# Thread: How well does relativity hold up under impossible reference frames?

1. ## How well does relativity hold up under impossible reference frames?

The whole point of relativity is that it allows easy interpolations between reference frames. Whether I designate one based on a point at rest equivalent to me, one going 100 mph compared to me, or one going at .999c. 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. Would all the imaginary numbers for real things cancel out and give us recognizable results, or would the equations just look at us like we're nuts and only spit out nonsense?

2. ## Re: How well does relativity hold up under impossible reference frames?

I think the math suggests that time starts going backward after 1C. Which is why a lot of time travel has you moving faster than light.

3. ## Re: How well does relativity hold up under impossible reference frames?

If you naively try to apply the Lorentz transform to move from a normal reference frame to a hypothetical inertial reference frame traveling at 2c, you get imaginary numbers for some of the new coordinates. I don't know of any possible physical interpretation of this. But I'm surprised at this question coming from somebody who is answering questions about black holes in another thread.

4. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by DavidSh
If you naively try to apply the Lorentz transform to move from a normal reference frame to a hypothetical inertial reference frame traveling at 2c, you get imaginary numbers for some of the new coordinates. I don't know of any possible physical interpretation of this. But I'm surprised at this question coming from somebody who is answering questions about black holes in another thread.
I'm a smart layman who knows my limits.

And I know that from a perspective that causes us to go at 2c we wind up spitting out imaginary numbers. I also know that hypothetical tachyons tend towards infinite speed as their energy tends towards 0, so sitting at "just" 2c is pretty energetic. What I'm wondering is if the earth spits out imaginary numbers because it's going ftl, and the moon is also spitting out similar imaginary numbers, do the imaginary numbers and backwards-in-timiness cancel out to give something that looks at least generally normal-ish.

5. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Anymage
What I'm wondering is if the earth spits out imaginary numbers because it's going ftl, and the moon is also spitting out similar imaginary numbers, do the imaginary numbers and backwards-in-timiness cancel out to give something that looks at least generally normal-ish.
I don't actually know for certain, but I'd expect the answer is yes. I think for this to not hold there would have to be a property of math that's used in the formulas and does not generalize from the real number line to the complex plane, and as far as I know relativity does not use anything that qualifies.

6. ## Re: How well does relativity hold up under impossible reference frames?

Imaginary numbers are not numbers. They have a lot of the properties of numbers but they're not numbers.

7. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
Imaginary numbers are not numbers. They have a lot of the properties of numbers but they're not numbers.

Complex numbers are not real numbers, is what I think you mean to say. 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.

In mathematics, they are both considered number systems. Also, in the spirit of what I think you were trying to say, we must always be careful to specify what number systems we are using, and be careful not to make assumptions based on number systems that the set of axioms we are working with may not be compatible with.

For example, we may have some explicit or implicit dependence on ordering of numbers in the axioms for a particular theory. If so, then complex numbers can't be naively plugged into that theory, since there isn't a general notion of less than or greater than in the complex plane.

8. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by gomipile
Complex numbers are not real numbers, is what I think you mean to say. 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.
It is difficult to describe mathematics in English. The math, however, is very clear.

Imaginary numbers are based on the square root of -1. This is not a number. Complex numbers have imaginary numbers inside them, so they too are not numbers.

Do not confuse the name "imaginary numbers" with what they are. "The map is not the territory."

9. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
It is difficult to describe mathematics in English. The math, however, is very clear.
Yes, indeed. The maths is very clear. i (or j) is the number that, if multiplied by itself, gives -1.

The confusing part is the use of "real" and "imaginary" to denote the difference between complex and normal numbers.

Complex numbers are not real numbers - as in "not a member of the set of numbers on the real line" - but are still none the less actual numbers. You would not be able to do mathematical operations on complex numbers if they were not actually numbers.

10. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Manga Shoggoth
You would not be able to do mathematical operations on complex numbers if they were not actually numbers.
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:

You cannot apply the math operator multiple to complex numbers.

11. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
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:

You cannot apply the math operator multiple to complex numbers.
So? I can't apply the Log operator to a negative number. By your logic that means that negative numbers are not numbers either.

You are confusing the usage of the word "real". In this sense "real" simply means that set of numbers that encompases Integers, Rational numbers and Irrational numbers.

