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  1. - Top - End - #31
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by BeerMug Paladin View Post
    The operators in the set of complex numbers take two ordered pairs of numbers in the reals and maps them to an ordered pair of numbers in the reals:
    I don't think you can dismiss this by stating that complex numbers are an ordered pair - becasue they are not.

    For one thing, are they ordered? = 5+4i = 4i + 5.

    For another 2 e^(2*i*pi) is a complex number that happens to have a real value, it is not an ordered pair even if it can be expressed as one..

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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by shawnhcorey View Post
    You are not doing math operations. math ops are only defined for real numbers. What you are doing is different ops that have the same name as math ops.

    A complex number is a vector (a,b) written in its own nomenclature: a+b𝑖. If you multiple two vectors together, you get the dot product: (a,b)∙(c,d) = ac+bd. But if you multiple two complex numbers together, you get something different:

    (a+bi)(c+d𝑖) = ac-bd+(ad+bc)𝑖

    You cannot apply the math operator multiple to complex numbers.
    Complex numbers are a proper field and all.
    Also, just because they can be written as a vector space of the real numbers doesn't make them less valid. After all real numbers are a vector space of the rational numbers and way more funky than just "add another dimension".

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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Khedrac View Post
    I don't think you can dismiss this by stating that complex numbers are an ordered pair - becasue they are not.

    For one thing, are they ordered? = 5+4i = 4i + 5.
    If they weren't an ordered pair, then you'd be saying that 5 + 4i = 4 + 5i. Which you clearly didn't do. The nature of the complex numbers as an ordered pair comes from the fact that there's a part representing the real component and a part representing the imaginary component. You (usually) can't interchange those values without changing the complex number you're talking about. (Except when the two numbers are the same anyway, so a transposition makes no change.)

    Quote Originally Posted by Khedrac View Post
    For another 2 e^(2*i*pi) is a complex number that happens to have a real value, it is not an ordered pair even if it can be expressed as one..
    Not sure why you're asserting this. Because the operators involved in that calculation have no meaning in the reals. The operators only mean something in the complex number set, which also, only returns complex numbers. To get to the real numbers, you need to introduce a mapping operator that removes the fundamental nature of the number as an ordered pair. It's pretty trivial to do this, but whether or not it's trivial, it still needs to be done to transition from the complexes to the reals.

    In a coordinate system, the coordinates (5,0) is not equal to the number 5. If you just described your coordinates as 5, it would be inadequate.
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by BeerMug Paladin View Post
    If they weren't an ordered pair, then you'd be saying that 5 + 4i = 4 + 5i. Which you clearly didn't do. The nature of the complex numbers as an ordered pair comes from the fact that there's a part representing the real component and a part representing the imaginary component. You (usually) can't interchange those values without changing the complex number you're talking about. (Except when the two numbers are the same anyway, so a transposition makes no change.)
    This one may just be a matter of unfamiliar jargon, but to me the natural meaning of "ordered pair" is that the order that they are listed in matters, not just which is which. 5 + 4i and 4 + 5i are different because which number is associated with i is different, not because they are written in a different order.

    Quote Originally Posted by BeerMug Paladin View Post
    A naive, direct interpretation of what I see when considering the argument "the reals are a subset of the complex numbers" is the statement:
    (x,0) = x
    That doesn't look like the same thing to me. One of those is an ordered pair and the other is not.
    That looks like a difference in notation without difference in meaning, or alternatively like a statement of definition, to me.

    Quote Originally Posted by BeerMug Paladin View Post
    Quote Originally Posted by Khedrac View Post
    For another 2 e^(2*i*pi) is a complex number that happens to have a real value, it is not an ordered pair even if it can be expressed as one..
    Not sure why you're asserting this. Because the operators involved in that calculation have no meaning in the reals. The operators only mean something in the complex number set, which also, only returns complex numbers. To get to the real numbers, you need to introduce a mapping operator that removes the fundamental nature of the number as an ordered pair. It's pretty trivial to do this, but whether or not it's trivial, it still needs to be done to transition from the complexes to the reals.

