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Calculus in physics?

  1. Jan 20, 2016 #1
    I can't understand some aspect of using calculus in physics.Here,i explain what i mean as an instance.Please give me your points on it.
    Let T to be temperature of a cubic solid,and any point of the solid could have different temperature.Suppose by experiment and other ways we know that if T be constant over the solid then the internal energy e of the solid is equal to f(T).V,in which f is a continuous function of T and the capital V is the volume of the solid.Now to calculate the e when T is not constant over the solid any more,physicists would claim that the e of an infinitesimal room would be very close to f(T).dV in which dV is the volume of the room,so the e of the whole solid is equal to "SSS f(T) dv",in which "SSS" is triple integral notation.If i was right so far,please tell me why do physicists think that the e of an infinitesimal room would be very close to f(T).dV?
    As in calculating the area under the graph of some positive function f(x), and above x-axis in calculus,by just taking a look at the graph we realize that the area of the region above a small line segment "dx" on the x-axis, and below the graph is very close to the area of the rectangle which area is "f(x)dx",but i'm wondering to know how in the internal energy example and many other such examples in physics,we could assume that such a thing holds.

    Best,
     
  2. jcsd
  3. Jan 20, 2016 #2

    jambaugh

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    There is a critical assumption that the temperature (or whatever physical quantity one is dealing with) as a function of position is continuous. This then allows that everywhere in that small "infinitesimal" volume or any volume for that matter, the temperatures all lie between a maximum and minimum value and that the difference between these max and min values approach zero as the volume's extent approaches zero. Note by extent I don't just mean the measure of the volume but also the longest dimension. (You can't just slice thin, you gotta dice it up.)

    It is this continuity assumption that makes it work. Then that energy function you're talking about is between f(Tmax)*V and f(Tmin)*V which as the volume's extent approaches zero approaches f(T(x)) dV... but more importantly the sum over all those volumes becomes the total energy. You should look at the construction of the limit of Riemann sums in a college level calculus textbook.
     
  4. Jan 20, 2016 #3

    boneh3ad

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    I think perhaps you are running yourself in circles trying to use the area-under-the-graph interpretation of an integral. it makes much more sense (to me, anyway) to view it as an infinite sum of infinitesimal pieces. So, in your example, the internal energy, ##e##, can be defined
    [tex]e = f(T)V[/tex]
    for a constant temperature. However, if the temperature is not constant, it is useful to picture the integral as adding up all of the tiny little individual pieces of internal energy, ##de##, that are contained within all of the tiny little piece of the volume, ##dV##. So, based on the previous equation, an infinitesimally small volume ##dV## contributes an infinitesimally small amount of internal energy to the total,
    [tex]de = f(T)\;dV.[/tex]
    Now all that remains is adding up all of the possible pieces of ##de## to come up with ##e##,
    [tex]e = \int_V f(T)\;dV = \int\int\int f(T)\;dx\;dy\;dz.[/tex]
     
  5. Jan 20, 2016 #4
    Thanks!"that energy function you're talking about is between f(Tmax)*V "The continuity gives us
    that f(Tmin)<f(T)<f(Tmax) so f(Tmin)V<f(T)V<f(Tmax)V,but how did you derive the statment above on internal energy?!
     
  6. Jan 20, 2016 #5

    A.T.

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    This visualization still works fine for a double integral (over a surface), where it is the volume under the function-surface. But it obviously fails for higher dimensional integrals, because we cannot think 4D dimensionally very well.

    Try imaging computing a 1D integral as going though a line of small boxes with different amounts of something in them (the function value), and adding those amounts up. 2D and 3D integrals are then just 2D and 3D arrays of such boxes.
     
  7. Jan 20, 2016 #6
    Non sence!
     
  8. Jan 20, 2016 #7

    boneh3ad

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    So wait, you don't want people to help you?
     
  9. Jan 20, 2016 #8

    jambaugh

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    Actually I should be a bit more careful in my statement but the continuity implies that for the small volumes the maximum of f(T(x)) and minimum of f(T(x)) and the actual energy per volume will all converge to the same value when your volumes' extent approaches zero.

    You should probably start with understanding single variable integration, how the Riemann sum approximates the "area under the curve" which itself will express the physical quantity you want in a given application. You then need to see how the error in that approximation goes to zero faster than the number of slices increases. Note that while we speak of infinitesimals and infinite sums, the modern version of calculus has dispensed with these as formal concepts (in the orthodox school of standard analysis).

    Differentials are not infinitesimal quantities but rather asymptotic ones.
     
  10. Jan 20, 2016 #9

    boneh3ad

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    Right, but from the physical perspective, I'd argue that this concept is still helpful in determining how concepts like integration can be used to solve physical problems. At least, I find it simpler to visualize the problems that way in a physical sense.
     
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