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Why is energy flat?

  1. Mar 1, 2010 #1
    I was thinking about the equation

    e = mc^2

    and started to get seriously confused.

    The confusion didn't come from this equation, but from the dimensionality of it, and then
    the dimensionality of energy itself.

    So a standard formula for energy for the most part.
    e = kg (m/s)^2
    e = kg * m^2 * s^-2

    Breaking this up... we have a scalar... kg. We have an acceleration s^-2 curve. Then we have this Plane m^2 for distance. My initial reaction is shouldn't this be m^3 ???

    This definition for energy (a very common one) seems to be missing a dimension. Energy is 3 dimensional right? An explosion of force or energy explodes in all directions, and not just on a plane.

    Just so you follow me on this...

    A one dimensional unit will be like (m).
    A 2 dimensional unit, a plane, will be like (m^2)
    A 3 dimensional unit, a sphere, will be like (m^3)

    a volume for a sphere is 4/3 pi * r^3
    volume for a cube is m^3

    So I am confused. I have looked at other equations that do this same thing.

    E = m c^2 .... E = M * (constant * constant) *(m^2)*(s^-2)

    so Energy is flat?

    I am confused. Help me out here.
    -Robert Powell
  2. jcsd
  3. Mar 1, 2010 #2


    Staff: Mentor

    No, energy is a scalar, but I don't think that has anything to do with the units. Torque has the same units as energy but is a vector. So the number of dimensions of a quantity and its units have little to do with each other.
  4. Mar 1, 2010 #3
    The number of spatial dimensions has nothing to do with the dimensionality of energy. Energy would have the same dimension as you derived above if space was 100-dimensional. The fact that energy involves the square of speed (m/s) is the reason you get an (m/s)^2.

    Energy is not a geometrical object, so it makes no sense to ask if it is flat.

  5. Mar 1, 2010 #4
    I don't think you understood the question.

    Picture this for a moment, you can see the mathematics of an equation.
    A single pulse from a singular source = PI * (r^2) * s^-1

    A pulse that accelerates or has a non linear curve will have some other s^-1. like s^-2.
    Much as the curve 1/r or 1/r^2 are different.

    So break down the components of the dimensionality for a minute, you have a propogation, a mass, and a planer surface. unless the planer surface is spherical, it makes no sense.
  6. Mar 1, 2010 #5
    I figured it out. Energy does propogate in all directions. But not as a solid sphere, but as a planer surface wrapped around a sphere that is hollow.

    Another example would be to look at a point in the center of a circle. You would think that the circle would represent an area, so you would need some distance squared... but if the circle is hollow, then all you need is the line around it, which dimensionally is just a distance.

    Einstein always said he wanted to have everything explained as a geometric function, and this does explain some things.

    If you read e = mc^2 from the left to right you see energy = mass over some moving surface. If you read it from right to left though. you see that a moving surface of mass = energy.

    got it, move along.
  7. Mar 1, 2010 #6


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    Robert, as Dale and torquil said, units have nothing to do with any geometrical representation, like a surface or a sphere or whatever. Energy doesn't have to "propagate". Potential gravitational energy of a stone on a mountain is just a number. The heat content of hot water is just a number. It is not "flat", or "hollow", or whatever.
    You could just as well say that US dollars are "flat" or "hollow" and "have to propagate".
  8. Mar 2, 2010 #7
    What follows is an aside from the regular thread, but interesting in and of its self.

    I have to totally disagree.

    I am a computer scientist. This is an under utilized talent actually that has been taught to me by many professors. Every part of an equation means something.

    If you look at a variable equation:
    y = x. you should see a line. In 3 dimensional space, a plane.
    y = x^2 you should see a parabola. In three dimensional space it is an infinite plane along z, but a curved plane none the less.
    When you see (1/sigma * sqrt(2pi)) * e ^ ((-1/2) * (x - Mu / sigma)^2)
    You should see a normal distribution. What the shape of the distribution is, is determined by the values of the various parts of the equation, but irregardless of the actual values... that generally will look like some sort of a normal distribution.

    1 = x^2 + y^2, you should see?

