• Physics Discussion
    973 replies, posted
Closer than the moon!
[QUOTE=JohnnyMo1;44751116]Closer than the moon![/QUOTE] What if it hits the moon :o
[QUOTE=Falubii;44751796]What if it hits the moon :o[/QUOTE] Then there would be like a crater... [I]on the moon[/I] I dunno if I could handle that
[IMG]http://puu.sh/8E0gf.png[/IMG] Spent an hour making a fancy vectorized thing for my thesis, love doing stuff like this. [URL="http://dl.dropboxusercontent.com/u/17216535/rotatie.svg"]SVG[/URL] There's still one imperfection in there, the endmarkers look weird. Edit: oh man I keep finding more imperfections. I edited this post like 5 times already.
Anyone have general career/undergrad advice? I'm sure some of you know some things now that would have been useful earlier in your studies.
[QUOTE=Falubii;44795079]Anyone have general career/undergrad advice? I'm sure some of you know some things now that would have been useful earlier in your studies.[/QUOTE] If there's something you want to do/try doing as a career then try to get in early with summer placements/internships and the like. Some areas like banking/finance are crazy competitive and it's much easier to get a foot in the door if you've got some experience under your belt. Same applies to research I guess, if you're thinking about academia/researching then ask around your department. Couple of my friends did that and helped out one of our postdocs for a month or so, even got paid reasonably well for it! Not sure if it's the case elsewhere, but that's the impression I get in the UK.
[QUOTE=Lord Pirate;44800766]If there's something you want to do/try doing as a career then try to get in early with summer placements/internships and the like. Some areas like banking/finance are crazy competitive and it's much easier to get a foot in the door if you've got some experience under your belt. Same applies to research I guess, if you're thinking about academia/researching then ask around your department. Couple of my friends did that and helped out one of our postdocs for a month or so, even got paid reasonably well for it! Not sure if it's the case elsewhere, but that's the impression I get in the UK.[/QUOTE] Definitely true in the US. I didn't start undergrad research until senior year and now I desperately wish I had been doing it since sophomore year or so. Doing good research can actually be more important than GPA.
(This is for homework due today, but I've looked through a whole script and half the internet and still can't find a good explanation, so I hope this is OK.) I'm supposed to show some things about the translation operator in for quantum mechanics homework, but the definition I have to find and examine is [img]https://upload.wikimedia.org/math/1/b/0/1b05716fe126463722bdff72412835fd.png[/img] and I honestly have not the slightest idea where that comes from or what an operator in an exp function even means. I think I'm slowly getting the hang of this, I solved the rest of the problems without much difficulty at least. This one is worth about a fourth of the points though, so it would be great if I could get a few pointers in the right direction.
exp(x) is simply alternate notation for e^x. Makes it easier to write when you would normally have a bunch of stuff cramped up into a superscript.
[QUOTE=JohnnyMo1;44803617]exp(x) is simply alternate notation for e^x. Makes it easier to write when you would normally have a bunch of stuff cramped up into a superscript.[/QUOTE] Yes, that much is obvious. I just don't know what to do with that operator in there or how to manipulate the expression. I found something on Wikipedia to check invariance against it by looking at infinitesimal changes (which gets rid of the exp function), but I don't think that's enough to find eigenfunctions. I think I could guess them, but that's hardly what I need if I don't know how to verify it. [editline]14th May 2014[/editline] Does it have something to do with the Fourier transform or at least something where I integrate over p so that the e-function operates on usual complex numbers? Actually, maybe I can solve it by Fourier-transforming the spatial representation into a momentum one and back... I still have no clear picture though, pretty far from it.
Oh, for some reason I read your above post as: "I honestly have not the slightest idea where that comes from or what an exp function even means." My bad.
