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A compressed spring has more mass

  1. Mar 8, 2012 #1
    My textbook says that a compressed spring has more mass than when it is not compressed. I'm assuming that this is because of E = mc2. A compressed spring has more energy and hence more mass. still, i have a real tough time wrapping my head around that. the spring has the same number of atoms. where does the extra mass come from?

    I put his in classical physics because it concerns springs which were described by Hooke in the 17th century, although Einstein is mentioned i think this belongs more to classical than modern physics.
  2. jcsd
  3. Mar 8, 2012 #2

    Ken G

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    Mass is not just a function of the number of atoms, it is a function of the energy, as you say. There's nothing more to "get" there-- what you were told (that mass is just about the number of atoms) is simply not correct, but it is so close to correct that we usually don't worry about it.
  4. Mar 9, 2012 #3
    I forgot that as an atom approaches the speed of light it increases in mass. Even though I don't understand why, I'll just accept it on faith for now. But in a compressed spring the atoms of the spring are not moving faster then when it's not compressed. Or maybe I'm wrong. Help me out here.
  5. Mar 9, 2012 #4
    You're assumption of this relating to E=mc2 is correct. An uncompressed spring has no potetnial energy, that is no ability to uncompress any more, whereas a compressed spring has some potential energy which was given to it by an external source, aka your fingers compressing it. This (potential) energy that has been given to the spring is equivalent to mass through E=mc2. The reason we usually don't worry on such things is that the mass equivalent to this amount of energy is so small, it is basically 0. Take 1Joule of energy, that is equivalent to 1E-17 kg!

    The question of why particle gain mass at high speed is the same principle. The faster a particle travels the more kinetic energy it has and this kinetic energy has mass.

    The difference between the spring and particle is one has potetnial energy and one has kinetic.

    Hope it makes sense!
  6. Mar 10, 2012 #5
    But I don't see why the particles in the spring are gaining speed if the spring is compressed. The only reason I can think of is due to the same reason why the particles in a gas in a container gain speed if the container is compressed, they start hitting each other more and the more they hit each other the faster they fly. I suppose that is the answer but I want to be sure.
  7. Mar 10, 2012 #6
    The particles in the spring aren't gaining speed. The spring is gaining potential energy not kinetic energy. Sorry for the confusion, I explained the speeding particle idea as you asked about it, but it is not applicable to a spring problem.
  8. Mar 10, 2012 #7
    I'm still confused. How about this. Say I have a ball, then I do some work an put it atop a 100m high ledge. It now has more potential energy. Are you saying that that ball now has more mass? If yes, then were is that new mass located in space. If no, then why does a compressed spring have more mass?
  9. Mar 10, 2012 #8
    Yes, the ball on top of the ledge has more potential energy, and therefore more 'mass'. This mass is not 'located' anywhere as such, mass is a form of energy. By giving the ball potential energy, you're not giving it more atoms, you're giving it energy which is equivalent to mass.

    It like asking where the kinetic energy of a speeding ball is located in space. It's not really located anywhere, its energy.

    Hope that doesn't confuse even more :-)
  10. Mar 10, 2012 #9
    you are not alone....more physics usually requires more ways to think about things.

    Einstein found that mass and energy and pressure and stress all affect gravitational attraction.....so we loosely say they affect 'mass' since all are linked via E=mc2
    That is not intuitive and is why it took an 'Einstein' to figure it out. Nobody knew that for thousands of years.

    You can consider that twisting and compressing a spring alters the relationship of the electrons to each other...their orbits get 'twisted' and misshapen as the metal is deformed for example so they are no longer in their lowest energy configuration. We say their degrees of freedom have been altered ever so slightly. Left uncompressed, the electrons will arrange themselves in their lowest allowable energy orbitals [if any were altered] and lattice configurations.

    You can also think of the compression as creating heat....it does......compress almost anything and it tends to get hot...and heat is a form of energy and hence a source of increased gravitational attraction['mass']. In this instance, kinetic energy of the electrons is changed. So if you just put the spring in an oven, for example, it will get ever so slightly more 'massive'; that is, will contain more energy.