    As one professor explained one time when I asking about some complex formula in a book somewhere. "The easiest way to explain what is going on is mathematics. But most people don't understand how to 'read' math and understand that the representation of the mathematical form generally comes as a simplification or generalization of an observation so others can 'read' it. If someone explained that (1/sigma * sqrt(2pi)) * e ^ ((-1/2) * (x - Mu / sigma)^2) types of equations were just normal distributions, it would not be as impressive when writing papers."

    But unfortunately, this is not as common or simple until it is explained.

    I understand it does not work easily in every equation, a scalar value may just be a color or intensity, etc, when you have 100 dimensions, it gets tricky. But even then, I am constantly asked to "Visualize this" on a computer. So for simpler equations, these things should be simple, or at least understandable.
  9. Mar 2, 2010 #8
    I am not taking sides here but I have a question for Robert.

    What happened to time in your discussion?

    BTW you should not equate dimensions as understood by the subject of dimensional analysis in physics with geometrical dimensions as understood in some systems of mathematics.
    Last edited: Mar 2, 2010
  10. Mar 2, 2010 #9


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    Yes. I would think that most people here on these boards do "see" those things the way you describe them. Indeed, x^2 + y^2 = 1 is a unit circle. All that is nice and well. We can have simple geometrical representations of these simple relationships. A "plot", so to say. But not all relationships have to be seen as representing physically "their plot". The plot of a function can be an interesting tool to understand certain behaviours, but don't go and think of every plot as being a physical representation of the relationship at hand.

    If you have: E = 1/2 m v^2, you can make the plot of E as a function of m, for constant v, and have a straight line, and you can make the plot of E as a function of v, with constant m, as a parabola, or you can see E as a curved surface representing the function f(x,y) = 1/2 x y^2.

    But that doesn't mean in any way that these PLOTS represent anything "physical". Kinetic energy is not a kind of "parabolic trough" in (physical ?) space at all. Yes, the mathematical relationship, when transposed into a geometrical relationship in the variables m and v, is plotted that way, but there's not much use in thinking of that "physically". Of course, it can be helpful to think of that plot *mathematically*, to understand the behaviour of the relationship (for instance, that an increase of v by 10 m/s has more effect on E when we already have a large "starting" velocity, than when we start from standstill, explaining why it is harder to accelerate from 150 km/h to 160 km/h than from 10 km/h to 20 km/h for instance).
    But there's not much of a geometrical figure in space attached to this, simply because the quantities in this relationship are not spatial coordinates.
    Last edited: Mar 2, 2010
  11. Mar 2, 2010 #10


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    This might indeed be the key confusion here.
  12. Mar 2, 2010 #11

    Vanesch, I understand your point. But there is a huge amount of physical evidence that can prove you utterly wrong when you think kinetic energy is not a parabolic trough through space.

    While it is foreign to think of it that way I grant, that is exactly what happens when you fire anything through anything. A bleeding off of energy happens that initially is nothing but a parabolic trough.

    Seconds 34 and 35 in this video, show a perfect example of the parabolic trough.

    Another example.

    I get it though, not all things behave simply. But you can infer many things once something is known.
    Last edited by a moderator: Sep 25, 2014
  13. Mar 2, 2010 #12
    Robert, I would counsel beeing very very cautious when applying mathematics to physics.

    There are whole provinces of mathematics that have no known counterpart in the physical word.

    Equally there are some notions in the physical world that are not directly reflected in mathematics.

    In particular when we draw a graph ( I mean an x,y,z....plot here not another sort) in mathematics the axes are just numbers. There is no reason to favour one axis over another.

    In Physics these axes also have units, which makes them different and yields the rich panoply of relationships that makes up Physics. Sometimes these axes can be continued in a numerical sense, but make no physical sense (eg continuing the temperature axis to the negative numbers). But mathematically there are still numbers there on the axis.

    We can also achieve nonsense by asigning units to our axes. For instance plotting the number of shirts sold in Idaho between 1910 and 2010 against the weights listed in the British Standard "List of Weights of Building Materials", is a perfectly acceptable mathematical process - they are both just a list of numbers, but make no sense at all in Physics.

    Finally there are phenomena in Physics for which we currently have no Mathematics.
    the Hydraulic Jump is one such, owing to the current definition of a function.

