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shelton Benjamin
shelton Benjamin
3:11 AM
Does the force remain same in relavistic frames
 
skullpatrol
skullpatrol
3:30 AM
 
 
1 hour later…
John Rennie
John Rennie
4:52 AM
@sheltonBenjamin things get complicated in relativity. Generally we use a four dimensional force vector called the four force not the non-relativistic force.
In SR what Newton called the force becomes an observer dependent quantity.
 
shelton Benjamin
shelton Benjamin
Sir one more question why is internal kinetic energy measured com frame?
I mean any special reason to define it this way?
 
 
4 hours later…
Slereah
Slereah
8:41 AM
Some Mathjax problems this morning
 
ACuriousMind
ACuriousMind
9:03 AM
MathJax on strike
 
Slereah
Slereah
quite the opposite
it's back and bigger than ever
 
skillpatrol
skillpatrol
a strike back
Right to Over Render
 
skillpatrol
skillpatrol
10:01 AM
 
 
2 hours later…
John Rennie
John Rennie
11:54 AM
Underground lakes on Mars. Yes, liquid water on Mars, though very salty liquid water.
 
Emilio Pisanty
Emilio Pisanty
the internet is a magical place :')
@JohnRennie hmmmmmm. Doesn't say what kind of salt. Knowing Mars, it is probably perchlorates (i.e., toxic) instead of chlorides.
 
John Rennie
John Rennie
@EmilioPisanty I think a healthy scepticism is required, but we can always hope :-)
 
Emilio Pisanty
Emilio Pisanty
ah, there we go
> We suggest that the waters are hypersaline perchlorate brines
from the abstract of the primary paper
Martian soil
Martian soil is the fine regolith found on the surface of Mars. Its properties can differ significantly from those of terrestrial soil, including its toxicity due to the presence of perchlorates. The term Martian soil typically refers to the finer fraction of regolith. So far, no samples have been returned to Earth, the goal of a Mars sample-return mission, but the soil has been studied remotely with the use of Mars rovers and Mars orbiters. On Earth, the term "soil" usually includes organic content. In contrast, planetary scientists adopt a functional definition of soil to distinguish it from...
perchlorates
↑ probably the thing that will make human life on Mars impossible
 
danielunderwood
danielunderwood
12:46 PM
Neat, never considered the soil on Mars to be toxic, though I suppose it makes sense. Is it known why concentrations are higher than on Earth?
Wikipedia suggests that the compounds could be formed by UV light, but it sounds like there are also a number of organisms on Earth that could process them
 
Emilio Pisanty
Emilio Pisanty
@danielunderwood I don't know, but it's a good question.
tbh, the structure of perchlorate itself makes little sense to me
but apparently it is indeed possible to oxidize chlorine
now
given that Earth's environment is highly oxidizing
and that there's tons of chloride out there in the ocean
I don't understand how and why that chlorine doesn't end up getting oxidized to perchlorate
@EmilioPisanty (or possibly what happens in perchlorate is that the central chlorine has oxidized the outer oxygens. Oxidation is bizarre when halogens are involved.)
 
Superfast Jellyfish
Superfast Jellyfish
@EmilioPisanty huh? What are the conditions for that to happen?
 
Emilio Pisanty
Emilio Pisanty
@SuperfastJellyfish no idea
but this is a thing:
Perchlorate
A perchlorate is a chemical compound containing the perchlorate ion, ClO−4. The majority of perchlorates are commercially produced salts. They are mainly used for propellants, exploiting properties as powerful oxidizing agents and to control static electricity in food packaging. Perchlorate contamination in food, water, and other parts of the environment has been studied in the U.S. because of its harmful effects on human health. Perchlorate reduces hormone production in the thyroid gland. Most perchlorates are colorless solids that are soluble in water. Four perchlorates are of primary commercial...
 
Superfast Jellyfish
Superfast Jellyfish
@EmilioPisanty seems like an outright electron ripper
 
 
2 hours later…
Yashas
Yashas
2:37 PM
I am having a bit of trouble trying to relate the uncertainty principle with measurement. The uncertainty principle reflects the variation in the wavefunction, right? A device that measures, say position, will report the position to some significant figures. Can I say that the position reported by the measuring devices (to whatever precision) is the position of the particle?
 
ACuriousMind
ACuriousMind
The uncertainty principle doesn't really have anything to do with measurement
 
Yashas
Yashas
So if I have a measuring device that can measure up to Planck's length, I can in principle get a very accurate value of the position? Now this would imply the momentum wavefunction now has a large standard deviation, right?
 
