@NeuroFuzzy Sadly quality is not measured by votes and fools are fooled by pretty pictures. I still maintain that your answer does not even begin to address OPs concerns.
@KyleOman Linux Mint 17 is pretty stable & based on Ubuntu, so you get all the benefits & none of the kinks
> Our test campaign can not confirm or refute the claims of the EMDrive but intends to independently assess possible side-effects in the measurements methods used so far
Look around that message for a brief discussion of that "news"
@vzn What it seems they're doing is ignoring that aspect as a systematic error and attributing some fantastic & unknown mechanism to producing an otherwise impossible thrust
The researchers from Dresden don't make any strong claims - they just say they've measured the same "thrust", but can not exclude e.g. feedback from the power supply lines or other causes for it.
@vzn IMO, it's funding hype. Claim some preposterous thing is real, get boat loads of public interest and the financial support follows (same thing happened with global warming research)
ok, found a thorough debunking link posted by Kyle, but it seems like there would be at least mild incentive to definitively disprove these false claims via some setup...
@KyleKanos ~shrug~ it's a clear demonstration of the phenomenon and gives a good reference. I'm not spending much time on a 0 vote 0 answer 1 sentence question.
@NeuroFuzzy It's a clear demonstration of light echoes. Sadly, the question was not actually about light echoes. At best, it's a so-so "answer"
Actually, it's a so-so demonstration of LEs. It shows a single image and provides no information as to what it is or how/why it works, only a reference
It's asking if light travelling through a cloud, perpendicular to LOS, is visible
Light echoes require specific geometries to be matched, so it's actually a somewhat rare event
We see clouds adjacent to stars that are being illuminated by the star (perpendicular fashion), so OP wants to know how we can see those clouds
Your answer says, "I'm going to ignore what you said, show you a flashy image, and gain 330 rep in the process"
More a statement of the lack of knowledge of the general public, since they're not knowledgeable enough to know what OP is asking and how you've entirely avoided answering it.
Ray emitted from supernova. Ray scattered off of cloud. Ray reaches a telescope providing an at-later-date record of where the supernova light had reached. Ray of light detected while passing through the cloud. That's what the OP's question was about.
If some of the light is reflected off the dust at such an angle that it is diverted to reach the observer, the observer will see that light. However, those specific photons reaching the observer will not reach B (unless they are reflected there by the observer). Similarly, unless the observer is...
I'm not sure whether this is a question about light echoes, or a fundamental misunderstanding of what it means to see something. — Lightness Races in Orbit18 hours ago
...and I'm completely unsure why that question has gathered so many upvotes.
The phrase "to see the light while it travels" in the question lets me suspect that OP doesn't actually understand what "seeing" is, and that the fact that there are light echos and whatnot in astronomy is actually wholly unrelated to what OP wanted to ask, regardless of how well they are explained or not.
@KyleKanos upvoted. If I can give a great relevant reference which plausibly answers the question, or let a question sink into oblivion, I'll do the former!
@obe : the physics isn't outdated. Don't be so quick to dismiss Einstein out of hand. Read the digital papers instead. Then when some textbook paints a totally different picture whilst appealing to Einstein's authority, put your thinking cap on.
@skill patrol : there does seem to be some kind of issue wherein textbook authors somehow "stray from the science" and students never read the original material. Then they grow up and write textbooks.
@JohnDuffield Indeed, and the "thinking cap" gets lost along the way.
The "original material" is distilled and refined to the point that even the original author can't recognise it @JohnDuffield as Einstein once said to Lorentz about special relativity.
Is the value of momentum an invariant?,
specificly for instance the momentum value $\mathbf p_{\text{lab}}[~\Lambda^0~]$ of a $\Lambda^0$ baryon (drifting from the (actual) interaction point of a collider experiment towards the beam pipe wall) with respect to suitable(1) constituents of the "lab"...
user37496: "If I measure the speed of a particle in the lab and then write down in my notebook the value" ... as real-number multiple of $c$ ... "an observer in a different inertial frame reads [...] the same" -- Correct. Real numbers are presumed unambiguous; they can be copied; +1. "(although [...])" -- Their graph structure (or topology) should remain distinctive enough. "context that led you" -- Foremost this: "speed itself is a coordinate system dependent concept"; then that. — user122622 days ago
He must just be trolling. The topology of $\mathbb{R}$? lol!
And if it transforms under various gauge groups, you get things like "It has components which are components which are components", if you want to write it out explicitly
And I just found out I can complete my Bourbaki collection :D Springer books for free is the best thing that happened to me since... Not really, but it's pretty nice :P
"I have fermented cabbage to alcohol. The largest problem with cabbage is its very small amount of sugar and its reaking smell when cooked down to a higher sugar level. If it is boiled for a long time it can be concentrated to the point that makes it worth while. "
"The ending result was stilled in a pot still. I had to run it a couple of time to get the cabbage flavor down to a level I could tolerate. "
@ACuriousMind @ACuriousMind $\mathrm{Real World Application}$ is an operator $R$ such that $R:\mathrm{GDP}\mapsto\mathrm{GDP}+\epsilon$ for some $\epsilon> 0$.
@ACuriousMind I don't; that would probably break something.
@Kyle Kanos : no it isn't a new low. I'm not misrepresenting observations. The coordinate speed of light is zero. Einstein would have called it the speed of light. Next time you see something moving faster than light be sure to tell the guys in Stockholm.
@0celo7 : no. Go and look it up. @Kyle Kanos : no, I'm saying c=0 because a black hole is black. The vertical light beam can't get out because at the EH the speed of light is zero. Please don't tell me you think it's can't get out because space is falling inwards.
I don't care what you believe when it comes to theories, but when you start actively rejecting experiments/observations because it conflicts with your theory, I think you should refrain from calling yourself a scientist because that really isn't science
I don't understand what all this talk about coordinate speed is for. Light can't leave a black hole because no light-like geodesic starting at the EH points outward. This is a completely coordinate-free statement. That the coordinate speed in any particular coordinate system should carry any significance at all is not obvious.
So E=MC2, which means that energy is equal to mass times the square root of the speed of light. Is it really necessary to add the words when the equation says the same thing?