    In a coordinate system, the coordinates (5,0) is not equal to the number 5. If you just described your coordinates as 5, it would be inadequate.
    Suppose, for the sake of discussion, that the following things are true:
    • The notation "4" is defined to mean the complex number 4 + 0i.
    • The notation "5i" is defined to mean the complex number 0 + 5i.
    • The real numbers are defined as the subset of complex numbers with imaginary part 0i. Tweak the wording to remove cyclic definition references if necessary.
    • The imaginary numbers are defined as the subset of complex numbers with real part 0, again worded to not be cyclic self-referencing.
    • Each actual-real-number math operation is replaced with the same-named complex number operation.

    What differences do these definitions make in how real number math operations work relative to whichever definitions you consider to be the actual ones? As far as I know, this is the full list:
    • The complex numbers do not have an equivalent of "less than"/"greater than" ordering without first converting them to real number magnitudes.
    • The complex version of an operation, performed on real numbers (by the definition I listed), may in some cases produce a non-real complex number where the actual real numbers version of that operation with the corresponding inputs would produce some equivalent of "does not exist", "not applicable", "inputs not in valid domain", etc.

    If there are other differences, what are they? If not, why is this set of definitions not considered valid?
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  5. - Top - End - #35
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Douglas View Post
    This one may just be a matter of unfamiliar jargon, but to me the natural meaning of "ordered pair" is that the order that they are listed in matters, not just which is which. 5 + 4i and 4 + 5i are different because which number is associated with i is different, not because they are written in a different order.
    The generalized way to write a complex number is:
    a + bi = (a,b)
    Which one is first denotes which component is the real part and which is the imaginary part. Thus the order matters.

    The reason why I'm using an ordered pair is because the construction of multiplication shown earlier utilizes pairs of real numbers to represent a single complex number. Every complex number has an imaginary component and a real component. Even if that component happens to be zero, it is still a part of the complex number.

    Quote Originally Posted by Douglas View Post
    That looks like a difference in notation without difference in meaning, or alternatively like a statement of definition, to me.
    I did intend it to be an accurate portrayal of the modern algebraic definition of a complex number. As far as I recall at least, I didn't look anything up in a textbook.

    Quote Originally Posted by Douglas View Post
    Suppose, for the sake of discussion, that the following things are true:
    • The notation "4" is defined to mean the complex number 4 + 0i.
    • The notation "5i" is defined to mean the complex number 0 + 5i.
    • The real numbers are defined as the subset of complex numbers with imaginary part 0i. Tweak the wording to remove cyclic definition references if necessary.
    • The imaginary numbers are defined as the subset of complex numbers with real part 0, again worded to not be cyclic self-referencing.
    • Each actual-real-number math operation is replaced with the same-named complex number operation.

    What differences do these definitions make in how real number math operations work relative to whichever definitions you consider to be the actual ones? As far as I know, this is the full list:
    • The complex numbers do not have an equivalent of "less than"/"greater than" ordering without first converting them to real number magnitudes.
    • The complex version of an operation, performed on real numbers (by the definition I listed), may in some cases produce a non-real complex number where the actual real numbers version of that operation with the corresponding inputs would produce some equivalent of "does not exist", "not applicable", "inputs not in valid domain", etc.

    If there are other differences, what are they? If not, why is this set of definitions not considered valid?
    One trouble with our language here might be that the complex number set is defined in terms of the reals, not the other way around. I'm not certain how it might be possible to go about things in the other direction, it seems it will inevitably be a circular definition. Although I seem to recall hearing that any system of operators and elements with all the algebraic properties of the complex numbers is in fact isomorphic to the complex numbers or a subset of the complex numbers... If true, that's pretty neat, but it's a discussion to be had beyond this level of understanding. And also not relevant to this particular discussion.

    One good distinction is that when you're considering physical models (I apologize for being on-topic), complex numbers aren't really a valid domain to engage with. There is no ruler that will return a complex number as an outcome. Likewise, returning a complex number from a physical model means you broke the math.

    Like I've said before, converting from a set of special-case complexes to the real numbers is a fairly trivial matter. Just set the imaginary component to zero and you're essentially just working with the real numbers. It's so trivial that people can casually think of the real numbers as being an actual subset of the complex numbers and they won't get in trouble. They're isomorphic to a subset of the complex numbers, but they're not actually a subset of the complex numbers. Isomorphic is an important word to understand.

