timo

Many measurement are made by some kind of measurement device, not by human senses. I do not know if [save for production of neutrinos on the subatomic scale] there are any objects that just disappear. People would start to wonder where they went if that happened often.

Err, well .. that was more or less my point, wasn't it?

Transform to cartesian, normalize the vectors, sum the entries for x-, y- and z-coordinate , convert back to angular coordinates. You supposedly don't want to get an angle of 179° as a bisection for two vectors with 0° and 358°, respectively.

He believes in provocation?

I remember being told that the original term is the only possibility. A quick reminder how the geometry-side of the Einstein equation can be [roughly] justified: 1) The matter-side, the energy-momentum tensor is divergence-free (=> local conservation of energy and momentum). 2) So should be the term representing the geometry side if you want to have a differential equation. 3) The geometry side shall somehow include the 2nd derivatives; the curvature. The Ricci tensor contains those (dunno to what extent using the Ricci tensor is required) but is not divergence-free. 4) Explicitly subtracting the divergence of the Ricci tensor then makes the geometry side a divergence-free function of the 2nd derivatives. 5) But: Since the metric tensor itself is divergence-free, adding any multiple of the metric will not spoil divergence-freeness of the geometry term while thought-wise it can still be considered part of the geometry. So what would that mean for your Lambda: - It should be a divergence-free tensor of rank 2. - It should not be some kind of particle content or otherwise it would logically belong to the matter part T. Substracting parts of T from both sides of the Einstein equation and calling that -Lambda probably isn't what you were after. - It should, by your constraint, not depend on the metric - whatever that exactly means. I would assume that "should not be some particle content" might be the killer constraint for you.

Perhaps you should also try to compile my example (copy&past, replace the filename with any jpeg on your HD) and see if you still have the same problem (i.e. a different output than the pdf I attached).

I cannot reproduce your problem. \documentclass[12pt]{article} \pagestyle{empty} \usepackage{graphicx} \begin{document} Below, we see a picture of Snail: \begin{figure}[h!] \centering \includegraphics[width=0.80\textwidth]{SnailsAvatar.jpg} \end{figure} It was not possible to find out what the picture is supposed to represent. \end{document} Then pdflatex Snail.tex Results in the pdf attached. I do remember having had a lot of trouble with pictures myself, though. For example, I cannot compile the source above using "latex" as command but only with "pdflatex". Perhaps you should try a different destination file format and see if the issue with the pathname prevails. Sidenote: I somehow overread the "at work" part so sry about that. Didn't mean to actually be rude in either case. Snail.pdf

You should definitely supply your TeX-code. Haven't you learned from last time? As TeXnicCenter is just a text editor, displaying the path is most probably not related to that. I've never had the path being displayed when embedding plots. My first guess is that you possibly do something like \begin{figure} \caption{NameOfTheJPGEG} \includegraphics{NameOfTheJPGEG} \end{figure} i.e. you explicitly put the filename as description of the figure.

I have a follow-up question on polarization of light. Assume I want to polarize microwaves with a metal grate. Assume the grate is oriented such that the bars of the grate are vertically aligned. Staying in the definition above: What will be the polarization of the beam of microwaves after the grate - and why?

I do! Because you have to show that the limit exists. Hint: Explicitely try n=0, n=1 and n=2 for the 1st derivative, first. You should already see a pattern from there.

If you have anymore questions, I'd advise having a look at Learn C++(Chapter 9.11), or alternatively the Learn C++ forums[/url']; it does an excellent job of explaining not only this. I know it is nitpicking but the part you quoted is funny in the context of "the page I have it from does an excellent job of explaining": Of couse, not every constructor taking an object as an argument is a copy operator .

Supposedly that you tried to compile the program with a C compiler (or linker) rather than a C++ one. That's just a guess, though.

For the first one consider a triagle with side-lengths 1, u, [math] \sqrt{1-u^2}[/math]. For the 2nd one use the result of the first one. Be sure to check the ranges, btw. Sidenote: I did not try them myself but these two approaches seem to be the obvious ones.

Since you have the distribution why would you not simply calculate the variance from it?

