when I try to explain a concept, usually after having been asked, and I get the roll eyes from someone who is ignorant and untrained because they don’t have a grasp on the subject and it doesn’t make sense to them or it’s counterintuitive.
If we can make some reasonable assumptions about the statistical average information content of a Kindleful of e-books, how much heavier does it make your brain once you read them all? Significantly, I would imagine. If so, since mass is conserved, all that weight has to come from somewhere. This suggests that the professor must be right – but as you absorb knowledgeable mass from the Kindle while reading it, that must suck all that weight out of the device. It’s all, uh, relativistically clear.
I read the article in the Times and thought about then. I think he’s right (qualitatively, at least; I wouldn’t know how to do the computation) although the explanation in the Times was pretty poor. Here’s my take on it. A Kindle with no books has its memory in a random state, let us imagine. (All bets are off if all bits are off.) That is a state of maximum entropy. To store a book in it requires that you expend energy. So I guess it is in a slightly higher energy state after than before.
But even if they couldn’t, it’s pretty clear from the story that this guy’s idea about the weight of a full Kindle is a theoretical calculation based on particular understandings of how matter reacts in certain situations, and on theories about the relationship between mass and energy.
You realize, i assume, that he didn’t reach his conclusion by filling his Kindle with ebooks and then placing it on his bathroom scale to see if he could detect any difference in its weight?
Apparently yes, but that is to measure an absolute mass of an attogram. Measuring a change in mass of an attogram in an object that weighs over a hundred grams is likely impossible at this point.
Take a look at the original article in the NY Times, which is actually pretty good (the one in the OP is complete shit). The explanation given there is basic physics:
Granted, you do need to know a little bit about the inner workings of solid state storage devices to come up with this, but given that the guy’s a full professor at Berkeley, I’m inclined to think he knows his stuff.
Thanks, that’s a lot clearer than the mangled version in the Telegraph. It also states the difference is a fraction of that caused by charging and discharging the battery.
Interestingly, the erased Flash is much more ordered than full Flash. Erased Flash is all 1’s, where a Flash full or books would appear pretty random, if you didn’t know what the data represented.
So if erased flash is all ones (do you have a cite for that?) then, for Prof Kubiatowicz to be correct, zeroed bits would have to weigh more than bits set to one, right?
That’s not impossible. You can represent 1s and 0s any way you want as long as you’re consistent and the hardware acts accordingly.
I’m not sure that that follows. Simply expending energy can’t always increase the weight. You expend energy to write a book to memory, but you also expend energy to erase it (or at least to remove it from the file index, bits are necessarily zeroed when a file is “erased”). So if expending energy increased weight, the device would gain weight every time you did anything with it!
Thinking about this some more; every time you store a new book, you expend energy and decrease the charge in the battery to do so. That means that, even if you are increasing the weight of the RAM, you’re simultaneously decreasing the weight of the battery.
Storing a book requires the expenditure of energy in tasks beyond just flipping bits in RAM. It has to power the radio receiver and transmitter, for example. Even if all of the energy expended went to just flipping bits, nothing is 100% efficient, so you’d still expend more battery energy then you’d end up storing in the memory.
So storing a book will decrease the immediate weight of the device.
However, if it is true that the weight of the ram increases as more information is stored, then the devices weight at X level of charge will be higher with more information stored than it will be with less.