    One way to see this more clearly might be to consider how you might prove the isomorphism.

    What we've got here is three things. A mapping from a subset of the complex numbers to the real numbers, a notion of a complex operator and a notion of a real operator.

    Consider m(x) to be the mapping from the complex numbers to the real numbers.

    Consider * to be the multiplicative operator in the reals and c* to be the multiplicative operator in the complexes.

    Then what you have to demonstrate to get one step towards proving this isomorphism is:

    m(a c* b) = m(a) * m(b), for all possible choices of a and b within the chosen subset of complex numbers
    That is, you have to show that it doesn't matter when you undergo the mapping, the result will end up the same.

    Either way you decide to go, you're still engaging with a mapping from one mathematical object to another. No matter how you go about it and no matter how trivial it seems, the mapping still transpires.
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    Default Re: How well does relativity hold up under impossible reference frames?

    Distinguishing between isomorphic objects is sometimes useful, and sometimes just an act of obscuration. Which real numbers are we talking about here, anyway? Dedekind cuts, equivalence classes of Cauchy sequences, what?

    Might as well spend our time arguing whether 0 is a natural number.

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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by BeerMug Paladin View Post
    One trouble with our language here might be that the complex number set is defined in terms of the reals, not the other way around. I'm not certain how it might be possible to go about things in the other direction, it seems it will inevitably be a circular definition.
    Even if there's no other way, you could always paste the definition of the reals in place of all references to them in the definition of complex numbers. Speaking of which, I may have approached this from the wrong direction.

    Defining complex numbers as points on a coordinate plane may simplify some aspects of reasoning about them, but it is absolutely not the definition I first learned, which I believe is also the definition that the concept originally started from. Specifically, "4 + 5i" is not just a notation for one way to represent a complex number in writing, but is literally the definition of the number - "the sum of 4 and the product of 5 and i".

    Starting from the reals and their operators and functions:
    • Add the axiom that there exists a number, written and referred to as i, with the property that i * i = -1.
    • Add the corollary (another axiom?) that the only ways of manipulating algebraic expressions that can remove i from the expression are substitutions using either i's definition or an expression that has a fixed constant value, such as 0 times anything or a (non-zero) number divided by itself. In all other ways, i is subject to the same algebraic manipulation properties as real numbers.
    • Define the complex numbers as the superset of the reals, i, and the values of all addition and multiplication expressions that can be formed from them.
    • Expand the definition of all operators and functions with a clause that, to the greatest extent possible, their domain is expanded to include non-real complex numbers, with output values for such inputs that satisfy all the same algebraic manipulation properties as for the reals, along with i's definition.
    • As an extension of the previous point, for each invertible operator or function that has a restricted domain in the reals, expand that domain to include all reals that the previous point added to the range of the inverse.


    To the best of my knowledge, this is a fully working and complete definition of the complex numbers and their operators and functions - all complex number operators and functions that correspond to ones for the real numbers can be, and I'm pretty sure originally were, derived from these points. For example, I think the equation exi = cos(x) + i * sin(x) can be derived from the algebraic manipulation properties of exponentiation by considering how those properties can be applied to reach a point where substitution with i's definition is applicable.

    By this definition, the real numbers are a subset of the complex numbers because the complex numbers are defined as a superset that includes them. Additionally, the complex number operators are derived generalizations of the real number operators, which arguably makes them not actually different operators.

    Quote Originally Posted by BeerMug Paladin View Post
    One good distinction is that when you're considering physical models (I apologize for being on-topic), complex numbers aren't really a valid domain to engage with. There is no ruler that will return a complex number as an outcome. Likewise, returning a complex number from a physical model means you broke the math.
    Not necessarily. It may instead mean that your model needs to define how a complex number maps to physical phenomena, or that your inputs represent a physically impossible scenario.
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    Default Re: How well does relativity hold up under impossible reference frames?