[math] v = \frac{2 \pi r}{24 \text{ h}} \approx \frac{6 \cdot 6 \cdot 10^3 \text{ km}}{24\text{ h}} = \frac{36\cdot 10^3}{24} \frac{\text{km}}{\text{h}} = 1500 \frac{\text{km}}{\text{h}}.[/math]. So if "tangential speed" is defined the way I assumed, then it seems you are right. However, seeing that I did some approximations it is well possible that the actual value would be ~1700 km/h. In that case the college book might have screwed up the units (or you misread them).

What specifically went bad? That obviously seems to be a good starting point for advices.

No, unless you put additional constraints on your question that lead to a yes. For example, in electron-positron annihilation conservation of energy and momentum prohibits e+ e- to react into a single photon. It does not forbid the reaction into three photons, though. The probability is just a little lower. I am even tempted to claim that there are no sensible constraints that leads to a yes - on the level of Feynman diagrams constraining to an even number of photons looks impossible, but cancellation of diagrams might do the trick. The typical example of a single photon being emitted is when an excited atom decays to an energetically lower state emitting light. That's mostly single photons.

I was almost thinking usage of $ might be the problem but it just did not fit to the fact that no tutorial said anything about it. Anyways \documentclass[10pt]{article} \usepackage{amsmath} \begin{document} \section*{Version 1 - The wanted newlines} \boldmath To determine the implicit solution of...$nothing to see here$ we proceed as follows. For the left hand side... \\ $nothing to see here also$ so... \\ $more nothing to see here$ \\ So the implicit solution is $no answer here, go learn some math$ \section*{Proper math} A simple equation or -as in this case- two subsequent simple equations are usually written in an equation-environment: \begin{equation} a=b \end{equation} \begin{equation} c=d \end{equation} The equationarray-environment is useful very often, too: \begin{eqnarray} E &=& \frac{p^2}{2m} \\ &=& \frac 12 mv^2 \end{eqnarray} \end{document}

I do not understand what you are looking for. Since you gave an example of what you want it to look like, perhaps you should also state what it currently actually does look like. That would probably help understanding your question.

The modus I knew/expected is asking Capn for some completely unnecessary change. Wait for Capn to say "I think I could do that". Wait for Capn to implement it. Wait for you to realize the change, and posting an announcement about it. I didn't know someone told you about the change. I consider that cheating.

For example iron that was previously magnetized. But: The magnetization must be stable enough not to change when you switch on your electromagnets (which is what I called "magnetic hardness"). So magnetizing the iron with the electromagnets before the experiment seems to be ruled out as an option. Sadly, I do not have any other ideas except perhaps placing additional electromagnets under the plate and have them running all the time.

Being a theoretician the following idea might be completely impractical/impossible, but: - Can't you just just start with a magnetized substance instead of a (initially unmagnetized?) steel plate? Iron is what I know as the typical ferromagnet. The property for the material you also need -the initial magnetization of the plate not flipping because of your shoes- is called "magnetic hardness", I think. - Or put other electromagnets under the plate (and perhaps switch to a plastic plate, then). Some other piece of advice: Have you looked at the estimated numbers? Do you expect a half-ways decent acceleration force with a current you still want to have running through your shoes, for example?

To answer this question I would need to know more about "mathematical physics" than I currently do. All I currently know about it is that a) The term exists, b) ajb said he is doing it and c) that the terminology in his posts is much more formal and abstract than that of the average theoretical physicist. The adjective "mathematical" for me is a hint that mathematical physics uses more mathematical rigor. I wouldn't be surprised if mathematical physics was what non-theoreticians expect theoretical physics to be.

I would think the difference is that contrary to popular belief theoretical physics is usually not closely related to mathematics whereas mathematical physics probably is.

The other way round is relatively interesting, too: [math] mg + \frac{mv^2}{r} = mg + \frac{mv^2}{r} \frac{g}{g} = mg\left( 1+ \frac{v^2}{rg} \right) [/math] (to be read as successive steps from the left to the right). The first step I did here is writing 1 in a complicated -but useful for the following steps- manner and using that [math]x = 1x[/math]. Multiplying things with a complicated 1 or adding complicated zeros to terms is something you see very often (at least in physics), hence this post highlighting it.