/sci/ - Science & Math

>>5226135

This is a *drastic* oversimplification of quantum computing. If it was as simple as 3-state logic we'd be able to implement a "quantum computer" efficiently on classical computers, no problem.

Quantum computers operate on qubits, which you can think of as 2d vectors in the plane (really they're complex 2d vectors, but real works fine for visualizing). The X axis represents "yes", and the Y axis represents "no". So qubits can really be any arbitrary mix of "yes" and "no" at the same time.

This would still be easy to emulate on a classical computer, after all it's only 2 numbers per bit. The *real* complexity comes in because *** every qubit in a quantum computer can be entangled with every other qubit, arbitrarily ***. This means that measuring one qubit can affect *every other qubit in the computer*. Storing this inter-relationship needs around 2^n floating point numbers for n qubits. This means that (naively) modern computers can simulate a single 32-bit integer quantum computer before they simply run out of memory.

To answer op's question... well, quantum computers. Being able to build dependencies and ambiguities into and between digital bits.

OP, as far as I can tell everyone in this thread has cited equations and said "this is true, therefore you can't go faster than light." I'd imagine that's not quite what you're looking for.

In the geometry that space and time have, the idea of "going faster than light" makes as much sense as "being closer than 0 feet". Intuitively, we expect there to be speeds faster than c, but it's just as wrong as assuming that you can keep heading in one direction and always end up closer to someplace.

>>4484984

You're assuming the potential only depends on position, and you're ignoring that there's any part of the state that *isn't* position. OP's works for the entire universe, yours works for a single spinless particle in a potential.

Besides, <3 dirac notation.

My submission, that neatly sidesteps the question of whether the bad bird would go to hell, with the added bonus of only containing concrete questions about the world itself:

Are you a bird, or is this the door to heaven, but not both? Or, if english had such a word: Are you a bird xor is this the door to heaven?

I believe (though others will surely disagree) that this counts as a single question since it only has one answer, and it doesn't use hypotheticals any more than "are you a bird *and* is this the door to heaven?".

If you use "and", then "yes" means you've got the right door, but "no" is ambiguous. If you use plain "or", "no" means you've got the right door but "yes" is ambiguous. Using "xor", "no" means you've got it and "yes" means use the other door.

Boolean logic wins again!

>>4009300

If you think so. But the inner products bit is really nice -- it's a notational reminder that the inner product in that vector space isn't commutative, something that's easily lost with other notation. That, and the identity with respect to an eigenbasis can be written

<div class="math">\sum_{j, k} | j \rangle \langle k |</div>

which is pretty easy to remember because it has such a striking form.

The notation is also nice when you're referring to eigenvectors of an operator with a degeneracy: in this case you have two or more eigenvectors with the same eigenvalue, and you need more than one index to uniquely identify the vector. With Dirac notation, you just stick the other index inside the ket: <span class="math"> | i, j \rangle [/spoiler]. In extreme cases, the number of indices is too unwieldy for the usual subscript index notation.

tl;dr; -- there's a reason it's like this. Dirac knew his shit.

>>3875072

Usually in relativity discussions we assume that when you see the light from an event, you can then backtrack and find out exactly *when* an event occured in your frame, because the speed of light is constant and you know where the light came from.

So in relativity when we say "event A occured at this time", we don't mean "we saw the light from event A at this time", we mean "it *must* have happened at that time in order for us to see it when we did". There's a similar thing for velocities. You pretty much ignore the light lag altogether.

So, when we say "you never see anything moving faster than <span class="math">c[/spoiler]", we literally mean nothing in your frame will move faster than <span class="math">c[/spoiler], whether you can actually see it with your eyes or not.

>>3773053
>>3772863

or rather, to use an oft-repeated analogy:

Imagine the universe is like the surface of an inflating balloon. There isn't a single point of expansion on the surface: it appears like every point is expanding from every other point. If you had to point to a center, it would be the center of the balloon, but people living only on the surface have no way of seeing it.

It's often easier to deal with 4 dimensions by dropping one of the spaces.

You're getting a lot of trolls saying NO under all circumstances, ignoring black holes and gravitational lensing. Ignore them.

You're also getting a lot of counter-trolls that are saying no because light always travels in a straight line, but the space it travels through may itself be bent due to General Revativity yada yada yada

Look.

Light will change direction in a gravitational field. It will not travel faster or slower -- light is funky like that. But it will change direction. This is acceleration, in the way you were using the term, and it's likely exactly what you wanted to know.