    Yes. The work you did to lift the ball requires energy and that is now 'stored' in the ball. Release it and it can do more work that it could when closer to earth. What you have done is to separate two masses, moving the ball away from the earth. Separating uncharged masses takes work....since gravity is always attractive.

    Get ready for other challenges to everyday thinking: maybe to 'wrap your head around' the fact that 'gravity is like acceleration'.....and that clocks tick at different rates in different gravitational potentials and when moving at different relative speeds.

    Enough of these ideas have been experimentally verified, measured, to convince us we are on the right track...but not every single phenomena has been measured as some effects are so far to small.
  11. Mar 10, 2012 #10
    Bob, that ball and its 'potential energy' is a relation. The relation you discuss is the one relative earths gravitational field but assume a meteor hitting it on the ledge. As it hasn't developed the relation relative the ground, more than 'potentially', it will not express this added gravitational 'energy' when interacting with that meteor, as I think of it at least :)

    The 'potential' stuff is just that, 'potential'. And it's the relation and final interaction that will define the outcome. So you could say that this 'potential energy' is 'spread out' in space, potentially there, but only as a result of and in a interaction.

    The spring though has been compressed, which is an interaction, and a outcome. There's nothing 'potential' about that compressed spring.

    You can also think of it in form of the conservation laws. You impart a 'force' by compressing it, and using some 'lock' you force it to keep the 'energy' that compression brought with it. If you kick something so that it start to move, as that ball, the 'energy' it gets by your kick will 'bleed off' in time expressed as friction etc. But the spring can't bleed it off, although it should sooner or later rust, or otherwise decompose, :) and so disappear as a spring.
    Last edited: Mar 10, 2012
  12. Mar 10, 2012 #11
    Hello Bob

    The essential point is that energy has mass. The discovery of the equivalence of energy and mass was a fundamental one – and, despite its importance, it’s anything but obvious. As far as we know, that's just the way nature is built.

    Adding energy to something (whether by kicking it to make it move, heating it, or raising it up and putting it on a shelf) therefore adds mass to it. Basically, every single constituent part of the object is now a bit heavier.

    Compressing a spring involves doing some work (energy). The atoms in the spring are held in place by electromagnetic forces. These are as much part of the spring as the atom itself. Pushing the atoms closer together, or twisting groups of them out of their normal position, effectively stores energy in the electromagnetic fields between the atoms. And since the mass of the spring is comprised of the masses of everything in it – particles and fields – the mass of the spring is greater.

    As KenG has said, confusion can arise because some teachers and textbooks use the words ‘mass’ and ‘matter’ interchangeably.

    The basic particles of ordinary matter can, if it helps, be imagined as chunks of 'frozen' energy (this is just a rough way of thinking). And since energy has mass, the familiar particles of matter have mass. This mass, if the particle isn’t moving, is called its ‘rest mass’. Not everybody would like this description, but the basic idea is that there is no distinction between the energy that manifests itself as the rest mass of a particle, and other manifestations of energy, such as thermal energy or light energy. They are all energy.
  13. Mar 10, 2012 #12
    "Adding energy to something (whether by kicking it to make it move, heating it, or raising it up and putting it on a shelf) therefore adds mass to it. Basically, every single constituent part of the object is now a bit heavier."

    I've seen similar statements before :) And had some discussions about it elsewhere. What I really would like to see is a experiment proving it. It should be possible to measure the gain in rest mass by elevating something. That heat transfers energy and that this will increase the mass as long as it is heated is correct, but the ball?

    To my eyes the potential energy gained for the ball is in the earth(gravity)/ball system, not in the ball itself.
  14. Mar 10, 2012 #13


    Staff: Mentor

    This is correct. More precisely, to actually measure the gain in energy (or "mass"), you have to look at the system including both the Earth and the ball; just looking at either one in isolation will *not* show any increased energy (or "mass").

    For example, suppose I measure the mass of the ball by taking a very small object and putting it in orbit about the ball, measuring the orbital parameters, and applying Kepler's Third Law. Let's suppose that I can find a suitable test object for doing this, and that I can find an orbital radius for the object around the ball such that the effect of any other gravitating body (such as the Earth) on the test object's orbit about the ball is negligible. (For a real ball anywhere near the real Earth, these conditions probably can't be realized; but we're doing a thought experiment here, and in principle there will be *some* conditions that meet these requirements.)

