Recent content by Paulpaulpa

The ##I_i## are the intensity of the rays, in other words energy per surface units per radians by seconds. The d##\Omega## are the solid angles The equation p75 isis what I don't understand. I suppose that each side represent the energy going and out of the surface dS but I don't understand...

Since ##\vec{F}_{total}.\vec{e}_y = 0##, we have $$\vec{F}_{N} = mg \vec{e}_y$$ $$\vec{F}_{Friction} = - \mu mg \vec{e_x}$$ and $$\vec{F}_{drag} = - C v^2 \vec{e}_x$$ As you did for your answer you then take ##\vec{F}_{total}.\vec{e}_x = 0## adn resolve the equation for ##\mu##.

You probably made an error with signs. Remember that ##F_{resistance} \propto - \vec{e_x}## while ##F_{driving} \propto \vec{e_x}##.

Thank you very much this is much more clear. I always noted total derivative ##d \mathcal{L}## so the ##\partial _{\nu}\mathcal{L}## confused me a lot. Your explanation of Noether's Current is more understandable than the one in my lecture.

Thanks for you answer. The first term is zero because of the equation of motion and the fact that ##\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) = 0 ## since ##\mathcal{L}## do not depend on x. Is that right ? But why is ##\partial_\mu \mathcal{L}## different...

In Sydney Coleman Lectures on Quantum field Theory (p48), he finds : $$D\mathcal{L} = e^{\mu} \partial _{\mu} \mathcal{L}$$ My calulation, with ##\phi## my field and the variation of the field under space time tranlation ##D\phi = e^{\mu} \frac{\partial \phi}{\partial x^{\mu}}## ...

Hello guys, I am a 22 years old french physics student. I am doing a master in Theoretical Physics and want to have a PHD. I love fundamental physics and this is what I want to do for the rest of my life. Because I am an idiot, I never really studied hard or listened during lecture so I have a...