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The Weight of an e-Book 243

Posted by samzenpus
from the because-he-can dept.
whoever57 writes "According to Prof Kubiatowicz from Berkeley, each time an additional book is downloaded to an e-reader, the mass of the e-reader increases. The effect doesn't really make the devices more difficult to carry: the professor calculates that 4GB of books would increase its weight by a billionth of a billionth of a gram— about the mass of a single virus or DNA molecule."
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The Weight of an e-Book

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  • by mug funky (910186) on Monday October 31, 2011 @02:30AM (#37891860)

    in other news, ipods get heavier as you fill them.

    maybe "the singularity" will happen when the internet gets so heavy the Earth collapses into a black hole?

  • Re:Not true at all. (Score:4, Informative)

    by rdebath (884132) on Monday October 31, 2011 @04:40AM (#37892400)

    Flash memory cannot be 'quick formatted' each block has to be properly erased before use because writing can only turn 1's into 0's. (Obviously, the filesystem on a flash device can be quick formatted; but a recent OS will tell the flash about this, using the "TRIM" command, and the flash will erase all the blocks anyway.)

    Some flash drives even understand the NTFS filesystem well enough to erase unallocated blocks without help; but that seems a little dangerous to me.

    BTW: What Kubiatowicz seems to be saying is that pulling electrons from the substrate into the gates of a flash drive makes it heavier. So erasing the blocks, ie shorting them to ground, makes them lighter. So while downloading a book could make the device lighter, erasing the device will make it lighter still.

  • by xded (1046894) on Monday October 31, 2011 @06:29AM (#37892832)
    ... (which the editors should've linked to []), it states:

    “Although the total number of electrons in the memory does not change as the stored data changes,” Dr. Kubiatowicz said, the trapped ones have a higher energy than the untrapped ones. A conservative estimate of the difference would be 10^(-15) joules per bit.

    As the equation E=mc^2 makes clear, this energy is equivalent to mass and will have weight. Assuming that all these bits in an empty four-gigabyte Kindle are in a lower energy state and that half have a higher energy in a full Kindle, this translates to an energy difference of 1.7 times 10^(-5) joules, Dr. Kubiatowicz calculated. Plugging this into Einstein’s equation yields his rough estimate of 10^(-18) grams.

    Of course Kubiatowicz also says that:

    [10^(-18) grams] is only about one hundred-millionth as much as the estimated fluctuation from charging and discharging the device’s battery.

    Which is a far better comparison than the one obtained from The Guardian where Graeme Ackland of Edinburgh University stated:

    "If Prof Kubiatowicz is really struggling with the extra weight, he is welcome to come to Edinburgh where it's cooler, and the lack of thermal energy in his Kindle will more than compensate."

    Slashdot, home of crowdediting.

  • Re:Not true at all. (Score:4, Informative)

    by rgbatduke (1231380) <.ude.ekud.yhp. .ta. .bgr.> on Monday October 31, 2011 @06:44AM (#37892898) Homepage
    Just to be picky -- the default state is 1's, not 0's. NAND in particular has to start out all 1's and then "writes" turn some bits in a block into 0's. And the issue is whether or not the 1/default state is the ground state of any given bit (first) and whether or not there is some interbit interaction energy (second) and what the sign of that interaction energy is (third). One could in principle write a very simple model hamiltonian for the system that looks like:

    H = - \sum_i (A b_i +/- \sum_j B_{ij} (b_i - 1/2) (b_j - 1/2))

    where the first term represents the additional energy gained turning a 1 into a 0 and the second one (probably summed only over nearest neighbors) the energy gained or lost when neighboring bits are the same or different states. b_i is bit state, 1 (default) or 0.

    The real question, then, is whether or not A is zero or if it should be e.g. A(b_i - 1/2) (symmetric) and whether B_{ij} is positive, negative or zero. If I didn't make any algebra mistakes in writing this down. If A and B are known, of course, one can easily estimate the cost of writing 4 GB of data, and I'm guessing that's what TFA does (without reading it, of course, what would be the fun of that!).

  • Re:Harddrives? (Score:3, Informative)

    by TheRaven64 (641858) on Monday October 31, 2011 @08:04AM (#37893210) Journal
    He seems to be under the impression that the storage devices have three states: one, zero, and undefined. This is not the case. There is no undefined state, when flash is erased all of the bits are set to one, when it is written some are set to zero. There is no difference in energy state between the a block that is erased and a block that is storing all ones. It is possible that the zero energy state has more mass than the one energy state, but that's not what he is claiming. Expect to see this show up in The Guardian's Bad Science column soon...

My problem lies in reconciling my gross habits with my net income. -- Errol Flynn Any man who has $10,000 left when he dies is a failure. -- Errol Flynn