timo

Wonder if adding such a "you are this intelligent" scale to homework exercises would increase the time people spend on them.

Morally: Yes (though I disagree in this case). Legally: Barely, especially not with the argument above. Considering the violation of the forum rules: For the mods to decide.

You can probably think of it as a background in which the physical objects (spacetime not being counted as a physical object for the sake of this argument) and processes are embedded. Like a piece of paper on which you draw a painting, except that the painting ususally doesn't influence the paper and that you usually view a picture at once, not line by line (<= time by time). Energy is a property of matter and -strictly speaking- also of non-matter objects (e.g. electromagnetic waves are usually not considered matter). It is, however, a property that plays a role in the gravitational influence similarly to the role that mass plays in classical Newtonian gravity where twice a mass exerts twice a gravitational force on other objects. "Does energy have a gravitational force" does not seem like a correct statement; it seems akin to asking whether charges have an electromagnetic force. In standard physics, there are three dimensions of space. However, there is no "the space" and "the time". Taking the piece of paper again: The piece is two-dimensional. Any line on the piece of paper is one-dimensional. There is, however, no "the line" as you can draw arbitrary lines on the piece of paper. You can decompose the piece of paper into sets of parallel 1D lines (ignoring that a piece of paper has finite size, here). Yet, there is not "the set of parallel 1D lines" since their orientation is still arbitrary. Similarly, you can cast a non-curved (flat, no gravity) spacetime into a set of spaces (each representing a different time, then). While there's some restrictions that have no analogy with the piece of paper example, the main statement remains the same: There is no "the set of spaces", you have some arbitrariness in the definition. No.

Sadly, I don't know what a "deck" is. What you assumed is that it's two rectangles being on one of the longer and one of the shorter sides, respectively. I could imagine it's either four of them (one on each side of the pool) in which case you are missing two addends for the total area (two additional "decks") or kind of a ring around the pool in which case you must add four additional "corner-rectangles" to fully enclose the pool. Except for that possible problem, your approach seem fine so far. Regardless of the actual term you use to calculate the total area (i.e. regardless of what additional area you decide to take for the "decks") from what you have you can solve the problem in the manner I explained earlier: Substitute either the lengths of the pool or the width of the pool with the respective other variable (note that xy=2500) to get expression for the total area that only depends on one of the two variables. Then, find the minimum total area as a function of this variable, either by differentiation or by drawing the function.

Reduce the two variables to one leftover variable by substituting one of the two equations you get into the other. Then find the remaining variable such that it minimizes the leftover function for the total area. Not completely sure if that works (I did not actually try the problem) but it should. If you have any intermediate results yourself and just got stuck, please show/post what you already tried and where and why you got stuck.

Free object (no potential energy): Definition of rest/invariant/proper mass: [math]m^2 := p^\mu p_\mu = E^2 - |\vec p |^2[/math] modulo some factors of c that I don't want to look up. It's what physicists call an invariant or a Lorentz scalar. Definition of kinetic energy: [math] E_{\text{kin}} := E - mc^2 = \gamma mc^2 - mc^2[/math]; obviously not a Lorentz scalar. => [math]| \vec v |(E_{\text{kin}}) = c\sqrt{1-\left( \frac{1}{E_{\text{kin}}/mc^2 + 1} \right)^2} [/math], [math]\lim_{E_{\text{kin}}\to \infty} | \vec v | = c[/math], [math]\frac{\partial | \vec v |}{\partial E_{\text{kin}}} > 0[/math]. Note that you can substitute the kinetic energy for total energy if you prefer that. Above is the standard definitions for kinetic energy and especially mass. Relativistic mass is merely the energy or an object expressed in different SI units (due to being divided by a constant factor c²).

Perhaps you can see that your reply was not an answer to my question. If not, then let's talk numbers: What's the density of an object with a mass of 300 g? Perhaps you might consider the idea that I asked the question for a reason (albeit I start to have doubts in the reason considering it has been said that density is mass divided by volume in almost every post in this thread, by now).

