Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Favorites infavorites:mine
infavorites:1234
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with Search options user 73067

Scalar and vector potentials in electromagnetism. The scalar potential is potential energy per unit charge. For potential energy, use the potential-energy tag.

1
vote
0answers
Wikipedia lists the gravitational potential as $$V(\vec{x})=-\int_{\mathbb{R}^3} \frac{\rho(\vec{r})G}{|\vec{x}-\vec{r}|} dv(\vec{r}) $$ with $dv(\vec{r})$ the volume element, G the gravitational …
asked Oct 9 '17 by Harald
1
vote
1answer
Let $\Phi(0, t)$ be the gravitational potential at the origin and $t$ the clock ticking at infinity, i.e. not influenced by any gravitation. Let a mass $m$ approach the origin with constant velocity … gravitational influence travels not with $c$ in empty space, but with a $c_{\Phi}$ that depends on the gravitational potential or (even worse) the metric along the path which, for the observer at …
asked Nov 25 '18 by Harald
1
vote
2answers
Suppose a cylinder of length $\ell$, radius $R$ with a bore hole through its long axis of radius $r$. How can the gravitational potential (not the gravitational field) inside the cylinder be derived … . I thought of using Gauss's law of gravity, but it only talks about the gravitational field and if $\ell$ is large enough, the field goes to zero. But this does not mean the potential is zero, its just constant. Is there a simple formula for it? …
asked Jan 20 '18 by Harald
1
vote
$, an outer radius of $R$ and a bore hole radius $r$. For a start lets concentrate on the gravitational potential for a point $\vec{x}$ in the middle of the bore hole. The volume integral to be solved … $ and $L=1000m$ should have generate a potential along the cylinder axis like this: One can imagine, that the longer the cylinder gets, the closer comes the potential to a constant along the axis of the cylinder. …
answered Jan 28 '18 by Harald