OK. I'm 10 ly from Earth "facing" Earth (x axis oriented at Earth), looking at Earth 10 years ago. This is {10, -10} as a x,t vector. When a spaceship passes me at 0.8c, we synchronize clocks. For that spaeceship, Earth is 6ly away and it's seeing light from 6 years ago, in other words {6, -6}. However, the only Lorentz transform that converts {10,-10} to {6,-6} is for v = -8/17, not v = 0.8
I have been thinking recently what will happen if one uses the energy momentum tensor of the Dirac field as a source in the Einstein Field equations. It is well known that in this case
$$
T_{\mu\nu}=i\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi
$$
What fallows however causes my confusion. T...
@barrycarter you can't make an automatic migration, but what you can do is create a chat room, comment to ask the other participant(s) to join you in that chat room, and then stop responding in the comments. (People often forget that last part)
Does anyone which one is bigger for an airplane? lift or drag coefficient? I guessed lift coefficient is bigger but in laboratory our numbers say drag coefficient is bigger
puu.sh/o4FBH/30db0a52cd.png anyone able to translate this in english? I read it as $F^n$ is equal to the set where the list of $x_f$ which is equivalent to $(x_1,...x_n)$ is an element of F. I don't get the rest of it though.
OH my monitor sucks I read that as $x_f$ when it's $x_j$ nvm i'm good
Magically, no, but you can manually give it that name. I think the template is just "Discussion between [username1] and [username2]" and the description is "Imported from a comment discussion on [link]"
If you ask me, the default name isn't particularly great, and you can probably come up with a better one.
And actually, you can also ask people to bring the discussion to this room, if it's not especially important to you to have a separate room for the discussion.
Yeah, that's what tends to happen when you (or I) ask people to move a discussion to chat. A lot of times, they keep going in the comments. But refusing to respond in the comments is remarkably effective ;-) They just wind up talking to themselves and it looks a little silly after a while.
But the worst part is that he's "slightly wrong" to the point he can confuse people, not blatantly wrong (like that other guy) where I feel free to ignore him.
The other thing you can do is flag (one or a few of) the comments as obsolete. Any time there's a comment asking to take a discussion to chat, it renders all other comments in that discussion obsolete, so we'll generally delete them and leave only the link to the chat room.
I think you all are still correct but some of us have some personal problems due to which we come to sites like this. I think we should be treated well.
I suspect the poster of that question thinks of good treatment as getting their questions answered.
3
@barrycarter it's probably stuff you know by now. Just the basics. You don't really need to. (But you never know, you might learn something, and there's a badge for it too I believe)
I'm actually frustrated that we don't answer homework questions or at least close them with: here's a link to some homework helper site resources. Or even migrate the question to math.SE or another site which DOES accept homework questions.
puu.sh/o4Hv2/7a15a64999.png Sorry for linking a puush again but it's not just tex this time. What does he mean by defining the sum of an element of $F^n$ and 0? How can you define such an elementary operation o_o and why does 0 have to be a list?
@barrycarter Well... the thing is, we really really don't want to be a homework help site. We want to ensure that when students are seeking the quickest path between them and the answers to their homework questions, that path does not lead through this site. So in some ways, we go out of our way not to be helpful to people who come here looking for that kind of help.
@DavidZ OK, we disagree on the definition of "homework help site". To me, that means guiding the student to the correct answer, making him do most of the work.
@Obliv So, F^n is the product of a set (field) F n times?
@dmckee Oh, I was referring to @DavidZ comment 'when students are seeking the quickest path between them and the answers to their homework questions, that path does not lead through this site'.
I just don't want to do that here. Once you admit that kind of question is inevitably becomes the bulk of the business of the site. Just look at Mathematics.
@barrycarter I loathe it. This is clearly a disagreement about goals. I suppose you could take it up on meta again. It's been a while since we went around on that issue.
@JohnRennie You know, I suggested in an earlier message that changing t=0 to something other than whatever I set it to in the problem would help. You're saying setting t=0 to 2022.5 would help?
@barrycarter - Hi--so on the broader conceptual point, would you agree that all statements about what distant events were simultaneous with a particular clock-reading in my past are just so much "bookkeeping", depending on what coordinate system I choose to use, with no frame-independent physical reality?
To a first approximation we can ignore the effect of the acceleration on the twin's clock because the acceleration period is so short (1 sec vs several years).
@barrycarter - I did, I think he was making pretty much the same point, assuming by "Fargo is to my right, and always has been" he meant "Fargo is in the direction I currently call right, and always has been" rather than "Fargo has always been in the direction I called right at the time". Might have been clearer if it was stated in terms of coordinates, like "Fargo is at a greater x-coordinate than my own current x-coordinate, and always has been".
("always has been" in the coordinate system I'm currently using, that is)
When the twin is stationary they are in the same inertial frame as the Earth, so they and the earth can compare clocks using any one of the clock synchroniastion schemes
@JohnRennie OK, I think you lost me there. If you ... crap, I have to go. I think your point is that choosing a different t=0 means we never have to go into unoccupied reference frames.
@Hypnosifl I take issue with your last statement... that's where I believe the confusion is.
@barrycarter Why do you take issue with it? Say that in the x-y coordinate system I am currently using, Fargo is at x=100, y=0, and I started at x=0, y=0, drove along the x-axis for 50 miles to x=50, y=0, then turned and started traveling along the y-axis, so I am now at x=50, y=10. Is it not true that in the coordinate system I am currently using, "Fargo is at a greater x-coordinate than my own x-coordinate, and always has been"?
That would be true even if on the first stage of the journey, I had been using a different coordinate system where I started at x=0,y=0 while Fargo was at x=0,y=100 in that coordinate system, and in this coordinate system I was moving along the y-axis. The point is that the statement in quotes above doesn't refer to this older coordinate system, so it's irrelevant to evaluating its correctness.
Hmmm, why did a post that was last edited last November show up in the reopen queue as "This question was edited after it was closed. Should it be reopened?"
I am in high school now and I want to get the basic idea of what relativity. Can anyone suggest me a book or website for it? I am also curious about the mathematics behind it. Is it possible for me to understand relativity in mathematical terms while still in high school? (Let's assume I know eve...
@dmckee Yeah, but the review shouldn't show "This question was edited after it was closed." if a non-review reopen vote and not the edit pushed it into the queue, should it?
@0celo7 I'm not an expert on the book policy because I don't like it much, but I think you are expected to explain why each book fits. Perhaps grouping them would be acceptable.
@ACuriousMind So is that a general property of gauge theories, that the fermions transform in a specific rep, whereas the gauge bosons transform independently of the fermion rep?
@Bass Yes, the transformation of the gauge field does not depend on the matter fields at all (there needn't even be matter fields, a pure Yang-Mills theory makes perfect sense).
@Obliv What operation is it supposed to denote? $\text{scalar}\cdot\text{vector}$ is defined to be scalar multiplication, but $\text{vector}\cdot\text{scalar}$ is not usually defined at all.
I have been thinking recently what will happen if one uses the energy momentum tensor of the Dirac field as a source in the Einstein Field equations. It is well known that in this case
$$
T_{\mu\nu}=i\overline{\psi}\gamma^{\mu}\partial^{\nu}\psi
$$
What fallows however causes my confusion. T...
I think you all are still correct but some of us have some personal problems due to which we come to sites like this. I think we should be treated well.