As for what happens inside a black hole... well. It's black. It's the blackest thing in the universe; no useful information (that we know of) comes out. So all we have is theory, and most of these say that space acts exactly the same on the inside as it does on the outside, except that any direction you travel in will bring you closer to the center. Alien geometry, for sure -- but light likely keeps it's peculiar constant speed.

Can still change direction tho.

>>3210343

Yes, but if there's some insight into an underlying process, better models can be constructed. Saying that finding the underlying reality isn't a part of physics is a little heavy-handed, even though I get why you'd say it.

Also, just wanted to let you know you're not the only one around here who's heard a few lectures in QM, even if it was only a freshman honors course for me. Next year's the more in-depth treatment -- I'm looking forward to it.

>>3149900

the first quadrant you described is the set of points <span class="math">r e^{\theta \pi i}[/spoiler], where <span class="math">r > 0[/spoiler] and <span class="math">0 < \theta < \frac{\pi}{2}[/spoiler]. If you square such a number, you have <span class="math">r e^{2 \theta \pi i}[/spoiler], which has <span class="math">0 < 2 \theta < \pi[/spoiler], so it's on the upper-half plane.

The inverse of such a mapping would be the *positive* square root. While the square mapping is usually two-to-one (not one-to-one), if you restrict the domain to the first quadrant and the codomain to the upper-half plane, it's bijective.

unless you need the function to be biholomorphic everywhere, and not just on D -> U.

squaring maps first quadrant to upper half plane, bijectively and holomorphically. Then you can use linear fractionals (I think you might call them Mobius transformations, but I'm not sure) to map the upper half plane to any other circle you want, such as the unit disk.

Complex analysis, bitches.

*sign*

Assuming x is an element of a ring on which addition is defined (otherwise, the notation itself is nonsensical)...

<span class="math">x - 9 = 19 + x[/spoiler] is equivalent to <span class="math">-9 = 19[/spoiler], so it's unsolvable if <span class="math">-9 \neq 19[/spoiler] (like in the reals, complexes, integers, and basically every other familiar ring).

However, if it happens that <span class="math">-9 = 19[/spoiler] in the ring we're working in, then <span class="math">x[/spoiler] can be any element in this ring. An example of such a ring would be the mod 2 ring, with elements 0 and 1.

<div class="math">0 + 0 = 0, 1 + 0 = 0 + 1 = 1, 1 + 1 = 0</div>

but this is a little pedantic, since it's unlikely we'd write -9 and 19 instead of 1 and 1.

>>3092688

I should clarify, I mean this thread will soon be full of trolls. Amazingly this hasn't happened yet. :D

>>3092690

As mentioned before, air resistance is a *force* that's a function of cross-section, velocity, and the density of the air. This air resistance force is the same for both spheres (if they're at the same velocity), in this case, since the cross section is the same and they're both in the same type of air.

However, forces cause different accelerations in different masses (F=ma, if you recall), so the more massive object will accelerate slower (due to air resistance) than the more massive object. Since the lead sphere slows down *slower*, it hits the ground first.

I know it's a troll thread, but I'm getting in on the ground floor with the correct answer.

Lead will hit first. With no air resistance, both would hit at the same time. They both accelerate due to gravity at the same rate. However, the air has less of an effect on the lead sphere, as the lead sphere will have a greater total momentum to overcome.

inb4 both hit at the same time, or lead hits first because it's heavier

>>2932219

*cough* my screwed-up example was
<div class="math">
(10^{\frac{1}{2}})^2 = 10^1 = 10 \Rightarrow 10^{\frac{1}{2}} = \sqrt{10}
</div>

>>2931622
>>2931559

whoah, entirely correct answer right off the bat! I'm proud of you, /sci/.

An easy way to remember how this works is to remember arithmetic rules for exponents, such as
<div class="math">
(a^x)^y = a^{xy}, a^x a^y = a^{x+y}
</div>

So, in your example,
<div class="math">
(10^{\frac{1}{2}})^2 = 10^1 = 10 \implies 10^{\frac{1}{2}} = \sqrt{10}
</div>

These rules are only a trick, though, and it breaks down a little for complex numbers (where <span class="math">a^b[/spoiler] has many values), but for almost anything else you'll encounter they work.

The definition given in the third post is a true *definition*, and will always work wherever <span class="math">e^x[/spoiler] and the logarithm are defined.

