Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
kwinse wrote:
Warp wrote:
So the original problem could be made slightly "harder" by phrasing it as: "Prove that every natural number can be written as the sum of unique, non-consecutive fibonacci numbers."
Am I missing something obvious or is that not possible?
5 = 2 + 3
6 = 1 + 2 + 3
8 = 1 + 2 + 5
8 = 3 + 5
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
rhebus wrote:
Bobo the King wrote:
I'm very interested, however, in where you draw the line. What is something that you consider useful but not "real"?
I'm going to suggest an answer: conventional current. That is, the classical notion that current flows from a positive terminal to a negative terminal. We now know that the current-carrying particles are themselves negatively charged, and they flow from negative to positive1. But you can get a long way in predictive power with a theory that assumes current flows from positive to negative. It's incredibly useful -- but is it real? The fact that we now know of the existence of electrons puts an interesting spin2 on it. Suppose we hadn't discovered electrons: would conventional current be more real than it is now that we have discovered them?
1Yes, you also get holes in p-type semiconductors; but really it's still electrons carrying the charge
2A spin of ½ (ho ho).
I think that's a very good response. I'm tempted to go in two directions with it.
First, I could say, "Good example. But it's real." After all, we don't question forces in physics, even though nature seems to heavily favor potential energy and action. Force is a construct that's most useful in the realm of classical mechanics (and even then, we can formulate the entire theory based on action). Why not do the same with conventional current?
Second (and this is more my inclination), I could try to invoke Occam's razor. If Benjamin Franklin had happened to decide that amber gains a positive charge and glass is left with a negative charge when they are rubbed together, we wouldn't have this problem today. We flipped a coin a few centuries ago and got the "wrong" result. Conventional current is a way of sticking to the old, flawed system. Nothing would be lost, however, if we called electrons positively charged and nuclei negatively charged. We could use positive current to signify the flow of the electrons. In a purely conceptual and practical sense, changing the sign of the charge carriers would be the "simpler" of the two theories. It's not that the theory isn't useful, it's just that we have an equivalent theory (that we refuse to use because of convention) that's equally useful and, in certain situations, more useful. Because the theories are largely equivalent, counter-conventional current would be the preferred one.
I think the best analogy I can come up with is in trigonometric notation. It's low on my list of pet peeves these days, but trigonometric notation is screwy and has always bugged me. For a few reasons, it would be convenient to swap the names of the sine and cosine function (the cosine is, in some sense, "more fundamental"). This would also "properly" align the secant function with the reciprocal of the sine and the cosecant function with the reciprocal of the cosine. And if it were up to me, we'd do away with "sin" and "cos" functions entirely and just draw little circles with horizontal and vertical radii, reminding us what axis we're projecting onto and illuminating the fact that derivatives just take you on a clockwise tour around the circle. But convention stuck and now we have to deal with it. The fact that the cosecant is associated with the sine function and the secant with the cosine function in no way implies that trigonometry is wrong or inconsistent. It just means it's confusing. Likewise, conventional current is not "fake", but counter-conventional current would be conceptually better.
I'm not sure if any of what I just said makes a whole lot of sense.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Truncated wrote:
Warp wrote:
Truncated wrote:
1) If virtual photons (or other virtual particles) cannot be observed directly or indirectly, is there any reason at all to assume that they exist?
Electromagnetic force is quantized, and it propagates at c in vacuum.
Yes. I don't see how that is relevant to the quoted section, or what I posted about in general. Can you elaborate?
Bobo the King wrote:
1) But virtual photons can be "observed" indirectly. The perturbative expansion (using virtual photons) of certain physical processes gives the most precise measurements in all of science; we know the fine structure constant out to ten or eleven digits, which is equivalent to measuring the distance from Los Angeles to New York to within the thickness of a hair. You might argue from a philosophical standpoint that virtual photons aren't real, but as I previously said, virtual photons are useful, and so they are in some sense real.
2) Your proposed claim that "virtual UFOs really exist" fails this test because virtual UFOs make no predictions about how physics should work.
3) As for the magnetic (or electric...) force having infinite range, the long range interactions are mediated by the "real" photons. They are directly observable. It's a little bit like the pressure pushing down on you right now. You're only feeling the air around you locally. Do you know that the upper atmosphere exists? Strictly speaking, no, but you can certainly feel its effects! And we might even make broad distinctions about lower atmosphere air and upper atmosphere air, but in truth, we know those distinctions are arbitrary.
