timo

1) Chose a magnetic field; I recommend [math] \vec B = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) [/math] (I'll skip units; you can use any units in principle). 2) Put in a particle with charge 1 at the origin with initial velocity [math] \vec v(0) = \vec 0[/math] 3) The acceleration due to the magnetic field is [math] \vec a = \vec F /m[/math]; I posted the force (Lorentz force) above. 4) From knowing the acceleration, the initial velocity and the initial position, try figuring out how the particle moves, i.e. what it's position at some time t (or time t=5 if that is easier for you) is. 5) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) [/math] and do the same again. 5) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) [/math] and do the same again. You'll want to look up circular motion for this part. 6) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right) [/math] and do the same again. You should find that this case can be reduced to the previous cases. 6) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} v_x \\ v_y \\ v_y \end{array} \right) [/math] and do the same again. You should find that this case is just a generalization of the previous case. 7) Start having ideas yourself (this point can also come sooner, of course).

Why so? I'm not quite getting your point here. We probably agree that the electromagnetic field exists and interacts with charged particles. So what is special about gauge invariance? The particle physicists tend to consider their Standard Model a great success and are very proud of it. Where do you see the big fail?

Congratulations. When will it go to mass production and be available on the market?

Strictly speaking that is not what comes out of this thought experiment. What it does say is that if there was such a rod (and if it was light enough that you can still push it) you could pass information faster than with a light beam. Saying that the assumptions made are not compatible with reality is a different statement, not a result of the thought experiment presented.

It's been an unwritten (maybe someone actually took the time to read it down) rule on SFN to start thread by not simply throwing in a link. Perhaps you can add a bit more to start a discussion, e.g. what you think of the video, what you think the message is, maybe if that guy is you or not, what you find interesting about the video, which interesting questions it raises, why you think anyone here might be interested, ... to options are almost endless. The current opening post looks a bit loveless and not very thought-through, as if a model Internet kid was chatting with his buddies.

I think you should look up the rule to see if it exists at all. Then check what conditions are set on the functions, e.g. if they have to have a real(-valued), non-zero limit. The way you stated it, it is obviously wrong.

I have not looked into the documents but: Have civilian deaths really been covered up to a large degree? I am fully aware that civilians do die in Afghanistan. Let's face it: The news that three civilians were killed in Afghanistan would not leave me sleepless for a week (and I don't think I am a bigger asshole than the rest of you guys). And news tend not to report "just like the last 51 times we reported, for the 52nd time now civilians have been killed - we now go live to our reporter in Afghanistan". In fact, I remember reports about some biggies considering killing civilians, in particularly blowing up a wedding ceremony (or was that Iraq? guess that already shows how much I do care about people far away dying). So my point is: do civilian casualties need cover up by the military or does the lack of interest (if I don't hear about it then I don't have responsibility) of the average North American or European already take care of that?

In my opinion it does not help. Neither for helping you to understand grad-school physics (that's what before-grad-school courses are supposed to do B) ), nor to impress people (except maybe fellow students), nor for communicating with others or understanding standard arguments (the language and way of thinking and expressing ideas is surprisingly different in math and physics ). But of course I do not know the US system, only a bit of physics and math. I wonder who recommended that? Did the professors really say people should have a degree in math or did they just mean that they'd be happy if their average PhD student did understand the work he does a bit better (for which not pursuing a math degree seems helpful due to the additional time to learn the math required in physics)?

I think North Korea attacking US territory is not the point. EDIT: @John: My original comment was possibly a bit offensive. I hope that you can at least understand that people might become pissed when one completely misses the point (like the lives of 50 million people living in South Korea) over some rather ridiculous self-centered scenario. North Korea probably was not supposed to play the role of the Russians in ajb's analogy.