How would you determine the density from the weight?

>> glEnable(GL_LIGHT_0); wtc+conspiracy -> 300k hits. No need to worry >> glEnable(GL_LIGHT_1); z+boson+lep -> 80k hits ....

What's unreasonable with [math] \frac{\partial \tau }{\partial t} [/math] i.e. change of eigentime with coordinate time ?

Given it's supposedly about invariance, I suspect M to be a rank-2 tensor and the dx to be infinitesimal shifts in coordinates that can be considered Lorentz-vectors in this case. The result of the product then is a scalar, i.e. a number that has the same value in all coordinate systems. I assume what Schutz is about to do is showing that if the objects dx and M are expressed in a different coordinate system (where they could be written as [math]M'_{ab}, dx'^a [/math]) the value of the product remains the same i.e. s'=s.

A simple e.o.m including mass would be that for an electrically charged object with charge q in an electric field E: [math]\ddot x = qE/m[/math].

I don't understand what you are saying.

I doubt it. Depends on what you mean by mass. If you mean energy, then, apart from the infinity problem, that's trivially true. If you mean mass as in "mass of a particle" (i.e. rest mass), then at least in the given scenario it's wrong. Surely, increasing the kinetic energy of a massive particle increases its speed. |v|(E) is a strictly monotonous rising (although limited) function except for massless particles.

I have doubts that you really need to integrate something. I am still not sure what you really want to calculate. I have the feeling you don't exactly know that yourself, so let me give you an offer of what you could calculate; maybe it comes close to what you try to do: Assume you wanted to compare the energy required to accelerate a car from stop to a velocity of ~17 m/s (ignoring any resistance due to friction) with the "biological energy" of a Snickers (or alternatively calculate the number of Snickers whose bio energy equates it): - The amount of energy required to accelerate the car equals the energy of the car after acceleration minus the energy of the car at rest. - The amount of "bio energy" in a Snickers in SI units is obtained by converting the calories into joules. - The number of Snickers whose "bio energy" equals the energy needed to accelerate the car equals the latter divided by the energy of a single Snickers.

I don't think it's correct. - The amount of energy required to cross 5 km in 5 minutes with a car is not unique. As you probably know, fuel consumption of a car depends on driving style. Notably on the driving speed. Just race close to the destination and then take a break until time ran up. You'll have spent more fuel than if you had been driving slower and -due to having a break- still can need the same time. Did you possibly mean the energy required to accelerate a car to a speed of 5 km per 5 minutes? That would greatly simplify your calculation and make the part where you seem to have made a huge error unnecessary. - You forgot the factor 1/2 when you calculated the energy. - Except for the missing factor of 1/2 you correctly calculated the energy required to accelerate a car to a velocity (ignoring energy losses which in practice can probably be quite large) of 5 km per 5 minutes. I see no reason why you suddenly take that number and divide it by 1 second. I don't see what the result is supposed to represent. I think it's plain wrong. The average power required is the energy required/demanded divided by the time that the power will be applied. I'm not sure you need power at all. - I've not read and checked the Snickers part; one problem at a time (in reality I'm just too lazy at the moment). - Very nice description of what you tried. It's way too rare that people bother mentioning units and intermediate calculation steps .

Dunno the story. But since density is mass divided by volume, knowing either the mass (by weighting) or the volume (by measuring the amount of water it displaces) alone will not result in knowing the density. You need both values to determine the density. A guess of mine would be that the weighting is not mentioned because it's kind of simple while measuring the volume via displacement in water is a pretty smart and not-so-straightforward idea.