ACuriousMind
ACuriousMind
What a measurement does is that after the measurement the system is in an eigenstate of the measured operator.
You can of course then apply the uncertainty principle to that eigenstate, but there's no real connection between that and the act of measurement
 
Yashas
Yashas
So if my position measurement device is very accurate, the momentum wavefunction becomes so scattered, that it's possible that the electron I measure on one side of the Earth ended up on the other side by chance?
Does the probability of taking some insane extreme momentum value become bigger the more accurate the position is measured?
 
ACuriousMind
ACuriousMind
position and momentum are annoying observables because they have continuous values
So you don't really ever measure a specific value, but a range of values. The behaviour of real position measurement is tricky to describe
but a naive (and simple to work with) idea would be that the resulting position wavefunction is some Gaußian centered at the measured value, and the corresponding momentum wavefunction is the Gaußian with the corresponding width by the uncertainty principle
 
Yashas
Yashas
2:43 PM
Is it right to say that by measuring quantity X more accurately, the more I disperse the wavefunction of quantity Y? (assuming that X and Y do not commute)
 
ACuriousMind
ACuriousMind
@Yashas not in general, but for position and momentum (or more generally canonically conjugate quantities) it is true
 
Yashas
Yashas
So do I increase the chances of my test electron being teleported to Mars by measuring its position with very very very high accuracy?
 
ACuriousMind
ACuriousMind
@Yashas Why would you - if you just measured its position it's definitely not on Mars!
(unless you're on Mars)
 
Yashas
Yashas
sorry, I meant measuring its momentum with very very very high accuracy
what I am coming to is that if I just measured the momentum to be near zero with very high accuracy, the probability that it is in Mars is higher (compared to a less accurate momentum measurement) but my momentum measurment apparatus is on Earth
it feels like by simply measuring momentum, I have teleported a particle which is very weird
 
ACuriousMind
ACuriousMind
well, note that all this only works like that for a free particle - a particle inside a box still can't leave the box just because it's in a state of definite momentum
(note also that momentum states inside a box are discrete, and applying the uncertainty relation is tricky)
In reality, most particles are bound, inside a box or otherwise not free. The absurd result that a particle of very definite momentum can be in a very large area doesn't work on scales like "Earth-Mars" because for one the particle won't be "free" on that scale and for the other the precision necessary for the apparatus to force it into that definite momentum state is impossible
note also that this is not "teleporting" - the position of the particle is uncertain. It's not "on Mars", neither is it "on Earth".
and once you want to ask whether you can instantaneously teleport things like this or what someone on Mars could measure you would need to do relativistic quantum mechanics instead where the whole idea of position is more complicated still
 
BioPhysicist
BioPhysicist
3:27 PM
I feel like there are users who are actively trying to fight site policies about homework questions, and I feel like it is bothering me more than it should haha
 
B. Brekke
B. Brekke
Just got a Mac and I don't know if I like the Mac keyboard for writing in latex. I find curly brackets, backslash and other symbols a bit unnecessary complicated. So I want to ask other Mac users, did you get used to it, or did you do something more drastic?
 
BioPhysicist
BioPhysicist
3:46 PM
0
Q: Can anyone help me in these question:(related to Work Done)

Hitman A7 Can work done by Kinetic Friction on an object be positive? Zero? Can Static Friction do Non-zero work on an object? Can Normal Force do Non-zero work on an object?

homework-and-exercises newtonian-mechanics work
Like this one... why is it getting upvotes, answers, and even comments defending the post?
 
ACuriousMind
ACuriousMind
@BioPhysicist I'm not a fan of the question but I don't think it's a HW question as we commonly understand it. It's a bunch of entirely unmotivated questions that likely originate from actual homework, but it's not the "solve this problem for me/do this calculuation"-type question I'd consider HW-like (that's why the name "homework policy" is terrible, but, well...)
 
BioPhysicist
BioPhysicist
That is a good point. I guess I somewhat recognized it as this since I VTC as needs more focus. But even then does that mean other users should be posting solutions to it? Perhaps I am being too harsh on this question?
 
ACuriousMind
ACuriousMind
It's not a good question for the reasons you name in your comment, but I don't think it's off-topic.
@BioPhysicist User can be posting solutions to whatever they want :P We do not police answers outside of the rather narrow "providing near-complete answers to HW-like question". Maybe they don't think the question lacks focus. Maybe they don't care. Maybe they think helping the OP is more important than not answering to encourage better questions (a viewpoint I disagree with but have sympathy for).
 
