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Fourth Integral ?

  1. Mar 3, 2006 #1
    Here I have a simple, yet strange question.
    On evaluating a double integral, we get area. On evaluating a triple integral, we get volume.
    But what we get on evaluating a fourth order integral i.e.

    (intergral)f(x,y,z,t)dx dy dz dt
    Is it space time? I cannot understand what the answer is telling us?
     
  2. jcsd
  3. Mar 3, 2006 #2

    TD

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    Double and triple integrals do not necessarily represent areas and volumes, you can compute both area and volume (in some special cases) with a single integral!
    In your analogy though, an n-dimensional integral generally represents and n-dimensional volume, of which 2 is area and 3 our 'classic' volume. From 4 on, I believe the term 'hypervolume' is used, cfr hypercube, hypersphere etc.

    Of course, this is purely mathematical. I suppose you can see the 4th dimension as time if you'd like a physical interpretation.
     
  4. Mar 3, 2006 #3

    dextercioby

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    It might be time, if it wasn't, then they would have written


    [tex] \iiiint f\left(x_{1},x_{2},x_{3},x_{4}\right) \ dx_{1} \ dx_{2} \ dx_{3} \ dx_{4} [/tex]...

    Daniel.
     
  5. Mar 3, 2006 #4

    HallsofIvy

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    Yes, and what
    [tex]\iiiint f\left(x,y,z,t\right) dx dy dz dt[/tex]
    "means" depends upon what the function f means.
     
  6. Mar 3, 2006 #5
    Everyone else has answered the question quite nicely, but they have left off a slight technicality.

    You (Excellence) mentioned space-time. The fourth integral COULD be time and you COULD be integrating over a space-time boundary. The problem is that you probably haven't been taught the technique to do such an integral. (Heck, I've got a Master's degree in Physics and I don't know how to do it. I've seen them done but that's far from saying I understand them.)

    The problem with a space-time integral is that it isn't being done in Euclidean 4-space, thanks to the Minkowski metric (a kind of version of Pythagora's theorem in space-time.) That adds a lot of complexity to the integration. All of the integrals mentioned in the previous posts are for integration in Euclidean 4-space, which is fairly straightforward.

    -Dan
     
  7. Mar 3, 2006 #6

    Hurkyl

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    The previously mentioned integrals aren't Euclidean space: they're just generic multidimensional integrals. There's no metric involved: it's just a differential form. (Or even merely an iterated integral, depending on how you write it!)
     
  8. Mar 4, 2006 #7

    benorin

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    Were we to stipulated that the metric used is Eucilidean, then triple integrals of any non-constant integrand, or quadruple integrals (of anything, even constants), ..., or generally n-fold multiple integrals can be interpeted as being the "content" or "hypervolume" of some region lying in 3,4,.., n-dimensional Euclidean space (as appropriate). And, just as 3-d solids (e.g. a cube) have volume, surface area, and "edge length," n-D objects (e.g. a hypercube; there is a better word for n-D object) have n-D content, (n-1)-D content, ..., volume, surface area, and edge length. The 4-D hypercube whose edge length is, say x, is a good example: it has 4-D content [tex]x^4[/tex], it has eight cubes which comprise its facets (facets are like faces for polytopes, which are n-D generalizations of polygons and polyhedra) and thus has 3-D content (e.g. volume) [tex]8x^3[/tex], and has then 24 squares which comprise its faces and thus has 2-D content (e.g. surface area) [tex]24x^2[/tex], and so on...
     
    Last edited: Mar 4, 2006
  9. Mar 4, 2006 #8

    HallsofIvy

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    We really should point out that mathematics is not physics! The simple fact that you have a function depending on 4 variables does not mean the variables are spacial coordinates and time.
    We can't tell you what any function means "physically" unless you tell us what you are applying it to.
     
    Last edited: Mar 16, 2006
  10. Mar 13, 2006 #9
    Ok HallsofIvy, I apply my question to the physics. Now, can u explain the interpretation of the outcome.
     