1) OK! I disagree with the sentiment "if it's useful, it exists", since I can think of a lot of analogies/ways of thinking which are useful, but does not make them real.
2) Well, that is easy to fix... but I don't think I'll go down that road.
3) If something is sending out "real" photons, it will lose energy. Magnets do not lose energy keeping up their magnetic field. I thought this was the very problem one tried to solve by using virtual photons?
Sticking with the numbering system you've introduced...
1) Again, I think your disagreement is based on philosophical grounds. In more familiar terms, we might ask if chairs exist. "Of course they do!" Well, what qualifies a chair? Is it something on which you rest your butt? Then what is a butt? Is a bean bag chair a chair? It has no specific solid form. How about a stump? That isn't man-made. What if I begin carving a seat into that stump? At what precise instant does it cease to be a stump and become a chair?
I find these arguments absurd and pointless. If you want to go that route, you quickly find yourself in the realm of essentialism. Philosophically, I tend to lean the other way, toward existentialism, which would appear to support and expand on your view. It isn't just that virtual photons don't exist, no human constructs can be meaningfully said to exist! We're all one big gelatinous mess of a wavefunction attempting to derive meaning out of subsets of ourselves, never able to grasp the big picture.
But I'm also a scientist. If I throw a ball into the air at a very specific speed and it consistently returns to my hand in two seconds, do I really need to question the existence of the ball itself? No. I like to get things done. In terms of their predictive power, virtual photons seem to be more real than balls or frictionless surfaces or blocks on inclined planes or other more familiar notions.
Your point is that other notions are useful but not real. Maybe, maybe not. It depends on what you mean by useful. For a scientist, numerical predictions trump all. What about non-mathematical notions? Is love real? It seems to be a construct of the mind, yet we seem to have a decent understanding of how love affects people. Again, this is a mostly philosophical question of where to draw the line and I begin to lose interest when we leave the realm of science and enter philosophy.
I'm very interested, however, in where you draw the line. What is something that you consider useful but not "real"?
2) I should have been more clear. Virtual UFOs make no testable, falsifiable, or substantiated predictions about how physics works. You said this is easy to fix, so what's your proposal? We don't have to discuss it at length.
3) You lost me here. Admittedly, particle physics isn't really my forte, but your claims seem fishy to me. It does take energy to create a magnetic field and a magnet will lose energy if it radiates some if it away. You may be confused by the fact that the magnetic force can do no work, but that can be reconciled (tediously) by entering a moving reference frame in which the magnetic field is transformed into an electric field. Historically, quantum field theory was developed to reconcile special relativity and quantum mechanics. Practically, the subject produces predictions of things like the magnetic dipole moment of the electron and scattering coefficients. I'm not sure how that ties in with the magnetic field energy.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Truncated wrote:
Bobo the King wrote:
Photons do mediate the electromagnetic force. ... There is no definite distinction between real and virtual photons, although we can make a few arbitrary distinctions:
• "Real" photons are observable, virtual ones are not. This criterion is the closest we can come to a definite distinction between the two, since it's what we mean (in practical terms) when we say a photon is real/virtual. If you observed its effects directly, it's real. If neither you nor anyone else observed it directly, it was virtual.
• Real photons last a (relatively) long time, virtual ones pop in and out of existence quickly. That's why we are able to observe them-- they last long enough to reach our eyes.
• Real photons travel (relatively) long distances, virtual photons travel short distances before disappearing. This can be deduced by using the speed of light and the previous point.
1) If virtual photons (or other virtual particles) cannot be observed directly or indirectly, is there any reason at all to assume that they exist? (What's stopping anyone from saying "virtual UFOs really exist, they just cannot be directly or indirectly observed"?)
2 and 3) The magnetic force has infinite range. How can it be mediated by virtual particles, if they last a short time and do not travel infinitely far?
But virtual photons can be "observed" indirectly. The perturbative expansion (using virtual photons) of certain physical processes gives the most precise measurements in all of science; we know the fine structure constant out to ten or eleven digits, which is equivalent to measuring the distance from Los Angeles to New York to within the thickness of a hair. You might argue from a philosophical standpoint that virtual photons aren't real, but as I previously said, virtual photons are useful, and so they are in some sense real.