BioPhysicist
BioPhysicist
The OP was the one who included the homework-and-exercise tag, so to me it seems like an answer to this question would be providing solutions to homework. Of course if the problem is more conceptual, like this one, it is usually harder to identify it as homework. So then if the question is conceptual then answers will not be deleted for providing solutions, since in general it is harder to prove that it is someone just wanting homework solutions?
 
ACuriousMind
ACuriousMind
The OP likely tagged the question with that because this is an actual exercise from their homework
they're new, new users essentially never apply that tag correctly :P
 
BioPhysicist
BioPhysicist
3:59 PM
So we are fine with helping people cheat on homework as long as they are conceptual homework questions?
 
ACuriousMind
ACuriousMind
Yes, it is the content of the question, not its origin, that matters for its status as a HW-like question. At least that is my understanding of the status quo, and I think this is sensible because we cannot really determine the origin of a question
if the user had asked the question and not tagged it, one might have guessed it might be from homework but we wouldn't really have a criterion except for gut feeling to say so
 
BioPhysicist
BioPhysicist
That does make sense I suppose. I guess I just feel weird about it when the OP has essentially indicated that it is homework
 
ACuriousMind
ACuriousMind
Personally, I don't really care so much about the "cheating" aspect. If it's a conceptual question it's usually pretty hard to copy the answer 1:1 from here since answerers tend to talk to OP as "you", go on tangents, insert pictures or link to Wiki articles, etc. In order to usefully pass this off as your own work you'd need to exhibit some rudimentary form of understanding it to rephrase it
 
skullpatrol
skullpatrol
With the advent of take-home-exams which allow Internet use it is no longer "cheating" imho.
 
ACuriousMind
ACuriousMind
for the calculation-type questions you can usually just copy the steps from the answer and don't need to do anything to it, so I don't think it leads to any increased understanding in many cases and also the solution to a specific exercise is of questionable use to anyone else not trying to solve the exact same exercise
i.e. these questions are both easier to use for mindless cheating and not useful to many other people, which is why I find it much more important to reject them
 
JMac
JMac
4:16 PM
@skullpatrol Geez all my exams were open book and that alone basically saved me on some exams. You would need to come up with some really clever questions to make an exam with internet access much of a challenge IMO. I'm curious how they are handling that.
 
BioPhysicist
BioPhysicist
Those are good points. Thanks @ACuriousMind
 
ACuriousMind
ACuriousMind
@JMac It's almost as if the skills tested by traditional exams are not the most relevant skills in the information age...
 
BioPhysicist
BioPhysicist
@JMac I suppose to at least prevent copy and paste searching one could take a problem and put a story with it
so that way the model is the same, but the relation to the problem is different
Like instead of "What is the acceleration of the block down the incline" you make up a problem about a pig going down a slide? I don't know haha
 
JMac
JMac
@ACuriousMind I'm just saying it seems like it would be hard to design a good test on understanding when people have access to the internet. Most of my exams were open book and basically tested problem solving on questions that weren't given examples in the book, and honestly it made them quite easy. I just figure it must be pretty hard to come up with good questions when the students have the full internet.
 
ACuriousMind
ACuriousMind
@JMac yeah, and I'm saying maybe if one can answer the questions just by having access to the internet they don't necessarily test abilities relevant in a world where we have constant access to the internet!
On the other hand you shouldn't underestimate how hard the work of transferring a given solution method in a book/on the internet to a new example is - it was easy for you, but was it easy for everyone?
that in itself - understanding enough of the stuff in the book/on the internet in a short period of time to use it to answer a slightly different kind of question - is pretty important and not automatic (it's not for nothing that IT people always say they solve their relatives' computer problems just by googling them :P)
 
JMac
JMac
4:31 PM
I'm sure it wasn't easy for everyone, but I'd honestly be worried for students like me. I relied on open book as a huge crutch and my actual education took a hit from it. Being able to google things directly instead of flipping through appendices and stuff would have made it super easy at very least. If they aren't heavily changing up test formats from the type I was used to, I think some people could really suffer, so that's why I'd be interested in how schools are handling it.
 
