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Does E really = MC squared?

  1. Jun 3, 2003 #1
    Have a running debate with a friend of mine that says if I run 1 mile then I will have burned the same energy when I walk a mile. I say that since E=MC2, then the ammount of energy I burn increases exponentially as I increase my rate of travel. Does anyone have a difinitive answer to this question? Could you possibly quote sources or provide a URL to an essay or article that explains this?
     
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  3. Jun 3, 2003 #2

    Integral

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    Your body must convert chemical potential energy to kinetic energy in order to run. Assume a flat track. Since kinetic energy is

    Ek=.5mv2

    since your your mass is pretty much constant, we only need be concerned with velocity, clearly, your velocity is higher if you run, therefore it requires more energy to run a mile then to walk.


    E=mc2 does not enter into the problem.
     
    Last edited: Jun 3, 2003
  4. Jun 3, 2003 #3
    Thanks

    Ok that makes sense. What about if two Cars travelled the same distance, but one went twice as fast. Would E=MC2 come into play then? If not, would you still use the Ek=.5mv2 equation to calculate the ammount of energy it takes, and if so wouldn't that have the same answer (yes the faster car would use/take more energy to get it from A to B)?
     
  5. Jun 3, 2003 #4

    chroot

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    E=mc2 does not come into play in any of these scenarios.

    The faster you move through air, the more drag you suffer -- and drag (wind resistance) depends on the square of the velocity. If you move twice as fast, you experience four times as much wind resistance. You have to provide the power to balance that wind resistance and continue moving at your desired pace.

    I will say that muscles are not very ideal physical things. For example, if you hold a 20 lb weight over your head in mid-air, exactly in position, you are doing no work -- in the physical sense. The position of the weight is not changing, therefore no work is being done. If you put the weight down on the table, you'll see that the table doesn't seem to have to do anything to support the weight -- the table is investing no energy. However, to hold the weight over your head is a very difficult thing, and it will eventually get you to huff and puff and feel tired. Why? Because your mucles are not rigid! The individual cells constantly contract and then release and then contract again. Each contraction takes further energy.

    So it's difficult to calculate how much energy you'll burn in running a mile -- very many physiological things come into play that make it a hard problem. For an ideal physical system, though, moving an object a mile down the road uses E = (force used to overcome friction) * (distance). The drag forces and so on are lumped into the friction force.

    So what does E=mc2 mean? Well, c is a velocity -- but a special one. It's the "natural velocity," or the only velocity in the universe that seems to have special qualities. Those qualities are:

    1) Anything moving at c (light, for example) appears to be moving at c for any observer who cares to look at it.
    2) c appears to be the ultimate limit of velocities. No matter what you do, you'll never be able to concoct a situation in which an observer will measure something going faster than c.

    E=mc2 unifies the concepts of mass and energy -- that's all. In many cases, people take c = 1 by using (for example) the second as a unit of time, and the light-second as a unit of distance. If you make this choice of units, then you just have

    E = m.

    Energy and mass then, in fact, are one and the same. The consequences of this relationship are far too numerous to list, but here are a few:

    1) An atomic bomb weighs a little less after it explodes.
    2) Magnets weigh a little more when they're separated than when they're together.
    3) The Sun is slowly losing mass, as it converts it to light.

    - Warren
     
  6. Jun 3, 2003 #5
    argh

    Ok Warren. Thanks for the post, but I'm not sure this really settles the bet. Would object X, consume more, less, or the same ammount of energy when travelling from point A (let's assume these objects are moving through a vacuum - drag was not intended to be part of the equation), to point B than object Z if object Z travelled twice as fast?
     
  7. Jun 3, 2003 #6

    chroot

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    If there is no friction, then no work is done at all in moving a body between two points at the same gravitational potential. For example, if you have an air-hockey table, it takes zero energy to move a puck from one side of the table to other. It takes energy to accelerate the puck, but the energy could be (in principle) reclaimed as you slow it back down.

    - Warren
     
  8. Jun 3, 2003 #7

    chroot

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    And the settlement of the bet is this: he's wrong. The energy used to run vs. walk a mile is certainly different, because of the physiological things I already mentioned, as well as the increased drag when running.

    However, your E=mc2 argument is just based on misconceptions.

    As a result, you win the bet, but not because you correctly proved him wrong.

    - Warren
     
  9. Jun 3, 2003 #8
    So I guess I didn't and still don't really understand what E=MC squared since I had previously believed that equation would apply to any object in motion. Energy = Mass x Velocity (squared). In other words I had previously believed that as the speed of an object increases the energy required to propel it also increases exponentially. By that logic, if you travelled the same distance at twice the speed, the energy required to get there would have increased significantly, but I guess this is wrong? I'm still unclear on this. Please advise.
     
