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Using differentials in the derivation of equations

  1. Apr 1, 2012 #1

    This is a hard question to ask, because it's so vague . . . I have real trouble getting my head around using differentials to derive equations. Stuff like the fundamental differential equation of hydrostatics, eulers conservation of momentum equation in fluid mechanics, and bernoulli's equation. All those ones where they say things like 'consider a rectangular element with horizontal sides of unit length and an infinitesimal height dh'. What's the purpose of using dh? why can't we just use h? what's the difference between dx and ∆x?

    I'd just like someone to give an uncomplicated answer as to how you use things dh, dx, dV, for example to derive an equation. What are the rules?

    I'm very familiar with calculus and know all about finding the slope of an equation with limits etc, this stuff just seems to get me and perhaps someone could give me a new perspective. Is there a book specifically on this stuff?

    Thanks :)
    Last edited: Apr 1, 2012
  2. jcsd
  3. Apr 1, 2012 #2


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    Staff: Mentor

    Can you write out a simple example of such a derivation, or link to one, and tell us exactly which steps you're having trouble with? Otherwise people are going to have to try to read your mind long-distance. :uhh:
  4. Apr 1, 2012 #3
    Yeah of course, sorry about that. I've attached something that's been giving me trouble. How do you decide what what becomes an infinitesimal increment? what is the purpose of deciding something is an infinitesimal increment?
    I understand this is probably basic stuff, but I seem to lacking an understanding in the fundamentals of it.

    Attached Files:

  5. Jun 7, 2012 #4
    As another newby to thermodynamics, and also moderately well trained in mathematics, I share the confusion.

    I read equations such as dU = TdS + PdV written without a clear definition or explanation.
    I understand Delta(U) = T Delta(S) + P Delta(V) as a meaningful statement, but I don't understand that dU = TdS + PdV has any mathematical meaning.

    Has anyone responded to your question? These kinds of mathematical (or at least mathematical appearing) statements seem to pervade geoscience textbooks. Are these statements and the differentials better explained and defined in Thermodynamics texts?
  6. Jun 7, 2012 #5
    A lot of physicists essentially think of things like "dU" or "ds" as arbitrarily small changes in these quantities. Mathematically, you can think of expressions like "dU = TdS + PdV" as describing the 1-form dU on, say, a manifold with all other variables on which U depends held constant (such as N). In this case, treating U as a 0-form, T = [itex]\partial[/itex]U/[itex]\partial[/itex]S, and P = [itex]\partial[/itex]U/[itex]\partial[/itex]V
    (although really P = -[itex]\partial[/itex]U/[itex]\partial[/itex]V, since internal energy increases when work is done on the system -- that's really just notational though)

    A lot of introductory fluid dynamics textbooks are really for engineers instead of physicists, and engineers are even less (explicitly) rigorous than physicists. Seeing stuff like "ds/2" is usually a way of justifying a formula in short-hand with minimal formalism. Maybe it's also meant to get students thinking about numerical techniques.
  7. Jun 7, 2012 #6
    Typically, [itex]\Delta x[/itex] (or any such quantity) is used to denote a real, finite quantity, whereas [itex]dx[/itex] is used to mean a quantity that is infinitesimally small--i.e. that can be integrated.

    For example, the energy equation for thermodynamics [itex]dU = T \; dS + P \; dV[/itex] really means that, between two states, the following integral equation is obeyed:

    [tex]\int_{U_1}^{U_2} dU = \int_{S_1}^{S_2} T \; dS + \int_{V_1}^{V_2} P \; dV[/tex]

    One has to consider that T,P may be functions of S,V.

    It's important to understand that differentials are used instead of finite differences because in general, T and P are functions of S and V, so when you do complete the integration, you get extra terms beyond just [itex]\Delta U = T \Delta S + P \Delta V[/itex]. When written in terms of differentials, we neglect all these higher order terms because infintesimals are treated like they square to zero. That may sound like mathematical voodoo, but it all has to do with the limiting process inherent to calculus.

    The use of infinitesimals allows you to use a limiting process in this manner and say that functions which should vary in space and time are constant over a very small length or duration.

    Example: deriving the area of a circle. The circle has radius [itex]R[/itex], and if we take a tiny sliver of the circle with angle [itex]d\theta[/itex], we can approximate the area of this sliver as a triangle. This triangle would have area [itex]R^2 \sin(d\theta)/2[/itex], but [itex]d\theta[/itex] is infinitesimal, and a Taylor series expansion would leave only [itex]\sin(d\theta) \approx d\theta[/itex] as the leading term. So you get

    [itex]A_{\text{circle}} = \int_0^{2\pi} \frac{1}{2} R^2 d\theta = \pi R^2[/itex]

    Calculus uses infinitesimal differentials to approximate complicated objects (like a circle) as a series of simpler objects (a bunch of triangles). We use infinitesimals because expressions that should vary over a finite interval can be treated as constant over an infinitesimal one, and this is often an essential step to being able to set up an integral.
  8. Jun 7, 2012 #7
    This reply is principally for pondhockey in reply to his PM

    For a discussion of total differential look at this thread, in particular post #8


    For a discussion of the points raised by Muphrid look at this thread and in particular post #13


    Unfortunately the formulae in this second thread no longer display correctly since the forum Tex change.
  9. Jun 8, 2012 #8
    I want to thank couchyam, murphrid and studiot for helpful replies. I'll spend some time with them.