The series expansion or the limit expression for exp() helps in understanding it. When we reached generators and stuff like that we did it the other way round, i.e. the momentum operator generates translations or some shit like that, and then you ended up with an exponential containing that operator. If they just throw that expression at you I can understand it looks weird as an exponential seems like a scalar function with a dimensionless argument. Or at least that's the first thing that I think about... I think we also used [URL="http://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#An_explicit_Baker.E2.80.93Campbell.E2.80.93Hausdorff_formula"]this thing for a couple of proofs[/URL].
[QUOTE=Number-41;44806409]The series expansion or the limit expression for exp() helps in understanding it. When we reached generators and stuff like that we did it the other way round, i.e. the momentum operator generates translations or some shit like that, and then you ended up with an exponential containing that operator. If they just throw that expression at you I can understand it looks weird as an exponential seems like a scalar function with a dimensionless argument. Or at least that's the first thing that I think about... I think we also used [URL="http://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#An_explicit_Baker.E2.80.93Campbell.E2.80.93Hausdorff_formula"]this thing for a couple of proofs[/URL].[/QUOTE] Thanks, I'll look at the link when I'm not completely tired any more. What we were supposed to do was exactly what you wrote I think. (The exponential wasn't given.) I still haven't understood it quite 100%, but now I have the solution from the exercise session so I can review it after sleeping. It turned out to be really short in the end, I'll probably put it here eventually so there's a solution to the question I posted. My problem is often that I don't know how I can manipulate a given equation so I hit a ton of dead ends that I can't look up online because googling maths doesn't work very well usually (and sometimes I don't even know something exists like in this case).
Oh now that I reread your post I think I know what they want you to do: derive the translation generator using the momentum operator. So first you have to think how you would write a translation in terms of a momentum operator. Remember that it is the derivative with respect to position and some complex-h prefactor stuff. You can then try to write down an expression for an translation delta_x using that derivative (series expansions, first order, stuff like that). So you find an expression that describes translation over a small distance delta_x thanks to the momentum operator. Then you try to write down an expression for a translation over some distance lambda=N*delta_X. You will find something that looks like the limit expression for exp() but without the limit, and then you take a limit for N to infinity and that conveniently ends up in an exponential. If I told it with maths it'd just be the solution :v: Oh and that link to that identity will probably be useful for relations between conservation and commutation and stuff like that. Don't think it's relevant to the current exercise. Edit: I don't think you even need to use the momentum operator at the start, just the derivative and then in the end you can rewrite it with the momentum operator.
[url]http://halo.wikia.com/wiki/Magnetic_Accelerator_Cannon[/url] There's some solid physics on that page I tell you what
[IMG]http://i.gyazo.com/5c6b22fcf2290ce0bec8c51d7f3c2b7c.png[/IMG] I'm supposed to derive the formula in the image, and I was able to get that the triangle with 'd' and the triangle in the medium were similar but other than that I had no idea where to go from this.
Anyone know of a good method for finding mean distances in a system of clustering particles due to interaction?
[QUOTE=gangstadiddle;44847724]I'm supposed to derive the formula in the image, and I was able to get that the triangle with 'd' and the triangle in the medium were similar but other than that I had no idea where to go from this.[/QUOTE] Snell's Law actually doesn't help. This is all trig. [t]http://s28.postimg.org/4hrsea2t8/Untitled_Note_May_19_2014_1_18_PM_Page_1.jpg[/t]
I'm kind of confused about nuclear binding energy. Everything I've read states that the mass of the nucleus is always less than the sum of its parts, and that the discrepancy in the mass represents the binding energy via energy-mass equivalence. My confusion is, how are they measuring the mass of the atom? Einstein's equation tells us that inertial mass (how its momentum changes under a force) is determined by the energy content of the system. If that were the case, how would you measure any difference in the mass of the atom, given that the only difference in the two situations is that it contains binding energy instead of exclusively mass?
I don't understand my physics book. It has loads of small errors and sometimes the answers don't match even though I'm 100% sure I've calculated them right. Doesn't really make learning by yourself easy :/
[QUOTE=Hattiwatti;44908986]I don't understand my physics book. It has loads of small errors and sometimes the answers don't match even though I'm 100% sure I've calculated them right. Doesn't really make learning by yourself easy :/[/QUOTE] Give an example and we can check.
