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Cartesian to Spherical co-ordinates (x,y,z) = (∞,∞,∞) | φ,θ are different.

  1. Jun 10, 2012 #1
    (This is NOT homework) just my personal interpretation,

    because these are the formulas as you already know:

    r = √(x^2 + y^2 + z^2)
    φ = arctan(y/x)
    θ = arccos(z/r)

    using (x,y,z) = (∞,∞,∞)

    I come across a bit of a sinister problem:

    r = √(∞^+∞^+∞^) = √(3∞^2)

    which is right because if we just said that ∞ was x, then it would be:

    x^2 + x^2 + x^2 = 3x^2 right? (just the small proof)

    that means that (using surd rules)

    √(3∞^2) = √3 x √∞^2 = ∞x√3 = ∞√3

    r = ∞√3
    φ = arctan(∞/∞) = arctan(1) = 45° or ∏/4 radians.
    θ = arccos(∞/∞√3) <---- (z / r) ||| arccos(∞/∞√3) = arccos(1/√3) - because can I please say that we shouldn't forget that the ∞'s cancel out(???) so:

    θ = 54.73561032° <--- This is what I don't understand, why is it that if you use x,y,z as infinity, why is the angle of θ NOT 60°? (I assume it's because r is in equidistance from x,y,z as they are all infinity, and φ is 45°, it shows r is equidistant from x and y) -

    ^ That is my thesis, I don't know why θ doesn't equal φ if all the lengths are equal, (again because r is equidistant from the line x, line y and line z) they SHOULD be the same no? (again, I assume it's because of the 1/√3 - but there is 3 infinities (to find out r) so it should be √3∞^2 no?)

    Thank's for reading, I was just reading through the spherical co-ordinate theories and formulas, it's not homework, it's actually something I thought other people would want to see.

    Thanks for reading.
    Eggman100

    Oh, -.-, I think I forgot that Phi is the angle with respect to r in the x and y axis, I think when I said about the equidistance, then I think I forgot that, "respect to r in the x and y axis" - I think I used it so that its the angle from the x axis and r which of course would have to implement a 'z' which ofcouse phi is only the x/y plane angle, sorry.
     
    Last edited: Jun 10, 2012
  2. jcsd
  3. Jun 10, 2012 #2
    Didiving and multiplying infinities isn't generally defined this way, unfortunately.
     
  4. Jun 10, 2012 #3
    How comes? I thought x,y,z could be any values? - including infinity?

    how would you generalise working with infinity in spherical co-ordinates?
     
  5. Jun 10, 2012 #4
    Infinity is not a real number. If you work with infinity like real numbers, you'll end up with odd situations such as 2 = 1.

    Here on PF there is a thread about it.
    https://www.physicsforums.com/showthread.php?t=507003 [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Jun 10, 2012 #5
    yeah, infinities do not cancel out like that. If you're trying to set up the limits of an multiple integral, you may try to do it "intuitively". first, think of the "grids" in the spherical coordinates, which are spherical shells, cones and planes (try some books to understand these). then, decide what point corresponds to what.
     
  7. Jun 10, 2012 #6
    Why don't you interpret the three spherical coordinates as limits?
     
  8. Jun 10, 2012 #7
    yeah, actually that's what i was trying to do. the (r,[itex]\phi[/itex],[itex]\theta[/itex]) possible coordinates are,
    r=∞,
    [itex]\varphi[/itex]=any number ≠ 0,n[itex]\pi[/itex]/2,
    [itex]\theta[/itex]= any number ≠ 0,n[itex]\pi[/itex]/2 (n integer)

    check them.
     
  9. Jun 10, 2012 #8
    Abstractly thinking, the three axes x,y and z split the three dimensional space into eight different regions. I think this question has to do with expressing the region [itex]D=\{0\leq x,y,z\}[/itex] in spherical coordinates.

    Thinking abstractly once again, we know the distance from the origin has to be positive and has to cover an infinite region, so it can take any value that is a non-negative real number. Phi and theta are right angles, thinking with the same sense. So our region is [itex]D=\{0\leq r,\,0\leq\varphi,\theta\leq\pi/2\}[/itex].

    However, to actually use the formulae you gave, we need to interpret them as limits. I hope you can handle that on your own.
     
  10. Jun 10, 2012 #9
    if anything's wrong, you can say it CLEARLY cause i'm just a high school student and i'm lousy at the "abstract" thing. As far as I can get from your comment, there seems nothing wrong in my answer.
     
  11. Jun 10, 2012 #10
    I didn't say there is something wrong with your answer, I just implied it lacks a proof.
     
  12. Jun 11, 2012 #11
    I read through your explanation, but you say that Phi and Theta are right angles, arent they not [itex]\pi[/itex]/4? (45°) rather than 90° because if you imagined the X,Y,Z grid (Example here: ) http://upload.wikimedia.org/wikipedia/commons/thumb/4/4f/3D_Spherical.svg/558px-3D_Spherical.svg.png

    Can you not imagine r how it is, but then think about it, if all the sides x,y,z are infinity, ∞ would have to be the same as x,y,z because they're all constant (they all have the value infinity - whatever value infinity is, that is x and y and z) so should the angle Phi and Theta not be 45°, because r is equidistant from each line, so r would be 45° between x and y, and would be 45° between the x/y plane and the z axis? or am I wrong?
     
