this post was submitted on 29 Nov 2024

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324

Fuck geometry (lemm.ee)

submitted 6 months ago by [email protected] to c/[email protected]

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[–] [email protected] 84 points 6 months ago (7 children)

This triangle is impossible.

If the distance between B and C is 0, B and C are the same points. If that is the case, the distances between A and B and A and C must be the same.

However, i ≠ 1.

If you want it to be real (hehe) the triangle should be like this:

    C
    | \
|i| |  \ 0
    |   \
    A---B
     |1|

Drawing that on mobile was a pain.

As the other guy said, you cannot have imaginary distances.

Also, you can only use Pythagoras with triangles that have a 90° angle. Nothing in the meme says that there's a 90° angle. As I see it, there are only 0° and 180° angles.

Goodbye, I have to attend other memes to ruin.

[–] [email protected] 22 points 6 months ago (1 children)

Context matters. In geometry i is a perfectly cromulent name for a real valued variable.

[–] [email protected] 4 points 6 months ago (1 children)

Oh shit, he used the word cromulent. Every one copy off this guy.

[–] [email protected] 2 points 6 months ago

That wouldn't be cromulent, would it?

[–] [email protected] 12 points 6 months ago

Mad mobile drawing!!

[–] technocrit 5 points 6 months ago* (last edited 6 months ago) (3 children)

As the other guy said, you cannot have imaginary distances.

Incorrect. There are complex valued metric spaces

And even if we assume real valued metrics, then i usually represents the unit vector (0,1) which has distance real 1.

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[–] [email protected] 2 points 6 months ago (2 children)

This is clearly meant to be a right triangle. And the distances between the points are the same (because the squares of the coordinate differences are the same), just the directions are different.

If you move 1 unit forward, turn the correct 90 degrees, and then move i units forward, you will end up back where you started.

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[–] [email protected] 47 points 6 months ago (3 children)

1 • 1 + i • i = 1 + (-1) = 0 = 0 • 0

Pythagoras holds, provided there's a 90° angle at A.

[–] [email protected] 28 points 6 months ago

this is why it is still a theorem

[–] [email protected] 22 points 6 months ago (4 children)

I'm so angry at people who think that distances can be imaginary.

[–] lugal 26 points 6 months ago

Never been together with people and still felt alone?

[–] [email protected] 24 points 6 months ago (3 children)

When talking about AC power, some of the power consumed doesn't actually produce real work. It gets used in the generation of magnetic fields and charges in inductors and capacitors.

The power being used in an AC system can be simplified by using a right triangle. The x axis is the real power being used by resistive parts of the circuit (in kilowatts, KW). The y axis is reactive power, that is power being used to maintain magnetic fields and charges (in kilovolt-amperes reactive, KVAR). And the hypotenuse is the total power used by the circuit, or KVA (kilovolt-amperes).

Literal side note: they're all the same units, but the different sides of the triangle are named differently to differentiate in writing or conversation which side of the power triangle is being talked about. Also, AC generator ratings are given in KVA, so you need to know the total impedance of your loads you want to power and do a bit of trig to see if your generator can support your loads.

The reactive component of AC power is denoted by complex numbers when converting from polar coordinates to Cartesian.

Anyways, I almost deleted this because I figured your comment was a joke, but complex numbers and right triangles have real world applications. But power triangles are really just simplifications of circles. By that I mean phasors rotating in a complex plane, because AC power is a sine wave.

[–] [email protected] 8 points 6 months ago

By that I mean phasors rotating in a complex plane, because AC power is a sine wave.

I read the entire thing as Air Conditioning and it made me think my tired ass had forgotten something important and then here comes like whiplash when it clicked that you were talking about Alternating Current.

More coffee needed.

[–] [email protected] 7 points 6 months ago

Please be careful with two different things. Complex numbers have two components. Distances don't. They are scalars. The length of the vector (0,1) is also 1. Just as a+bi will have the length sqrt(a^2 + b^2). You can also use polar coordinates for complex numbers. This way, you can see that i has length 1, which is the distance from 0.

The triangle in the example above adds a vector and a scalar value. You can only add two vectors: (1,0) + (0,1) which results in (1,1) with the proper length. Or you can calculate the length/distance (absolute values) of the complex numbers directly.

[–] [email protected] 1 points 6 months ago

Its another classic case of Euler's Identity

[–] [email protected] 16 points 6 months ago (3 children)

They're about as imaginary as numbers are in general.

Complex numbers have real application in harmonics like electronics, acoustics, structural dynamics, damping, regulating systems, optronics, lasers, interferometry, etc.

In all the above it's used to express relative phase, depending on your need for precision you can see it as a time component. And time is definitely a direction.

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[–] technocrit 5 points 6 months ago* (last edited 6 months ago) (1 children)

You're mad at mathematicians for constructing complex valued metrics? It's all just formalism, nothing personal.

[–] [email protected] 1 points 6 months ago

Omg, yes. This is horrible. :)

[–] [email protected] 4 points 6 months ago

But that's not the definition of the absolut value, I.e. "distance" in complex numbers. That would be sqrt((1+i)(1-i)) = sqrt(2) Also the triangle inequality is also defined in complex numbers. This meme is advanced 4-4*2=0 Works only if you're doing it wrong.

