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Yech. While some people do use s2=(ct)2-(x2+y2+z2) rather than s2=x2+y2+z2-(ct)2, it is a very very bad idea to mix conventions. Stick with one. Assuming you are using the convention s2=x2+y2+z2-(ct)2, that is kind of right. Remember that this is an expression for the square of the interval. This makes a time-like interval imaginary rather than negative.

The problem is that you appear to be misinterpreting that meaning. [math]s^2=x^2+y^2+z^2-(ct)^2[/math] is the "distance", aka "interval", between the origin and some point in four dimensional spacetime. This can easily be extended to get the "distance" between two points in spacetime: [math]s^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2-(c(t_2-t_1))^2[/math], or more simply, [math]s^2 = \Delta r^2 - c^2\Delta t^2[/math]. This is very similar to a distance, but that it can take on negative values means it does not satisfy one of the key mathematical concepts that characterize "distance". Nonetheless this is an extremely useful concept, so it is handy to give it a name. That name is the spacetime interval. Your misunderstanding stems from your thinking that a light-like interval, one in which [math]s^2=0[/math], is *the* spacetime interval. It is but one example of a spacetime interval.

You asked why [math]ds^2=dx^2+dy^2+dz^2-c^2dt^2[/math] is called the spacetime interval. I answered that question. If you wanted the answer to be "because that's what its called" then why didn't you just say so?

Depends on what you mean by "weight". Colloquially throughout the English-speaking world, and legally in the US, "weight" is a synonym for mass. If you bring a physicians beam scale to the Moon and step on it it will register a very slight increase. (While a beam balance measures mass, it is subject to buoyancy effects.) On the other hand, in much of physics and engineering, "weight" (aka "actual weight") is a synonym for "gravitational force". With this definition in hand, your weight ("actual weight") on the Moon is about 1/6 of your actual weight on the Earth. Ignoring that the Earth and Moon aren't quite round, this is given by Newton's law of gravity: [math]F=G M_p m_b / {r_p}^2[/math]. (G is the universal gravitational constant, Mp is the mass of the planet, mb is the your body's mass, and rp is the planet radius.) Since the Moon's mass is 0.0123 Earth masses and its radius about 0.273 Earth radii, your actual weight on the Moon is about 0.123/0.2732=0.165≈1/6 of your Earthbound actual weight. On last definition of weight, called "apparent weight" or "scale weight", is that quantity measured by a spring scale. In magnitude this is very close to gravitational force. As a vector, scale weight points away from the planet's surface rather than into it. Unlike gravitational force, scale weight is a measurable quantity. Your scale weight on the Moon is about 1/6 of your scale weight on the Earth.

Euclidean distance is but one measure of the distance or "interval" between two points in some space. One generalization of the concept of distance is based on a metric, a function of two variables that is symmetric, positive definite, and obeys the triangle inequality. (There are other generalizations such as the taxicab norm that cannot be expressed in terms of a metric). The Lorentz metric is not really a metric because it is not positive definite. It is instead merely non-degenerate. It perhaps would have been better to call it something else. However, it does share many other things in common with a metric, so for lack of a better concise term it is still called a metric and the result of applying the Lorentz metric is still called an "interval" (aka distance).

Right. I guess that exemplifies why nobody teaches the non-standard michel123456 math. What you will learn in standard math is that In vicinity of the origin of the half plane x>0, xy takes on all values from 0 to infinity. This means from the perspective of a limit (or continuity), 0^0 is indeterminate. However, it is extremely convenient if we define it to be one, at least in certain contexts such as the binomial expansion and power series. --------------------------------------------------------------------------------------------- Does this theorem have a name? 0^0 is indeterminate. http://www.wolframal.../input/?i=0%5E0 As far as finding such a pair of functions, here is one such pair: f(x)=x, g(x)=ln(a)/ln(x). Note that with f(x) and g(x) defined in this manner that f(x)g(x) is identically equal to a for all x>0. Note also that both f(x) and g(x) approach zero as x→0+ as required.

That is but one view of determining the value of 0^0. Let's try a couple more. 1. For all positive values of x, 0^x=0. Thus [math]\lim_{x\to 0} 0^x = 0[/math]. So perhaps we should have 0^0 be zero? 2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x0 and such that [math]\lim_{x\to x_0} f(x)^{g(x)} = a[/math]. Bottom line: 0^0 can only be defined in terms of limits, but different limits can be found that let 0^0 take on any value whatsoever. It is an indeterminate form. That said, it is convenient to define 0^0 to be 1 as an abuse of notation. This lets us write power series such as [math]f(x) = \sum_{n=0}^{\infty} a_n x^n[/math] Without this convenient abuse of notation, this series would not make sense at x=0.

