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When you solve a problem like this, you are using a system of reference (actually you use one in all problems, but here it is very explicit). In this case, the easiest one is y in the vertical and x in the horizontal. Almost all the forces are already in one of these 2 directions. Namely, you have all the weights pointing downwards, so in the -y direction ...

1

How about this for a more "physical" definition: two points in space-time are time-like separated if and only if a massive particle starting at one could, if subjected to appropriate finite forces, reach the other. Replace "massive" with "massless" to get the definition of light-like separation. If neither is possible the the points are space-like ...

1

Let $\lambda, \mu, \nu$ be functions on the reals to points (events) in spacetime. Let these be "straight" curves, in the sense that $\lambda', \mu', \nu'$ each all have the same direction for all values of their parameters. For example, $\lambda(u) = \lambda_0 + lu$ is a simple case, as $\lambda'(u) = l$. The vector $l$ is the vector along the direction ...

2

First, let's assume we are working in Minkowski space just for simplicity (altough this can be argued too in a curved space-time using curves that connect the events/space-time points). This space has an affine structure, you can think of it as vectors connecting pairs of points. This vectors are geometrical objects and we don't need to specify a basis for ...

0

For example: From "Gravitation and Spacetime" via Google Books Added to address a comment below: My question was intended to be addressed without referring to any coordinates; including no coordinates (!) such as "clock times T, T1, T2" or somesuch. But these clock times aren't coordinates. The reading of a single clock is the proper time ...

1

Here's a contrarian opinion: there's no such thing as seeing a fully rounded earth! Standing at the top of a high building, you look out from the center of a circular (fully round?) disc (with a hump from the spherical earth). The edge of that disc is the horizon, the farthest point you can see in any direction. Since you are above that rim, you are ...

5

This will be limited by our field of view (FOV). I couldn't find a better source, but Wikipedia says the vertical range of the field of view in humans is typically around 135° and the horizontal range around 180°. So for the Earth to be entirely within your field of view it will be limited by your vertical range. And by using a little bit of geometry you ...

0

I see the problem you have here, change the r*r drdϕdz there to z*r drdϕdz, then you should get the correct answer

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Your question is a bit muddy, but I believe the answer to everything is "yes, but..." Here's what's described in the referenced article: The laser produces a small spot on the target and the lens in the camera creates an image of that spot on one (ideally) pixel. By using simple lens equations, or by similar triangles, if you know the lens focal length ...

2

"What is the probability that a particle's speed lies between $u$ and $u+du$.?" Short answer The probability that a particle's velocity has an $x$-component between $u_x$ and $u_x+du_x$, and similarly for $y$ and $z$, is proportional to the volume taken up by that small cubic chunk of velocity-space, $du_xdu_ydu_z.$ Given these three components of the ...

2

The intuition is that $4\pi u^2$ is the area of a sphere of radius $u$, and now to find the volume of the thin shell between radius $u$ and radius $u+du$, you multiply the area of the surface of the shell by the thickness of the shell and find that its volume is $4\pi u^2du$. We can make this intuition more precise in a number of ways. Here are two: ...

1

When doing surface integrals like you say, you would always normalise by $R^2$. So, if you give a ray's direction by the spherical co-ordinates $\theta,\,\phi$, and you want the solid angle subtended by a bundle of such rays (e.g. a light field) converging on some point, then it is by definition the area of the part of the sphere "pierced" by the bundle. ...

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John Rennie's answer seems fine to me (+1). I'll only add the relevant pieces of the BIPM brochure (PDF, p. 118). BIPM rules.

6

The solid angle is defined as the area on the unit sphere subtended by the angle divided by one unit area. It's a ratio so it's a single dimensionless number. I see why you think it should be a 2D quantity, because the surface of a sphere, and any patch on it, is a 2D manifold and you need two quantities (traditionally $\theta$ and $\phi$) to map it. When ...

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