[–] [email protected] 24 points 6 months ago

You didn’t really expect an imaginary triangle to behave like a real one, did you?

[–] [email protected] 24 points 6 months ago (1 children)

I get it, it's projected on a comlplex sphere. B and C are the same point

[–] [email protected] 1 points 6 months ago

please stop making it make sense

[–] [email protected] 16 points 6 months ago* (last edited 6 months ago)

C and B have a wormhole between them

[–] [email protected] 11 points 6 months ago

In the complex plane each of these vectors have magnitude 1 and the distance between them is square root of two as you would expect. In the real plane the imaginary part has a magnitude of zero and this is not a triangle but a line. No laws are broken here.

[–] [email protected] 10 points 6 months ago (1 children)

i = 1 is the only logical choice

[–] [email protected] 5 points 6 months ago

Literally, this is one of those questions where they're testing logic and your understanding that the figures aren't necessarily representative of physical reality.

[–] [email protected] 9 points 6 months ago

I kept seeing this pop up recently, and I finally understand it: it's an introductory problem in Lorentzian general relativity.

AB is a space-like line, while AC is a time-like line. Typically, we would write AC as having distance of 1, but with a metric such that squaring it would produce a negative result. However it's similar to multiplying i to the value.

BC has a distance of 0, but a better way of naming this line would be that it has a null interval, meaning that light would travel following this line and experience no distance nor time going by.

I'm sure PBS Spacetime would explain all of this better than me. I just woke up and can't bother searching for the correct words on my phone.

[–] [email protected] 8 points 6 months ago

It gets worse once you start doing trig on it

[–] technocrit 6 points 6 months ago (1 children)

Maybe the problem is constructing a metric that makes this diagram true. Something like d(x,y) = | |x| - |y| | might work but I'm too lazy to check triangle inequality.

[–] [email protected] 3 points 6 months ago

Triangle inequality for your metric follows directly from the triangle inequality for the Euclidean metric. However, you don't need a metric for the Pythagorean Theorem, you need an inner product and, by definition, an inner product doesn't allow non-real values.

[–] [email protected] 5 points 6 months ago* (last edited 6 months ago) (1 children)

Wish me luck for I'm doing trig test with radians (2 pi rad = 360 ?)

[–] [email protected] 8 points 6 months ago (2 children)

Radians are the objectivly better way to do angles tho. Just remeber π=180deg and ur right. Btw here is a another brain fuck the units radians/second is just Hz

[–] [email protected] 3 points 6 months ago (2 children)

Thank you for reminding me!

Btw, Radians/sec = Hz? What is this, physics?

[–] [email protected] 2 points 6 months ago

Engineering unit maths. Cos angles are unitless so radians/second =1/second=Hz

[–] [email protected] 2 points 6 months ago* (last edited 6 months ago)

The easiest way to think about it is that 1 full rotation (2*pi radians) in 1 second makes 1 Hz.

The number of rotations made in a second corresponds to Hz in the same way that the number of sine wave periods that fit in a second also represents Hz. This gif does a really good job of showing how rotation relates to sine/cosine waves, which just so happens to help visualize the rad/s -> Hz <- periods/s relationship:

[–] [email protected] 1 points 6 months ago

Radians are the objectivly better way to do angles

Yes, and tau is objectively better than pi. Just remember tau = 360°. Which is a full circle, which easier to work with than half a circle.

[–] [email protected] 4 points 6 months ago* (last edited 6 months ago) (1 children)

funny Interpretation: in the complex plane, the imaginary axis is orthogonal to the real axis. so instead of the edge marked with i (AC), imagine an edge of length 1 orthogonal to that edge. It would be identical to AB, so ~~AC~~ CB is 0.

[–] [email protected] 1 points 6 months ago (1 children)

But then CB couldn't also be 0; wouldn't it be cos(1 + i)? Or something like that.

[–] [email protected] 3 points 6 months ago

oh I mixed up the points, I meant to say CB is 0 in the end

[–] [email protected] 4 points 6 months ago (1 children)

You can make something like this properly by defining a different metric. For example with metric dl^2^ = dx^2^ - dy^2^ the vector (1, 1) has length 0, so you can make a "triangle" with sides of lengths 1, -1 and 0.

[–] [email protected] 2 points 6 months ago (3 children)

That's not a metric. In any metric, distances are positive between distinct points and 0 between equal points

[–] [email protected] 1 points 6 months ago* (last edited 6 months ago) (1 children)

It depends which metric definition are you using. The one I wrote is a pseudo-Riemannian metric that is not positive defined.

Normally physicists use that generalized metric definition because spacetime in most cases has a metric signature of (-1, 1, 1, 1). Points with zero distance are not necessarily the same point, they just are in the same null geodesic.

[–] [email protected] 1 points 6 months ago

You're talking about a metric tensor on a pseudo-Riemannian manifold, I'm talking about a metric space. A metric in the sense of a metric space takes nonnegative real values. If you relax the condition that distinct points have nonzero distance, it's a pseudometric.

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