Activity for celtschk
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #292713 | Initial revision | — | about 2 months ago |
| Answer | — |
A: Should (lone) black holes emit gravitational waves? To start with, black holes in the GR model don't radiate. At all. Even Hawking radiation isn't predicted by GR. Rather, it is predicted by doing quantum mechanics in curved spacetime. That is, you take the GR result as background to your quantum calculations (not unlike how you calculate the atom sta... (more) |
— | about 2 months ago |
| Edit | Post #287186 | Initial revision | — | about 2 years ago |
| Answer | — |
A: maxwell equation in 1d Yes, that would the obvious interpretation of that equation in one dimension. Note also that in that case, the divergence also reduces to the ordinary derivative. In other words, in one dimension, the electric field is constant wherever there is no charge, and if we additionally demand that the el... (more) |
— | about 2 years ago |
| Comment | Post #286310 |
If the lighthouse bean fails to regularly illuminate a ship that's still in safe waters, it is either badly built or out of order. The ship's lights don't need to illuminate the submarine; it only has to reach the periscope. If the submarine doesn't have the periscope sticking out of the water, the p... (more) |
— | over 2 years ago |
| Comment | Post #285628 |
Judging from [this Wikipedia article](https://en.wikipedia.org/wiki/Planet_Nine) things are not quite as clear-cut as you claim. (more) |
— | almost 3 years ago |
| Comment | Post #285586 |
Or we could have defined the spring constant the other way round, so that you'd get the displacement as product of the constant and the applied force. (more) |
— | almost 3 years ago |
| Comment | Post #285554 |
Another place where the relativistic mass concept can cause confusion is that people are tempted to insert it into Newton's equation $F=ma$. This does not work in general, but it *does* work if the force is orthogonal to the velocity. In general, the acceleration is not even in the direction of the f... (more) |
— | almost 3 years ago |
| Comment | Post #285323 |
@#53922k The formula for kinetic energy you used is only for non-relativistic particles. It definitely cannot be applied to photons. Photons do have kinetic energy (indeed, all of their energy is kinetic energy), but it isn't given by that formula. (more) |
— | almost 3 years ago |
| Edit | Post #285321 |
Post edited: Escaped dollar signs to prevent them being interpreted as math delimiters |
— | almost 3 years ago |
| Suggested Edit | Post #285321 |
Suggested edit: Escaped dollar signs to prevent them being interpreted as math delimiters (more) |
helpful | almost 3 years ago |
| Edit | Post #285239 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: How can the kinetic energy equation be intuitively understood? Change in momentum is force times time. On the other hand, change in kinetic energy is force times distance (more accurately, the component of the force along the movement times distance; a force perpendicular to the movement doesn't change the energy). Now imagine you apply a force for a given ti... (more) |
— | almost 3 years ago |
| Comment | Post #285145 |
@#53922 Exactly, space and time (times c) are different coordinates, and as vectors you simply have separate components. The minus sign in the scalar product mathematically comes from the Minkowski metric tensor, which tells how to form the scalar product. (I couldn't answer earlier; my internet acce... (more) |
— | almost 3 years ago |
| Edit | Post #285145 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: Clear up confusion on Minkowski signature As you probably know, one of the postulates Einstein's Special Relativity is based on is that all observers see light in vacuum go at the same speed $c$. Now consider a lamp at rest relative to John being switched on. The light is going through vacuum from then on, until after some time it arrives... (more) |
— | almost 3 years ago |
| Comment | Post #285133 |
Is it specifically on Minkowski's thoughts (in which case I can't answer, as I didn't read Minkowski's original work), or is it just about how you can derive it (independent of whether it's how Minkowski did it)? (more) |
— | almost 3 years ago |
| Comment | Post #285133 |
My question was an “or” question. So which of the two alternatives is it? (more) |
— | almost 3 years ago |
| Comment | Post #285133 |
I'm not entirely sure what your question is: Is it about why the time dimension is handled differently from the space dimensions, or is it about the difference between putting the $-1$ in the metric and putting a factor $i$ into the time coordinate? (more) |
— | almost 3 years ago |
| Comment | Post #284645 |
No problem. Maybe I should have been more explicit that I was talking about what I see *in the video.* That would probably have avoided the misunderstanding. (more) |
— | about 3 years ago |
| Comment | Post #284645 |
Where do you read that statement? I only see: “*Entropy* can't be destroyed but it can be created.” And while for energy that statement would clearly be false, for entropy it's true. Except in equilibrium, of course, because there the entropy is already maximal. But then, if the world as a whole were... (more) |
— | about 3 years ago |
| Comment | Post #284434 |
BTW, I now notice that you incorrectly inserted. If you insert $t=5\\,\mathrm s$ into $ut$ you do *not* get $5u$, but $u\cdot 5\\,\mathrm s$. Also, $t^2$ should become $(5\\,\mathrm s)^2$. As a result, in your initial equation, $S$ equals a length, while in your final equation, $S$ equals a velocity.... (more) |
— | about 3 years ago |
| Edit | Post #284448 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Should I always write units in equation no matter if it looks like variable? In short: Yes. And there are standard ways to distinguish variables from units. Let me explain in detail. In physics, we deal with physical properties of objects and systems. Those quantities can be split down into quantites, that is, properties that can be quantified. To quantify basically means ... (more) |
— | about 3 years ago |