Your proposed claim that "virtual UFOs really exist" fails this test because virtual UFOs make no predictions about how physics should work.
As for the magnetic (or electric...) force having infinite range, the long range interactions are mediated by the "real" photons. They are directly observable. It's a little bit like the pressure pushing down on you right now. You're only feeling the air around you locally. Do you know that the upper atmosphere exists? Strictly speaking, no, but you can certainly feel its effects! And we might even make broad distinctions about lower atmosphere air and upper atmosphere air, but in truth, we know those distinctions are arbitrary.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
I happened to tune back in just as they passed that horrible part they were stuck at on Route 9. According to one comment, they'd been stuck there for 11 hours.
Edit: Calling it now: they're going to get knocked out before they reach the next PokeCenter.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Warp wrote:
Fun fact: Albert Einstein is best known for the theory of relativity. Much less known is that the concept of photons was also first invented by him. In fact, it was Einstein who first introduced the whole concept of quanta (and his Nobel prize was related to that. For some reason he was never awarded a Nobel for his theory of relativity, even though it's one of the most fundamental theories of physics today.)
To continue the story, for much of the rest of his life, he campaigned against quantum mechanics. This is where the apocryphal (but accurate in spirit) quote, "God does not play dice with the universe," came from. What I think is even funnier is that every effort he made to poke holes in quantum mechanics just illuminated new facets of the theory. He may be second only to Niels Bohr in importance to quantum mechanics' formulation, despite his distaste (or outright disgust) with the theory.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Tub wrote:
Aren't photons the force-carrying particles of the electromagnetic force anyway, and thus a force themselves?
I've never understood hawkins when he talked about "virtual" photons (that are supposed to carry the force) and "real" photons (that somehow don't?).
Photons do mediate the electromagnetic force. It's an abuse of language, however, to say that photons are the electromagnetic force. As for real versus virtual photons, both can exert a "force" (better put, both can deliver an impulse). There is no definite distinction between real and virtual photons, although we can make a few arbitrary distinctions:
• "Real" photons are observable, virtual ones are not. This criterion is the closest we can come to a definite distinction between the two, since it's what we mean (in practical terms) when we say a photon is real/virtual. If you observed its effects directly, it's real. If neither you nor anyone else observed it directly, it was virtual.
• Real photons last a (relatively) long time, virtual ones pop in and out of existence quickly. That's why we are able to observe them-- they last long enough to reach our eyes.
• Real photons travel (relatively) long distances, virtual photons travel short distances before disappearing. This can be deduced by using the speed of light and the previous point.
• Real photons travel at speed c=3*10^8 m/s, virtual photons have a nonzero amplitude to travel faster or slower than that. Over long distances/times, the amplitudes for traveling faster or slower start to "cancel out", blurring the distinction between the virtual and real photons. (Don't think of this as the photon traveling "sometimes faster, sometimes slower". The cancellation is more fundamental than that.)
• Real photons satisfy E=pc, where E is the energy, p is the momentum (magnitude), and c is the speed of light. Virtual particles can have more or less energy than this.
I'm sure there are other ways to distinguish between the two, but that's all I know.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Warp wrote:
RGamma wrote:
Gravity
Gravity isn't a force, as far as we know.
Light passing through a transparent solid causes it to change direction. This is caused by a really complex quantum-mechanical effect (it has something to do with the light taking all possible paths and interacting with the particles of the object, or something along those lines), but I don't know if it could be said to be caused by a force since I know next to nothing about QM. (I suppose interaction with other particles could be considered a force?)
Warp is close to the crux of the issue. Applying classical notions like force to quantum mechanical and relativistic (decidedly non-classical) entities like photons is a stretch at best.
Do photons interact with "stuff"? Yes, they definitely do. So forces act on photons? Maybe. Photons can be deflected in many different ways. If your definition of a force is something that causes a change in an object's velocity, then sure, forces act on photons. But photons fail to follow Newton's second law, F=ma, which is the most useful thing we can do with forces. Calling a force something that changes an object's velocity is a weaker notion than F=ma (or F=dp/dt, the more powerful version of the second law). The whole question is kind of built on a fallacious notion of forces.
