# Tag Info

0

Other answers being good, i'll add 2 more cents here. Inertia as quoted in question (btw, this is nice, to see actual historical sources), is (defined as) the tendency of matter to continue in its current state, unless changed. This is inertia, the term "force of inertia" is just another way to state the same thing but using Newton's second law, $F=ma$, so ...

3

The question of mass has arguably been one of the two most important issues in physics (the other being the electromagnetism). Physics has tried to uncover the true nature of mass for hundreds of years, to no avail so far. Not surprisingly, its description is somewhat circular: “In physics, mass is a property of a physical body which determines the body's ...

4

The concept of inertia is indeed useful in two ways. I think your notion of it as a technical promotion of the everyday word "sloth" (without the baggage given it by the Roman Catholic translation of the "deadly sin" Ἀκηδία) as extremely close to the mark. In physics the notion of "inertia" has two, very alike uses: The first is practical, through a weak ...

0

It boils down to a matter of convention. Nothing stops you from choosing the annihilation operator to be $a^\dagger$. Still, in quantum field theory, you decompose e.g. a scalar quantity in plane waves $$\Phi(x) = \int \frac{d^4 k}{4 \pi} \left( a(k) e^{ik\cdot x} + a^*(k) e^{-ik\cdot x} \right)$$ where obviousely the second part is the c.c. of the first ...

2

I'll give this a shot though I'm uncertain if this will clarify or cloud the issue. electrons are people, OK current is the number of people, No, that's not the correct analogy. Current is a flow and a people (electron) current is a flow of people (electrons) and the amount of current is the number (Coulombs) of people (electrons) passing a ...

1

Since your question is kind of pictorial, i will try to stick with this and avoid any deeper mathematical / physical explanations. electrons are people ok with that current is the number of people no, that's a mistake. Actually current is is this case the number of people crossing a line on the street per time. Like Steeven already mentioned in ...

1

But what is potential in this example? In your people-in-the-street analogy, the electric potential would be a shop with sale in one end of the street, or maybe an accident or something else interesting. People tend to move towards the interesting end (less potential) and away from the other and less interesting end (higher potential). Just like a stone ...

1

In mathematics, complex numbers and rotation are intimately related. After rotation $\theta$ in the complex plane, the number $1$ becomes $e^{i\theta}=\cos\theta + i\sin\theta$. Although $e^{i(2n\pi)} = 1 = \cos(0) + i\sin(0)$, you actually moved by $\theta$ and not by $0$. If you rotated something by $\pi$, you could say the angle is now $-\pi$ - but you ...

1

What you can say is that you have a distribution of the results you are going to get (be it a discrete or continuous random variable), and when you calculate the average of a large sample, you are adding the random variables and multiplying by a constant. The addition of random variables translates into a convolution of the probability density functions, ...

3

In general, we can say nothing about finite $n$, but most of the time, we can safely assume some "niceness" of the distributions in question. If, for example, we assume a finite Variance $\sigma^2$ (a quite common feature), we could use Chebyshev's inequality for a rough error estimation of the form $$P(|\bar{X_n} - µ| > \alpha) \leq ... 6 This is the exact reason why we do statistical hypothesis confidence testing. In essence, the confidence interval we get from this test is a quantitative measure of "how far we've converged". For example, consider an experiment to test whether a coin is imbalanced or not. Our null hypothesis is that it is not: in symbols, our hypothesis is {\rm Pr}(H) = ... 2 You know Newton's second law:$$\vec F = m \vec a~.$$Now you multiply both sides by "\vec r \times~", then you get$$\vec r \times \vec F = \vec r \times m \vec a = \frac{d}{dt} (\vec r \times \vec p)~.\tag{1} (If you carry out the time derivative, the first term from the product rule is $\vec v\times\vec v$, which of course is $0$.) This is very ...

Top 50 recent answers are included