Recent content by Alpha&Omega

Would it be an integral of an expression for the force with respect to time? In my book it has two expressions for relativistic force. One is F=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}ma if it's perpendicular to the x-axis and the other is F=\frac{1}{(1-\frac{v^2}{c^2})^\frac{3}{2}}ma if the...

The momentum divided by c? So if momentum is MeV/c then in this equation we have MeV/c2 which is the mass! (I know it would have saved trees, but for some reason I decided to do it this long and complicated way =D). So I have to work from this equation to the equation posted in the...

magic! This may sound weird, but how did you know to use \frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}?

Ah, darn. Let me see: \frac{{\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}{c}}{\sqrt{1-\frac{(\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}})^2}{c^2}}} \frac{{\frac{c^2}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}}{\sqrt{1-\frac{(\frac{c^2}{{1+\frac{m_0^2c^2}{q^2E^2t^2}}})}{c^2}}}...

I felt insecure. XD Do you mean as a physical quantity? I would just do this: \frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} \sqrt{\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}} \sqrt{\frac{1}{1-\frac{v^2}{c^2}}-1 Is this what you meant? =S

1. Consider the decay of particle A with rest mass M_{0A} into two particles, labelled particle 1 and particle 2. The energy of particle 1 is denoted by E_1 and the rest mass by m_{01}, similarly for particle 2 the energy is E_2 and the rest mass m_{02} . i). in the rest frame of particle A...

1. A particle of charge q and rest mass m_0 is accelerated from rest at t=0 in a uniform electric field of magnitude E. All other forces are negligible in comparison to the electric field force. Show that after a time t the relativistic expression for the speed of the particle is...

Oh, I thought this was an actual physics term =S If m< 1.67E-27 it would imply that the least amount of energy that both the electron and proton have is less than the rest mass energy of the proton. This can't happen!

YAY! BYW? m=\frac{1}{c^2}\sqrt{E^2-(pc)^2} m=\frac{1}{c^2}\sqrt{(1.67 \times 10^{-27}c^2+6.63 \times 10^{-19}c)^2-(6.63 \times 10^{-19}c)^2} m=3.18 \times 10^{-27}kg (sorry about the units, I worked out everything in kg =S).

So the CM energy momentum 4-vector of the system is \rho_{CM}=(1.67 \times 10^{-27}c^2+6.63 \times 10^{-19}c, 6.63 \tims 10^{-19},0,0) (it's probably simpler using it in this form). That's a little weird! Slowly but surely I'm starting to get a better grasp of this question. I meant...

Okay, so the CM energy-momentum 4-vector is (E, \rho_x, \rho_y, \rho_z). Since the proton is stationary relative to the electron, it's CM energy-momentum 4-vector is (1.67 \times 10^{-27}c^2, 0,0,0). In the case of the electron, i'll assume it's traveling along the x-axis towards the...

Right, I have a few questions. Firstly, what is a "CM energy-momentum 4 vector"? Does this have 4 dimensions so I should be including time as well? Secondly, is there any reason why it's "<E,E>" in the invariant mass equation? Here's my go at finding the invariant mass...

Ah, I understand what you mean. For part ii) you can use E=\rho c. Therefore \rho=\frac{E}{c} . There's also the equation for De Broglie wavelength (it won't let me put it here for some reason). Hence if you make the energy larger the wavelength of the electron decreases. A smaller...

Define f(x) such thatf(x)=x^3-2. f(x) is continuous because it is a polynomial (you can prove this if you like). f(1)=1-2=-1<0 f(2)=8-2=6>0 By the Intermediate value Theorem, \exists \ x_0 \in \mathbb{R} \ s.t \ f(x_0)=0. Hence x_0^3-2=0 so x_0^3=2.

Protons and neutrons have a diameter of around 10^{-15}m. Despite this small size, physicists were able first to probe inside these particles in the 1970s, by scattering high energy electrons off them. i). explain briefly why it is important that the electrons have high energy. Since...