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Why is squaring so significant when it comes to basic physical laws?

  1. Jun 16, 2013 #1
    Why is "squaring" so significant when it comes to basic physical laws?

    Let me begin by saying I do not have a background in physics, or science for that matter! I just have an interest in things like cosmology and particle physics because I like to understand things.

    So my question is this: why does taking a number and multiplying it by itself so significant, and pops up all the time in fundamental physical laws?

    Gravity and electromagnetism operating on an inverse square law. Kinetic energy increases proportional to the speed squared of a given object, mass-energy equivalence, etc....

    Is it something that just "is" and science doesn't question it? Does it have something to do with spacetime being geometric, so forces/fields/vectors have to obey some "shortest path" rule, and that path happens to be a "square/inverse square" path? Why aren't these physical phenomena cubed?

    Thanks! I look forward to reading your responses!
     
  2. jcsd
  3. Jun 16, 2013 #2

    mfb

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    Cubes appear in laws, too - higher orders are just less frequent than lower orders. Most values are not squared, some of them appear squared, a few values get cubed, and so on.

    The square in the laws of gravity and electrostatics has a geometric reason: It is the number of spatial dimensions (3) in our universe minus one.

    Some examples of very high powers:
    Lennard-Jones potential with the 12th power of the distance
    Triple-alpha process with the 40th (!) power of temperature
     
  4. Jun 16, 2013 #3

    Bandersnatch

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    As far as the inverse sqare law goes, that's the result of the world being (spatially) three-dimensional.
    It's easy to see why, when you begin by considering a source of whatever you want(e.g., radiation, force) in one-dimensional space.

    So, we've got a line(1d space) with a point-like source located somewhere on it. The source is causing some sort of interaction to propagate from it in all directions. In 1d space, that means all the output is divided equally between the two directions. Whichever point on the line you choose, no matter how distant from the source, the strength of the interaction measured there will always be the same as at any other point. There's nowhere for the interaction to dissipate.
    So in such a space, e.g. the gravitational force would not have [itex]\frac{1}{R^2}[/itex] in it, as the strenght is distance independent.

    Now let's take a 2d space(a plane). Now the interaction propagates in the form of expanding circles centered around the source. If you pick one point somewhere on the plane, you'll find out that the interaction has been spread thinner, as the same amount of it needs to cover more space. At any given distance R, there are [itex]2\pi R[/itex] points that equally share the original interaction between them. The force of gravity would have the factor [itex]\frac{1}{R}[/itex] in it.

    Three-dimensional space adds another dimension into which the interaction must spread, so that at any given distance R there is [itex]4\pi R^2[/itex](concentric spheres) points sharing what the source had produced. Hence the [itex]\frac{1}{R^2}[/itex] factor in force and flux equations.
     
  5. Jun 16, 2013 #4

    Doc Al

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  6. Jun 17, 2013 #5

    anorlunda

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    Don't forget that squares of real numbers are always positive. (Squares of complex numbers p, if defined as x times conjugate x, are also always positive.

    If you accelerate a mass to the right, it gets kinetic energy proportional to v squared. Ditto if you accelerate it to the left. If energy were proportional to v, one would be positive and the other negative. V has direction, v squared does not.

    The same applies to gravity and the inverse square law, the force between earth and the sun remains the same even if you swap their positions.

    The surface of a sphere also increases with the square of radius.

    These are partial answers to your question. There may be hundreds more answers.
     
  7. Jun 17, 2013 #6
    Damn. Thanks mfb, bander, and anorlunda. You guys are awesome!

    Doc Al, I tried to read the wiki article before posting this question, but I couldn't register the "why" aspect of my question. The posters above described it in more layman's terms, and it's clear now.
     
  8. Jun 17, 2013 #7

    Jano L.

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    There is great number of formulae in physics. As there is an infinite number of integer exponents, some of them have to occur less frequently. It seems natural that the higher the power is, the less frequently it will occur.

    So, I would guess that exponent 1 is more frequent than 2 and that is less frequent than 3 etc.

    The idea that squaring turns coordinates into rotation independent quantities is also quite convincing.

    I do not understand the comment about the gravity. The explanation of 2 in the Newton gravity law "number of dimensions - 1 = 2" does not seem much better than the explanation "first prime = 2" or "2 is the power used in the Pythagorean theorem". In other words, there is no immediate connection. Can you elaborate on this idea with gravity?
     