  11. Mar 13, 2006 #10

    HallsofIvy

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    Since there was no physics in your post, no I can't. If you gave a physics problem in which such an integral appeared, I might be able to. But an integral of any kind- without a physics meaning assigned to it- is just that- an integral without any physics meaning.
     
  12. Mar 13, 2006 #11
    What is 'the physics'. There are many physical systems with more than 3 variables. The standard 4 dimensional integral is 3+1 dimensions in space-time but it could be a 4 spacial dimensional integral, which is something you might find in string theory, which deals with more than just 3 space dimensions.
     
  13. Mar 13, 2006 #12

    Hurkyl

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    Or, it could be 4-dimensional because you're integrating over two spatial coordinates of two different particles. (All spatial dimensions)

    Or, it could involve parameters that aren't related to geometry at all!
     
  14. Mar 17, 2006 #13
    Integrating in 4 dimensions

    A quadruple integral will give you the content of a region in 4-space. Content is the generalization of length, area, volume (1-, 2-, and 3-D content respectively) to any number of dimensions.

    Minkowski spacetime has a pseudoEuclidean metric (Lorenzian), not a Euclidean metric - so directed integration in spacetime requires one to use a metric such as t^2 - x^2 - y^2 - z^2, which according to Special Relativity should be the same for all observers even though the actual durations in time and spacial distances might be different for different observers.

    Therefore, the quadruple integral as you have written it expresses a 4-content in Euclidean 4-space, not in Minkowski spacetime.
     
  15. Mar 18, 2006 #14

    HallsofIvy

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    And, of course, the Euclidean 4-space could be given many different interpretations in different physics problems!
     
  16. Mar 24, 2006 #15
    Now, the replies are going over my head.
     
  17. Mar 24, 2006 #16

    HallsofIvy

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    All you need to know is that mathematics is not physics! It makes no sense to talk about a physical meaning of an integral, or any other mathematical process, without stating a specific application.

    Yes, we use mathematics in physics. We also use grammatical symbols. Asking about the meaning of an integral in physics is like asking about the meaning of a semi-colon! Show me the sentence in which it is used, and it might well have significance- but not out of context.
     
    Last edited: Mar 25, 2006
  18. Aug 5, 2009 #17
    A dimension mathematically can be a reference to all sorts of things physically, not just space. Or rather, as long as you're referring to anything with any sort of variable intensity, you can plot it as a dimension. Like the loudness in decibles of a sound could be your x axis, ability to concentrate your y axis, and you could have z and a be your position in the room, to map the relationship between standing in a particular spot on a flat surface and being able to concentrate while a band is playing next door. Or in my case, a celtic dance studio... although I have no idea what the hypervolume of the fourth integral of functions of those would measure... maybe plot z and a in respect to time t as parametrics, and have it be something like the total measure of how much I've been able to concentrate in this room?
     
  19. Aug 5, 2009 #18
    The notion of length, area, and volume are generalized in a branch of mathematics called measure theory. All three are instances of a "measure" on a set, which intuitively gives a sort of "size" to each set (but a "size" that is distinct from cardinality).

    In some books, term volume is used for any standard measure in R^n. So, the length of a line segment is a volume, as is the area of a square. It's just a terminology.

    If you were to come up with a more physics-y word for the standard measure in R^4, you'd probably call it hypervolume. This kind of naming follows suit with the terms hypersphere and hypercube. You'd probably also use hypervolume for R^n where n > 4.

    It's just a naming thing though.
     
  20. Aug 5, 2009 #19
    I think this may be a common mistake. In single-variable calculus, I was taught to equate the definite integral of a function with the area under its curve. In multi-variable calculus, I erroneously assumed that integrals of additional variables represented the "area" under the "curve" in the additional dimensions. HallsofIvy corrected me on a previous thread.
     
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