| Edit | Post #283921 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Find equation of motion using. Lagrangian given equation is $L'=\frac{m}{2}(a\dot{x}^2+2b\dot{x}{y}+c\dot{y}^2)-\frac{k}{2}(ax^2+2bxy+cy^2)$ Your mistake is that you did a second derivative of $L$, taking the derivative according to both degrees of freedom together. Instead you need to make a separate equation for each degree of freedom. Since you have two degrees of freedom ($x,y$), you get two equations: \begin{align} \frac{\math... (more) |
— | about 3 years ago |
| Comment | Post #283695 |
@#53922 As I wrote, $\delta q$ is arbitrary (within limits). For the integral to be zero for *arbitrary* $\delta q$, the only possibility is for the prefactor to be zero. As an analogy, if you know that the equation $ax+by=0$ holds for *arbitrary* $x$ and $y$, then you know that both $a$ and $b$ must... (more) |
— | about 3 years ago |
| Edit | Post #283695 |
Post edited: Fixed a misleading formulation |
— | about 3 years ago |
| Edit | Post #283695 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Prove differential form of Lagrangian Let's first remember how you check that you are at an extremum (minimum, maximum, saddle point) of a normal differentiable function. You do so by checking that the first derivative is zero (plus further tests that are irrelevant for the current question). Now what does the derivative being zero me... (more) |
— | about 3 years ago |
| Comment | Post #283410 |
The reason why I think it doesn't belong to the Problems category is that it doesn't contain your attempt to solve it, but a question about a solution given in a text book. But then, my understanding on what this category is for might be wrong, which is why I only commented with a question. (more) |
— | over 3 years ago |
| Comment | Post #283410 |
Doesn't this question actually belong on Q&A? (more) |
— | over 3 years ago |
| Comment | Post #283420 |
… in your next line (the last one of calculating $F$), the result would be $9506\\,\mathrm{N\\,m^{-3}\cdot V}$. (more) |
— | over 3 years ago |
| Comment | Post #283420 |
However you did it a bit wrong: $V$ is a quantity, therefore it also includes the units (otherwise your result could not be $V = 20.6185\\,\rm m^3$). Remember, units are not something you tug on, it is an integral part of your quantity (you don't have a length of 5, you have a length of e.g. 5m, whic... (more) |
— | over 3 years ago |
| Comment | Post #283420 |
You've got a number of intermediate steps with numbers but without *any* units (and the units don't match up if we assume true unit-less numbers). It is important to *always* take the units with the numbers, or else your intermediate steps are essentially meaningless (and easy to slip in unit mistake... (more) |
— | over 3 years ago |
| Comment | Post #283393 |
By solving the differential equation. How to do that is *highly* dependent on the exact force involved. Indeed, the main topic of a classical physics lecture is how to find and solve the equation of motion of physical systems. (more) |
— | over 3 years ago |
| Comment | Post #283392 |
Sometimes you'll walk at constant speed, sometimes you'll accelerate. And sometimes you'll do both at the same time, namely when turning at constant speed. (more) |
— | over 3 years ago |
| Comment | Post #283395 |
@#54107 While you cannot make photons go faster or slower, you *can* change their direction. A change of direction is also an acceleration. Or said differently: While the *speed* of a photon is the same in all inertial frames, the *velocity* is not. And acceleration is a change in velocity, not neces... (more) |
— | over 3 years ago |
| Edit | Post #283396 |
Post edited: Fixed a typo |
— | over 3 years ago |
| Edit | Post #283396 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Why we can't find a particle accelerating unless there's some other particle accelerating somewhere else? If you are accelerating while running on Earth, actually you are also accelerating Earth in the opposite direction. However for a given force, the acceleration is inversely proportional to mass, therefore when some $60\\,\rm kg$ person accelerates by, say, $1\\,\rm m/s^2$, then Earth, which has a mas... (more) |
— | over 3 years ago |
| Edit | Post #283393 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283393 | Initial revision | — | over 3 years ago |
| Answer | — |
A: How to find position of a particle at a time given a position dependent force In general, the only way to do it is to solve the equation of motion. In simple cases, that can be done analytically (that is, you can find an explicit formula, but in most cases (outside problems given to students) you have to either make approximations (that is, essentially find a sufficiently clos... (more) |
— | over 3 years ago |
| Comment | Post #283304 |
The Photography & Video community has a Gear Recommendation tab. That seems to be a quite similar to the book suggestions idea. So any experience with that one should be valuable input on whether if could work on Physics.
I'm not active on that site, so I can't tell how well it works. The fact tha... (more) |
— | over 3 years ago |
| Comment | Post #283009 |
What exactly do you mean with “the time the particle spends in the magnetic field”? Since the magnetic field of the wire is non-zero in the entirety of space (except inside the wire, but the particle won't go there anyway), the time the particle spends inside the field is infinite. Do you have a spec... (more) |
— | over 3 years ago |
| Comment | Post #283251 |
If the initial acceleration is zero, then the initial force is zero, and therefore the particle has infinite distance from the central mass (or in other words, there does not exist any place in space where the acceleration is zero — unless we take other gravitating bodies into account). More general... (more) |
— | over 3 years ago |
| Edit | Post #283215 | Initial revision | — | over 3 years ago |
| Answer | — |
A: What is the meaning that the universe is flat? To understand this, it helps to look at a dimension less. Imagine, you're having a dispute with a flat-earther about whether the earth is a sphere or a flat plane. Moreover, there's a third person who holds the position that the earth is actually shaped like a saddle. As an additional problem, ... (more) |
— | over 3 years ago |
| Edit | Post #281722 | Initial revision | — | over 3 years ago |