/sci/ - Science & Math

>>5755769
Also, I do realize you're a troll. My response was intended for people who might believe otherwise.

>>5755769
http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment#Subsequent_experiments
http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment#Recent_experiments
http://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment#Recent_optical_resonator_experiments

As far as I can tell, Tesla didn't like GR because he simply didn't understand it. As far as his "theory" of gravity goes, it's complete nonsense. Non-Lorentzian aether theories were falsified by the MM-experiment and countless other similar experiments, and Lorentz-invariant aether theories don't generalize easily or simply into gravity theories unlike Special Relativity.

Put simply, aether is erroneous crap that was proved unnecessary over a century ago, and attempts to revive it are conducted exclusively by crackpots who have no real knowledge of physics. Key words like "conspiracy" are dead giveaways.

>>5755670
>Maybe after about 7 years you will have accumulated enough knowledge to actually have a layman's understanding of QCD.

I don't think that's too accurate. You can certainly develop enough knowledge to begin working on QCD after ~2-3 years from your first intro physics course if you're interested enough and willing to put in the effort. I know from experience because I started reading QFT textbooks my sophomore year in college, only 2 years after taking my first intro physics course.

>>5755490
>>5755637
As much as I hate to admit it, these anons are absolutely correct. Whereas you can give a simplified version of General Relativity or basic Quantum Mechanics, there just isn't a boiled-down version of QCD that would be satisfying. It's too subtle and intricate a subject, with very little analogy that can be drawn to everyday phenomena. I.e. there's simply no bullshit "it's like a bowling-ball on a trampoline" analogy that will satisfy you.

You have to learn basic QFT first, including QED and how Feynman diagrams are used/interpreted, then you can get into more complicated stuff like QCD.

Measuring the speed of light with a bar of chocolate and a microwave?

>>5751700
In the case of a single particle, the operator would be the S-matrix operator with:

<div class="math">|\langle f|S|i \rangle|^2=|\langle i|S|i \rangle|^2=0.5</div>
where "i" represents the multiparticle state where the significant particle has not decayed and "f" represents where it has. The actual physical process isn't usually explained in the thought experiment, but it would probably be something like neutron decay, and then calculating the S-matrix requires lattice methods and a whole bunch of other shit that gets really complicated.

>>5751687
>http://www.youtube.com/watch?v=0NkdEfvmr5g
That didn't seem very accurate to me. They didn't even explain what the thought experiment was, they just had some nonsense quantum-mysticism crap about "making your own reality." Not even slightly related to Schrodinger's cat.

>can manipulate vectors
That doesn't make any sense, because vectors are just abstract mathematical objects. They're not physical "things" can you can touch or manipulate.

The function:

<div class="math">f(x)=\frac{10-x}{10-x}</div>
is equal to one everywhere, except at x=10 where there's a hole and it's undefined. So you could set f(x)=1 if you were simply trying to evaluate the limit as x->10, but the function itself is not defined there.

>>5746085
They behave exactly like fractions for all practical purposes.

>>5746072
<div class="math">ds=\sqrt{dx^2+dy^2}=\sqrt{1+\left (\frac{dy}{dx} \right )^2}dx</div>

>>5746060
Also, mathematicians seem to have a stick up their ass when it comes to hand-wavy explanations that involve multiplying or dividing by differentials, and I've never really understood why, considering that it *does* work.

>>5746060
If it works it works. There's no need for rigor when there's a more pedagogical explanation that yields the right answer.

>>5746027
Use [*eqn][*/eqn] tags instead of [*math][*/math]

>>5746017
First one you just multiply ds by 1=dt/dt, then take the denominator inside the square root.

Second one comes from the fact that:
<div class="math">x=rcos\theta </div>
<div class="math">y=rsin\theta </div>
So, with basic product rule:
<div class="math">dx=cos\theta dr-rsin\theta d\theta </div>
<div class="math">dy=sin\theta dr+rcos\theta d\theta </div>
When you plug that into the Pythagorean theorem, everything simplifies so that you're left with:
<div class="math">ds^2=dr^2+r^2d \theta^2</div>

>>5745945
Stop being a semantic prick.

Yep. In flat 2D space, if you move an infinitesimal distance in the x-direction, dx, and an infinitesimal amount in the y-direction, dy, then the total distance you've moved is given by the Pythagorean theorem:

<div class="math">ds^2=dx^2+dy^2</div>
The length of some arbitrary curve C is then an integral of ds over the curve:

<div class="math">L=\int_C ds=\int_C \sqrt{dx^2+dy^2}</div>
Then all you need to do is parametrize the curve and evaluate the integral. Here's a simple example: a basic parabola y=x^2 from x=0 to x=1. If we parametrize with respect to x and get the integral in a more useful form, it becomes:

<div class="math">L=\int_0^1 \sqrt{1+\left (\frac{dy}{dx} \right )^2}dx</div>
Now, dy/dx=2x, so we simply plug that in:
<div class="math">L=\int_0^1 \sqrt{1+4x^2}dx \approx 1.479</div>

>>5745022
>But when you stretch the rubber band, the dots move farther apart.
That's pretty much what I said. I'm not sure what the rest of your post has to do with anything.

Are you comfortable with a bit of calculus? Say we take an elastic band, and we mark each point on that band with a coordinate; call it "x". Now, assuming each point on the band is equally spaced, the distance between two arbitrary points is given by what's called a "metric." In general, two points infinitesimally close together are separated by a distance "ds", where:

<div class="math">ds=kdx</div>
where k is some constant relating coordinate separations to distances. Now, instead of an unchanging band, what if we consider a band that's being stretched or compressed over time? The coordinates don't change, because they're simply labels that we drew on the band. But distances certainly do change, and they change as a function of time. So the metric in this case would look something like the following:

<div class="math">ds=f(t)dx</div>
This is, essentially, what's happening with the universe. You ascribe a coordinate to each point in space, and the distance between coordinates is increasing over time. The exact function f(t) which dictates how the distances change is found via the Einstein field equations, and are related to the energy density in spacetime.

>>5743608
Same shit. Let x=u.

It's 2/3. The way I did it was with L'Hopital's rule, but I sincerely doubt he'd expect you to know that in day 2 of your calc class.

A non-rigorous way to show it would be to get a calculator, plug in x=1.9999999, then plug in x=2.0000001, and see that they're both ~0.6666667.

Nothing strange happens when you cross the event horizon. It's just a sort of imaginary line in the sand where you can't ever escape. Like traveling down a river with a waterfall at the end - at some point the current becomes fast enough that you wouldn't be able to swim against it, so no matter how hard you fight you're going to topple off the end.

Some of the classics would be "derive Snell's Law from Fermat's Principle," or "given two points 1 and 2, with 1 higher above the ground, in what shape should we build a frictionless roller-coaster track so that a car released from point 1 will arrive at point 2 in the shortest possible time?"

Calc of variations is absolutely invaluable in more advanced physics too, because virtually everything is represented as a Lagrangian. This includes gravity, quantum field theories, etc. And the most convenient, condensed way to ensure a new theory is local, has a Newton's third law analogue, certain symmetries, etc., is to express it in terms of a Lagrangian.