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x(x-y)

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  1. I suspect the problem stems from using 0.5º as [latex]\sigma (\theta)[/latex]. You should be using radians.

     

    Once you've made this mistake once, you tend not to forget it.

     

    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.

  2. 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?

  3. 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.

  4. 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.

  5. 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)!

  6. 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.

  7. 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...

  8. 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".

  9.  

    Meteors compress the air in front of them and that generates a great deal of heat. I forget what they call it, but this can cause them to burst before they hit the ground.

    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.

  10. For some reason, here in the UK anyway, it is shameful not to be able to read or write, but to be mathematically illiterate is fine. It is even joked about.For example, when people know I "do maths", they either chuck hard sums at me or say "I cannot do maths".

     

    Yes, agreed, I can relate to that (being in the UK studying a Physics degree myself). It really does irritate me when I hear people say "I'm terrible at maths" followed by laughter (and most people tend to laugh and agree with them). When has it ever been acceptable to be illiterate? So why should it be acceptable to be innumerate?

     

    Ugh, I could rant all day. Anyway, I guess part of it stems from the fact that people just think of maths as "doing big sums with lots of numbers" - hence the reaction which you stated of people chucking hard sums at you.

  11. C++ is good for beginners. But MATLAB is my current favourite - it's a high level computing language which is very nice to use and very satisfying to program with, also it can be used in conjunction with Simulink to create both simple and advanced models for physics, climate, biology, economics etc... All in all, MATLAB is a good choice for many different people with varying goals (and especially good for physicists and mathematicians).

  12. I agree - however, fortunately, I am fairly competent in mathematics (and I have studied basic quantum mechanics thus far on my degree, but not GR - looking forward to it though). Anyway, if you just keep on practising the maths and build yourself up from the very basics, then you have a fair shot at being able to understand the likes of QM and GR eventually; however, of course it does help to have some sort of lecturer/teacher/tutor.

  13. I hate cheese

    You must leave now.

     

    Anyway, there aren't many foods that I really don't like - I suppose I don't like mushrooms, nor am I keen on duck meat; I like pretty much every other type of meat and fish.

     

    Oh, and I much prefer margarine on toast and in sandwiches than butter.

  14. I was wondering what the implications of the failure of Einstein's original prediction of gravitational waves would be - how far would this invalidate the theory of general relativity? Several professors and researchers at the university I read physics at are heavily involved with international collaborations to detect gravitational waves, and so this lead to me to think of how "bad" it would be for current established physics if it turns out that they do not exist*.

     

    I don't mind if you include mathematics and fairly advanced physics in your replies, as I am currently studying physics at degree level - so I should be able to handle some of the stuff you throw at this thread! Although, I realise that some "physics experts" here have PhDs in different fields of physics so obviously I can only handle so much!

     

     

     

     

    * Indeed, I am aware that they are incredibly difficult to detect (or so it seems) and so even if we do not detect them within the next 100 years that doesn't necessarily mean that they don't exist.

  15. Gravity doesn't actually couple to mass as far as I know, it actually couples to energy and momentum as shown by a rank two stress energy tensor in general relativity; gravity is neither a vector nor a scalar field, rather it is a tensor field. Whereas the Higgs field is a scalar field with a non zero potential at origin state, unlike say the electric field. A scalar field does not generally interact with a tensor field, and so the Higgs field and gravity are not really intertwined inherently.

  16. Indeed - as swansont stated - the dielectric material is an insulator (typically air, glass etc used in between capacitor plates) which will be partially polarised when an electric field is passed through it; a partial dipole is created. If you'd just spent a long time charging up the plates of a capacitor and then someone came along and put an intrinsically polar material (i.e. a conductor) between the plates then, of course, you wouldn't be a happy person.

     

    Note that the capacitance C of a capacitor is given by

     

    [latex]C = \frac{Q}{V}[/latex]

     

    where Q is the charge and V is the potential difference, and one can express the latter as V = Ed, where E is the magnitude of the electric field (strength) and d is the distance between the plates, yielding

     

    [latex]C = \frac{Q}{Ed}[/latex]

     

    Thus, by decreasing E you will act to increase the capacitance C assuming Q remains constant.

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