That's because it's wrong. What Mr. Skeptic probably had in mind is that [math] \lim_{x\to 0} \frac{a}{x} [/math] (with any a>0) does not exist (approaching 0 from the left gives [math]-\infty[/math], approaching from the right gives [math]+\infty[/math]) while [math] \lim_{x\to 0} \frac{a}{|x|} = +\infty [/math] formally seems to exist (if you accept that positive infinity was a result). That is, however, not the same as the question whether [math] \frac{a}{|0|} [/math] does exist. Mathematically, neither does the existence of a limit when approaching some point P imply the existence of the function at this point P, nor does the existence of the function a point P imply that the limit exists or equals this value. For me, the most sensible meaning of division is purely algebraic. Algebra is something that is not taught in schools (despite parts of school math being called "algebra", there). In algebra, you define multiplication as a mapping of a pair of elements (numbers) on a single element (new number, product). Division there is the multiplication of an element with what is called the "inverse element". There is an inverse element for every element except one element which is called "zero" (the existence of such an inverse would violate the standard axioms). Hence, the multiplication with the inverse of zero does not exist hence the division by zero does not exist. Calculus, in particularly dividing by something else than zero, then taking limits and claiming that one did divide by something similar to zero so the result must be similar, strictly speaking has nothing to do with the question about division by zero (though in practice one often also calls something zero when one actually means something very small). That is just the slightly more elaborate version of what was said before: Division by zero does not exist by definition. I'd say that pretty much every contrary statement seen on forums comes from the poster not knowing algebra (which is nothing to be ashamed of - it really is told in university courses, only), not from strokes of ingenuity. (Shouldn't we have some FAQ about this topic? Probably occurs even more often than 0.999999... = 1)

How about using the standard books used in school?

I'd just approach them directly: Look around the Pixar (or some other suitable company) homepage to get some idea about the company, then send a mail to some suitable person there asking for information. I don't think they are too angry if a suitable employee contacts them asking for information.

I don't know what he means (can guess it, though). But assume a homogeneous magnetic field (i.e. the field lines are all parallel lines into one direction). If the initial velocity of your ion is parallel to these field lines then the ion will fly parallel to the field lines and be unaffected by the magnetic field. If the initial velocity is perpendicular to the field lines, then the ion will fly in circles perpendicular to the magnetic field because the field affects it via the Lorentz force (the [math]q \vec v \times \vec B[/math] from above which was zero if [math]\vec v[/math] and [math]\vec B[/math] were parallel). If the initial velocity is neither purely parallel nor purely perpendicular then it is a mix of both. The two possibilities become superimposed, causing a screw-formed path of the ion. TonyMcC probably meant "have a velocity that is not parallel to the magnetic field lines" when he said "cut across the magnetic lines of flux". Three quick comments: - Despite being a construct for visualizing it is uncommon to call the magnetic field lines "imaginary". - The speed of an ion in a magnetic field does not change, see my previous post. The velocity can change, of course. More precisely, the velocity component perpendicular to the magnetic field lines changes; the component parallel to the field lines remains unchanged. - If you are interested in understanding it, I suggest you sit down and try to calculate the motion of an ion in a magnetic field - you will be able to answer a lot of possible questions yourself after having done that. Start with some special cases and then see how far you get in generalizing this. This is a topic where a lot of people on sfn are competent enough to help you with the math.

Surely you meant [math] 0.25 \text{ kg} \cdot 5 \frac{\text{m}}{\text{s}^2} = 1.25 \text{ N}[/math] - the units absolutely belong to the calculation in every step. This result should be correct. 1) 1250 N is wrong, your answer should be fine. 2) Check if the answer sheet possibly said [math]1250 \frac{\text{g m}}{\text{s}^2}[/math] which might be easily mistaken for [math]1250 \frac{\text{kg m}}{\text{s}^2}[/math]. 3) While it is a good idea to do all calculations in SI units it is not strictly necessary. As long as you are carrying around your units in every step the actual choice of units does not matter. But just staying in SI is also fine if you have trouble understanding the reason.

I don't think an analytic solution for [math]\int_a^b e^{-x^2} \, dx[/math] is known. But the integral [math]\int_{-\infty}^\infty e^{-x^2} \, dx[/math] exists and is well known (except to me who never remembers such things) and various ways to calculate is should be available online. The term you want to google for is probably "gaussian integral".

Right.

Was that question serious? The probability is roughly [math]\left(\frac{355}{356}\right)^{21310} = 10^{-26} = 0[/math], of course.

My old wishes: - (Option to have) Avatar back on the left. - Automatic ignoring of threads (full threads, not only the posts; and also on the forum search level) started by people on my ignore list (or whatever the current equivalent for people on my ignore list is).

