A Water clock used in ancient Greek is designed as a closed vessel with a small orifice O.The time is determined according to the level of the water in the vessel. What should the shape of the vessel be for the time scale to be uniform. find mathematical equation governing curve AOB.
question ...
@barrycarter there's always a grey area with homework, which that question obviously is. If I think I can usefully discuss the concepts involved then i will gnerally try. In this case I pointed the OP to the physics they need, but I didn't actually give an answer and I don't disagree with the close decision.
@barrycarter The review queues are rather slow here, giving people who either disagree with the close reasons or don't care about on-/off-topicness plenty of time to answer questions
I've actually got a bad case of fatigue on this. I'm too tired to think and write carefully about it. It's clear that the current policy is too nebulous and not well understood by all active members much less new askers, so I suppose it still needs doing, but ... uhg! — dmckee ♦20 hours ago
The metric components in a two-dimensional spacetime are given in terms of the coordinates $(t, x)$ by$$ds^2 = -\cosh x\,dt^2 + dx^2.$$Consider a particle that is "held in position" at $x = 1$. What is the acceleration of this particle, i.e., if the particle has unit mass, how much force must be ...
@barrycarter again it's grey. My view is that someone who just wants their homework done won't be interested in me explaining the concepts involved so they will consider they haven't got an answer and hopefully not come back.
Anyone who is genuinely interested will pursue the points I make and hopefully learn something.
Let me ask that in a more specific way. Do you believe there are any questions that should be both closed and answered, in the SE specific definition of answer?
But if we leave the question open long term someone is sure to come along and (in good faith) provide exactly the sort of explicit answer the homework cheat is looking for.
In my perfect world high rep users could immediately and unilaterally close a homework question, but could then if they wished add a conceptual answer.
I'm proposing the Axiom of Choicelessness: either you believe a question should be closed, or it should be answered, but you can not hold both beliefs at the same time. Do you agree with this axiom?
@barrycarter in that case I disagree with your axiom. You should not provide a complete and explicit answer, but it's OK to provide an answer that discusses general principles.
@barrycarter yes. Under no circumstances should anyone answer specifically. That just encourages the cheaters and the lazy and pollutes our hallowed halls.
The hard liners would criticise me for even providing a discussion of concepts.
In the water clock example presumably the OP felt my discussion of concepts was enough to help them understand the problem and that's why they accepted it. If so they learned something, which is fine by me.
@JohnRennie Oh, I wasn't talking about specific examples. To me, you close a question because it's not valid and you only answer a question if it is valid. I can't see holding both views at the same time.
originally the Physics SE was intended to be a research community like the Math Overflow but that didn't work out. Instead it is basically a learning resource i.e. physics students (in the loosest sense of *student*) can come here to get help with concepts that puzzle them.
So my view is that if I can leave the OP knowing more about physics than they did before that justifies an answer.
Just doing their homework teaches them nothing of course.
You need to calculate the proper acceleration at the point $x = 1$, and you do this using the geodesic equation:
$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} $$
This is simpler than it looks since the only non-zero component of the fou...
Is this an okay proof that : If $f:A \to B$ is injective, there must exist a left inverse $g:B \to A$ such that $g \circ f: id_A$. Suppose f is injective. Define a function $g$ where for every $b \in B$ $g(b) = a$. thus for every $a$, $g\circ f (a) = a$ and so $g\circ f(A) = id_A$
@JohnRennie: Physical intuition: A geodesic is a free-fall motion. The acceleration an object in free fall feels is zero (that's what "free fall" means, after all) - the geodesic equation precisely determines the curves along which proper acceleration is zero.
For the question you linked, you have to take the worldline described by the question (particle with fixed spatial position in those coordinates) and compute its proper acceleration.
@JohnRennie "You" don't exist when talking about proper acceleration. Proper acceleration is an invariant (proper vector), you don't need any comparisons to compute it.
Ok but in this case we have an observer held stationary at x=0 and a freely moving object momentarily stationary at $x=0$. My contention is that the equation I gave does give you a vector whose norm is the proper acceleration of the stationary observer.
The gravitational force on your body, called your weight, pushes you down onto the floor.
$$W=mg$$
So, what is the weight equation through general relativity?
yeah I think short term memory aka working memory lets you score higher on iq tests. it's related to 'fluid' intelligence directly. thats what I got off googling it anyway
let $f$ be the map $A \to B$ and $g$ be the map $B \to A$. The proof to this claim: $f$ is bijective only when there exists an inverse $g$ such that $g \circ f = id_A$ and $f \circ g = id_B$. Suppose for all $b \in B$ the image through $g$ is $g(b) = a$ and for all $a \in A$ the image through $f$ is $f(a) = b$. then for all $a \in A$ $g \circ f (a) = id_a$, so $g \circ f(A) = id_A$. For all $b \in B$ $f \circ g (b) = id_b$ and so $f \circ g(B) = id_B$. Is this right?
so based on that definition, there has to be a $g$ where the image of $B$ through $g$ is equal to every single $a$ that corresponds to $b$. It's just given I thought
@barrycarter Actually, you have to say that there are roots that can't be expressed instead of saying "they" can't be expressed. Many fifth-order and higher roots can still be expressed, there's just no general formula for them
You need to calculate the proper acceleration at the point $x = 1$, and you do this using the geodesic equation:
$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} $$
This is simpler than it looks since the only non-zero component of the fou...
The gravitational force on your body, called your weight, pushes you down onto the floor.
$$W=mg$$
So, what is the weight equation through general relativity?
