# Weight difference between an empty and a full memory stick

1. Dec 8, 2009

### Rebu

Problem:
"What is the weight difference between an empty memory stick and the same memory stick when it contains data?"
Can the answer to the previous question be proofed?

2. Dec 8, 2009

### Staff: Mentor

Just a quick note to the General Physics regulars. This is a continuation/retry of a thread from earlier tonight. The OP has clarified the question that he wants to ask, and to me, it's actually an interesting physics question that I'm curious about the answers to as well.

In addition, from my experiences with flash memory technology, some flash drives store information in differential cells, so there is no difference in stored energy, but there is a difference in stored information. I don't know if that makes a difference in the final physics/relativity answers, but I think it's an interesting angle to the OP's question.

So if there is a difference in the energy stored in information storage on a flash memory stick or chip, does that change the mass of the device? What if there is a difference in the information stored, but not in the amount of stored energy it takes to encode that information?

3. Dec 9, 2009

### sportsstar469

why was my reply deleted =p.

4. Dec 9, 2009

### Staff: Mentor

It looked like you were saying that you expected this thread to be locked as an unauthorized repost of the other problematic thread. I've been working with the OP to morph this into a valid thread where we can all learn something. That's why. Oh, and txt speak is not permitted here ;-)

5. Dec 9, 2009

### rcgldr

Seems it would depend if all 0 bits or all 1 bits would be heavier than some specific pattern.

6. Dec 9, 2009

### Stonebridge

It's an interesting question and the answer will come from first defining exactly what you mean by "information".
If all the bits were set to "0" would that be information? How much? If they were all set to 1 would that be information. If half are 0 and half are 1, is that information and how much more or less information is it compared to the other cases?
What is held in the memory stick may be information to one person, but not to another.
Information only has meaning or existence in conjunction with the one who is observing or decoding it.
From a consideration of these questions, it could be possible that the memory stick could hold different amounts of information to different people.
Information is not a form of energy, and as such, the information has no "weight". On the other hand, if storing the 1's and 0's involves a gain in stored energy, then of course there is a change in weight.
Just my thoughts.

7. Dec 9, 2009

### Rebu

Well one of the memory sticks would be unpartitioned and the other would contain a FAT32 partition with some music files on it.

Can the change in weight, if there is one, be proofed?

http://en.wikipedia.org/wiki/Flash_memory

8. Dec 9, 2009

### Galap

Hmmm... I guess this would take a little more info. does a memory device with '000000000000000000000000000000000' stored store more energy than '1010101001011110101010110101010101'? If so, by E=MC^2, it should be more massive.

9. Dec 9, 2009

### Andy Resnick

This is an interesting question, but the question should refer to 'mass' rather than weight.

If there is a difference in energy required to have a '1' or a '0', then the answer is clear (yes, because of E = mc^2)

If there is no difference in energy, there is still be energy associated with the entropy: kT ln(2) per bit of information, and again, there will be a change in mass.

So, my 8 GB memory stick at room temperature has a maximal information difference of 8*2^9*kT*ln(2) joules of energy (about 1*10^-17 J) which corresponds to 4*10^-17 fg. Not much.

10. Dec 9, 2009

### mikeph

Can you explain the energy associated with entropy? And how you do relate entropy with information on a microscopic level?

11. Dec 9, 2009

### Stonebridge

How do you define a change in amount or quantity of information?
In a single memory byte location, which of these has the most "information"?
00000000
11111111
00001111
11110000
If you claim to be able to allocate "mass" to information, which state has the most "mass"?

12. Dec 9, 2009

### uart

Ok I just measured it at about 10^(-17) kg heavier with music. That's for a 4GB memory stick. I'm not sure of the measurement error though as I used my kitchen scales.

13. Dec 9, 2009

### Andy Resnick

14. Dec 9, 2009

### Andy Resnick

I'm not an expert in information theory, but IIRC, a string of identical bits has *zero* information (and zero entropy), because of the way information is encoded in a signal. See the way entropy of a signal is defined here:

http://en.wikipedia.org/wiki/Information_theory

15. Dec 9, 2009

### IMP

If all 00000000's has the same mass as all 11111111's, then any combination in between should have the same mass I am guessing.

16. Dec 9, 2009

### kote

"The entropy, H, of a discrete random variable X is a measure of the amount of uncertainty associated with the value of X."

According to that link, entropy is a function of your expectations. If your options are 1 and 0 and you don't know which you're going to get, you have maximum entropy. The values that you actually get, 00000, 01101, etc, don't (typically) change the entropy level.

If you know that the sum of your bits is odd, you have 5 total bits, and you know what the first 4 are, then you can predict the final bit - there is no entropy or uncertainty associated with it. It is only when your bits are correlated like this that the actual series of bits has anything to do with total entropy.

This type of entropy doesn't have anything to do with physical E=mc^2 energy though. If it did, gaining knowledge about what's on my flash drive would change its mass. If my drive has all 0s and I expect a random sequence of 1s and 0s, it is at maximum entropy. If my drive has all 0s and I expect it to have all 0s then it has no entropy. You don't need a physical change to change your information entropy, so it can't be related to mass.

There's no inherent "amount of information" in a string of bits, no matter what they are. It all depends on what you are expecting - what algorithm you are using to encode or decode your bits.

The mass would change if the drive physically has more electrons on it when storing 1s or 0s and your density of 1s and 0s changes. That's the only way I can think of it changing though.

Last edited: Dec 9, 2009
17. Dec 9, 2009

### IMP

18. Dec 9, 2009

### kote

Interesting, but not quite related . The theory is that the universe is inherently discrete and made of some physical equivalent of bits. We don't exactly have the technology to make flash drives that store information at the qubit level (if there is such a level). We can't just count how much mass it would take to get 8gb worth of qubits (is there even a straight conversion?). Even if we could, we would have to know the mechanism by which the 1s and 0s are encoded into qubits. Which particle property corresponds with a 1 and which corresponds with a 0? Does the particle mass change when we change between these expressions of properties? Is compression used?

The real questions still are, how exactly are 1s and 0s physically encoded, and does the physical 1 have more mass than the physical 0.

On another note, a formatted flash drive and a full drive will typically have about the same random amount of 1s and 0s. Flash bits can only be flipped so many times, so the drive won't be reset to all 0s when it is formatted. The key file system data will be reset, and the rest of the bits are just left randomly set to whatever they are. This is the same as doing a "quick format" vs a full format in Windows.

The only exception is when you first get the drive from the manufacturer, and then it might all be 1s or 0s. Even then, smart drives will use all of the empty space before going back and resetting the used space, so all of those initial values are relatively quickly overwritten.

Last edited: Dec 9, 2009
19. Dec 9, 2009

### IMP

You could build a "flash drive" out of wood. It could have 1000 levers. Any levers in the up position are 1's, any in the down position are 0's. Why would the position of the levers change it's mass? All down, all up, any combination in between should have the same mass. Now the 32GB one is rather large...

20. Dec 9, 2009

### uart

Initially I made a guestimate based on typical structure of modern flash memory of about 10^3 electrons per floating gate (that is, per bit). Stupidly I took the mass increase to be the mass of these electrons (approx 10^(-27) kg per bit). That's nonsense of course, as each cell remains overall charge neutral and the electrons are just redistributed from one plate of the capacitor to the other.

Looking at it again I'll say approx $n\, q_e\, V\, /\,(4\,c^2)$ kg per bit. So based on n approx 10^3 electrons per bit and assuming V is a few volts, I get about 10^(-33) kg per bit as a serious guestimate for modern flash memory.

Last edited: Dec 9, 2009