[QUOTE=Falubii;44865663]I'm kind of confused about nuclear binding energy. Everything I've read states that the mass of the nucleus is always less than the sum of its parts, and that the discrepancy in the mass represents the binding energy via energy-mass equivalence. My confusion is, how are they measuring the mass of the atom? Einstein's equation tells us that inertial mass (how its momentum changes under a force) is determined by the energy content of the system. If that were the case, how would you measure any difference in the mass of the atom, given that the only difference in the two situations is that it contains binding energy instead of exclusively mass?[/QUOTE] I'm not entirely sure what you're asking? But experimentally, if you used a mass spectrometer to measure the mass of a bound nucleus, then you get a result that should be equal to the rest masses of the constituent protons/neutrons minus a 'mass' which corresponds to the binding energy (using E=Mc^2). To measure the mass of the nucleus 'without' binding energy, you'd have to measure the rest mass of an individual proton/neutron and multiply the result by the number of protons/neutrons in the nucleus. Sorry if that's not what you were asking! [QUOTE=Hattiwatti;44908986]I don't understand my physics book. It has loads of small errors and sometimes the answers don't match even though I'm 100% sure I've calculated them right. Doesn't really make learning by yourself easy :/[/QUOTE] Post them here or try to get a later edition of the textbook? The errors normally get ironed out.
Nah I think I figured it out. Either I'm dumb or introductory physics texts handle binding energy in a confusing way.
[QUOTE=JohnnyMo1;44909217]Give an example and we can check.[/QUOTE] Well I had my exam today, was actually fairly easy. But yea the book has lots of small errors, some problems have wrong numbers (fixed by our teacher), wrong units, typos, answers at the back can have a) b) c) while the actual problem only has a) and b) etc... And sadly this is the latest print. And then there's the RCL-circuit problems. In the book and teacher notes it says that with a coil the voltage reaches its maximum before the current does and for a capacitor it's the other way around, but in all the problems the current always reaches its maximum first no matter what components were used. [t]http://4st.me/rV0Vk.png[/t] "Calculate the phase angle" Book and notes say that phase angle φ = dt/T * 360°, or at least I can't find any other way to calculate it. From the graphs time between maximums is 5s and T is 20s, so φ = 5/20 * 360° = 90° and the right answer is 60°
Definitely looks like a 90 degree phase angle, your book may be shit.
That may be the case :P Oh well, got an 8 (grades from 4 to 10) from the course, which is fine by me since it's good enough to not lower my average of 9. Completed it at the last possible moment too, since graduation is on Saturday.
Freezing point of hydrogen is -434.5°F. Freezing point of oxygen is -361.8°F. If water is just one hydrogen atom and two oxygen atoms, why does it freeze at only 32°F when the atoms that make up a water molecule freeze at a significantly lower temperature?
Had an exam about Markov Chains and Stochastic Differential equations, didn't go too well. I hate professors who ask 3 self-contained questions about the entire course.
[QUOTE=cqbcat;44999127]Freezing point of hydrogen is -434.5°F. Freezing point of oxygen is -361.8°F. If water is just one hydrogen atom and two oxygen atoms, why does it freeze at only 32°F when the atoms that make up a water molecule freeze at a significantly lower temperature?[/QUOTE] Condensed matter physics is pretty much all about how condensed matter behaves differently than the sum of its parts. It's not so simple. I believe the answer has to do with water being a polar molecule, meaning it can form stable solids just from electrical interactions. Hydrogen becomes a liquid at such low temperatures because hydrogen can only bond with itself through van der Waals forces. If the temperatures are too high the hydrogen atoms are moving too quickly to bond through such a weak force.
Oh wow, my representation theory course was fun until they introduced the fucking [I]physics[/I] crystals can fuck right off
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