It reminds me of a chapter from Surely You're Joking, Mr. Feynman, in which an acquaintance asked him, "Is electricity fire?" Of course, Feynman immediately knew that the answer is "no", but simply couldn't convey it to this man. The man had a notion that fire something that burns. "Logically", he scaled this down to something that burns just a little bit in a small area, which seems to be what happens when you rub your socks on a carpet and then touch a doorknob. All he was doing, however, was extrapolating something intuitive and familiar to him into a field in which it isn't applicable.
I think the same thing is going on with the question at hand. A lot of us (especially those in introductory physics classes) are familiar with forces, how you can feel and measure them, and how they affect objects. However, we struggle with notions of photons, which travel at a constant speed and appear to have wave-like properties. But wouldn't it be neat if we could connect these two ideas? Forces on photons?
Nope. Can't be done.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Scepheo wrote:
Bobo the King wrote:
There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless.
2n3 + 3n2 + n =
(2n2 + n)(n + 1) =
n(2n + 1)(n + 1)
I think it's obvious that either n or n + 1 will be divisable by 2. Then remains to see that at least one term is divisable by 3. There are three cases:
1) n = 3k
2) n = 3k + 1
3) n = 3k + 2
In the first case the first term (n) will be divisable by three, in the last case the third term (n + 1) will be divisable by three. In the second case, we can show that
2n + 1 = 6k + 2 + 1 = 3(2k + 1)
which is again, divisable by 3. As at least one term is divisable by 2, and another by 3, the result will be divisable by 6. The proof is rather long, but I find the result to be at least somewhat intuitive, looking at the numerator.
Well when you put it that way, it seems so obvious! But I swear there was a nontrivial question in there somewhere. There are other integer-coefficient polynomials that are not factorable (I think...) but are divisible by six when you plug in an integer. There are others for all other integers. I think I proved something many years ago about the polynomials divisible by prime numbers using Fermat's little theorem, but I got stuck on nonprime numbers.
Basically, I wanted a way to produce the set of all polynomials that are always divisible by some integer, n. I quickly deduced the following:
• If some polynomial in x is always divisible by n, then that same polynomial plus n*xk, for any k, is also divisible by n. Therefore, the coefficients only needed to range from zero to n-1, inclusive.
• If a polynomial is always divisible by n, any integer times that polynomial is also divisible by n. The coefficients should have no common factors.
• The sum of two polynomials divisible by n is also divisible by n. These last two criteria taken together mean the set of polynomials always divisible by n form a basis set of a vector space.
• If a polynomial is always divisible by n, then it will still be divisible by n if it is multiplied by x.
• Since we are free to multiply by x, we have a strong incentive to look only for the polynomials up to order n-1. Our vector space will then have dimensionality n and if we can find n linearly independent polynomials that are all divisible by n, we should be able to reproduce all others by multiplying by x, multiplying by integers, adding the polynomials together, and/or taking the coefficients mod n.
I wrote up a Matlab program that would find integer coefficient polynomials. I believe I also had suspicions that non-integer coefficients could work as well. Would anyone like to pick up where I left off? I think a good start would be listing all third-degree polynomials divisible by 4. If they're all factorable, then I clearly overlooked the obvious.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
FractalFusion wrote:
Bobo the King wrote:
(2n2 + 3n + 1)/6.
How interesting! The average is always an integer! Who'd've thunk it!?
What? No.
The sum is always an integer. Doesn't mean the average is.
Argh, you're right. It's been a rough day and I'm juggling four or five things at once. I'm actually a little surprised I didn't screw up the broader proof.
Anyway, I do think it's interesting that the sum is always an integer, despite being divided by six. There is nothing in the numerator of that expression that is a clear indication that it will be divisible by six, yet it is regardless. Once upon a time, I was really curious about that property and wanted to find other polynomials that, with integer arguments, were sure to be divisible by a specific number (in this case, 6). I seem to remember finding the general property at some point, but I can't remember what it is. It's somewhere in the study of modular arithmetic, rings, and ideals. Can anyone derive it? How can one quickly determine what polynomials up to order n-1 are always divisible by n?
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Warp wrote:
So is there a simple formula for calculating the average of all squares between 12 and n2 (inclusive)?
Yes, I calculated it while I was writing up my previous post.