  13. Jun 11, 2012 #12
    You are wrong. We are defining the region, not a single point.
     
  14. Jun 11, 2012 #13
    No the question I gave was for that when x,y,z (the point) is equal to ∞,∞,∞ <--- find r, phi and theta, i'm not talking about that, i'm just wondering why the angle for theta isn't 45' like phi is, because it's the same length for x,y,z all the angles (phi and theta) should be 45' like in an example

    (5,5,5)

    r = √75
    phi = arctan(1) = 45'
    theta = arccos(5 / √75) = 54.735' (rounded)

    ^ *sighs* okay don't worry. <--- however I don't understand why theta is different, if all the lengths are the same and it's a regular cubic shape (if you was to draw lines from the axes to show the cube-shape the angles i would have though would have been the same unless i'm expected to just do trig, and even if I do that means the answer for the theta angle should be the same for any length because the Adj/Oppo/Hyp would be the same length so that would be my answer =/ - although i'm not happy having to trig infinity.... (so i'll use 5 first - then use that as my answer)

    Okay, this is what I did:

    S=O/H <----> I have r (which is my Hypotenuse) and z (the length, not the actual axis - as the length being 5) which is O

    Sin^-1 (5 / √75) = 35.26438968' - now this is the angle inside the triangle where it sits on the x/y plane so take away from 90' (because the Z axis of course is at 90')

    so 90 - 35.26438968 = 54.73561032 = 54.7', so if the lengths are ALL the same, theta would have to always be 54.7' because of the trigonometry giving the proof: again with the value 8:

    r = √192
    phi = arctan(1) = 45'
    theta = arccos(8 / √192) = 54.7' (again)

    and thus:

    trigonometry:

    (Z length is 8 units), r length is √192

    so S=O/H <---> sin^-1 (8 / √192) = 35.26438968 <---> take away from 90' so that = 54.7' again, (with trig)

    so am I safe to say that because infinity is always itself, and if x,y,z (my point) always has lengths (∞,∞,∞) then using the fact that if the lengths are always the same number, then theta always has to be 54.7' (unless one of the values are different etc) then that proves how it is 54.7' and not 45'.
     
    Last edited: Jun 11, 2012
  15. Jun 11, 2012 #14
    Oh, you mean that. Well, next time you ask a question, keep in mind that there is no such point with a coordinate as infinity.

    You have r right. With phi, the angle made with the x-axis by the corresponding component of r, we have pi/4 (45 degrees.) For theta, the angle made by the line to the xy-plane, we have a right triangle with edges 5 and 5√2 Hence, theta becomes arctan(1/√2), not 45 degrees. You just need to think abstractly.
     
  16. Jun 11, 2012 #15
    You're making the mistake of thinking of infinity as a single point.

    If your xyz coordinates can be parameterized as (t,t,t) for [itex]t\to \infty[/itex], then you get one answer, but (t,2t,t) will give you different results for the limits, yet this also goes to [itex](\infty, \infty, \infty)[/itex]. There is no difference between [itex]2\infty[/itex] and [itex]\infty[/itex].
     
  17. Jun 11, 2012 #16
    There is no point equal to (∞,∞,∞)

    There is no such thing.

    Why are you convinced you can treat ∞ as a real number? It is not a real number. It is a concept.

    From the very first line of the wikipedia article:

    "Infinity refers to something without any limit, and is a concept..."

    It is a concept. Not a number. You can not add, subtract, divide, take the arctan of concepts. These operations only work on numbers.

    With a higher level of understanding of mathematics you will learn the proper way to treat and to deal with infinity. But what you have written is just gibberish.

    For example when you write:

    ∞x√3

    You are multiplying a concpet and a number.

    That is like me saying, "love x 5 equals 22". WTF?!? that doesn't make any sense.

    I hope you can see that.
     
  18. Jun 11, 2012 #17
    Ohhhh okay, reasonable enough explanation =) ^^
     
  19. Jun 11, 2012 #18
    ya, that's what i meant when i said "i'm lousy at the abstract thing". just check my answers. it's correct and i think that's enough reason it should be correct. i can't explain more. i thought of those points because they just occurred to me.
    furthermore, this proves that you can't treat ∞ as a real number. if it would be real number, then you certainly get just one answer. but because it's not a real number, the point (∞,∞,∞) in Cartesian can mean a lot of points i provided. got it?
     
  20. Jun 11, 2012 #19
    Yeah, I can see you like that. I was only trying to figure out my own personal problem, sure it is a concept, but don't forget that if "Love x 5 = 22" everyone knows that love would have to be 4.4 - it's just saying love the emotion is counted as X.

    When I get to a higher level sure, but I'm going to PM you something that I think is important before you actually think something that's wrong.
     
  21. Jun 11, 2012 #20
    You can deal with infinity as it is the part of the real subset, but it is certainly not a real number. It just belongs to what we call the extended real number line. In the extended real number line, infinity and negative infinity are defined as follows: [itex]\infty=\mathrm{sup}\{\mathbb{R}\},\,-\infty=\mathrm{inf}\{\mathbb{R}\}[/itex]. However, these are certainly not real numbers.

    In general, with an operation like infinity times 2, you take infinity as a divergent sequence. If in a certain operation you end up with different answers with different divergent sequences, we say that it is undefined. The same happens when the limit does not approach the same value from both sides.
     
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