Yashas
Yashas
what is a measurement in QM
are all interactions measurements?
 
vzn
vzn
huh, 34 refs to "measurement" on wikipedia pg on HUP en.wikipedia.org/wiki/Uncertainty_principle
 
BioPhysicist
BioPhysicist
@Yashas You should make sure there aren't any cats in the chat before you ask about QM
 
ACuriousMind
ACuriousMind
@Yashas No. See physics.stackexchange.com/q/43406/50583. Essentially it is an interaction that decoheres the systems being measured, but this is a subtle question that touches on different resolutions to the measurement problem and what it "really means" is firmly in the realm of quantum interpretation
 
skullpatrol
skullpatrol
::Meow:: @BioPhysicist :-)
 
Yashas
Yashas
4:43 PM
Why did Schrodinger think of live and dead cats of all possible scenarios? :(
There were hundreds of different ways to construct that experiment
 
ACuriousMind
ACuriousMind
the cat thought experiment was intended to illustrate the absurdity of the measurement problem w.r.t. to macroscopic entangled systems
it was never meant to become the canonical way of teaching people about QM :P
 
BioPhysicist
BioPhysicist
And yet that is typically the only thing most people are exposed to about QM
 
skullpatrol
skullpatrol
especially in pop-sci
cargo cult anyone :-)
 
BioPhysicist
BioPhysicist
In addition to Schrodinger's cat being both alive and dead, it is also located at every single point in the universe at once
Although it can only have certain amounts of energy; it is in a box after all.
 
skullpatrol
skullpatrol
throw in a ball of string theory for the cat to play with
 
Physics Meta
Physics Meta
5:32 PM
0
Q: Understanding the voting process

BuraianI'm quite puzzled as to why this answer I wrote got three downvotes (here) The op's question indicates they had a fundamental misunderstanding and probably got really lost while self studying the subject. Something which I can relate to as I self study as well. So, due to that I thought of writin...

support
 
skullpatrol
skullpatrol
With voting you don't understand anything, you just get used to it.
 
Yashas
Yashas
6:25 PM
Consider a particle in a box system. Suppose I measure the energy to be E. Now in another measurement I measure X that is incompatible with E. Now since the E and X are incompatible, the state of the system can be expressed as linear combination of energy eigenstates. So if I measure E again, shouldn't I be getting the previously measured energy eigenstate with unit probability because there is only one state with energy E and energy must be conserved?
 
ACuriousMind
ACuriousMind
@Yashas No, the measurement is an interaction with the system. After that the energy of measurement apparatus + measured system is conserved, but you can't say anything about the energy of the system alone before/after measurement
it's like a collision
 
Yashas
Yashas
so the measuring apparatus should have energy that is more than the zero point energy of the system being measured (otherwise, the system being measured cannot take enough energy to go higher if it were already in the ground state)
 
vzn
vzn
hmmm, huh! "collision" sounds like something outside the copenhagen interpretation. aka "ontological"... or even more dramatically/ taboo, a "hidden variable"...
 
ACuriousMind
ACuriousMind
@Yashas All you can say about energy conservation is that the expectation value of energy of system + apparatus is constant.
 
Yashas
Yashas
if the system was in ground state with energy E and the expected value of the energy of the apparatus is less than E => a measurement is impossible?
 
ACuriousMind
ACuriousMind
6:36 PM
sure, the apparatus probably needs "enough energy" if you're trying to measure high energies but the very definition of "measurement apparatus" means that it can for ce the system into eigenstates of all possible eigenvalues of the observable it measures.
but usually the measured system in this context is microscopic compared to the "classical" measurement apparatus so this isn't really a concern
I'd advise not to think too much about the finer points of measurement until you firmly understand the rest of QM. In many situations you don't need to think about the measurement apparatus at all and "a measurement" is just a magic word for applying the Born rule and getting an eigenstate of an observable out without any specific underlying physical apparatus in mind
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
Hey guys, anybody here to help me with number of significant digits? Im totally puzzled
 
Yashas
Yashas
@FarhadRouhbakhsh what's the question
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
Determine the number of significant digits: 16.264 – 16.26325
actually 16.264 – 16.26325 is equal to 0.00075
but I think it has 5 significant digits
because of our first number 16.264
@Yashas Am I wrong?
 
Yashas
Yashas
You retain the number of decimal places in subtraction/addition. So 16.264 - 16.26325 is 0.
I don't really recall the motivation for this rule but this is what I was taught in high school.
You retain significant digits in multiplication and division and decimal places for addition and subtraction.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
how did you retain? I guess 16.26325 retained is 16.263
 
Yashas
Yashas
6:47 PM
16.264 - 16.26325 = 0.00075 but you have to retain only three decimal places (because you have three decimal places — the least number of them — in 16.264), so you round it off to three decimal places in the result => 0.001 (so it's not zero, sorry)
 
ACuriousMind
ACuriousMind
I've always thought that significant digits are a silly concept. If the values are exact then you shouldn't throw any digits away and if they are not exact you should figure out proper error bars instead of throwing away random digits :P
 
Slereah
Slereah
It is silly, yes
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
@Yashas so the answer will be 1?
 