Assume it takes the form of the following polynomial
p(n) = a1n + a2n2 + a3n3.
We will show it does indeed take this form, but you should be led to try that solution because the integral of x2dx is 1/3 x3.
This formula works for n=0, the base case. (I altered your statement to all squares from 0 to n2. Clearly, this makes no difference.) Now we need to use induction to show that the formula works for n=k+1, assuming it works for n=k.
We require that p(k+1) = p(k) + (k+1)2 (the sum up to k, plus the k+1th term). This is the condition that will pin down the coefficients for us.
We start with...
p(k+1) = a1(k+1) + a2(k+1)2 + a3(k+1)3
and after a little expanding of terms (left to you), we find
p(k+1) = p(k) + (a1 + a2 + a3) + (2a2 + 3a3)k + 3a3k2.
Now apply the condition that p(k+1) = p(k) + (k+1)2 and we see that the polynomial coefficients of (k+1)2 must match up with the coefficients of the "leftover" powers of k from the induction step. This means
a1 + a2 + a3 = 1
2a2 + 3a3 = 2
3a3 = 1
which is just a linear system of equations. Solve this system of equations any way you like to obtain
a1 = 1/6
a2 = 1/2
a3 = 1/3
(or just verify that those are the correct values needed to solve the system of equations).
Putting this all together, the sum from 0 to n of i2 is equal to
1/3 n3 + 1/2 n2 + 1/6 n = (2n3 + 3n2 + n)/6.
Try it out!
Edit: As RGamma points out below (and I see from reviewing your question), you wanted to know the average, not the sum. Just as RGamma suggests, divide the above result by n to obtain
(2n2 + 3n + 1)/6.
How interesting! The average is always an integer! Who'd've thunk it!?
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Warp wrote:
Given the classical problem "what's the sum of all integers from 1 to 1000" (or whatever number), most people have memorized the formula that gives the answer without understanding it.
I, however, like to think of the solution as: "The average of all the numbers multiplied by their amount." It's relatively easy even intuitively to understand why that gives the correct answer.
The great advantage of thinking about it like that, rather than blindly memorizing a formula, is that it can be generalized to other situations as well. For example, what if the problem were "what's the sum of all even numbers from 2 to 1000?" Many people would not know how to resolve the problem, but it becomes easy when you understand that it's, again, their average multiplied by their amount. (Their average is (2+1000)/2 and their amount is 500.) The same principle can be used for "all numbers divisible by 5" or whatever.
Now the problem becomes calculating the average of such a sequence. For a linear sequence it's trivial, but what about non-linear sequences? For instance, what if the problem is: "What's the sum of all n2, where n goes from 1 to 1000?" There are 1000 integers there, but what's their average?
Is there a generic way of calculating the average of np (where n is a linear sequence and p is a constant)? How about other possible functions?
To expand on RGamma's response, I like to solve these kinds of problems using a little induction.
First, note that the finite sum of consecutive integers to the mth power is closely approximated by taking the integral of xm. Therefore, we expect it to be a polynomial of degree m+1. If it had a higher or lower degree, this approximation would surely diverge for large upper bounds.
(Incidentally, the coefficient you get by integrating matches the coefficient in the finite sum. For example, the sum of integers (m=1) up to n is equal to 1/2 n2 + 1/2 n. Likewise, the integral from x=0 to x=n of xdx is 1/2 n2. This fact isn't necessary for the remainder of the proof, but will cut down on your work in later steps.)
The rest of the process proceeds by induction. Assume a polynomial solution of the form
p(n) = a0n0 + a1n1 + ... + am+1nm+1
For the base case, we surely know the value when n=0. All that remains is to write the k+1th term in terms of the kth term and match all of the coefficients. You'll end up with m+1 equations and m+1 unknowns (the a coefficients). I'll leave that up to you to try out.
The utility of my method is that it allows for series such as the sum of squares of even numbers.
Edit: Changed my xs to ns.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
As long as we're on this topic and I'm putting off work, I'd like to ask a fairly simple question.