  9. Jun 17, 2013 #8
    I assume you are referring to the post by mfb. Did you read the succeeding post in which Bandersnatch elaborates on the point in question with a convincing and enlightening argument? I have quoted it for your convenience.

     
  10. Jun 17, 2013 #9

    atyy

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    If one takes gravity to be the solution of Poisson's equation, which is an equation that is meaningful in arbitrary dimensions, then one gets the inverse square law in 3 spatial dimensions.

    One can also take Poisson's equation as the slow motion weak field limit of the Einstein field equation which also exists in arbitrary dimensions.

    In both Poisson's equation and the Einstein field equations, it is the second derivative that makes the inverse square law in 3 spatial dimensions. The Einstein field equations are the unique equations for the spacetime metric (ie. spacetime geometry) which give covariant energy conservation and have at most second derivatives (I'm not sure I got that exactly right - but I'm thinking of Lovelock's theorem)

    Then one might ask why only second derivatives. In the most general case, if we only take it that the gravitational field should be the spacetime metric (ie. spacetime geometry), and "general covariance" then we allow many higher order derivatives. In the quantum version of the theory, one can show that all the higher order corrections are negligibly small for all experiments done so far. These are ideas from the "effective field theory" treatment of quantum general relativity. http://arxiv.org/abs/1209.3511

    Very famous example of exponents that are not "simple ratios" are the critical exponents in low dimensions. Some examples are given in http://en.wikipedia.org/wiki/Ising_critical_exponents, where you can see that the exponents below 4 dimensions are not "simple ratios". These exponents are relevant near criticality, which is where the distinction between liquid and gas ends: :)

    The interesting thing is that although they are not simple, they are not material dependent - so they reflect "universal properties". They depend only on the symmetry and dimensionality of the underlying material. In fact, the theory that explains these universal exponents near the critical point is the same theory I mentioned above about the effective field theory treatment of quantum general relativity - just applied to fields with different symmetries. The theory uses what are called "renormalization group" ideas, which can be considered to generalize the central limit theorem, which is an explanation for the "universality" of bell-shaped distributions.
     
    Last edited by a moderator: Sep 25, 2014
  11. Jun 17, 2013 #10
    The focus so far has been on the inverse-square law. But I would just like to suggest that if one has a physical quantity (which one believes is a scalar, i.e. kinetic energy) that they think should depend in some way on a vector quantity, then the dot product of the vector with itself incorporates the magnitude of the vector and produces a scalar in the process. You can't have a scalar be linearly proportional to a vector.

    There are more sophisticated ways of talking about this in terms of requiring Lagrangians to be isotropic in space and Galilean invariant. If you decide at some point to really build your background in physics, then I would suggest Landau's "Course of Theoretical Physics" Vol. 1. There is a good discussion of what I mentioned in the first sentence of this paragraph.

    Interesting question, btw.
     
  12. Jun 17, 2013 #11
    As pointed out, there are several reasons why squares pop up in physics, such as the inverse-square law, which has to do with the area of spheres in 3 dimensions, and the fact that the dot product of vectors is frame-independent.

    There is one I would like to add though. In statistics, we often take as an assumption that random variables follow a Gaussian distribution, and we calculate p-values using this assumption. However, unless one can legitimately invoke the central limit theorem, this is just an assumption--why shouldn't we assign "standard scores" and "p-values" based on a distribution that decays exponentially away from the mean rather than one which is a gaussian centered on the mean? i.e., why do statisticians always use e-(x-x0)2 instead of e-|x-x0|/σ? In fact, some statisticians have studied the latter distribution, and this is an alternative kind of statistics. The problem with this approach, though, is that the function is not differentiable at x=x0, so it is just easier mathematically to use the gaussian. In other words, x2 is a function with some nice properties (always positive, always concave up, infinitely differentiable and integrable, etc.) that is also very easy to manipulate.

    But then again, the square does show up quite naturally due to the Central Limit Theorem.
     
  13. Jun 17, 2013 #12

    atyy

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    I think this is a very important point. The renormalization group, which explains why even if we put in higher order derivatives in our theory, they don't show up at low energies, can be considered a generalization of the central limit theorem.
     
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