    If your physical model doesn't make sense of imaginary inputs, then it is either wrong, or it is doing the same thing by a different name. In electromagnetics imaginary numbers crop up all the time, but we could replace them with a "potential energy" variable and just have to do ten times the work to get the same numbers. Lots of other places where they crop up work the same way.

    And my favorite way to teach imaginary numbers is to draw a number line, ask people to define points on the line, and then ask someone to define a point not on the line.
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Anymage View Post
    But if I give this imaginary point a speed of 2c, any by extension give myself and every real thing in the universe an equivalent FTL speed, I'm curious how the numbers would pan out.
    The numbers would be more or less consistent, but they would make actually interpreting the numbers a huge headache (even compared to how relativity normally is) because cause and effect are now disassociated from the time axis.

    If you have the entire universe moving at 2c you have no real change but are using a lot of weird numbers to describe things. If you have things moving at 2c relative to each other then you'll have all the problems of a subliminal object.

    In my opinion, one should start from the perspective that the proper numbers (proper distance, proper velocity, proper acceleration) are metaphysically real and the coordinate numbers are abstractions. Put in any real numbers in for the proper values and you always get sense sensible values back out for the coordinate values.
    Quote Originally Posted by gomipile View Post
    The awkwardness of your choice of how to use the language is highlighted by the fact that you call imaginary numbers imaginary numbers in the same sentence that you say they are not numbers.
    The language is fine. An <adjective> X may or may not be a X (depending on the adjective and noun); for example a former teacher isn't a teacher. I'd also disagree with shawnhcorey about "number" as a mathematical term, but his description of his position is fine.

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    Default Re: How well does relativity hold up under impossible reference frames?

    So how does that work when space itself is expanding? As I understand it, the farther away something is from us, the faster it is moving away from us (due to the way the universe is expanding), so that galaxies that we can currently see will disappear from our sight a mere few billion years from now (they will be moving away from us at faster than c, so the light they emit will never reach us).
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Quizatzhaderac View Post
    If you have the entire universe moving at 2c you have no real change but are using a lot of weird numbers to describe things. If you have things moving at 2c relative to each other then you'll have all the problems of a subliminal object.
    I was idly wondering about tachyons too, in that there's no guarantee that they all have the same direction of motion. But I also have a hunch that they'd be more complicated to work out than a pointless forum post would warrant and that I wouldn't be able to understand all the ugly math of people who do work all the complex stuff. Still, thanks for covering my pointless math curiosity after other people's geeking out about the complex plane.

    Quote Originally Posted by Lord Torath View Post
    So how does that work when space itself is expanding? As I understand it, the farther away something is from us, the faster it is moving away from us (due to the way the universe is expanding), so that galaxies that we can currently see will disappear from our sight a mere few billion years from now (they will be moving away from us at faster than c, so the light they emit will never reach us).
    There are places we can see* where space is expanding faster than light. Inside the event horizon of a black hole. Usual disclaimer here how quantum gravity might throw off what we think we know now, but the way things get really loopy beneath an event horizon is pretty well covered if you bother looking around. It's just that by definition, we can't interact with or be affected by this region.

    Out past the particle horizon is probably very similar. It'd look very weird and time and space would look messy if we could somehow see them, but the word "horizon" means that it's impossible. The big difference is that the black hole's pulling of spacetime is defined by the central singularity (and thus, it can be said to occupy a specific region of space and its location is largely independent of the observer), the particle horizon is defined by being far enough away from you that the expansion of space accumulates to superluminal speeds. Even if you could teleport instantaneously, the particle horizon would be too far away to matter. And if in the future it gets close enough to be meaningful, you'd be looking down a big rip sort of situation and are pretty much screwed.

    *(Edit: For certain definitions of "see". Someone's going to be pedantic that we can't see inside if I don't make a note of it first, but we can still see the general region of space and point to where it'll be.)
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Douglas View Post
    To the best of my knowledge, this is a fully working and complete definition of the complex numbers and their operators and functions - all complex number operators and functions that correspond to ones for the real numbers can be, and I'm pretty sure originally were, derived from these points. For example, I think the equation exi = cos(x) + i * sin(x) can be derived from the algebraic manipulation properties of exponentiation by considering how those properties can be applied to reach a point where substitution with i's definition is applicable.