Why do we care about irrational (or transcendental, or definable...) numbers? I'm a physicist by training and I've never cared if the number I am working with is rational or not (much like I've never cared whether I use a less-than or less-than-or-equal-to sign). There are some really elegant theorems concerning different classes of numbers, but at the end of the day, is any of it important? I genuinely want to know if anyone uses this stuff for even moderately practical purposes.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
thatguy wrote:
That's right Bobo the King. There is one bit of neatening I can do to the proof. For the section where you discuss Pythagorean triples, you don't even need to invoke the formula for generating them. All you need to do is reason thus:
If x^2 + y^2 = z^2, then either:
x and y are both even, z is even
x is odd, y is even, z is odd
x is even, y is odd, z is odd
And in each case x+y-z is even.
Actually, my post originally said, "By a quick parity argument, it can be seen that either all three sides are even, or one leg is even while the other leg and hypotenuse are odd," which matches what you've written.
Unfortunately, we both overlooked the possibility that x and y might both be odd while z is even. Looking back, however, I see this doesn't affect the proof, so... uh... we're both right?
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Flip wrote:
A polygon has vertices with integer coordinates (in 2 dimensions) and integer length edges. Show that the perimeter is an even integer.
Define the following:
Pp: the parity of the figure's perimeter.
Px+y: the parity of the net displacement in the x coordinate plus the net displacement in the y coordinate.
Lemma:
It can be shown that all Pythagorean triples can be generated by the following equations:
a = k*(m2-n2)
b = k*2mn
c = k*(m2+n2)
where k, m, and n are all integers. Because all three are integers, b is surely even because of the factor of 2 in its expression. Therefore, either all three sides are even or one leg is even while the other leg and the hypotenuse are both odd.
Proof:
Begin with Pp even and Px+y even (i.e., start with even perimeter and even x+y displacement-- plausible for the trivial polygon consisting of a single point).
Iterate over all diagonal sides of the polygon. Because its vertices are on integer coordinates and the sides are integer length, a right triangle with integer sides can be constructed such that its hypotenuse coincides with the diagonal side of the polygon under consideration and the right angle is at integer coordinates. In other words, the right triangle's sides comprise a Pythagorean triple.
With each side, the additional perimeter contributed is c. If c is even, then Pp is unchanged and because both legs are even (by the lemma), Px+y is also unchanged. If c is odd, then Pp is changed and because one of the legs must be even while the other is odd (again, by the lemma), Px+y is changed as well.
Because both Pp and Px+y change concurrently, after all diagonals have been iterated through, both are surely either 0 or 1.
All diagonal sides have been iterated through, so all that remains are the vertical and horizontal sides. For this to be a closed polygon, the net displacement after iterating over all sides must be zero. Of course, the parity of the net displacement must be zero as well. If Px+y is odd, then we have to take an odd number of horizontal and/or vertical integer "steps" on the grid, which will add an odd number to Pp (which was odd before considering horizontal and vertical sides). If Px+y is even, then we have to take an even number of horizontal/vertical integer steps, which will add an even number to Pp (which was even before considering horizontal and vertical sides). In either case, we end with Pp being even, completing the proof.
I'm not a mathematician and although I'm confident my proof is correct, I'd like someone to clean it up for me. I imagine there's a more elegant way to put all this.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
MUGG wrote:
Also, please tell me why your lua script will help find the hitbox. Isn't it just another way to search the memory? I could simply use VBA's memory search for that..
Uh... I'm kind of at a loss for words...
Your first post asked how to find hitboxes. I offered an easier way to find hitboxes and now you say the Lua script is "just another way to search the memory". Yes it is, but isn't that what you were asking for in the first place?
What baffles me is that it sounds like you've considered going about it by disassembling and/or tracelogging, which is complete overkill. I get the impression that you want advice on a very different topic from what's been discussed so far.
Smart RAM Search is especially handy for determining the addresses corresponding to (nearly) continuous parameters that are graphically displayed. That means it's ideal for things like hitboxes. Based on the Super Mario Land example I posted on the Smart RAM Search topic, it looks like we can confidently say that Mario's y-extents are stored in addresses C00C and C014 (or possibly C010 and C018).
I ran your script (if anyone is wondering, it draws hitboxes on the screen for all sprites except Mario), so it's clear that you have some experience finding hitboxes. What do you need to do in Kirby (or other games) that you haven't done there?