    PS I ask again

    What happened to time in your musings?
  14. Mar 2, 2010 #13


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    You seem to be trying to equate energy to a physicality that does not exist. Energy does not have to have any geometric analog. As previously mentioned, a static system like an object raised in a gravitational potential has energy but no geometric properties to this energy can be attributed. Likewise, if I have two molecules with dipole moments, then rotating one of the molecules about its axis will change the energy of the system. I can just as easily make arguments about the flow of energy being in one, two or three dimensions. In addition, shouldn't the flow of energy have an added set of dimensions to represent it's speed of propagation? In such case you would be talking about power, not energy. Finally, dimensional analysis can be a misleading tool as it is in this case. Why separate the components out into mass, length and time? Why not use just mass and speed? If we took mass and speed it would be more analogous to the familiar situation of the kinetic energy of a moving object. But this is misleading because there is also potential energy stored in static systems.

    EDIT: I also fail to see how your linked videos make any evidence of energy bleeding off as a "parabolic trough" as you put it. Those videos are not measuring energy, they are simply footage of ballastic eye candy. If you are you purporting that the cavities created by the shockwave somehow indicate a parabolic trough then you are mistaken. The cavities are created from the pressure wave created by the passing projectile. You would first need to prove that there is a linear relationship between the size of these cavities to the amount of energy that has been given up.
  15. Mar 2, 2010 #14
    I am in a bar ATM and on an I phone I will think of how to make time's relationship clearer when I get home.

    Btw I misspoke earlier when I said Vanesh was "utterly wrong" that was me getting caught up in the discussion. His arguments are valid and appreciated.
  16. Mar 2, 2010 #15
    I don't have this worked completely out, but here is an idea...

    Since we know what it is supposed to look like already from what I have described above.

    a planar surface over a circle or sphere.

    couldn't you do this instead... kg * m^2 * s^-2

    using the identity r^2s = x^2s + y^2s

    or kg * m^2 * (x^2s + y^2s)^-1

    so you would have kg * (m^2 / (x^2 + y^2)s)

    Something like that.
  17. Mar 2, 2010 #16


    Staff: Mentor

    Again, energy is a one dimensional scalar, the units are irrelevant. As I mentioned previously, torque has the same units as energy but is a three dimensional vector, so the number of dimensions of a quantity is not related to the units. This is a simple counter-example that shows clearly that your assumption (that the units imply some specific dimensionality) is wrong. It is no use proceeding with your analysis when your assumptions are demonstrably wrong.
  18. Mar 2, 2010 #17

    Why do you think that the Kg/m/s system has some sort of fundamental priority in modeling the laws of nature. What happens to your "mass over moving surface model" if you work with the imperial system. E=(lb*(ft/s)^2)/ft/s^2=lb*ft. Now how do you visualize this and which model is correct?
  19. Mar 3, 2010 #18
    I think you again are misunderstanding the concept.

    Physics is not only about understanding the value of some answer... that is a desireable aim. But to do so, you have to understand what is going on.

    To quote John Von Neumann... "There's no sense in being precise when you don't even know what you're talking about. "

    So, to understand what you are talking about you have to be able to build a conceptual model for it. Once you have the conceptual model... you can build a mathematical model on the conceptual model and begin to test its validity.

    However, if you are working with proven theories, or theories that have been experimentally justified again and again... you may be able to almost Data mine.. the equations for other nuggets of insight into the conceptual model.

    I will leave you with another quote by Von Neumann.. “The sciences do not try to explain, they hardly even try to interpret, they mainly make models. By a model is meant a mathematical construct which, with the addition of certain verbal interpretations describes observed phenomena. The justification of such a mathematical construct is solely and precisely that it is expected to work.”

    The fact that it works only tells you it works... not why it works. And if you are able to glean why it works... THEN you have something.
  20. Mar 3, 2010 #19
    The visualization of this problem is this.

    An expanding hollow sphere (a 2 dimensional object) with some mass distributed over the area of the surface of the sphere... intersecting with a line on that sphere... so the distribution of the mass would be over that line. Which if you visualize it correctly is just a circle.

    That is one visualization, you could do another as a flat planar surface and not a circle resulting in a line.. what ever is easier to think of.

    When you take electomagnectics and study gaussian surfaces, you realize, that a distribution not only of charge but of mass, and other things can be over many different shapes and throughout many objects.
  21. Mar 3, 2010 #20
    So? You've lost me completely.

    And you are still side stepping my question about time.
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