4 ms·

There are some great notes in there and also things that you will almost certainly not know already. Beware the SI's broken definition of Hz. You should treat

There are some great notes in there and also things that you will almost certainly not know already. Beware the SI's broken definition of Hz. You should treat the radian as being correct, as a fundamental dimensionless property of the universe that falls out of pure math like the Taylor series for sin[x], and you should treat the Hz as being a fundamental property of incompetence by committee. [...] Either way, if I ever develop a time machine, I'm going to go back and knock both groups' heads together. At a frequency of about 1 Hz. Or better yet, strap them to a wheel and tell them I'm going to spin one group at a frequency of 1 Hz, and the other at 1 radian/s and let them try to figure out which one of those stupid inconsistent definitions means what. Hint: It'll depend on which time period I do it in, I guess, thanks to their useless inconsistent definition changes. The rest of the diatribe on Hz ends up amusingly spittle flecked with some decent frothing in evidence. I always knew that the kilo was bollocks (lump in a jar) but that barely gets a nod here. I had no idea about the state of some of the other units (like Hz)

repsilat 8y ago
Hmm, I guess it would have worked if a radian was defined as 1/(2pi) instead of 1. On the downside, you'd have to redefine things like `sin` to take multiples of "whole rotations" to make `sin(2pi radians)` (i.e., `sin(1.0)`) do the right thing.
- andrepd 8y ago
  If you define a "radian" like that you lose all the nice properties of sin like sin(x)' = cos(x)
  - vlasev 8y ago
    Yes. The radian is exactly what it HAS to be. Nothing more, nothing less. Just like the natural constant e.
    - repsilat 8y ago
      It has a clear and fixed definition, but that definition isn't the numeric constant 1. Associating it with a numeric value is a convenience, and different choices give different conveniences. In physics "c" is a constant. In some circumstances it's most convenient to define it as 1, and in others we define it as 3e8m/s. You can argue that 1 is more convenient for radians than 1/2pi because of identities with natural logs and derivatives of other transcendental functions, but that's an aesthetic call -- you can still get similar identities, just with different scaling factors. More to the point, the derivative of `sin(x radians)` will still be the derivative of `cos(x radians)`, so I'm not actually sure that argument holds water -- another way of saying "you have these scaling factors" is to say "you have to use the right units." Maybe that's still too big a pill to swallow, but I'd be wary of the endowment effect here.
      - deleted 8y ago
        [deleted]
      - vlasev 8y ago
        I think you're wrong. Let's investigate the trigonometric functions from the point of view of differential operators. Those are essentially functions that take in functions and spit out functions. The operators we'll look at are made up of derivative operators like d/dx. Let's say we have a differential operator D. We might want to find properties about it. One good place to start is to find its kernel (all functions u such that Du = 0) and find it's eigenpairs (all pairs (k, u) such that Du = ku for k a constant. We'll focus on the latter problem Let's look at the simplest D, that is D = d/dx. What function satisfies du/dx = ku? An exponential, of course, because functions of the type u = a^x have derivative u' = ka^x for some k. What'e the most fundamental such function? Well, it's the one where du/dx = u, i.e. k = 1. This is none other than the exponential function e^x with e as currently defined. Every other choice of a "fundamental" exponential function will run into a lot of inconveniences later on. What about another example. Let's look at D = d^2/dx^2. It's simple enough. What functions satisfy d^2u/dx^2 = ku. It appears the solution depends on whether k > 0 or k < 0. If k > 0, we can show that the hyperbolic trigonometric functions do the trick, so we can rehash the discussion above. If k < 0, however, we get into the usual trigonometric functions, namely what we now know as sin(rx) and cos(rx) for some choice of r, since for both u'' = -r^2u and you can find r from there. Again which r is the most fundamental? r = 1 of course, and this leads to the usual trigonometric functions. If you choose r to be something else, we'd have a problem as we'll no longer have the nice property that sin'(x) = cos'(x) and so on. Also, these two examples are really the same example since e^(ix) = cos(x) + i*sin(x). As a side note that should really be the main note...aesthetic calls are very important in mathematics. No one likes scaling factors to their identities. We like using fewer symbols to talk about common concepts. If we could make e = 1 or pi = 1 without inserting the opposite problem of complicating almost every single identity in mathematics needlessly, we would have done it a long time ago. I think the reason c = 1 works in physics is because you can change a few other constants around (usually to 1) and things works out well. I fail to see how we can do the same in mathematics where things need to work on many many different layers. Also, if you look at any "Natural units" systems, they all have pi in them. Why didn't they get rid of it? Because it's even more fundamental than c!
oldandtired 8y ago
Easy to fix Hz == 1 revolution / sec revolution == 2 Pi radians The problem of not having the revolution as the conversion unit is the massive oversight.
- Firadeoclus 8y ago
  That wouldn't work because revolution would simply be a dimensionless number which would define Hz as 2Pi/s. Which it isn't.