Put that into a diagram. What are you complaining about? My use of "this"? My use of "for all x not equal to zero"? Is this better: [math]x^0 = \frac{x^1}{x} = \frac {x}{x} = 1[/math] This result, x0=1, is valid for all x except for x=0. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. First year algebra in high school? If that is the case, perhaps you are learning things a bit out of order.

My answer wasn't simple enough? Sigh. Try this one more time. Suppose x is not zero. That means we can divide x raised to some power by x. For example, [math]x^4/x = x^3[/math], [math]x^3/x = x^2[/math] and [math]x^2/x = x^1=x[/math]. Note that [math]x^n/x = x^{n-1}[/math]. So what is [math]x^1/x[/math], both expressed as a numeric value and as a power of x?

So let's start with the basics, the integers. Exponentiation extends multiplication just as multiplication extends addition. Multiplication in the integers is defined as repeated addition. For example, 4*3 = 4+4+4. Recursively, the base case is[math] a*1 = a[/math] and [math]a*n= a*(n-1) + a[/math] is the reduction rule. Exponentiation is similarly defined as repeated multiplication. The base case is [math]a^1=a[/math], but now the reduction rule uses multiplication rather than addition: [math]a^n=a^{n-1}*a[/math]. This definition of raising a number to the power of a positive integer still applies when one is working in the rationals or the reals: [math]x^n = x^{n-1}*x[/math] So, let's naively substitute n=1: [math]x^1 = x^0*x[/math] Now solve for [math]x^0[/math], and use the fact that by definition [math]x^1=x[/math]: [math]x^0 = \frac{x^1}{x} = \frac {x}{x} = 1[/math] This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

Nonsense. The only problem here (other than the obvious language issue) is your lack of understanding. Period. There is a solution: Go to school.

Use of obscenities is strictly forbidden here, in any language. Ku-ku! I have got on a mathematical forum or to a psychiatric clinic? 1. Just because someone else made a mistake in their math does not mean you are right. 2. Insulting people is contrary to the rules of this forum. So, let's do this right. We are given that [math](1)\quad \frac{df(x)}{dx} = 2x[/math] [math](2)\quad F(x) = \int f(x) dx[/math] [math](3)\quad F(2) = 5[/math] Taking the anti-derivative (aka indefinite integral) of (1) yields [math](5)\quad f(x) = x^2 + c_1[/math] where [math]c_1[/math] is an as-yet undetermined constant. Integrating again yields [math](6)\quad F(x) = \frac{x^3}3 + c_1 x + c_2[/math] Note that we now have two constants of integration. Since (3) provides only one piece of information, we can resolve one, but not both, integration constants. I'll express [math]c_2[/math] in terms of [math]c_1[/math]. Applying (3) to (6) yields [math](7)\quad c_2 = \frac 7 3 - 2c_1[/math] Applying this to (6) and replacing [math]c_1[/math] with [math]a[/math] yields [math](8)\quad F(x) = \frac{x^3+7}3 + a(x-2)[/math] where [math]a[/math] is an arbitrary constant. It is always a good idea to double check one's work. Note that the second term in (8), [math]a(x-2)[/math], contributes nothing at x=2. Regardless of the value of a, F(2)=5 as required. Differentiating yields [math]f(x)=\frac{dF(x)}{dx} = x^2+a[/math] Differentiating again yields [math]\frac{df(x)}{dx} = 2x[/math] as required.

Yep, it is. Good that you recognized that. That's correct. The important thing is that a, b, and c have to be integers. Relax this constraint and there are an uncountable number of solutions. For example, [math]3^3+4^3 = (\sqrt[3]{91})^3[/math]. However, since the cube root of 91 is not an integer, it is not a solution.

What do you mean by inertia? "Inertia" is one of those words that physicists now eschew (exception: Moment of inertia). Without a qualifier, inertia has an ambiguous meaning. In some contexts it means mass, in others, momentum. Why use a word with ambiguous meaning when there are two perfectly good words that do not suffer that ambiguity? If you mean momentum, photons have non-zero momentum. Look back at what you wrote. That is exactly what you are doing. Then do it. Don't put the burden on us. If you are not using mathematics you are not doing physics. Period.

The numbers a, b, c, and n have to be positive integers. Without this restriction there are uncountably many solutions to an+bn=cn for any given position value of n. If you do not understand the basics of the problem you will not understand Wiles' proof. I'll venture further: you will never understand it. Very few PhD mathematicians understand it.