Slereah
Slereah
mostly they're just to not carry around too many digits, but physicists think they're more important than they are
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
@Yashas I mean 1 significant digit
 
Yashas
Yashas
6:48 PM
@FarhadRouhbakhsh I am not 100% sure but from I remember from high school, the answer is 0.001.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
how about the significant numbers of 20000?
Is it 5 or 1?
 
Yashas
Yashas
It depends on the context. It can be both 5 and 1. How many apples do you have? Say 1. This one has infinite significant digits.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
In the question it just says determine the number of significant digits in these numbers
Unfortunately it doesnt talk about any context :(
 
Yashas
Yashas
It's 5 then I guess
2.00 x 10^4 has three significant digits while 2.000 x 10^4 has four significant digits. Normally, trailing zeros are counted and leading zeros are ignored.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
37.76+0.132 is another one which is equal to 37.892
If you consider the least digits in your addition then the answer would be 3
But I myself think the answer is 4 :)
 
Yashas
Yashas
6:55 PM
@FarhadRouhbakhsh I think 37.89 is the answer but I am confused. Should there be three significant digits or four?
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
Im confused too
 
Yashas
Yashas
I think I'll check my high school textbook once. I need to recall the motivation for these rules too.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
I always hear they say in addition you should consider the number with least digits
 
Yashas
Yashas
"In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures."
" In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places."
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
So the answer would be 3?
because 0.132 has 3?
 
Yashas
Yashas
6:59 PM
Yes, I think the answer is 3.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
And for the 16.264 – 16.26325=0.00075 the answer is 5 by this rule
 
Yashas
Yashas
@FarhadRouhbakhsh It's two I think. The leading zeros are not counted.
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
but your rules say it should "retain" as many significant digits as there are in the least one.
 
Yashas
Yashas
Wait, it's one. Sorry. 16.264 – 16.26325=0.001 (which has one significant digit)
"least digits" and "least decimal places" mean different things
the former means the total number of digits and the latter means the number of digits after the decimal point
@FarhadRouhbakhsh that's for multiplication and division only
retain the number of significant digits in multiplication and division; retain the number of decimal places in addition and subtraction
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
does "retain" mean it should have the same amount?
or it could be less?
 
BioPhysicist
BioPhysicist
7:06 PM
@Yashas Interestingly, this question has been recently asked
0
Q: Since measurement collapses wavefunction and everything is constantly interacting, how does uncollapsed wavefunction exist?

John DoeInteraction causes change and that change can be measured, so shouldn't everything be measured constantly

quantum-mechanics
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
@Yashas If "retain" means it could be less then you are right. but if it means it should be the same amount then I am. Any comments?
 
Yashas
Yashas
@BioPhysicist I think I know the answer to this one. The notion of uncollapsed is ambiguous or relative. If you ask the observer in Schrodinger's cat experiment to raise a red flag if the cat is dead or a green flag if the cat is alive, then for the person who will be looking at the flag, the whole system (the first observer and the cat) are entangled. The wavefunction of the whole setup collapses when he sees the flag whereas for the first person might see the cat wavefunction collapse.
I think it's just the choice of interpretation.
but QM is complex that I have gone to the origin in the Dunning Kruger graph
there is so much objective and subjective stuff in QM that I keep losing track of things
 
BioPhysicist
BioPhysicist
I just thought it was interesting that you asked it here and then someone else asked a similar question on the main site
 
Yashas
Yashas
Oh, I never knew there was a "follow" question feature until now.
@FarhadRouhbakhsh I think it's just retain. Less or more shouldn't matter. I am not 100% sure.
 
BioPhysicist
BioPhysicist
@Yashas It's a somewhat new feature I would say. Maybe about 6 months old
 
Farhad Rouhbakhsh
Farhad Rouhbakhsh
7:18 PM
@Yashas Thanks for your help
 
Yashas
Yashas
Looks like the OP asked something else in the question and I interpreted it as something else
 
Slereah
Slereah
7:50 PM
user image
7
 
Superfast Jellyfish
Superfast Jellyfish
@Slereah brilliant!! xD
Where is this?
 

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