Redingold
Redingold t1_jadkevs wrote
Reply to How do measurement uncertainties propagate through calculations in an experiment? by slackslackliner
Generally this is not a straightforward task, however if the uncertainties on the variables and the correlations between the variables are small, you can use a formula:
For a function f(x, y, z...), where x, y, z... have uncertainties σ*x, σy, σz..., the uncertainty of the function σf* is approximately sqrt((∂f/∂x)^(2)σ*x^2 + (∂f/∂y)^(2)σy^2 + (∂f/∂z)^(2)σz*^2 + ...).
You can see for simple cases that this produces sensible results. For example, for the case of multiplying a value by 6 to convert between volume per 10 seconds and volume per minute, we'd have f(x) = 6x. This gives us ∂f/∂x = 6, so σ*f* = sqrt(6^(2)σ*x^(2)), or σf* = 6σ*x*. Multiplying a value by 6, therefore, increases the uncertainty on it by 6 as well.
For another example, consider adding two variables, so f(x, y) = x + y. Then, ∂f/∂x and ∂f/∂y both equal 1, so σ*f* = sqrt(σ*x^(2) + σy^(2)). This is only an approximation based on assuming x and y are uncorrelated, if they are correlated then this isn't quite accurate (if they're correlated then σf* = sqrt(σ*x^(2) + σy^(2) + 2σxy*)), but the nice thing about this formula is you can use it in a lot of different situations.
Redingold t1_j9jmpip wrote
Reply to comment by [deleted] in How did Paul Dirac predict the existence of a positron before the actual discovery? by bazongaenthusiast
I want to expand on the bit about negative energy there.
The issue with having negative energy states allowed is that, generally, particles try to occupy the lowest energy states available to them. Electrons in an atomic orbital for example, will emit photons and move onto lower energy orbitals if those orbitals are empty. Having negative energy states, and, in particular, having negative energy states with no lower limit (you can see that as p gets larger and larger, -sqrt(p^(2)c^2 + m^(2)c^(4)) gets lower and lower without bound) means that electrons should all be shedding as much energy as they can as they cascade down towards negative infinity energy. This doesn't resemble the reality we live in, so something else must be happening.
What Dirac reasoned was that you could make it work if all the negative energy states were already filled, all the way down to the bottom. The positive energy states would be free and available for all the conventional electron physics we already know about, and the negative energy states would be blocked, so electrons won't have room to go barrelling down towards negative infinity.
This has an interesting implication, though. There's a gap between the positive and negative energy states: when p = 0, the energy jumps from E = mc^2 to E = -mc^(2) (so the gap is 2mc^2 wide, or wider if p is not 0), and there are no states at all in this region. If, however, you supplied that 2mc^2 worth of energy (or more), you could make an electron jump up from an occupied negative energy state, into a free positive energy state. This would leave a hole in the negative energy states, and this hole would act like it had a positive charge (because when an electric field is applied, the electron next to it moves into that space, leaving a new space where it was before, and then the next electron moves into that space, and so on, like how a bubble rises in water by the water falling down and pushing the bubble up. The hole would move in the opposite direction of the electrons, and so would have the opposite charge, i.e. positive). So by supplying at least 2mc^2 of energy, you could make an electron, and a hole that behaves like a positively charged electron. That's a positron.
Nowadays, modern physics doesn't really use this Dirac sea model, as it's called. Quantum field theory lets us treat the ground state as a vacuum containing no particles, not an infinite sea of negative energy particles, and it treats positrons as real particles, not holes where an electron isn't in the negative energy sea. The Dirac sea model also struggles to explain why we don't see the effects of having infinity negatively charged particles everywhere in the universe (although the energy of the vacuum isn't a solved problem in quantum field theory, either, see the cosmological constant problem). That said, the two are mathematically compatible to an extent.
Redingold t1_j7zhgav wrote
Reply to comment by johndburger in Do tonal language speakers understand each other while whispering? by Paulix_05
No, they really are different. It's a smaller difference than voiced vs unvoiced, known as fortis and lenis. Voiced consonants in English are fortis, and are pronounced more forcefully, and unvoiced consonants in English are lenis, and are pronounced less forcefully.
Redingold t1_j40wuj3 wrote
Reply to comment by leb0b0ti in Were muslim armies harder to maintain in the field? by DJacobAP
Because Martin directly claims they're an amalgam of real historical cultures with only a dash of fantasy. He makes a direct claim to historical accuracy and it doesn't hold up in the least. Martin has consciously cultivated the appearance that his series is "how it really was" and that in turn distorts what people think about real history.
Redingold t1_ixcq4k3 wrote
Reply to comment by bellsandwhistles in How exactly is the "direction" of the flow of power measured in a alternating current system? by Landhund
So named because it points in the direction of the flow of power (not really, it's the Poynting vector, not the Pointing vector, named after physicist John Henry Poynting, but it's an amusing case of a sort of nominative determinism).
Redingold t1_jdvk4rf wrote
Reply to comment by [deleted] in eli5: I have a math/percentage question by StevieRaveOn63
That's not the right method to use here. The 28 grand the person paid is just the tax, just the 4.5%, not the tax plus inheritance. The inheritance is the amount they're getting, not the amount they're paying.