    By this definition, the real numbers are a subset of the complex numbers because the complex numbers are defined as a superset that includes them. Additionally, the complex number operators are derived generalizations of the real number operators, which arguably makes them not actually different operators.
    (bolding mine)

    Functions actually get a lot more messy in the complex numbers. Take a simple square root and see what happens if you stroll around the 0. In order to keep the function smooth we need to take a complex space made from cuting and cross-connecting of two complex planes. Things only get worse from here with irrational power functions or logarithms and become truly funky with for example square roots of for example x(x+1)(x+2), since the complex space defined by that function is homeomorphic to a torus instead of a sphere.


    That being told: as sets, real numbers are a subset of complex numbers. There can be no contest there, since for sets there are no defined operations on them. Whether the connection between the algebraic structures on complex and real numbers is as simple or not, that is a different question though. Right now, I do not feel good enough about it to give a solid answer. I do have some opinions though:

    I found a video on the construction of natural numbers with an interesting answer to the question we have here: whether two structures connected by an isomorphism are the same thing. Relevant part starts here.

    Actually, we can look at it this way: in any number set equality relation implies we have the very same number on both sides of the = operator. It is pretty much hardcoded in the definition of the basic number sets and is inherited by all the more complicated ones. Question of existence of an isomorphism (X = Y iff there exists an isomorphism between X and Y) between some algebraic structures fulfills all the criteria for an equivalence relation. If so, it works like regular = and it is up to us to decide if we form some meta-algebraic axiom that if there exists an isomorphism between two structures, then they are both merely different representations of the same structure. In the spirit of algebra my answer would actually be yes.
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Radar View Post
    (bolding mine)

    Functions actually get a lot more messy in the complex numbers. Take a simple square root and see what happens if you stroll around the 0. In order to keep the function smooth we need to take a complex space made from cuting and cross-connecting of two complex planes. Things only get worse from here with irrational power functions or logarithms and become truly funky with for example square roots of for example x(x+1)(x+2), since the complex space defined by that function is homeomorphic to a torus instead of a sphere.
    Sure, square roots, exponentiation, and logarithms are messy if you try to compute them using the a + bi form of a number, but they become almost trivial if you just switch to the aebi form instead (a and b are not the same numbers in each form).

    Square root of a complex number:
    sqrt(aebi) = (aebi)0.5 = a0.5e0.5 * bi

    Irrational, or even non-real, exponent of a complex number:
    (aebi)x = axexbi

    Natural logarithm of a complex number:
    ln(aebi) = ln(a) + ln(ebi) = ln(a) + bi

    Logarithm with a complex base:
    logx(aebi) = ln(aebi) / ln(x)

    I'll grant that fully simplifying that last one to get a result of either a + bi or aebi form takes a few extra steps, including transforming parts of the expression from one form to the other, but it's still a fairly straightforward sequence of substitutions.
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Douglas View Post
    Sure, square roots, exponentiation, and logarithms are messy if you try to compute them using the a + bi form of a number, but they become almost trivial if you just switch to the aebi form instead (a and b are not the same numbers in each form).

    Square root of a complex number:
    sqrt(aebi) = (aebi)0.5 = a0.5e0.5 * bi

    Irrational, or even non-real, exponent of a complex number:
    (aebi)x = axexbi

    Natural logarithm of a complex number:
    ln(aebi) = ln(a) + ln(ebi) = ln(a) + bi

    Logarithm with a complex base:
    logx(aebi) = ln(aebi) / ln(x)

    I'll grant that fully simplifying that last one to get a result of either a + bi or aebi form takes a few extra steps, including transforming parts of the expression from one form to the other, but it's still a fairly straightforward sequence of substitutions.
    It is not about calculations getting messy - it is all about the need to redefine the complex space in a non-trivial way in order to make those functions analytic. Using the exponential notation try how the square root behaves when you start from sqrt(1)=1 and keep increasing the argument (b in your examples) until you reach again sqrt(1). You need another full circle to get back to the original value of sqrt(1)=1.

