43,455 reputation
2080173
bio website ellipsix.net
location Wuhan, China
age 29
visits member for 4 years, 5 months
seen 5 mins ago

I'm a postdoc doing research in high-energy particle physics. I also have a hobby interest in computer programming.

You can find me on Twitter, or check out my blog and personal website! Or you can email me at stack@ellipsix.net.

Please don't contact me through any other channel about something that should be handled through this site. If you do, I'll probably just tell you to post it here.

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6h
comment What is a body's momentum really equal to?
I'm intentionally calling it outdated even though I know some people still use it - the implication is partly that it should be outdated, not that nobody uses it.
16h
answered What is a body's momentum really equal to?
23h
comment How to determine which object in a special relativity physics problem has proper time?
@Brionius the modern terminology, as I know it, is that what you learned to call "rest mass", formerly denoted $m_0$, is simply called "mass" and denoted $m$, and "relativistic mass", formerly denoted $m$, is now just "energy" or "total energy" $E$. Or more precisely, instead of talking about relativistic mass, we talk about energy, which differs by a factor of $c^2$. For example, the (total) energy of an object is its mass energy plus its kinetic energy, and the total energy increases with velocity, but the mass never changes. (My main complaint is with the use of $m$ for relativistic mass.)
1d
comment How to determine which object in a special relativity physics problem has proper time?
My one complaint about this answer is that it's not clear about the difference between relativistic mass (an outdated concept) and rest mass. Otherwise, very nicely explained.
Apr
23
revised Is This A Working Time Machine?
rolled back to a previous revision
Apr
23
comment How fast would you have to go to claim you saw the red light as green due to the Doppler Effect?
What have you done to try to figure this out yourself?
Apr
23
revised Particle moving under force $F=-cx^3$
edited tags
Apr
23
comment Why don't things get destroyed by gas molecules flying around?
Yes, but even intermolecular forces are strong enough to resist most collisions with air molecules. I mean, I don't know the numbers offhand, but it must work out that way, otherwise objects as we know them wouldn't exist ;-) So perhaps you only need a sphere of air the size of, I don't know, the moon to get one air molecule moving fast enough to break an intermolecular bond, instead of 16 Earth-orbit-sized spheres. And besides, breaking one molecule off something hardly counts as destroying it.
Apr
23
comment How do I choose the right value of $r$ to find where the electric field is zero?
Ah, well I left out the absolute value signs in my previous comment because we were only talking about a point in the region where $x>0$. The more general expression is that the distances are $|x|$ and $|x+2|$, in your convention, or $|x|$ and $|x-2|$ in the other convention. So if you use the other example's definition of $x$, which has $x>0$ to the right of the origin, then the distance from $x=-3$ to $Q_2$ is $|-(-3)+2|=5$. Note that $|x-2|=|-x+2|=|2-x|$, so you can use any of these interchangeably in this situation. (PS the fact that $Q_2$ is negative is completely irrelevant.)
Apr
23
comment How do I choose the right value of $r$ to find where the electric field is zero?
Your diagram shows that $x$ is defined as distance to the left of the positive charge, as I mentioned in the answer. So, using your definition, points to the left of the positive charge correspond to positive values of $x$. If you think about it a bit, you should be able to understand that this corresponds to your equation as well: take a point 3 units to the left of the positive charge, for example. Its distance from the positive charge ($x$) is 3 and from the negative charge ($x+2$, in your convention) is 5. If $x$ were $-3$, that wouldn't work out.
Apr
23
comment How do I choose the right value of $r$ to find where the electric field is zero?
@DevinCrossman That's how they defined $x$, not how you defined $x$ - at least, not according to your diagram or your equation.
Apr
23
comment special theory of relativity-time dilation
A web search for "time dilation" or "special relativity" or any of various similar terms would lead you to several online resources that explain this; therefore I don't think it's a good question.
Apr
23
revised special theory of relativity-time dilation
edited tags
Apr
23
answered How do I choose the right value of $r$ to find where the electric field is zero?
Apr
23
revised How do I choose the right value of $r$ to find where the electric field is zero?
edited tags and title, improved formatting
Apr
22
reviewed Approve Why is this spin expectation value a vector
Apr
22
comment How do I maniptulate the Fourier transform of a field using the delta function?
I'm voting to close this question because it's been cross-posted to Mathematics.
Apr
22
comment Units for displacement current density
You can't delete it because it has an upvoted answer. But in this case, it's your answer, so you can delete the answer and then you should be able to delete the question. That being said, I think this is a good question and answer - even though it was caused by a minor mistake, someone else could easily make the same mistake, and leaving this post here would help them. (Plus, free rep :-P)
Apr
21
revised Why do $S_x$ and $S_y$ flip up/down spin states but $S_z$ does not?
improve formatting and title
Apr
21
comment How do you become a theoretical physicist?
I'm voting to close this question as off-topic because it is about career advice.