Yep. Nope, unless you define mass to be energy (-> relativistic mass) which is kind of silly because you could as well call it energy, then. It's a bit like a dog and an animal. A dog is an animal but an animal doesn't have to be a dog. Kinetic energy is energy. Potential energy is energy. Mass is energy. But the reverse statement that energy is kinetic energy does not need to be true. Same with potential energy. Same with mass. Mass is a form of energy; you might think of it as a form of potential energy for calculational purposes. Even better, more precise and less prone to making errors: Mass is a term that contributes (or at least can contribute) to the energy of a system. Therefore... Energy, not mass. In this case, none of the energy comes from mass; it's purely kinetic energy.

http://www.nytimes.com/2008/03/29/science/29collider.html?_r=2&oref=slogin&oref=slogin Teh l33tspeak mod sucks.

Probably a joke by either the admins or the authors of the underlying forum software. A bit annoying, indeed.

http://askanexpert.web.cern.ch/AskAnExpert/en/Accelerators/LHCenvironment-en.html#3, not exactly the page I was looking for; there is a more explicit answer on the CERN homepage iirc. But probably more authorative than WP. There is no widespread doubts, at least not if "widespread" means "a lot of people". I've actually only heard those doubts twice: First time on the CERN homepage where I read a comment similarly to the one I linked above. It was directly after the question if CERN really has an X33 plane, I think. So I took it as kind of a joke. Second time I heard about it was on the german Wikipedia where someone wanted to add these concerns to the LHC article. The cited source was an interview with a researcher on non-linear dynamics (i.e. neither a particle physicist nor an astrophysicist or anything you'd expect for someone familiar with the topic) on an internet page for computer programmers. So in some respect that would indeed fit the term "widespread" . @Klaynos: We have detected evaporating black holes? Where and when?

Gravitons are supposed to be light-like. Virtual gravitons are not (in my eyes) gravitons. Depending on preferred view, either: - via virtual particles, that can cross the EH. - via the field, which clearly (at least for the gravitational field) exists outside the EH. No, but Google gives e.g. these results: - http://www.physicsforums.com/showthread.php?t=65583 : Didn't read it, but Janus has the habit of knowing what he sais. - http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980601a.html, http://www.fnal.gov/pub/inquiring/questions/blackholes.html : I somewhat dislike the heavy use of the term virtual particles that sounds like circumventing problems by usage of misleading but undisputed terminology. But Nasa and Fermilab should be relatively serious sources.

The event horizon is a boundary, where no light-like or time-like paths lead out. Virtual particles are not particles, merely terms in a Taylor expansion that look akin to particles. Interactions work via a source influencing a field (in this case energy creating the gravitational field via the Einstein equations) which then influences other particles, e.g. via [math] \ddot x^\mu - \Gamma^\mu{} _{\alpha \beta} \dot x^\alpha \dot x^\beta = 0[/math] which you (Obelix) seem to know, already. There is initially no reason why the field equation should be altered when you go from the non-quantized version to a theory of quantum gravity. As a matter of fact, reproducing the classical version for classical scales (in which case spacetime outside a BH is non-Minkowskian just as if there was a non-BH object with the same mass creating the field) is a requirement for a quantum theory of gravity. If you insist on interpreting the terms in the Taylor expansion, the virtual particles, as particles then those "particles" are not restricted to being time-like or light-like and hence you shouldn't have any problems with the EH. Sorry for being a bit vague. It's a bit hard to make comments on a non-existing theory. The bottom line is that virtual particles are not particles and especially not required to be time-like or light-like. Sidenote: The question is a classic. There should be dozens if not hundreds of good replies to the question "how can gravitons escape the event horizon" online.

What exactly is your question? How to understand the factors 2 and 4 from the pictures, what the pictures actually show, why the waves have to have the form sketched in the pictures or something else? You didn't find any latex reference or it wasn't mentioned? - A guide is here: http://www.scienceforums.net/forum/showthread.php?t=4236 - The TeX code for any capital greek letter is \Letter, the code for a small greek letter is \letter, e.g. \Lambda for [math]\Lambda[/math] and \lambda for [math]\lambda[/math]. Got a good explanation why there's a maximum amplitude for the open end, too?