655 reputation
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location United Kingdom
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visits member for 3 years, 4 months
seen 13 hours ago

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.


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?
Dec
21
comment Proper distance and embedding diagrams?
Thanks. I quite like that picture of the two concentric circles because I can then easily see how it builds into the kind of upside down witch's hat shape of an embedding diagram. But how does he calculate the circumference of the inner circle to be 0.99999999x2pi miles less than the outer circle? Does it involve some nasty integral of the proper distance equation I gave in my question? My definition of "nasty integral" is pretty all encompassing.
Dec
21
comment Proper distance and embedding diagrams?
thought I should point out that at my level nothing is trivial.
Dec
21
comment Proper distance and embedding diagrams?
Thank you. The context of the question was trying to understand a really simple embedding diagram in the form of a picture of a couple of 1 mile apart concentric circles around the Sun at pitt.edu/~jdnorton/teaching/HPS_0410/chapters/…. He says, "for each mile that we come closer to the sun, the circle does not lose 2π miles in circumference; it loses only (0.99999999)x2π miles". How does he work that out from the equation?
Dec
21
comment Proper distance and embedding diagrams?
Thank you. The context of the question was trying to understand a really simple picture of a couple of concentric circles around the Sun at pitt.edu/~jdnorton/teaching/HPS_0410/chapters/…
Dec
20
asked Proper distance and embedding diagrams?
Dec
19
comment Difference between coordinate and proper distance in Schwarzschild geometry
@genneth - thanks for that. I'm assuming there's no problem in measuring r with my ruler in the "flat space" circles at the bottom of the diagram?