    Given the specifications above, the "mass" of the ball, as measured by the orbital parameters of the test object, will be the same regardless of whether the ball is 1 foot above the Earth's surface or 1000 miles above it. The "potential energy" of the ball has no effect on this measurement.

    However: now consider a different measurement. Suppose I put a test object in orbit about the Earth-ball *system*; in other words, I put it in orbit far enough away from the Earth and the ball that the orbital parameters are determined by the combined properties of the system as a whole; any effects of the Earth alone, or the ball alone, are negligible. Then I measure the "mass" of the Earth-ball system the same way as I did for the ball alone: I measure the test object's orbital parameters and apply Kepler's Third Law.

    If I do this, I will find that the "mass" of the Earth-ball system is *larger* when I lift the ball to 1000 miles above the Earth, than it was when the ball was only 1 foot above the Earth. Here the "potential energy" of the ball *does* affect the measurement; but as yoron says, that means the potential energy is not really a property of the ball alone, but of the Earth-ball system.

    However, there is yet *another* wrinkle to this. The energy used to lift the ball from 1 foot to 1000 miles has to come from somewhere. Where did it come from? To really "balance the accounts" properly, we have to take this extra energy into account; and for the observed "mass" of the Earth-ball system to increase as I just said, the energy that lifts the ball has to come from *outside* the system.

    Suppose, for example, that I just throw the ball 1000 miles up. (I have *very* strong arms. :wink:) Does the "mass" of the Earth-ball system change? *No*; it does not. The energy I imparted to the ball came from me, and I am part of the Earth-ball system; so the energy stored in me counts towards the total mass of the Earth-ball system. In throwing the ball, I am not increasing that total mass; I am simply transferring some energy from me to the ball.

    But suppose that a small rocket engine, with fuel tank attached, is dropped to Earth from very far away. We attach the engine to the ball and fire it, and it turns out to have just enough energy to lift the ball from 1 foot to 1000 miles above the Earth. Then the total mass of the Earth-ball system will increase--but it will increase as soon as the rocket engine is received, *before* we lift the ball. The firing of the rocket, like my throwing of the ball, does not change the total energy of the system; it simply transfers energy from the rocket to the ball.

    Similar remarks apply to any system where "potential energy" can be stored and released; yes, the total "mass" of the system can increase when potential energy is stored (such as when a spring is compressed), but only if the energy that is being stored comes from outside the system. (If I compress the spring by muscular effort, the spring's mass increases because I am not counted as part of the spring "system". But if the spring is part of a larger system, such as a whole machine, and the spring is compressed by energy transferred from some other part of the system, the compressing of the spring does not change the mass of the total system.)

    I know this was a bit long-winded, but the issue is complex and deserves detailed discussion.
  15. Mar 10, 2012 #14
    Nope, not long winded at all. You went through it in a clear and logical fashion, and enjoyable too :) It all goes back to that scarlet pimpernel, doesn't it?

    'Energy' :)
    And what one mean by defining a 'system' of course.

    Both becomes tricky too me at last as I think of 'energy' as transformations primarily, even though I use the concept as freely as any other here :) And a 'system' as a rather open-ended definition if defined to reality, although not in theory. Even the concept of entropy seems to have problems with open ended systems.

    Btw: Thanks, it was a rather nice explanation.
    Last edited: Mar 10, 2012
  16. Mar 10, 2012 #15


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    Two quick comments - well, it morphed into more.

    The extra energy in the spring can ultimately be considered to be due to the electromagnetic fields that hold it together. Electromagnetic fields have energy, and you need to include it to get the total energy. I'm not aware of any good simple model of the electromagnetic field in a spring. If you try to consider the energy of an electromagnetic field in a point charge, for instance, you get infinity. So it's not clear exactly how to model it, though it is clear that's where the energy is stored.

    The second comment is that most of the above is a special relativity analysis, and the conclusions are about the special relativistic mass. It turns out there are at least three sorts of mass in GR, so the situation in GR is more complicated. If mass were as simple as it was presented, GR wouldn't need three different formulations of it.

