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x(x-y)

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About x(x-y)

  • Birthday 02/10/1994

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  • Interests
    Physics - especially quantum mechanics, particle physics, mathematical analysis of waves.

    Mathematics - especially calculus, complex analysis, group theory and vectors.
  • College Major/Degree
    Currently studying Physics at University of Birmingham (UK)
  • Favorite Area of Science
    Physics

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Meson

Meson (3/13)

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  1. I would've just edited the OP, but I can't now - anyway, an update to my system: CASE - Corsair 900D SuperTower Case (to be released next month) RAM - TeamGroup Xtreem LV 16GB DDR3 @ 2133MHz
  2. I see - that makes sense.
  3. Yeah, your answer seems to be correct - looks like an error in your textbook.
  4. Ah, yes, what an amateur mistake. That seems odd though, if I change everything to radians then the errors make sense - however surely if you were using degrees for theta then you'd stick with that convention all the way through your calculation? Well, obviously not I guess - I really should stop using degrees altogether, I work in radians most the time anyway. Thanks.
  5. I have a set of results from an experiment I did recently in the lab where one array of results is for the variable [latex]\theta[/latex]. In order to obtain a graph from which I can analyse the results and obtain a value for the electron mass I must take the cosine of this variable, to give me the following data: [latex]\cos \theta = [0.94, 0.77, 0.50, 0.17, -0.17, -0.50][/latex] Now, the errors on theta are plus/minus 0.5° and so using the standard error formula and assigning a variable y to the cosine of theta such that [latex]y = \cos \theta[/latex], I obtain the error on y (which will be the error on cos theta) as: [latex]\sigma (y) = \frac{\partial y}{\partial \theta} \, \sigma (\theta) = -\sigma (\theta) \sin (\theta)[/latex] But when I plug the values of theta and the error on theta into said error formula I obtain the following errors: [latex]\sigma (\cos \theta) = \pm [0.17, 0.32, 0.43, 0.49, 0.49, 0.43][/latex] As you can see, these errors are pretty large (nearly 300% in some cases) and so I get the feeling I am doing something wrong here. So, am I being stupid and missing something blindingly obvious?
  6. Yep that works - but you can skip a few of those last steps actually and get to the answer easily from [latex]\frac{1}{E_2} - \frac{1}{E_1} = \frac{1}{m_e c^2} \left(1-\cos \theta \right)[/latex] Anyway, thanks for the help.
  7. I see, thanks. So after searching for specific terms I can find the derivations just by searching for "compton scattering" generically. Great.
  8. Could anybody point me in the right direction as to how one would derive the equation below? [latex]E_{\gamma}'=\frac{E_{\gamma}}{1+\frac{E_{\gamma}}{m_0 c^2} \left(1- \cos \theta \right)}[/latex] I read it in my lab manual along with the Compton Wavelength equation (which I know how to derive) and I'm wondering where it comes from. Thanks.
  9. That'll work - I'll try to formulate a plan to study something to do with a simple Chua's circuit, maybe measuring the voltage across the capacitors on the oscilloscope and objserving the lissajous figures to identify bifurcations as the source voltage is increased. Something like that anyway.
  10. I'm just about to buy these components to build my own computer: CASE - Corsair 800D Full Tower Case CPU - i7-3930K @ 3.20GHz (obviously can be OC'd) GPU - GeForce GTX Titan 6GB GDDR5 RAM - Corsair Dominator Platinum 16GB 1866MHz HDD1 - Samsung PRO 840 SSD 256GB HDD2 - Seagate Barracuda 2TB 7200RPM COOLING - XSPC RayStorm D5 EX360 WaterCooling Kit MONITOR - Dell UltraSharp 27" U2713H 2660x1440 MOTHERBOARD - Asus Rampage IV Extreme Intel X79 (S2011) PSU - Corsair Professional Series 860 W '80 Plus Platinum' So, go ahead and post your current (and/or planned system specs)!