Something I'm sad to know is not taught in most schools: how to solve every classical physics problem ever.

1. Draw force diagrams for all the objects. All of them. And don't make any assumptions about movement yet.

2. Write F=ma from the force diagrams. Still, no assumptions!

3. Write down constraints. In this case, the acceleration of one mass must be the opposite of the acceleration of the other (a_1 = -a_2) because the string has a constant length. Also, the two tensions are the same.

4. You should now have a system of equations... now solve them!

5. Check your solutions with limiting cases that make sense. What happens when m1 goes to infinity? to zero? What about changing g? Also, check your units! Make sure they work out!

Most of all, keep all your inputs as VARIABLES! It is tiresome to work with numbers like "0.552 g" everywhere, and tiresome to grade. Use "g, T, m1, m2, ...".

The only thing that will ever change in this process is small tweaks. Eventually, you'll start getting systems of differential equations instead, for things like springs. Also, you'll add in conservation of energy and torque diagrams/angular momentum to help deal with trickier problems, but the process stays about the same.

>>2178334

most certainly a troll, but just so we're clear...

remember SETI@Home? Well, this is what happened to it.

>>2177892

this is the one that agrees the most with what I've been taught so far... and I like it! Here's my current understanding of how to construct the basics of math.

If you've already got mathematical induction, you can just say that N (the natural numbers) is a set containing 0 and 1, and that it is also "inductive" (that is, if n is in N, then n+1 is as well).

Then, extend N with additive inverses to get the integers Z (which is an "ordered ring", having additive inverses but not multiplicative inverses).

Then you can extend Z with multiplicative inverses by taking pairs of integers p, q and defining addition and multiplication carefully. These are the rational numbers Q, which is an "ordered field" (since it now has both inverses, and an obvious ordering).

You can get the real numbers R by completing Q with all the non-infinite infimums and supremums of all subsets of Q. This basically adds the irrationals to the rationals. One way to do this is with Dedekind cuts (http://en.wikipedia.org/wiki/Dedekind_cut).). This is a "complete ordered field". Completeness is *required* to do calculus in. At least, the naive calculus that's initially introduced... Technically, you need ordering too, but this is easy to work around for non-ordered sets like R^n.

If you want to go farther, you can extend R with the element i, defined as i^2 = -1. More rigorously, you can define this new set as pairs of reals, and then define multiplication carefully. Or you can use matrices... same thing really. This is the complex numbers C (a superset of R), which is an "algebraically-closed complete ordered field".

Algebraically closed fields have some very nice properties. For example, in C, every polynomial of degree n has exactly n (maybe non-distinct) roots! Not true of reals (see: x^2 + 1 = 0).

Fun!

Oh, and I forgot the best part: treating time as a dimension like space leads to the best part of relativistic physics. Mass, momentum, and energy all collapse into relativistic momentum. The first three components (the space components) are what we classically think of as momentum, the fourth component is what we classically think of as energy, and the magnitude is the mass. This is a *much* nicer way of dealing with these properties: believe it or not, relativistic collision problems are *easier* than classic collision problems.

>>2035394

Yes, it is. The kinetic energy that you know and love, (1/2)mv^2, is just the most obvious term in the taylor expansion for relativistic energy. This expansion has other powers of v in it, so that implies that v is unitless, if we want them all to add at the end of the day. So either distance is measured in seconds, or time is measured in meters. You pick, it's the same either way.

This has the added bonus that in such a system of measurement, mass, momentum, and energy all have the same units, and E=mc^2 becomes redundant: energy isn't *converted* to mass, it *is* mass.

>>2035359

Physics Majors basically go like this.

Freshman: Crash Course in all fields, learn calculus.

Sophomore: Write differential equations using F=ma, solve. If you get fancy, you use relativistic equations here too.

Junior: write differential equations using Maxwell's stuff, solve.

Senior: write differential equations using Schroedinger's Equation, solve.

It's way more interesting than it sounds. :P

As a starter, surprisingly few people know that (classical) conservation of energy, momentum, and angular momentum are direct results of F=ma.

>>2035576

technically, he is correct... we would write e as 10.

Of course, an irrational (and non-algebraic, no less) number base is TOTALLY USELESS for representing rational numbers, which is what we deal with daily.

Also, non-natural number bases are usually dumb.

>>2035796

yes. Technically, each term in the sum should be, but it's the same thing so whatever.