    In particular, if you have a stationary metric (roughly speaking - one that isn't time-varying), pressure in GR causes "extra" gravity, above and beyond the special relativistic mass increase. This extra effect of pressure reflected in the defintion of the Komar mass density, which is rho + 3P / c^2 in reasonably flat space-time. (THere are some other factors needed if you have significant gravitational time dilation anywhere).

    Defininig the Komar mass requires that you have a stationary metric - more exactly a timelike Killing vector. Without this property, the above formula isn't correct and the Komar mass isn't defined.

    IT's quite common to have a system that's either stationary or quasi-stationary. It's not quite clear how to compute the error when the system is quasi-stationary though, at least i've never seen an analysis.
  17. Mar 10, 2012 #16
    The extra mass comes from the energy stored in the electromagnetic field. That means there are more photons interacting with the charged particles making up the atoms of the spring. Each photon has energy = hf (h - Planck's constant and f is frequency of the photon). The photon is a boson that carries force between electrically charged particles. To whatever extent there is an effect inside the nuclei of the atoms, you could have other force carriers contibuting to an increase in energy (slightly more gluons carrying the color force between quarks, or the quite negligible effect of W+ and W- bosons carrying the weak force between associated particles).

    When people make reference to the need to include the complete system in the analysis, they probably have in mind the need to include the additional bosons being exchanged between the relevant particles.

    Goodison_Lad has already mentioned the storing of energy in the electromagnetic field in an earlier post.
  18. Mar 10, 2012 #17


    Staff: Mentor

    Good point, I should have clarified that my remarks only apply to a system for which a "mass" can be defined in GR. For the "almost Newtonian" case of the Earth and the ball, either of two definitions of mass in GR apply, and both give the same answer (at least to the level of approximation we are using): the ADM mass, which applies to asymptotically flat spacetimes, and the Komar mass, which applies to stationary spacetimes. For the spring, with proper idealizations it is essentially stationary, so the Komar mass could be applied. I'm not sure about asymptotic flatness since we normally don't consider springs to be alone in the universe :smile:.
  19. Mar 10, 2012 #18


    Staff: Mentor

    This should be done, yes (at least if you're working at a level of detail where you need to model interactions as exchanges of bosons, instead of just modeling them more simply as potentials).

    But it's not really what I, at least, was talking about when I talked about including the "complete system". The point I was making was that whenever you apply a force to an object (such as compressing a spring or lifting a ball from a lower to a higher height), the energy used to apply the force needs to come from somewhere. If you include that energy as part of "the system", then the total energy, or "mass", of "the system" does *not* change when you compress the spring or lift the ball; energy just gets transferred internally from one part of the system to another. The only way, in these situations (bear in mind my previous comment that all this only applies in situations where a conserved "total energy" or "mass" can be defined in the first place), for the "mass" of "the system" to change is to introduce energy from somewhere else, from "outside" the system.
  20. Mar 10, 2012 #19
    It is tricky, isn't it :)

    What defines a 'system' and where to apply 'cut offs', if I may express it that way.
    But I get your gist Peter, and I think it make sense. And komar mass is something I really need to look up. I've seen it used a lot now in describing 'gravity', but I'm not sure how to see it.
  21. Mar 10, 2012 #20


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    There's also no such thing as a compressed spring in isolation - if a spring is compressed, something else somewhere is under tension to keep it compressed.

    There are two cases: you include the object keeping the spring compressed, to have a closed system, or you don't.

    If you've got a non-closed system, it's possible in general for different observers to disagree about what the mass of the system is - this is true even in SR. This makes the concept of the mass of a non-closed system rather ill-defined, because you'll get results that depend on your observer or your coordinate choice.

    If you've got a closed system, you need to consider the tension in whatever is compressing the spring, which decreases the total mass. The cancellation of the tension and pressure effects is complete in flat space-time, though it's not necessarily complete in non-flat space-time.

    So if you have a closed system in flat space-time where both the Komar and the special relativistic concepts of mass both apply and are well-defined in an observer-independent manner, you'll find that they are equal, but can be and usually are distributed differently.
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