    As a consequence for example a circle with unit radius around a pole of a square root function will inevitably have a circumference of 4Pi and you cannot even draw a circle around the singularity of a logarithm at all. The impact on topology is quite significant. In fact, the key reason why the only closed loops with non-zero integral over them are those around singularities behaving like 1/x is that the antiderivative of 1/x (which is log(x)) is defined on a different complex space altogether, so the loop stops being closed when the endpoint values are evaluated. As far as I know, this is a unique feature of the 1/x function.
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Radar View Post
    It is not about calculations getting messy - it is all about the need to redefine the complex space in a non-trivial way in order to make those functions analytic. Using the exponential notation try how the square root behaves when you start from sqrt(1)=1 and keep increasing the argument (b in your examples) until you reach again sqrt(1). You need another full circle to get back to the original value of sqrt(1)=1.

    As a consequence for example a circle with unit radius around a pole of a square root function will inevitably have a circumference of 4Pi and you cannot even draw a circle around the singularity of a logarithm at all. The impact on topology is quite significant. In fact, the key reason why the only closed loops with non-zero integral over them are those around singularities behaving like 1/x is that the antiderivative of 1/x (which is log(x)) is defined on a different complex space altogether, so the loop stops being closed when the endpoint values are evaluated. As far as I know, this is a unique feature of the 1/x function.
    But square root having multiple valid outputs is a normal problem. Complex numbers make it trivial to see what is going on. As Douglas said, you can easily calculate the square (or any) root of a polar form of a complex number. You can even see all the possible repetition by realizing that angles repeat every 2 pi. Just because it disrupts your topology is no reason to get down on complex numbers. Lots of normally analytic functions have problems at zero. Do you think that zero is a problem that destroys the beauty of math?
    Quote Originally Posted by Wardog View Post
    Rockphed said it well.
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  16. - Top - End - #46
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    Default Re: How well does relativity hold up under impossible reference frames?

    Many mathematicians would say that these unique features of complex analytic functions are strengths of the complex number systems, not weaknesses.

    The fact that creating a domain where your function is analytic and entire requires extra topology beyond a flat sheet leads to Riemann surfaces, which are quite beautiful and useful in various areas of mathematics.

    Also, those pesky singularities lead to the Cauchy integral theorem, which was alluded to above. The usefulness of that theorem more than makes up for the initial learning curve of dealing with singularities of complex functions.
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    where is the atropal? and does it have a listed LA?

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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Rockphed View Post
    But square root having multiple valid outputs is a normal problem. Complex numbers make it trivial to see what is going on. As Douglas said, you can easily calculate the square (or any) root of a polar form of a complex number. You can even see all the possible repetition by realizing that angles repeat every 2 pi. Just because it disrupts your topology is no reason to get down on complex numbers. Lots of normally analytic functions have problems at zero. Do you think that zero is a problem that destroys the beauty of math?
    I do not get down on complex numbers or at least I did not intended for my posts to be viewed in such a light. I probably should have been more carefull with my adjective choices.

    All I was arguing is that there is a distinct and non-trivial difference between real spaces and complex spaces which makes comparing them a bit more tricky. Also that there is a huge difference between comparing sets and comparing spaces. I would not learn complex analysis if it were not useful
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    Default Re: How well does relativity hold up under impossible reference frames?

    A thought occurs. How do they know it takes an imaginary number as a value. If it's merely the square root of a negative number it could take any of the quaternion* or even sedenion constants as a coefficent

    *i^2=j^2=k^2=ijk=-1

  19. - Top - End - #49
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    Default Re: How well does relativity hold up under impossible reference frames?

    Quote Originally Posted by Bohandas View Post
    A thought occurs. How do they know it takes an imaginary number as a value. If it's merely the square root of a negative number it could take any of the quaternion* or even sedenion constants as a coefficent

    *i^2=j^2=k^2=ijk=-1
    I think it is about picking the least compliacted system that can describe a given situation. If it can be all written in terms of complex numbers and introducing quaternions would not give any particular advantage, then there is no reason to overcomplicate things.

    As a counter-example, many integrals in real space are way easier to solve in complex space, so it is better to move to a more complicated number system.
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