That's correct. And on even smaller scales it is overcome by the chemical bindings, the electromagnetic interaction, the nuclear forces, the fact that points stay points (for e.g. molecules, atoms, nuclei, elementary particles). Gravity is only a special case of a force countering the expansion. It can be considered special because it is the largest-scale effect and it interestingly comes from the same theory that predicts the expansion in the first place.

As some minor remarks: A field does not need to be an assignment of a vector to each point in space. It can also be other stuff, e.g. a rank-2 tensor, a scalar, a spinor. The issue is even a bit more complicated than it might seem now because the common meaning of the term "vector" is not consistent over all physics. Ajb said "value" (including the "") on purpose. I don't think this statement is correct or even sensible.

I find the question a bit strange because you use pretty specialized terminology but your question is either beyond my understanding or really basic. So I'll just write a few more general lines before going to a one-line answer: The electroweak symmetry breaking in the Standard Model (SM): In the SM there is the electroweak symmetry group. This symmetry is not seen in everyday processes. The idea is that the laws of physics do follow the symmetry but that our point of view is shifted so that the symmetry is obscured: The point around which the symmetry hold is not the point we consider as our (everyday) reference point. To achieve this a mechanism is added to the SM which makes the symmetry point a very unfavorable state for the universe to be in (in favor of our reference point, of course). This keeps the electroweak symmetry group in principle but causes nature to favor a reference point from which the full symmetry is not fully apparent. Subsets of the symmetry might still remain. Symmetries of Grand Unified Theories (GUT): GUT symmetries are symmetries that are larger than the symmetries of the SM (all symmetries of the SM must be contained somehow). It is not expected that in everyday processes nature obeys a larger symmetry group than what is already contained in the standard model. So this extended symmetry must be destroyed somehow (*). The actual answer: - It is not known whether GUT symmetries exist in nature at all. - If they do it is only straightforward to assume that the breaking of the symmetry works basically the same as in the breaking of electroweak symmetry. - I have not heard of any other idea of breaking GUT symmetries. I think GUT is such a speculative thing that you'd make it even more speculative by adding something completely new without the need for it. - There is the possibility that I simply didn't get the point in "spontaneous symmetry breaking", so take this reply with a grain of salt (e.g. if you consider all I said above as a wrongified baby-language version of what you learn in your grad-school courses then I am probably not the person that should answer questions for you). (*) Note with "destroying a symmetry" I do not mean that no hint of it remains; the symmetry just shows in different ways. The electroweak symmetry is broken to leave only electromagnetism. But the remnants of this not-a-symmetry-anymore (and the mechanism by which the reference point was shifted) are still detectable as the weak force and the Higgs-Boson.

The one in the "Notices" above works for me: http://www.scienceforums.net/forum/showthread.php?t=53357

I am not sure if air is a great example for a liquid<->gas phase transition so I'll just assume you have water vapor in your container. In that case, you'd expect a phase transition under cooling down. In statistical physics, the background field of thermodynamics, one considers the set of all possible particle configurations (positions) and weights them according to the Boltzmann factor. The lower-energetic liquid-like configurations get a higher weight than the vapor weights but the higher total number of vapor-like configurations can cause the system to still be vaporous. As you decrease the temperature, the relative numbers of configurations stay the same. But the weights are shifted towards the lower-energetic liquid states, eventually causing an overall favor of the system being liquid. Pressure is irrelevant for this calculation. (Of course you might in principle just skip the liquid phase and jump towards the crystal directly).

I think you'd have accelerations like in today's fast trains like the ICE or TGV (which is a soft acceleration, in case you've never been in a fast train) and only (if at all) a higher top speed. After all, you want the travel to stay comfortable for the passengers. You could try to look up the accelerations of said trains and calculate (an estimate for) the stopping process for a maglev train of your choice's top speed from that.

Yes, but the letters are fairly standard. So if for example you don't know what a vector or a derivative is, then you'd possibly better ignore my calculation. Anyways, the situation about what happens to the speed of an ion traveling through a magnetic field. - Arrows over objects indicate the object is an R³ vector. - F is the acting force. - p is the momentum of the ion. - q is the electric charge of your ion - v is the velocity of the ion. - B is the magnetic field. - m is the mass of the ion. - t a time coordinate. - | ... | is the length of a vector.