  10. Jun 3, 2003 #9

    chroot

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    Yes, this is wrong. c is not any velocity, it is a single particular velocity. c is a conversion factor between the metric units of mass (kilograms) and energy (joules). c is not a variable in that equation, it's a constant. Its only purpose is to relate those metric units.

    If you choose to use "natural units," like the second and light-second, the conversion factor is just one, and you're left with just E = m.

    - Warren
     
  11. Jun 3, 2003 #10

    russ_watters

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    I'm not sure anyone still really explained why e=mc^2 doesn't apply here. It doesn't apply because it is a description of the conversion of matter to energy and vice versa. Ie, it only applies to nuclear reactions. There are no nuclear reactions going on when you run/walk a mile.
     
  12. Jun 3, 2003 #11

    chroot

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    That is absolutely incorrect. It seems correct because only nuclear reactions demonstrate the effect strongly enough to be easily measurable -- but in fact, E = mc2 applies to a great many systems not involving any nuclei. I gave a couple of examples.

    - Warren
     
  13. Jun 3, 2003 #12

    no, in a sence you are absolutly correct. the problem is that the effect is in no way noticable to objects traveling at running or driving speeds. it is only as one increases near the speed of light that he would be burduned down by the effects of E=MC2.
     
  14. Jun 3, 2003 #13

    chroot

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    Uh... thanks Maximus. Unless you mean "the effects of relativity," you are wrong.

    - Warren
     
  15. Jun 3, 2003 #14
    the effects of relativity and e=mc2 are very similar and in some cases dependant on each other. as one approaches the speed of light the energy inherent in their movement (because E=MC2) adds to their mass. if one didn't know E=Mc2 they'd have a hard time understanding the effects of relativity.
     
  16. Jun 3, 2003 #15

    chroot

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    Uh, maximus... E = mc2 is one of the equations derived from the postulates of relativity. They aren't two unrelated things. I took offense to your wording: you don't experience effects from equations; you experience effects from phenomena which are perhaps described by equations.

    - Warren
     
  17. Jun 3, 2003 #16
    You mean, result from equation (due to equation).
     
  18. Jun 3, 2003 #17

    i am sorry. i, of course, meant no offense.
     
  19. Jun 3, 2003 #18

    Integral

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    Lets point this back to the original question.

    To summerize what has been said.
    1. This is not a problem in relativity, because we are assuming speeds which we encounter in our daily lives, walking, running and driving.

    2. We are not converting mass to energy so E= mc2, which it a statement the rest mass energy equivelent is not a factor and need not be considered.

    3. This IS a problem of Newtonian Forces and Energy. So the important relationships are.
    Kinetic Energy
    i) Ek, = .5mv2
    and Newtons 3rd (?)
    ii) F = ma

    As well as conservation of Energy (to a certian extent)

    We will so need to understand

    iii) v = at (assume stating from v=0
    in these equations
    v=velocity
    a=acceleration
    t=time
    m=mass
    F=Force
    Ek= Kinetic Energy

    So, let us start with some body at rest (v=0) To increase the velocity to something greater then 0 we must apply a force. By equation ii, this force will cause an acceleration, over time by equation (iii) the acceleration will increase the velocity.

    Now by (i) we can see that the Kinetic energy of a body depends on the mass and the velocity, so we can see that how much energy is required to accelerate a body depends on the final velocity. This of course is a MINIMUN number, if there is any friction in the system it represents lost energy.

    Now by the conservation of energy we must be converting some form of energy to kinetic energy, for the human body and the car this is chemical energy stored in our muscles or in gasoline, we must expend an amount of energy at least equivilent to the kinetic energy of the moving body.

    Our body does not regain such energy when we slow down, if fact just the opposite is true, slowing down is every bit as much an acceleration as speeding up so we must spend more energy to stop.

    The human body is a very complex chemical engine, it is better to learn the fundamental physics by considering simpler systems.

    FURTHER POSTS SHALL NOT MENTION RELATIVITY! THERE IS NO NEED, ONLY CONFUSION THERE!
     
  20. Jun 3, 2003 #19

    russ_watters

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    Well actually, you gave 2 examples of nuclear reactions and one I wasn't familiar with. Point taken though. Nuclear reactions are by far the most commmon use for e=mc^2.
     
  21. Jun 5, 2003 #20

    pmb

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    I doubt that what you're friend said is true. The human body is a pretty complicated machine and as such does not work like a simple machine. When jogging the legs are doing more work since there is a lifting action which is absent in walking. Then there is other effects like cooling and the energy require to evaporate water etc. Far too complicated to derive in simple equations. I have a text on exercise phisiology - I'll look this up later tonight and get back tommorow.

    E = mc^2 only means that the total amount of energy you do results in a loss of total rest mass by the amount m = E/c^2. The relationship between mass and energy is a linear one. I.e. A doubling of the mass results in a doubling of the energy.\

    Pete
     
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