Recent content by Carolina Joe

Ah, so the triangle-like quality of that expression is useful, but in an area of physics I've never studied. My texts are undergraduate level, so I guess if you're looking at graduate textbooks they go further into special realativity using Minkowski spacetime? Is that just a framework for...

Energy "Triangle" It seems to me that the total energy of a particle forms a kind of right triangle. E^2 = (m_0 c^2)^2 + (pc)^2 Where the total energy is the hypotenuse, and it's square is the sum of the other 2 sides, each squared. Does anyone know of any physical significance to...

Okay, I see that we're starting from rest, but just from a calculus standpoint, I have trouble seeing how to treat these limits -- especially if this hadn't been a physics problem, but just a calculus problem. When s=s, mv=mv seems to be true no matter what! I sense that I'm just...

I asked this in another thread, but I think this forum might be a better place for it (not trying to spam the same question). When deriving the formula for relativistic kinetic energy, we start with KE = \int_{0}^{s} \frac{d(mv)}{dt} ds = \int_{0}^{mv} v d(mv) So I figure that since v =...

What I'm asking is, given that we've rearranged the differentials, what rule of calculus is being used to determine the new limits of integration? When we change from \int_{0}^{s} \frac{d(mv)}{dt} ds to \int_{0}^{mv} v d(mv) we're changing the limits of integration, which seemed fairly...

Thanks! Thanks for the reply, Whozum. So you're using the chain rule to rearrange the differentials, with the goal of getting v by itself, and at that point the only differential left standing is d(mv)? I guess I'm wondering what the new limits of integration would be in a more complicated...

Hi Everyone, I'm reading in my modern physics book about relativistic kinetic energy. I'm a little confused about what rule of calculus allows the following statement: KE = \int_{0}^{s} \frac{d(mv)}{dt} ds = \int_{0}^{mv} v d(mv) I see that they must be saying KE = \int_{0}^{s}...