EDIT: Since I can't link my Maths textbooks (or even find them these days):

An "imaginary number" is a multiple of a quantity called "i" which is defined by the property that i squared equals -1. This is puzzling to most people, because it is hard to imagine any number having a negative square. The result: it is tempting to believe that i doesn't really exist, but is just a convenient mathematical fiction.

This isn't the case. Imaginary numbers do exist. Despite their name, they are not really imaginary at all. (The name dates back to when they were first introduced, before their existence was really understood. At that point in time, people were imagining what it would be like to have a number system that contained square roots of negative numbers, hence the name "imaginary". Eventually it was realized that such a number system does in fact exist, but by then the name had stuck.)

12. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Manga Shoggoth
You are confusing the usage of the word "real". In this sense "real" simply means that set of numbers that encompases Integers, Rational numbers and Irrational numbers.
I am not confusing anything. In modern algebra, a mathematical system is a set of objects that have an associate operation on them. Real numbers and complex numbers are different systems because they have different multiple operations. That means complex "numbers" are not numbers.

13. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
I am not confusing anything. In modern algebra, a mathematical system is a set of objects that have an associate operation on them. Real numbers and complex numbers are different systems because they have different multiple operations. That means complex "numbers" are not numbers.
Funny you should say that - those issues are answered in the three links towards the bottom of the page I quoted.

Real numbers are a subset of the complex number system (where the imaginary component is zero), just as integers, rationals and irrationals are subsets of the "real number" system.

14. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Manga Shoggoth
Real numbers are a subset of the complex number system (where the imaginary component is zero), just as integers, rationals and irrationals are subsets of the "real number" system.
No, integers, rationals, and irrationals are subset of real numbers because they use all the same operations. Complex numbers used different operations. They are not the same.

15. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
No, integers, rationals, and irrationals are subset of real numbers because they use all the same operations. Complex numbers used different operations. They are not the same.
This is an arguement that literally nobody is making here. Nobody is arguing that Complex Numbers are Real Numbers. We are arguing that complex numbers are numbers.

This is what I was taught in school (starting with A-Level Algebra). This is how we dealt with them when I was doing my Physics Degree. This is what every single on-line source that I have checked has said.

I suggest that you read section headed "Do complex numbers really form a number system?" (Hopefully I have the link right this time...).

16. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Manga Shoggoth
This is what I was taught in school (starting with A-Level Algebra). This is how we dealt with them when I was doing my Physics Degree. This is what every single on-line source that I have checked has said.
I suggest you study modern algebra before you claim you know what algebra is.

17. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
No, integers, rationals, and irrationals are subset of real numbers because they use all the same operations. Complex numbers used different operations. They are not the same.
You sound like the people who argued that negative numbers weren't really numbers back in the 1600s and 1700s. Including negative numbers requires modifying the definitions of the basic operations, after all. So your argument would apply equally well to exclude negative numbers from being numbers since the "negative times a negative is a positive" isn't required when working with only non-negative real numbers.

Unfortunately for your position, you're at least 150 years too late to the discussion about complex numbers being numbers.

18. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by gomipile
You sound like the people who argued that negative numbers weren't really numbers back in the 1600s and 1700s. Including negative numbers requires modifying the definitions of the basic operations, after all. So your argument would apply equally well to exclude negative numbers from being numbers since the "negative times a negative is a positive" isn't required when working with only non-negative real numbers.

Unfortunately for your position, you're at least 150 years too late to the discussion about complex numbers being numbers.
No, I'm talking about modern algebra. If you don't want to study that field of mathematics, that's fine but don't claim you know algebra if you don't.

19. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
I am not confusing anything. In modern algebra, a mathematical system is a set of objects that have an associate operation on them. Real numbers and complex numbers are different systems because they have different multiple operations. That means complex "numbers" are not numbers.
The multiplication operation for complex numbers is exactly the same as the one for real numbers. The only difference is that often you have to apply the distributive property of multiplication because a single complex number is represented as the sum of a real number and an imaginary number, and that's really a difference in the numbers being multiplied, not in the operation itself.

20. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
It is difficult to describe mathematics in English. The math, however, is very clear.

Imaginary numbers are based on the square root of -1. This is not a number. Complex numbers have imaginary numbers inside them, so they too are not numbers.

Do not confuse the name "imaginary numbers" with what they are. "The map is not the territory."
I just make a point to refer to them exclusively as "complex" outside the few limited cases where the subtle distinction between complex and imaginary numbers is relevant. For example, i by itself is a complex number, even though its real term is 0.