repsilat 8y ago

Hmm, I guess it would have worked if a radian was defined as 1/(2pi) instead of 1. On the downside, you'd have to redefine things like `sin` to take multiples of "whole rotations" to make `sin(2pi radians)` (i.e., `sin(1.0)`) do the right thing.

andrepd 8y ago
If you define a "radian" like that you lose all the nice properties of sin like sin(x)' = cos(x)
- vlasev 8y ago
  Yes. The radian is exactly what it HAS to be. Nothing more, nothing less. Just like the natural constant e.
  - repsilat 8y ago
    It has a clear and fixed definition, but that definition isn't the numeric constant 1. Associating it with a numeric value is a convenience, and different choices give different conveniences. In physics "c" is a constant. In some circumstances it's most convenient to define it as 1, and in others we define it as 3e8m/s. You can argue that 1 is more convenient for radians than 1/2pi because of identities with natural logs and derivatives of other transcendental functions, but that's an aesthetic call -- you can still get similar identities, just with different scaling factors. More to the point, the derivative of `sin(x radians)` will still be the derivative of `cos(x radians)`, so I'm not actually sure that argument holds water -- another way of saying "you have these scaling factors" is to say "you have to use the right units." Maybe that's still too big a pill to swallow, but I'd be wary of the endowment effect here.
    - deleted 8y ago
      [deleted]
    - vlasev 8y ago
      I think you're wrong. Let's investigate the trigonometric functions from the point of view of differential operators. Those are essentially functions that take in functions and spit out functions. The operators we'll look at are made up of derivative operators like d/dx. Let's say we have a differential operator D. We might want to find properties about it. One good place to start is to find its kernel (all functions u such that Du = 0) and find it's eigenpairs (all pairs (k, u) such that Du = ku for k a constant. We'll focus on the latter problem Let's look at the simplest D, that is D = d/dx. What function satisfies du/dx = ku? An exponential, of course, because functions of the type u = a^x have derivative u' = ka^x for some k. What'e the most fundamental such function? Well, it's the one where du/dx = u, i.e. k = 1. This is none other than the exponential function e^x with e as currently defined. Every other choice of a "fundamental" exponential function will run into a lot of inconveniences later on. What about another example. Let's look at D = d^2/dx^2. It's simple enough. What functions satisfy d^2u/dx^2 = ku. It appears the solution depends on whether k > 0 or k < 0. If k > 0, we can show that the hyperbolic trigonometric functions do the trick, so we can rehash the discussion above. If k < 0, however, we get into the usual trigonometric functions, namely what we now know as sin(rx) and cos(rx) for some choice of r, since for both u'' = -r^2u and you can find r from there. Again which r is the most fundamental? r = 1 of course, and this leads to the usual trigonometric functions. If you choose r to be something else, we'd have a problem as we'll no longer have the nice property that sin'(x) = cos'(x) and so on. Also, these two examples are really the same example since e^(ix) = cos(x) + i*sin(x). As a side note that should really be the main note...aesthetic calls are very important in mathematics. No one likes scaling factors to their identities. We like using fewer symbols to talk about common concepts. If we could make e = 1 or pi = 1 without inserting the opposite problem of complicating almost every single identity in mathematics needlessly, we would have done it a long time ago. I think the reason c = 1 works in physics is because you can change a few other constants around (usually to 1) and things works out well. I fail to see how we can do the same in mathematics where things need to work on many many different layers. Also, if you look at any "Natural units" systems, they all have pi in them. Why didn't they get rid of it? Because it's even more fundamental than c!

oldandtired 8y ago

Easy to fix Hz == 1 revolution / sec revolution == 2 Pi radians The problem of not having the revolution as the conversion unit is the massive oversight.

Firadeoclus 8y ago
That wouldn't work because revolution would simply be a dimensionless number which would define Hz as 2Pi/s. Which it isn't.