    Murphrid writes:
    For example, the energy equation for thermodynamics dU=TdS+PdV really means that, between two states, the following integral equation is obeyed:


    I would note that whether the infinitesimal approach is taken or not, the given equation did not, to me, imply the integral equation that murphrid contributed above. I would say that the integral equation does not follow mathematically from the original statement, and perhaps it is the integral equation that would be a better (clearer) statement of physical law.

    Again, thanks to all of you.

    [edit: I should probably have phrased the last comment as a question rather than an assertion, given my lack of authority on the subject!]
    Last edited: Jun 8, 2012
  10. Jun 8, 2012 #9
    Nice to see some polite people about.

    Keep the questions coming.

  11. Jun 8, 2012 #10
    The fog is beginning to lift; thanks again.

    It looks to me now that part of the notational confusion for me can be cleared up by the clue that dU is related to dS and DV, from which I can infer that the longer version of the statement would start out

    dU(S,V) = ...

    That T and P are pretty much along for the ride, and that murphrid's integral equation begins as a double integral on the left, an equivalent iterated integral on the right, reducing to the sum of two univariate integrals. This seems confirmed by couchyam's restatement of T and P as the missing parts of the differential of U.

    Presuming that this is all true, it was more than I was able to reconstruct from the simplified notation, so your help is invaluable. The reference to the thread by omberlo/studiot is very instructive.
  12. Jun 10, 2012 #11
    I'd like to follow up on one point that Murphrid makes:

    "Typically, Δx (or any such quantity) is used to denote a real, finite quantity, whereas dx is used to mean a quantity that is infinitesimally small--i.e. that can be integrated."

    This still seems to be an issue for me. First of all, the treatment of differentials that I'm comfortable with defines dx and dy as reals. dx is not much different than delta x, but while delta y is the actual increment of y corresponding to x + delta x, dy is the increment of a tangent line approximation to y.

    Defined in this way, the ratio of differentials has the appropriate derivative as a limit.

    treating dx and dy as "infinitesimals" was a frequent heuristic that was used in math and science, but is not something that I've ever seen made precise.

    In particular I don't follow a rigorous meaning for "integrating infinitesimals". Do you regard this as a heuristic for a more explicit formulation? And if so, does the heuristic always seem to work out for generating mathematical statements that can be proved another way? Are there rules for working with differentials as infinitesimals that result in correct mathematical statements?
  13. Jun 10, 2012 #12
    If you're having trouble with differentials, it's probably not the best to start looking at thermodynamics equations, although you can if you can think in abstract quantities.

    Physicists put equations in the form of differentials so that they're ready to be summed. Have you tried looking at probability distributions? Physics textbooks will write p(x)dx, and never just p(x), as the probability distribution. This is because we can integrate ∫p(x)dx, but we can not integrate ∫p(x), as you've probably already learned in calculus. For example, the Gaussian distribution is more or less a*exp(-b*x2). To denote that it is a distribution, we write f(x)=a*exp(-b*x2)dx, where we say f(x) is the probability. This is because it just doesn't make any sense to pick a point on the plot of a*exp(-b*x2) and say, "this is the probability."

    Another reason for using differentials is that it's the physicist's simplification of continuity. Math has a rigorous sense of continuity, but physics just takes it for granted. For example, often times, you'll see a book say something like "we observe what happens when x goes to x+dx." This really means "we observe what happens when x goes to x+Δx", but putting a dx there tells the reader that you can choose any value for Δx (ie. Δx is not quantized, which sometimes is the case).

    In any case, if you see a differential and you're trying to visualize it, you can replace the dx with Δx as a mental picture; just let the Δx be very narrow. For example, if you're viewing a graph, say, of f(x)=x2, then x2dx at x=2 is simply a vertical line drawn from the point (2,0) to (2,4). That's what it means by infinitesimal.

    The last thing I want to mention is that in most cases, when the differential is used, it's to make it dimensionally correct. Take, for example, kinetic energy.

    Ek = 1/2 mv2
    dEk = 1/2 (mv dv + v2dm)

    If dE has units of energy, then all terms on the right better have units of mass times velocity squared, and although this is sloppy math, it's only used for visualization purposes.
    Last edited: Jun 10, 2012
  14. Jun 10, 2012 #13


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    Differential forms are the way to treat dx properly.

    The dual concept from differential geometry, "tangent vectors", capture the notion of an infinitesimal displacement.

    Part of the problem is that in the sloppy heuristic notation, people use dx when they really mean a tangent vector, and this gets in the way of forming a consistent notion of the two ideas.

    The rules for manipulating differential forms are basically just implicit differentiation.

    [tex]d\left( e^t x^2 \right) = e^t \, d(x^2) + d(e^t) x^2 = e^t (2x \, dx) + (e^t \, dt) x^2 [/tex]

    then if you have a tangent vector, you can combine them to get a directional derivative:

    [tex]\nabla_v \left( e^t x^2 \right) = e^t (2x \nabla_v x) + (e^t \nabla_v t) x^2[/tex]

    Or, you could integrate your differential form instead, via what you learned in your multi-variable calculus class. (or even single-variable calculus, if all quantities are a function of, say, t)
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