Recent content by Rick16

Thank you, that is an interesting trick with the ratios. It is not likely that I would have thought of it. I suppose you know how to do it? Because I don't.

This question does not seem to evoke a lot of interest. Could somebody just tell me if it is legitimate to write ##\frac{d}{d\vec v}(\vec v \cdot \vec v)=2\vec v##? I.e. I wonder if I can take the derivative directly with respect to the vector as a whole or if I must take the derivative...

I have begun reading Landau's Mechanics. In chapter 1, §3 he writes ##\frac {\partial L} {\partial \vec v}=const.##, where ##L## is a function of ##v^2##: ##L=L(v^2)##. He then writes that from this it follows that ##\vec v=const.## I want to try to show formally that v is constant, but I am...

Thank you. This is it. It is not even new to me. I read about it probably more than once, but I had completely forgotten about it. This shows what happens when you just read physics texts without doing problems.

One more comment to show where exactly my problem lies: Potential energy is defined at a specific position. A System does not have a specific position. How then can I define/understand what the potential energy of a system would be?

Here is the solution: The kinetic and potential energies of the system are $$U_k=2\cdot\frac{1}{2}mv^2=mv^2,~~~U_p=-\frac{Gm^2}{2r}.$$ To show how these are related, apply Newton's second law to the motion of one of the particles: $$F=ma\Rightarrow \frac{Gm^2}{(2r)^2}=m\frac{v^2}{r}.$$ Multiply...

Here is problem 1.55 (a) from Schroeder's Thermal Physics: Consider a system of just two particles, with identical masses, orbiting in circles about their center of mass. Show that the gravitational potential energy of this system is -2 times the total kinetic energy. I did not know how to...

I made the substitution because without it my integral looked like this: ##\int d(\gamma v^x)##, which I found was a very odd looking integral. If instead I write it as ##\int \frac {d}{dt}(\gamma v^x)\,dt##, it does not look odd anymore and it all makes sense. Problem solved, thank you very much.

I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get...

I just came across a post in another thread, “Some thoughts about self-education“, where gleem wrote the following: “Sometimes, after many responses, the person may give a courteous thank you, but leave us wondering if our help was understood. One thing that I thought might be of value is for...

I cannot believe how simple this turned out to be. A long time ago, I had decided that Δm was the mass of the expelled gases and I stubbornly stuck with this misconception. That's why I always wanted to write m - Δm. Writing m + Δm seemed nonsensical to me, because in my mind this meant adding...

Thank you. This is a very instructive way to look at it.

I see. I have to be clear about what ##\Delta x## actually represents, whether it represents ##x_{final}-x_{initial}## or ##x_{initial}-x_{final}##. Everything else should then fall into place automatically. I had a rather vague idea of ##\Delta m## as simply the mass of the exhaust gases that...

When I wrote ##m-\Delta m## for the change of the mass of the rocket, I did not see this as ##m -(m_{initial}-m_{final})##. I rather considered ##\Delta m## the mass of the exhaust gases, which is positive, and I simply subtracted this positive mass from the total initial mass of the rocket. I...

There is one more complication. I learned the foundations of classical mechanics with John Taylor's Classical Mechanics, and I have now traced the origin of my misunderstanding about deltas and differentials back to this book, specifically to footnote 4 on page 58, where Taylor writes: As...