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E=MC squared

  1. Jul 15, 2003 #1
    E=MC squared? Assuming this is correct which I assume it is, a rechargeable battery should weigh more when charged, compared to un charged, has anybody actually put this to the test?
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  3. Jul 16, 2003 #2


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    Have you calculated how much mass would result from that increased energy? I don't believe it would be possible to measure that small a change.

    I believe it is possible to measure the increase in mass (via momentum) of elementary particles at high speed and the results have confirmed the mass increase due to relativity.
  4. Jul 16, 2003 #3
    What about a large battery then ? Ok maybe the amount of charge/mass would be to small to measure, but has anybody anywhere actually put this theory to a practical Physical test ?
  5. Jul 16, 2003 #4


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    Not by measuring batteries, but there have been many accurate tests.

    The problem with the battery idea is that it would be akin to measuring the weight of a mountain with a precision of an ounce.

    In order to be able to measure it, you eithr need very high speeds or energies. The effect can be clearly seen in nuclear reactors and in particle accelerators.

    In accelerators, for instance, you know what strength you need on a magnetic field in order to steer a proton (to keep it on a circular path); as the speed approaches that of light, much stronger fields are required. The reason is that it gained mass through E=mc2.
  6. Jul 18, 2003 #5
    Hey I just found this forum and decided to post to this thread as I think I can clear things up.

    I'm all sure were fond of the metric system and the mks system so here goes...

    The formula E=MC^2 gives you energy in joules when mass is in kilograms and the speed of light is in meters/second (approx 300*10^6)

    Now if I recall correctly some 9V batteries store somwehere around 10 Mj of electrical energy. we work this out to be 10,000,000=M*C^2
    Knowing C is approx 300,000,000 we divide: 10*10^6/300*10^6=0.0333... kg or.. 0.0000333.... grams. That is a whopping 3/100,000th's of a gram in one 9v battery... this of course is a direct energy to matter conversion, if one had half that much antimatter and reacted it with that much matter they would get an explosive reaction of 10 megajoules witch is a bit of energy to go off in your face spontaniously. Although remember now this is a chemical reaction not a nuclear annhilation one, the mass lost will be signifigantly less than that MUCH less... the energy that enters the chemical system as photons is stored as bond energy between atoms... you would have to work out the reaction to find out hom many Kj per mole were released and then convert the raw energy to it's mass equivalent.

    Keep in mind though this is excluding MANY things but I hope it should clear things up a bit.
  7. Jul 18, 2003 #6
    tar Peter_C and wow I,ll need some very accurate bathroom scales then
  8. Jul 18, 2003 #7
    One thing I may be forgetting in the above post is that the molecular energy is stored similar to kinetic energy... it is potential energy waiting to be released... For example a compressed spring holds no more mass than the same spring uncompressed. I think the mass of the battery might not vary at all. That is the total mass, including the gasses released by the battery. I think if you had such precise measuring equipment you would notice a big diffrence in the mass of the battery due to escaping gasses. Now say you have a pile of nuclear material undergoing steady decay, if you were to say measure all the energy released from the material you could find out how much (very little) mass was lost when the energy in the atomic glue (gluons) was released when the atoms underwent decay. Sorry for misleading you there for a min :-) hadn't slept in some time.
  9. Jul 20, 2003 #8


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    That's a bit high, the highest I could find stored around 1,924,560j

    Oops. You only divided by c here, not c² like you should have.

    By using the value I found, and dividing by c² we get 2.14 x 10-14 grams. A 9v battery masses about 30g so you would need a scale that was accurate to within 1 part in 1015
  10. Jul 20, 2003 #9


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    Not correct, as shown by the following example:

    Assume that you have two springs: A and B. The springs are compressed, so that when triggered, the two springs are pushed apart. As a result, spring B is pushed "backward" as spring A is pushed "forward" by the release of compression.

    Basic Parameters:
    Springs: 1kg each.
    Velocity each attains from the energy release: .01c

    We will also assume that there is a third reference C moving in the opposite direction of A at .99c relative to the two springs before were released (the original rest frame).

    First we analyse it using classical mechanics:

    For either mass to attain .01 c, it must have an KE of mv²/2 or:

    1(300000)²/2 = 4.5e12 joules

    Therefore, the total energy of the system as measured from the rest frame is 9e12 joules, either as stored energy before being triggered, or as the combined KE of both masses after being triggered.

    Now, let's look at things from the frame of C.

    At the start, both A and B are moving at .99c relative to C, so the total KE of the system at this point is:

    2(297000000)²/2 = 8.8209e16 joules
    plus the energy store of 9e12 joules equals

    8.8218e16 joules total energy.

    After the springs are triggered:
    A will have a velocity of .01+.99c =1c relative to C
    This gives it a KE relative to C of 4.5e16 joules.

    B will have a velocity of .99c-.01c = .98c relative to C
    This gives B a Relative KE to C of 4.3218e16 joules

    Adding the KE's of A and B together gives us a total energy of 8.8218e16 joules, the same value we got for total energy before the springs were triggered. Thus energy is conserved at all times from all frames, as it should be.

    Now by relativity:

    The Relativistic KE formula is KE = mc²(1/sqrt(1-v²/c²)-1)

    Therefore, in order for the masses to attain .01 c wrt to the original rest frame the total KE is

    (2)c²(1/sqrt(1-(.01c)²/c²)-1) = 9.0006750563e12 joules wrt to the rest frame.
    This is also the amount of stored energy needed for the compressed springs. Thus it is also the total energy of the system at any point during the test, relative to the rest frame.

    Again, at the start, both A and B are moving at .99c relative to C.
    So we just use the relativistic KE formula to find the KE relative to C.


    KE = 2c²(1/sqrt(1-(.99c)²)-1) = 1.095986169e18 joules

    Plus the 9.0006750563e12 joules of the stored energy equals


    In order to determine A and B's relative velocity wrt C after the springs are triggered, we must use the addition of velocities theorem W = (u+v)/(1+uv/c²)

    This gives us a relative velocity of A wrt C of
    (.99c+.01c)/(1+.99c(.01c)/c²) = .99019705c

    The KE of A relative to C using the Relativistic KE formula is:

    5.5434143392e17 joules

    B's relative velocity is found by
    (.99c-.01c)/(1-.99c(.01c)/c²) = .98979901

    B's relative KE wrt C is

    5.4170853919e17 joules

    Adding these two KE's together gives us

    1.0960499731e18 joules,

    Which is 5.48e13 joules more than what we got for the total energy of the system before the springs were released. We have more energy after than before, and energy is not conserved.

    The problem comes from not taking into account the mass of the stored energy of the springs. This works out to:

    9.0006750563e12/c² =.00010001 kg, which has to be added to the 2kg of our springs before we can calculate the KE of the system wrt C


    KE = 2.00010001c²(1/sqrt(1-(.99c)²)-1) = 1.096040974e18 joules

    Plus the 9.0006750563e12 joules of the stored energy equals

    1.0960499731e18 joules.

    Which equals the total energy of the system after the springs are released, and energy is conserved.

    Therefore, one can see that "stored" or potential energy does add to the mass of an object. If if didn't, energy wouldn't be conserved in all frames.
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