Recent content by Rick16

I understand what you mean, and yet I wanted to check the validity of Taylor's result using the formal differentiation process with the given variables, i.e. the traditional way: I first take the derivative of ##f## with respect to ##\alpha##: ##\frac{\partial f(y+\alpha\eta)}{\partial...

I withdraw the last part of my last post. I guess technically it should be ##\frac{\partial f(y+\alpha\eta)}{\partial y}=\frac{\partial f(y+\alpha\eta)}{\partial (y+\alpha\eta)}\frac{\partial (y+\alpha\eta)}{\partial y}##

I understand that if you have a function ##f## and you take its derivative with respect to whatever you want, you get ##f'##. If, however, you have the function ##f(x)## and you want to get ##f'(x)##, you have to take its derivative with respect to ##x##: ##\frac{df(x)}{dx}=f'(x)##. Derivatives...

I have read your insight. I don't see an exact parallel to how Taylor uses ##y## in his text, except for the ambiguity in ##y##. But as I see it, this ambiguity comes from the fact that Taylor uses ##y## in a rather "cavalier" way. Office_Shredder writes "this is very abused notation". So this...

I see. So ##\frac{\partial f}{\partial y}## actually means ##\frac{\partial f}{\partial (y+\alpha\eta)}## here. But this only works because ##y## is linear in the function argument. If the function were ##f(y^2+\alpha\eta)##, then ##\frac {\partial f}{\partial y}\neq\frac{\partial...

The problem with this is that ##y## does not depend on ##\alpha## and ##\frac {\partial y} {\partial \alpha}\neq \eta##.

Here are two examples: The first one is from Taylor's Mechanics, section 6.2: Taylor takes the function ##f(y+\alpha\eta, y' + \alpha\eta', x)## and differentiates it with respect to ##\alpha##. In the expression for the function, ##y## and ##\eta## depend on ##x##, ##\alpha## is an...

Thanks a lot! I constantly run into problems with minus signs. I think they make up 98% of all the mistakes that I make. It's very frustrating. And yes, the text says "the work performed on the rod". Sorry about the bad quality of the scan. I scanned the page several times and this is the best...

This is problem 3 in section 2.3.4 from Conquering the Physics GRE by Kahn and Anderson: And here is the solution from the book: The point of confusion for me is that they use the work-energy theorem in the form W = Tinitial - Tfinal, instead of the other way around. If I were to do this...

Are there currently professional physicists who work on Einstein’s quest to unify the gravitational field and the electromagnetic field (as classical fields), or has this idea been completely abandoned? Is it simply a hopeless endeavour?

I want to come back to my question about the meaning behind tensor contractions. I just finished Susskind’s theoretical minimum volume 4. On page 325 he writes: “I don’t know any particular physical significance or geometric significance to the Ricci tensor or the curvature scalar.” This then...

Thank you very much. This is basically the same approach as in #45. If I understand correctly, ##dm## in #45 corresponds to your ##dm_e## and ##dM## in #45 corresponds to your ##\dot{m}dt##. What I find confusing in your equation is the term ##m(t)## in ##[m(t)+\dot{m}dt]v(t)##. Shouldn’t this...

This is pretty great. At first sight your solution seems to be the same as Taylor’s, with the only difference that instead of writing ##dm## for the change in the rocket mass and ##-dm## for the change in the mass of the ejecta, you write ##dM## for the rocket and ##dm## for the ejecta, where...

Finally somebody understands my frustration. Thank you!

It means the same for me, but when I write ##m-\Delta m## and the result is less than the initial value, then ##\Delta m## is positive. If ##\Delta m## were negative, then I would subtract a negative value, which would mean that the result would be bigger than the initial value.