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location Baltimore, MD
age 30
visits member for 4 years, 10 months
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I'm a physics graduate student.


1d
comment Error propagation for exponents
This formula just assumes the errors are small, not that they're Gaussian.
Aug
20
comment How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
I suppose what you want to know is something along the lines of, "What motivated people to come up with that definition of work in the first place?" I don't know, but that's a different question than the one I was answering, which was an accusation that the logic is circular.
Aug
20
comment How can Kinetic energy formula and the Work formula be derived without assuming the other to be true?
I fixed a mistake in the derivation. I'm not sure what you're asking for with "how do we know the work formula"? I am taking that as the definition of work.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
No, that is completely wrong. Nothing in my answer says anything remotely like that. It is only because both position and velocity had constant magnitude and rotate at the same frequency that we can take their derivative in the same way. It is obviously not true for arbitrary functions.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
The answer does assume understanding some basic properties of derivatives, such as linearity.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
Yeah, I'm really not getting your point. Maybe you can think more about it, phrase it more clearly, and post it as a separate question.
Jun
22
comment Intuitive explanation for why centripetal acceleration is $\frac{v^2}{r}$
Sorry, but I can't understand your point at all. There's nothing about polynomials here. It's an argument based on geometry.
Jun
10
comment Size of a glass capillary for noticable capillary action
Downvoted because it's obvious that the material affects the height of the column.
Jun
9
comment Why are log scales so common?
see quora.com/Why-do-humans-perceive-logarithmically
Jun
3
comment Why is the light reflected at the same angle from mirror?
A single atom scattering isotropically violates momentum conservation.
Jun
1
comment Two people are holding either end of a couch, is one person exerting more force than the other?
This answer says the same thing as the upvoted answers; why is it downvoted?
May
30
comment Spin angular momentum conservation and entanglement
related: physics.stackexchange.com/q/4047
May
30
comment Black Body Golf Balls
Cool, thanks for following up.
May
29
comment Black Body Golf Balls
I find your fourth paragraph obscure. It appears that you think that 50% of the radiation from any patch of a hemispherical dimple will be re-absorbed, which I agree with. However, you then seem to ignore this fact and focus on "normals", the importance of which is unclear as "normal" refers to solid angle with zero measure and therefore zero radiation. Since 50% of the radiation from the dimple will be reabsorbed and its area is $2\pi r^2$, it emits the same power as a flat patch of $\pi r^2$, IE the same as a flat cap to the dimple would emit.
May
28
comment Black Body Golf Balls
Perhaps you are confused about the word "cross section". Your statement is simply wrong.
May
28
comment Black Body Golf Balls
No, the answer does not contradict itself. It discusses two scenarios: one in which the golf ball has exactly the same cross section as the sphere, and one in which the golf ball starts out the same as the sphere and has dimples added. Since those scenarios are different, they have different answers. The cross section determines what radiation is absorbed from the shell because it determines what lines of sight from the shell are blocked by the sphere. I don't understand how this isn't obvious.
May
28
comment Black Body Golf Balls
The cross section of an object determines what radiation it absorbs from the shell. I find this to be extremely obvious and do not see how it would be possible for you to fail to understand it. Your statement about "cross-sectional surface area (perimeter)" is very confusing. I don't know what you mean by that. I am referring to the cross-sectional area because it determines how much radiation is absorbed from the shell. I am not discussing the perimeter or the surface area.
May
28
comment Black Body Golf Balls
Based on your comment, you are clearly unwilling to read my answer carefully; I said very, very plainly in my original answer that if you start with a sphere and add dimples, the emitted radiation will go down very slightly. Your denial of that means you are not truly attempting to understand what I wrote.
May
28
comment Black Body Golf Balls
Suppose the sphere absorbs power $p_s$. Then, in order to remain the same temperature, it must also radiate power $p_s$. Because the golf ball is very nearly the same shape and size as the sphere, it absorbs very nearly power $p_s$, and so it emits very nearly $p_s$, so the golf ball and sphere emit and absorb very nearly the same power. The only difference is that adding dimples actually makes the golf ball's cross section very slightly smaller so it absorbs/emits slightly less than $p_s$. A very small sphere would absorb much less than $p_s$ and so would radiate much less.
May
28
comment Black Body Golf Balls
I will have to guess at what you're saying because it is extremely difficult to understand you. My best guess is that you have completely ignored the crucial part of my argument - that the golf ball and the sphere absorb the same amount of radiation as each other because they have the same cross-section. A small sphere clearly does not absorb the same amount of radiation as a large sphere, so the argument in my answer does not indicate that a small and large sphere would radiate the same amount as you appear to be implying.