Activity for celtschk
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| 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) |
— | about 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 years ago |
| Comment | Post #285133 |
My question was an “or” question. So which of the two alternatives is it? (more) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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) |
— | over 2 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 2 years ago |
| Comment | Post #283410 |
Doesn't this question actually belong on Q&A? (more) |
— | over 2 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 2 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 2 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 2 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 2 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 2 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 2 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 2 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 2 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 2 years ago |
| Comment | Post #280385 |
I think that “seconds and light seconds” refers to convention (T), but what I actually proposed was convention (M), where you'd use seconds and meters just as in non-relativistic physics, and the light is purely in the metric. BTW, if you set all three of $\hbar$, $c$ and $G$ to $1$, what you actual... (more) |
— | over 3 years ago |