Recent content by Thadriel

Anyway, I appreciate the responses. These will help me. But I have to spend some time to really think about what the math means.

So just divide F by m then. In easy math, g is the acceleration, which is ##F/m_1##, correct? But why is it ##F/m_1##? Is it because ##m_1## is negligible compared to the big mass? I guess if we divided by M, should you not get the same acceleration? I mean you can choose either one as the rest...

So you are integrating in three dimensions, correct? That is what the ##\mathbb{R}^3## means, right? And will you please explain why you have a gamma there? This isn't the same gamma as the Lorentz factor, is it?

Thanks. I'll think on this a couple of days. Since you brought up ##\vec{r}##, I'm wondering if this will look something like Coulomb's law: ##F = \frac{q}{4πε_0} \sum_{i=1}^N q_i \frac{r - r_i}{|r-r_i|^3}## which honestly is stretching my math. Am I on the right track there, that I'd need...

$$F=G\frac{m_1 m_2}{r^2}$$ is presumably for point masses. If the masses weren't a point masses, then wouldn't you need a version of the formula that sums up the gravity for each infinitesimal portion of the masses? And for my money, "summing up" in physics is integrals, right? So would it be...

:doh::headbang: Probably the closest you can get.

Convenience is probably the best answer I could have imagined. Like choosing your axes so that they line up with the acceleration in an inclined plane problem.

Why, may I ask, do you prefer this sign convention over the other one?

Yes I got plenty of great answers.

All good friendo. I expect that sort of attitude towards those weirdos who think they’re here to disprove all of science. If I came across that way, that was unintentional.

Yes I learned a great deal from that incredibly educational post, and even more so from the further mockery of a non-physicist daring to not already have a complete physics education.

But would that be stable, or a quasi-star so ridiculously massive that it just takes a long time to collapse? (according to a distant observer) The Wikipedia article says when they'd run out of fuel they'd "dissipate, leaving behind the intermediate mass black hole." I do not understand what...

So what I gather from this is that if a star begins to collapse into a black hole (or neutron star), there is simply no going back?

I’m sure that when a star is in the process of becoming a black hole, there must therefore be one inside it at some point during the process (correct me if I’m wrong on that). But if so, how long does that take? Could there exist a supergiant star that has a black hole inside it for a long...

I think maybe what that person is getting at is that when discussing quantum mechanics, it is not good to use every day experience as the baseline for “normal.” But apologies is that wasn’t the point. Obviously for materials that glow in the dark, glowing is quite normal behavior.