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Oct
15
asked Help understand article on thin shell formalism
Oct
10
awarded  Yearling
Oct
8
comment Why does the sea horizon line always seems to be at the same height as one's eyes?
@CarlWitthoft Ah, I didn't know. I only new the word in my language. So looked it up wikipedia in portuguese, looked the same article in english and thought that it was that the word I was looking for ^^ Thank you. I also learned linguistics today.
Oct
8
comment Why does the sea horizon line always seems to be at the same height as one's eyes?
Hum. I see your confusion. You're schema has one problem: scale. Your altitude will never be greater than the earth radius. Let me make a drawing and gather my comments as a proper answer. Hopefully i'll address your doubt.
Oct
8
comment Why does the sea horizon line always seems to be at the same height as one's eyes?
Well, no =) The triangle is just a simple visualization tool. Obviously you can pick any other kind of triangle to visualize the setup. The rectangular triangle has one advantage, though, at least for me. It's quick to prove my assertion mathematically if one uses the Pythagorean theorem and the definition of cosine.
Oct
8
comment Why does the sea horizon line always seems to be at the same height as one's eyes?
@CarlWitthoft English isn't my main language and in my main language we say "cateto" for cathetus. I don't know any other word for it, actually. Out of curiosity, since this comment will most likely be erased, what do you call a side of a triangle rectangle that isn't the hypotenuse?
Oct
8
comment Why does the sea horizon line always seems to be at the same height as one's eyes?
+1 for a very good question. My suggestion: draw a triangle rectangle with one cathetus horizontal and another vertical (imagine you're are on the top of that cathetus). Now take the horizontal cathetus to be very big. What happens to the angle between that cathetus (horizontal) and the hypotenuse (your line of sight)?
Oct
7
awarded  Popular Question
Sep
17
comment Differentiation in general relativity
The index $\mu$ is called a dummy index that's why it can be replaced be any other letter in the (implicit) sum. See Einstein notation.
Sep
17
accepted Why should any physicist know, to some degree, experimental physics?
Aug
27
comment Why should any physicist know, to some degree, experimental physics?
@MattReece In the sense that ultimately physics, as all sciences, must be about the real world. Sure we all have a sense of beauty and elegance and we like to transpose that into our theories but what really should matter is if those theories might or not explain the universe, as such be able to be falsifiable. So a "proper" theoretical physicist should always have that in the back of its head or it just dwells in the fantastic, and to me that is not doing science.
Aug
27
awarded  Popular Question
Aug
27
awarded  Nice Question
Aug
26
comment Why should any physicist know, to some degree, experimental physics?
+1 Nice examples. Your point was basically what I meant with my first point.
Aug
26
asked Why should any physicist know, to some degree, experimental physics?
Aug
14
comment Why is the Schwarzschild radius the radius of an event horizon?
I laughed so much. Thank you for your time. I'll check Physics SE later, then.
Aug
13
comment Why is the Schwarzschild radius the radius of an event horizon?
Btw, just to make sure I understand what you're saying: when you say that, for example, $U$ is spacelike you're saying that if I consider curve, say, $c$, parameterized by a parameter $\lambda$, given by the parametric equations $c(\lambda)=(c^v,c^u,c^\theta,c^\phi)$ whose tangent vector is $\dot{c}(\lambda)=(0,\dot{c^u},0,0)$ then $c^uc_u>1$?
Aug
13
comment Why is the Schwarzschild radius the radius of an event horizon?
I believe that your answer is by far the most complete so I'm inclined to set it as the correct answer. Nevertheless, I'm having an hard time figuring out how do you know that the coordinate $U$ is spacelike for $r>2M$ and timelike otherwise. Do you conclude that from the expressions for $U$? Because I can't see it. Another problem am having is figuring out why does the fact that $U$ is timelike in the interior region implies that the light cone always point to the singularity.
Aug
13
revised Why is the Schwarzschild radius the radius of an event horizon?
edited title
Aug
13
asked Why is the Schwarzschild radius the radius of an event horizon?