Could you be a bit more illustrative of what you are talking about? Computer language, operating system, application domain, perhaps an example of what you mean?

Oh please. Obviously a conjectured model of physics first and foremost must agree with already-accumulated evidence.

What a bunch of hogwash. Who is more qualified to remove a tumor from your brain, a trained neurosurgeon, or someone who has read the wikipedia article on brain surgery? Who is more qualified to argue a case before the Supreme Court, a practiced constitutional lawyer, or an amateur pundit who has some twisted view of what the constitution means? What we have here is one of the key problems of the 21st century, the silly notion that people with no training in a specialized field can add any value to that field. If anything, this internet cult of the amateur subtracts value from society. Moreover, this is not how science works. When someone throws out a bunch of excrement and someone knowledgable in the field calls BS, it is not our job to prove the person who made those excremental claims false. The burden of proof lies solely on the person who made the claims. In the field of physics, those claims need to be backed up with mathematics. Words alone are just armchair philosophizing, aka mental masturbation.

Yes and no. The distance driven by a taxicab in Manhattan (or any other city with a nice rectangular grid of streets) from point A to point B is not the same as the distance that a crow would fly in getting from point A to point B. The number of moves taken by a king on a chess board illustrates of yet another definition of "distance". There are an infinite number of mathematical definitions all of which describe in some way the concept of "distance". The number of choices drops dramatically when one requires that distance be generated by an inner product. In fact, ignoring isomorphisms, there is only one. Relax the constraint that the inner product has to be positive definite, instead requiring it to be non-degenerate, and the number of possibilities opens up again. However science is constrained to mathematics that matches reality. Given the experimental confirmations that space is locally Euclidean, that the equivalence principle is true, that the speed of light is the same to all observers, and that motion obeys certain physical laws. of the physics of motion, and of the the observed constancy of the speed of light, there just aren't many choices left. In fact, there is only one (to within isomorphisms). This is the Minkowski metric.

When something obviously has zero meaning, no, its not arrogant. Pioneer's post, like yours, was word salad. What, exactly, do you think energy is? What do you even mean by this? Energy is not a thing in and of itself. All of the forms of energy that we know of, except for gravity and dark energy, involve particles. Dark energy doesn't count (yet), because nobody has any idea what it is (yet). Gravity is in a sense in the same boat as dark energy. We don't know what causes it (yet) as a deep theory of physics. Developing a model that unifies quantum mechanics and gravitation is the Holy Grail of physics. We're not there yet. No, it's not. Where did you get this idea? This is word salad. You are traveling at 0.999999 c with respect to some object in the universe. Why haven't you become radiant again?

Just. Read. This. http://en.wikipedia..../Antiderivative Russian wikipedia entry: Неопределённый интеграл — Википедия

The derivative of a constant function is zero, and the derivative of sum of two functions is the sum of the derivatives. Thus if F(x) is an antiderivative of some function f(x), so is G(x)=F(x)+c. This is basic, basic stuff. Have you taken even one course in calculus?

The only misunderstanding here is yours. You are spamming multiple forums with this nonsense. Many people have told you many, many times over that you are omitting the constant of integration in the indefinite integral.

You asked a question that you could have answered yourself had you spent even a minute researching it. http://www.wolframalpha.com/input/?i=1+year+%2F+1+sidereal+month

Playing with this expression yields another explanation for why lunar tides are greater than solar tides. Designating [math]r_m[/math], [math]\rho_m[/math], and [math]\delta_m[/math] as the radius, density, and ngular diameter of the Moon, the above expression for the tidal force becomes [math]a \approx \frac {\pi\,G\,r_e} 3 \, \rho_m \, {\delta_m}^3[/math] There is nothing special about the Moon here. The same expression applies for Jupiter, Saturn, and the Sun. The tidal force at the points on the surface of the Earth along line connecting the center of the Earth and the perturbing object (Moon, Sun, etc) is proportional to the product of the object's density and the cube of the object's angular diameter. Note hat the angular diameters of the Moon and the Sun as seen from the Earth are nearly equal (0.518° for the Moon versus 0.533° for the Sun). This means the relative strength of tidal forces from these two objects depends primarily their average densities, which is about 3.34 grams per cc for the Moon but only 1.41 grams per cc for the Sun. ============================================================== What "lunar gap"? Gravity is not 5 times higher at the Bay of Fundy. Have you ever sung in the shower? Your vocal chords are not 5 times stronger when you are in a shower.