685 reputation
315
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location United Kingdom
age
visits member for 2 years, 10 months
seen Apr 12 at 19:25

Manual worker. Live in England. Formal scientific education "peaked" many years ago when I failed maths and chemistry A-level and scraped a bare pass in physics. Still enjoy pottering about in maths and physics.


Mar
5
comment Still trying to understand gravitational potential and Poisson's equation?
thanks, but this is way over my head. I confess I have no idea what a 3d Dirac delta distribution is.
Mar
5
comment Still trying to understand gravitational potential and Poisson's equation?
Thanks. Bear with me if I do this one step at a time. So$$r^{2}\left(\frac{\partial}{\partial r}\left(\frac{1}{r}\right)\right)=-1$$ $$\frac{\partial}{\partial r}\left(-1\right)=0$$ Therefore $\nabla^{2}\phi=0$ which is what it should be. Finally, did you get rid of the second and third terms of your first equation by assuming $\theta=\phi=0$?
Mar
5
comment Still trying to understand gravitational potential and Poisson's equation?
Thanks. What does IIRC mean? I've just found this in Wikipedia for inside a uniform spherical body$$\phi=\frac{2}{3}\pi G\rho\left(r^{2}-3R^{2}\right)$$which (yippee) I managed to change into your equation. Does all this mean that the equation for the gravitational field $$\mathbf{g}=-\mathbf{\nabla\phi}$$ is only true for points outside of the mass, ie for $r>R$?
Mar
5
accepted Still trying to understand gravitational potential and Poisson's equation?
Mar
5
asked Still trying to understand gravitational potential and Poisson's equation?
Feb
23
comment Why no basis vector in Newtonian gravitational vector field?
@Manishearth - Just showing my colossal ignorance! I'm a self learner and have only recently learned that $y=y\left(x\right)$ means $y$ is a function of $x$ so I was wondering why there's a $\mathbf{g}\left(\mathbf{r}\right)=$ for the first equation but only a $\mathbf{g}=$ for the second.
Feb
23
accepted Why no basis vector in Newtonian gravitational vector field?
Feb
23
comment Why no basis vector in Newtonian gravitational vector field?
@Manishearth Thanks, I didn't know that and it's a lot clearer now.
Feb
23
comment Why no basis vector in Newtonian gravitational vector field?
Also, why aren't there basis vectors in $\mathbf{g}=-\mathbf{\nabla\phi}=-\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right)$, for example $\mathbf{g}=-\mathbf{\nabla\phi}=-\left(\frac{\partial\phi}{\partial x}i,\frac{\partial\phi}{\partial y}j,\frac{\partial\phi}{\partial z}k\right)$. Is that incorrect for some reason?
Feb
23
comment Why no basis vector in Newtonian gravitational vector field?
Thanks. Afraid I'm still a little confused with the maths notation. Can we do this in baby steps? When you say $\phi\left(x,y,z\right)$ I assume you mean $\phi$ is a function of $x,y,z$. But when you say $\nabla\phi=\frac{Gm}{|r|^{3}}\left(x,y,z\right)$ I think you mean $\left(x,y,z\right)$ is a vector, which is equal to the vector $|r|e_{r}$. So $\mathbf{g}=\mathbf{g}\left(\mathbf{r}\right)=-\mathbf{\nabla\phi}$. So why does the author say $\mathbf{g}=-\mathbf{\nabla\phi}$ and not $\mathbf{g}\left(\mathbf{r}\right)=-\mathbf{\nabla\phi}$?
Feb
23
asked Why no basis vector in Newtonian gravitational vector field?
Feb
22
accepted Trying to understand Laplace's equation
Feb
22
comment Trying to understand Laplace's equation
@mtrencseni - Now I think I get it. I was making assumptions that I shouldn't (I was assuming I could ignore y and z and find the Laplacian of -Gm/x). I popped the real phi into the WolframAlpha calculator took second partial derivatives, added them all up and there was zero. Brilliant. Thank you very much.
Feb
22
comment Trying to understand Laplace's equation
@mtrencseni - thank you, but that doesn't answer my question. How/why does Laplacian equal zero when the radius doesn't equal zero. I'm thinking that if I differentiate phi twice wrt r I should be getting zero. Is that correct? How does that give zero? I'm a physics novice so don't worry about making your answer too simple.
Feb
21
asked Trying to understand Laplace's equation
Feb
5
awarded  Tumbleweed
Jan
31
awarded  Teacher
Jan
31
answered Getting started general relativity
Dec
21
comment Proper distance and embedding diagrams?
I actually found a useful reference to this at physics.ucsd.edu/students/courses/winter2011/physics161/…. I managed to input the integral equation into Excel and come up with the same results as the author. I then calculated the difference between coordinate height and proper height of Mount Everest, due to the Earth's gravitational field, is 0.0000062m. You'd hardly notice. Another productive day spent!
Dec
21
accepted Proper distance and embedding diagrams?