Originally Posted by shawnhcorey
No, integers, rationals, and irrationals are subset of real numbers because they use all the same operations. Complex numbers used different operations. They are not the same.
You're thinking of the hypercomplex numbers (the quaternions, octinions, and sedenions). Those, indeed, do not play well with arithmatic, the regular complex numbers use pretty normal arithmatic, you just add as if i was an unknown variable

so, for example, 2i+6 + 3i+23 = 5i+29

21. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
No, I'm talking about modern algebra. If you don't want to study that field of mathematics, that's fine but don't claim you know algebra if you don't.
I am a mathematician, and a glance through my algebra texts doesn't show any statements to the effect of complex numbers not being numbers.

22. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by gomipile
I am a mathematician, and a glance through my algebra texts doesn't show any statements to the effect of complex numbers not being numbers.
Dig deeper.

23. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
It is difficult to describe mathematics in English. The math, however, is very clear.

Imaginary numbers are based on the square root of -1. This is not a number. Complex numbers have imaginary numbers inside them, so they too are not numbers.

Do not confuse the name "imaginary numbers" with what they are. "The map is not the territory."
It's maps all the way down, my man. Algebra is abstract, there is no territory.

Originally Posted by shawnhcorey
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:

You cannot apply the math operator multiple to complex numbers.
That's... not right. Complex numbers aren't vectors, i doesn't have a direction associated with it, and nor does 1. They have some similarities, and it's sometimes helpful to think of complex numbers as being 2 dimensional, but they aren't the same.

Not that this actually matters, because, as has been noted elsewhere, that multiplication definition is the same as the one we use for real numbers. You can tell because it has identical properties.

Also, dot products aren't multiplication. The dot product is what's called an inner product: an operation that takes two vectors and returns a scalar. This means that the inner product isn't closed on all vector spaces (really just some trivial one dimensional vector spaces, and even then there's a difference between a 1D vector and a scalar so that's questionable), which excludes it from being the multiplicative basis for an algrebra. It is useful for defining different types of vector spaces, however.

Finding the angle between two vectors involves an inner product, but the fact that that inner product is distinct from multiplication of complex numbers doesn't really mean anything. Hell, you can define an arbitrary inner product, and it won't mean anything, because the inner product of a given vector space isn't multiplication, and really is only tenuously related.

Originally Posted by shawnhcorey
I am not confusing anything. In modern algebra, a mathematical system is a set of objects that have an associate operation on them. Real numbers and complex numbers are different systems because they have different multiple operations. That means complex "numbers" are not numbers.
I feel like it might be a language thing that's tripping people up, because there is the sense that complex numbers are a different field than real numbers. And that's a meaningful thing, because all a field requires is an additive and a multiplicative operation, and those operations are insufficient to go from the real numbers to the (non trivially) complex numbers. But at the same time, those operations are insufficient to reach most real numbers (the irrationals, transcandentals, non-computables, etc).

Originally Posted by shawnhcorey
Dig deeper.
Kinda feel like it's on you to support your argument at this point. A few people have explained the opposing side in several different ways, and your own source doesn't actually make the claim you're making.

24. ## Re: How well does relativity hold up under impossible reference frames?

Maybe it would help if shawnhcorey provided the definition he is using for the word "number".

25. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by crayzz
Kinda feel like it's on you to support your argument at this point. A few people have explained the opposing side in several different ways, and your own source doesn't actually make the claim you're making.
I am not going to present a 2nd-year university course here for anyone.

26. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
I am not going to present a 2nd-year university course here for anyone.
Just need a textbook name, my man. I'm sure if this is second year stuff there will be at least a few.

27. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by shawnhcorey
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:

You cannot apply the math operator multiple to complex numbers.
Consider the expression (a + bi)(c + di) and treat it as though i were any other number, while a through d are all real numbers.

Apply the distributive property of multiplication:
ac + adi + bci + bdi2

Apply the definition of i, which says i2 = -1:
ac + adi + bci - bd

Apply the commutative property of addition:
ac - bd + adi + bci

Apply the distributive property in the opposite direction on the imaginary part:
ac - bd + (ad + bc)i

There, I just got the correct result of complex number multiplication solely by applying properties of real number addition and multiplication and the definition of i, all in a generalized way that works for all complex numbers.

28. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Anymage
I'm a smart layman who knows my limits.

