Recent content by vladimir69

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    How Does Photon Energy Relate to Mass Loss in Particle Decay?

    I thought this question must use relativistic principles since it was asked in the special relativity chapter. The solution does it as \mathbf{P_{2}}=\mathbf{P_{1}}-\mathbf{K} \mathbf{P_{2}}^2=\mathbf{P_{1}}^2-2\mathbf{KP_{1}}...
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    How Does Photon Energy Relate to Mass Loss in Particle Decay?

    Homework Statement Set the speed of light c=1. A particle of rest mass m_{0} decays at rest into a photon and loses rest mass \delta in the bargain. Show that the photon energy is \omega=\delta(1- \frac{\delta}{2m_{0}}) in the particle's rest frame before the decay. Homework Equations...
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    2 spheres connected by wire, find tension

    Thanks for clearing that up, the bit at the end was what I was missing - the ceiling is pulling up with the same force.
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    2 spheres connected by wire, find tension

    OK thanks for that. I suppose one way of looking at it would be to consider the forces on one of the spheres, there is only the electric force and the tension, for the sphere to remain where it is these forces would have to be equal. How about the case when the force is different on each...
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    2 spheres connected by wire, find tension

    Homework Statement 2 conducting spheres of radius r=0.5cm are connected by a long thin piece of conducting wire of length L=2metres. A charge of 60 micro Coulombs is put onto one of the spheres, find the tension T in the wire. Assume the charge is evenly distributed along the surface of the...
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    MSc Preliminary at Monash University: Academic Guidance & Career Help

    OK Thanks for the tips I will take what you said into consideration. Even though a job in academia would be nice, a math job of any sort would be nice. I was thinking that after 3 or 4 years potential employers will think I am no longer interested in maths and physics so by doing some award...
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    MSc Preliminary at Monash University: Academic Guidance & Career Help

    I have completed a physics open yale course but that isn't official. Like I could just say I have watched every youtube clip on every subject in maths and physics but doesn't necessarily show how well I understood it and whether I actually did complete it or not. When I do an award course I...
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    MSc Preliminary at Monash University: Academic Guidance & Career Help

    Hello, This thread is probably half career and half academic guidance I am seeking. So here is the trouble: I completed a Bachelor of Science in 2006 in maths in Australia at UNSW, and since then have been helping out with the family business, which is totally unrelated to math and do not...
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    Electric dipole in an electric field

    Ok I see now, thanks for your help.
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    Electric dipole in an electric field

    I thought the units were ok in both the equations
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    Electric dipole in an electric field

    Homework Statement An electric dipole (something that has charge +q on one end and charge -q on the other end separated by a distance 2a) is in a uniform horizontal electric field of magnitude E. Initially the electric dipole is aligned horizontally until it is displaced slightly by an angle...
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    Electric field on a piece of wire

    Ahh damn, that was careless. I used \phi because it helps with getting the right direction for the unit vector, but screwed up the differential unfortunately. (actually I did a similar error the second time aswell) Thanks for the help.
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    Electric field on a piece of wire

    Homework Statement A thin piece of wire, with charge per unit length \lambda lies on the x-axis and extends from x=-L/2 to x=+L/2 so that it has a length L. Show that the electric field (at the point (x,y) = (0,D)) in the +y-direction has magnitude given by 2\frac{\lambda...
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    Electric field of semi circular piece of wire

    OK thanks for that. Glad I am not the only one having a bit of trouble with the unit vector.
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    Electric field of semi circular piece of wire

    ok here is what i managed to think of \mathbf{E}=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}d\mathbf{E} =\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} k_{e}\frac{\lambda }{R}\left( \begin{array}{cc} \cos\theta \\ \sin\theta \end{array} \right)d\theta=-k_{e}\frac{Q...
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