  11. I have a diode, an inductor (basically a coil) mounted on a breadboard which is connected to an oscillator from which I can alter the frequency, the dc offset and the source voltage, and it is also connected up to a an Oscilloscope. So I was wondering if anybody could come up with an interesting experiment to perform related to chaos in this electrical circuit and/or encoding/decoding signals. I have already done a simple experiment where I increased the source voltage and measured the points at which bifurcations in the voltage oscillations on the CRO display occurred (period-doubling bifurcations, windows of order etc). One idea I have is to connect another "chaos circuit" part (the inductor and diode) up to the other inductor-diode set up and try to determine whether the signal produced as shown on the CRO signifies true chaotic nature of the diode-inductor system - i.e. observing whether a sine wave output is returned to the screen indicating that the circuit is not exhibiting true chaos. Thanks.
  12. I do find that fascinating though that this bit of "rock" transferred considerably more energy to the atmosphere than the nuclear weapons we designed to explode with huge force. Just imagine if a Tunguska like event just so happened to occur directly over a major city, the consequences would be devastating - it would be tantamount to setting off a hydrogen bomb mid-air over said city. Extraordinary, for all our advances in technology and the supposed anti-missile systems which cover countries like the US - if one of these bits of "rock" penetrated our atmosphere and headed straight for a major city, there'd be nothing we could do to stop it in time. It's also interesting to note the key reason why the energy released from these objects is so high - the kinetic energy (classically) of a body is Ek = 1/2 mv^2, so it is clear to see here that v will have much more of an impact of the kinetic energy of the body than its mass. Suppose, for example, that we were able to accelerate a measly 1kg mass up to the 99% of the speed of light and then get this mass to impact a body - I'm fairly sure that the relativistic kinetic energy is given by [latex]E_k = (\gamma - 1)m_0 c^2[/latex] where gamma is the Lorentz factor and m0 is the rest mass - we are assuming that the mass will remain constant at 1kg, which it will not but just for simplicity we'll make such an assumption. Plugging in the values, we get that [latex]E_k \approx 5.4\times 10^{17}\, J[/latex] which is around 130MT of TNT equivalent, or 2.6x the energy of the most explosive nuclear weapon ever tested which was Tsar Bomba which exploded with an energy of 50MT TNT equivalent. In other words, this simple 1kg mass would be the deadliest, most destructive weapon developed by man if we could achieve that kind of speed - which, by the way, is extremely unlikely and entirely unfeasible as well as it would take a huge amount of energy to accelerate the mass to such a speed anyway! Anyway, that's enough of my ramblings...
  13. Makes sense. @Michel: The flash of light is essentially caused due to a very large amount of energy being transferred to the surroundings in a very short amount of time - due to increased relative surface area. Think about burning pieces of wood - if you set fire to a very large piece of wood (which has a high volume in relation to its surface area) then it will take a long time for it to burn and consequently the flame intensity (brightness) will be relatively low over a small amount of time. However, if you burn very small pieces of wood (which have high surface area to volume ratios) then the brightness of the flame per unit time will be relatively larger than the former scenario. It's the same situation with this meteorite - as it has exploded into smaller fragment, thus it's overall surface area to volume ratio will be larger than when it was a single rock, thus energy is transferred at a faster rate; hence the light "flash".
  14. It is called a bolide, but as far as I'm aware this event was not a bolide - the meteorite did not explode during descent, rather it fragmented and vapourised (hence the vapour trails) with only a few fragments reaching the ground. The "explosion" itself was the shockwave caused by the meteorite breaking the sound barrier as it entered the Earth's atmosphere; note that the shockwave was also due to the loss in kinetic energy of the rock as it entered our atmosphere, it was moving at some 8km/s before entering and slowed right down upon entry, that energy loss has to go somewhere. NB - An example of a Bolide event would the Tunguska Event in 1908, which was several orders of magnitude more powerful than this event. Also, note that the term "bolide" has many definitions - astronomers tend to use it to describe a magnitude -14 or greater event which explodes mid-air.
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