Anyway, I realize my post may look combative and I just want to say that is not the tone I intend. I'm genuinely curious about what you're looking for.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
One quick-and-dirty way to find hitboxes would be to use the Smart RAM Search script that FatRatKnight and I wrote. You'll find it here. You could put the bar over the enemy, then slide it up and down as the enemy moves. If its movement is complicated enough, the script should have no problem finding the enemy's y-position. From there, you'll need to look at which of those positions corresponds to the (a?) hitbox.
Unfortunately, I never got around to adding the capability to flip the bar on its side and measure things' horizontal positions. The good news is that the x-position should be stored somewhere near the y-position and you might be able to find it by inspection. Do this for a few different enemies and you may be able to deduce where all the hitboxes are stored (they are likely uniformly spaced throughout the memory).
If you decide to go this route, tell me what you come up with! I'm always eager to hear that my scripts are being used!
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
marzojr wrote:
For reference, HHS is absolutely correct -- potential energy is as real as kinetic energy, or chemical energy, or whatever other form of energy you want. Except for New Age bullshit "energy", but that is not relevant here.
But gist of the issue is that "potential energy" is a highly misleading name: there is a load of things that are very real that get assigned that name, such as electromagnetic potential energy, gravitational potential energy, and so on. These are only "potential" because we misleadingly call them that -- they are very real interactions of the objects we are studying with well defined and measurable fields. Where does the "kinetic energy" go when you climb up a gravitational field? Into the gravitational field. Where does the kinetic energy come from when you fall back? From the field. Where is the elastic potential energy stored in a spring? In the electromagnetic interaction between the atoms of the spring. And so on. The energy is real, and it is all there; it is so real, in fact, that general relativity says right out of the box that it generates gravitational fields. I can elaborate on that, if so desired, but I won't do it now.
I agree with marzojr, and he certainly has proven himself many times in this thread to be quite the physicist.
But maybe it isn't necessary to invoke gravitational fields or electromagnetic interactions...
I once asked my research advisor if virtual photons are "real". By that I meant that since they are a perturbative expansion on the action integral, can we truly imagine little unseen packets of light shooting back and forth between two particles? (Don't worry if you don't know what a virtual photon is. It's not central to my argument.) His response (after much badgering) was, "Is force real?" I realized that technically it isn't real because there is no quantum mechanical analogue to force. He said, "Yes, but does that stop you from using it?"
He then quoted a famous physicist (probably Steven Weinberg) as saying, "It's real if it's useful."
So that's my answer to this question. Potential energy is a useful concept, therefore it is real.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
Warp wrote:
Speaking of potential energy... I know this is basic high school physics, but could someone explain what it exactly means?
I always thought that it was just an abstract concept used to aid figuring out the math, rather than it being an actual physical thing that actually exists. If it really physically exists, and it's a form of energy, it feels like a rather... esoteric form of energy. Where exactly does it exist? Can you point out to some point in space and say "there are X joules of potential energy right here"? Can you measure it?
Your notion of potential energy is built right into its phrase. As I understand it, early physicists believed that energy is observed only in the motion of objects. They later discovered that by adding some terms for "bookkeeping", you could conserve energy for many more physical situations than just ideal elastic collisions. The manifest energy of motion became "kinetic" energy ("kinema" is Greek for "motion") and the "bookkeeping" energy became "potential" energy, as in the potential for real energy.
As for where potential energy resides, I'm no expert on the subject, but I believe you could build up a theory based on the fundamental particles that mediate forces. For example, if your system has electric potential energy, there must be photons being exchanged. My research advisor would probably go one step further and say that it's all somewhere in the quantum fields, but I don't have a good grasp on what those are so I won't offer it as my official explanation.
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
I said the one on the table remains hotter because the table will re-radiate back some of the lost energy. I think the question is ambiguous.
(Though I would be curious to see if we could determine which of the two effects dominates for "typical" materials and an ideal blackbody table.)
Experienced Forum User, Published Author, Player
(79)
Joined: 8/5/2007
Posts: 865
I see a lot of stuff posted here and I don't know if Warp yet accepts the answer already, but may I offer a simple analysis?
All games where you pick a car, the host will pick a goat (at random). Only half the games where you pick a goat, will the host pick another goat; in the other half, those trials are discarded because he shows a car. The net effect is that you are discarding some trials when you had picked a goat, when you would benefit from switching.
Sorry if all this has been said before.