And I know that from a perspective that causes us to go at 2c we wind up spitting out imaginary numbers. I also know that hypothetical tachyons tend towards infinite speed as their energy tends towards 0, so sitting at "just" 2c is pretty energetic. What I'm wondering is if the earth spits out imaginary numbers because it's going ftl, and the moon is also spitting out similar imaginary numbers, do the imaginary numbers and backwards-in-timiness cancel out to give something that looks at least generally normal-ish.
So--- as an experimental particle physicist (that's one of my masters, I have a doctorate in engineering as well)--- I think the original question was an interesting idle wonder, and I've read as well about the backwards in time interpretation of particles blindly using special relativity equations with V>c. But.

It somewhat defeats the purpose of science to extend the math of science beyond any physical data. It is no longer science, it's just math that doesn't mean anything.

To move any particle of finite mass to the speed of light takes infinite energy. Which means you cannot do it. (And, I should note, the power requirements to accelerate particles have been proven out to v=0.99999c and beyond --- special relativity is a VERY well verified field.) Yes, there are arguments that --- that energy barrier at c is symmetric and asymptotic apparently from the equations we have for v<c, so perhaps there are particles which we could detect ("tachyons") which somehow are beyond the speed of light. But we've wandered from "science" to "science fiction". It is supposition which is un-anchored in actual data--- interesting hypothetical particles. Cool to wonder about, but--- there are interesting things in this universe without having to extrapolate without data. Dark matter, heavy baryons and antimatter and black holes. As a scientist, I always chuckle at my brethren (cough, cough, gravitational theoreticians and string theory guys) who wander aimlessly and do math without real world data.

OK, *some* physicists are looking at massless excitations which may move faster than c. See: Chao, R, "What is known about tachyons, theoretical particles that travel faster than light and move backward in time? Is there scientific reason to think they really exist?", Scientific American, 21 Oct 1999.

Which--- ok, the original poster asked about reference frames which move above c, so that could be valid if we restricted ourselves to that. There are some recent works on wave phenomena above c--- see something like U.D. Jentschura and B.J. Wundt, "Localizability of tachyonic particles and neutrinoless double beta decay", Phys. J. C (2012) 72:1894, doi: 10.1140/epjc/s10052-012-1894-4.

29. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by RocketBoy11
So--- as an experimental particle physicist (that's one of my masters, I have a doctorate in engineering as well)--- I think the original question was an interesting idle wonder, and I've read as well about the backwards in time interpretation of particles blindly using special relativity equations with V>c. But.

It somewhat defeats the purpose of science to extend the math of science beyond any physical data. It is no longer science, it's just math that doesn't mean anything.
Thanks for the input. Always great to get more people who are smarter than me on the boards (so I can learn more)! And may I be the first to say: "Welcome to the Playground!"

30. ## Re: How well does relativity hold up under impossible reference frames?

Originally Posted by Manga Shoggoth
Funny you should say that - those issues are answered in the three links towards the bottom of the page I quoted.

Real numbers are a subset of the complex number system (where the imaginary component is zero), just as integers, rationals and irrationals are subsets of the "real number" system.
I see there is an argument around this statement. I just want to make what I feel is an important clarification.

A subset of the complex numbers can be mapped to the reals. Usually, people shorthand this statement as "the reals are a subset of the complex numbers". It's pretty easy to see why. And for most everyday understanding and usage of these ideas, it's not terribly important that the statement be 100% precise when people are first introduced to it. Pedagogical statements tend to be dirty, filthy lies.

Anyway, the real numbers are not a subset of the complex numbers. Because, among other things, the operators are different with the complex numbers.

Speaking of...

Originally Posted by Douglas
There, I just got the correct result of complex number multiplication solely by applying properties of real number addition and multiplication and the definition of i, all in a generalized way that works for all complex numbers.
This is called a construction. What you are doing is constructing a new operation within the complex number system out of two copies of the real number system and its related properties by introducing a new rule. In the reals, there is no variable substitution that behaves like i.

Operators are defined as mappings from one set to another. Let's consider the multiplication operator a bit more.

The operators in the reals take two numbers from the reals and maps them to numbers in the reals. So for multiplication, the mapping is:
a * b = c
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:
So, (a,b) (complex *) (c,d) = (e,f)
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. However, we can define the mapping:
(x,0) ==> x
There, now we are back to my original statement.

A subset of the complex numbers can be mapped to the reals.

This may seem like nitpicky and finicky little details that don't matter, but mathematics is the home for pointless pedantry. Get off the daggum lawn ya dern kids!

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