Okay, so I need to figure out how the number of stars per unit range of parallax (plx) changes with plx, assuming stars are uniformly distributed in the sky. Hmm, let's start by recalling what parallax means. Parallax is a measure of the apparent shift in a star's position in the sky when observed from Earth over six months. The closer a star is to us, the larger its parallax. So, parallax is inversely related to distance. 

Since the stars are uniformly distributed, their number density should depend on how they are spread out in space. But how does that translate to the parallax? Let's think about the distribution of stars in terms of distance. If stars are uniformly distributed, their number density in terms of distance (say, in parsecs) would follow a certain pattern. Wait, no, uniform distribution in space would mean that the number of stars in a spherical shell of radius r and thickness dr is proportional to the surface area of the shell times dr. The surface area of a sphere is 4πr², so the number of stars in a shell dr would be proportional to r² dr. 

But parallax is 1/distance, so if distance is r, then parallax plx = 1/r. So if r = 1/plx, then dr in terms of plx is dr = -1/plx² d(plx). But since dr is a small change in r, the negative sign might not matter here. So the number of stars per unit parallax range would be the number of stars in dr divided by the change in parallax. Let me think again.

The number of stars in a shell of radius r to r+dr is 4πr² n dr, where n is the number density (stars per volume). But we want the number per unit parallax. Since parallax is inversely proportional to r, the differential change in parallax d(plx) is related to dr. So let's express dr in terms of d(plx). Since plx = 1/r, then r = 1/plx. Differentiating both sides, dr = -1/plx² d(plx). The negative sign indicates that as plx increases, r decreases. But since we're looking at the absolute change, the magnitude is 1/plx² d(plx). 

So the number of stars in a shell dr is 4πr² n dr. Substituting r = 1/plx and dr = 1/plx² d(plx), we get:

Number of stars per d(plx) = 4π (1/plx)^2 n * (1/plx²) d(plx)

Wait, that would be 4π n (1/plx²) * (1/plx²) d(plx) = 4π n / plx^4 d(plx). So the number per unit parallax range is proportional to 1/plx^4. But wait, that would mean the answer is B. But let me check again.

Alternatively, maybe I should approach this using the concept of volume in parallax space. The volume element in terms of parallax would be something else. Let's think in terms of probability. The number of stars with parallax between plx and plx + d(plx) is proportional to the volume of the region corresponding to that parallax range. Since parallax is 1/distance, the region between plx and plx + d(plx) corresponds to distances between 1/(plx + d(plx)) and 1/plx. Let's compute the volume of this region.

The distance ranges from r1 = 1/(plx + d(plx)) to r2 = 1/plx. The volume of the spherical shell between r1 and r2 is 4π(r2³ - r1³). Since d(plx) is small, we can approximate r2³ - r1³ ≈ 3 r2² (r2 - r1). Let's see: using the approximation that (r2³ - r1³) ≈ 3 r2² (r2 - r1) when r2 ≈ r1. 

r2 - r1 = 1/plx - 1/(plx + d(plx)) ≈ (plx + d(plx) - plx) / (plx (plx + d(plx))) ) ≈ d(plx) / (plx²). So substituting back, the volume is approximately 4π * 3 r2² * (d(plx)/plx²). But r2 is 1/plx, so r2² = 1/plx². Therefore, the volume becomes 4π * 3 * (1/plx²) * (d(plx)/plx²) ) = 12π d(plx) / plx^4. 

Thus, the number of stars in this volume is proportional to this volume times the number density in terms of distance. Wait, but the number density in terms of distance is n (stars per unit volume). So the number of stars would be n * 12π d(plx)/ plx^4. Therefore, the number per unit d(plx) is proportional to 12π n / plx^4. 

But wait, the problem states that stars are uniformly distributed in the sky. Wait, uniformly distributed in the sky? That might mean that their distribution is uniform in the celestial sphere, but perhaps their depth (distance from Earth) is not considered. Hmm, maybe I misinterpreted the problem. If the stars are uniformly distributed in the sky, does that mean their positions on the sphere are uniform, but their distances could vary? Or does it mean they are uniformly distributed in three-dimensional space? The wording is a bit ambiguous. 

Let me check the original question: "Assume stars are uniformly distributed in the sky." The phrase "in the sky" usually refers to the celestial sphere, meaning their positions are uniformly distributed over the sphere. However, their distances from Earth would still vary. So the density in three-dimensional space might not be uniform. Wait, but if they're uniformly distributed in the sky (angular distribution), but their distances are such that the density per unit volume is uniform. Hmm, maybe not. Let me think again.

If stars are uniformly distributed in the sky (angularly), but their distances are such that for any given distance, the number per unit volume is uniform. Wait, that might not make sense. Alternatively, if the stars are uniformly distributed in three-dimensional space, then their density is uniform (n is constant). But the problem states "uniformly distributed in the sky," which might mean that their angular distribution is uniform, but not necessarily their density in three-dimensional space. 

Wait, maybe the problem is using "uniformly distributed in the sky" to mean that their positions are uniformly distributed over the celestial sphere (i.e., their angles are random and uniformly distributed), but their distances from us could be anything, possibly with some distribution. But the question is about how the number of stars per unit parallax changes with parallax. 

If the stars are uniformly distributed in the sky (angularly), but their distances are random, then the number density in three-dimensional space would not be uniform. The number density in three-dimensional space would be n(r) = (number of stars per unit area on the sphere) * (1/(4πr²)) * (number of stars per unit depth dr). Wait, that's getting complicated. Let me try another approach.

Alternatively, if the stars are uniformly distributed in the sky (angularly), but their distances are such that the number per unit volume in three-dimensional space is uniform. Then, the number density n is constant. But in that case, the number of stars per unit parallax would depend on the volume element in parallax space. 

Wait, maybe another way: the number of stars in a given parallax range is the integral over the corresponding distance range. If the stars are uniformly distributed in three-dimensional space, the number density is n = constant. The volume between r1 and r2 is 4π (r2³ - r1³). The number of stars in that volume is 4π n (r2³ - r1³). Since parallax plx = 1/r, then r = 1/plx. So plx1 = 1/r1, plx2 = 1/r2. The parallax range is from plx1 to plx2. The number of stars in that parallax range is 4π n ( (1/plx1)³ - (1/plx2)³ ). 

But wait, the question is about the number per unit parallax. So dN/dplx = derivative of N with respect to plx. Let's compute dN/dplx. 

N = 4π n ( r³ ) evaluated from r1 to r2. But r = 1/plx, so N = 4π n ( (1/plx)^3 - (1/(plx + d(plx)))^3 ). For small d(plx), we can approximate this as 4π n (3 (1/plx)^2 * d(plx) ), using the expansion (1/(x + dx))^3 ≈ 1/x³ - 3/x^4 dx. So the change in N is approximately 4π n * 3 (1/plx^2) d(plx). Therefore, dN/dplx ≈ 4π n * 3 / plx^2. Wait, that's proportional to 1/plx². But that doesn't match any of the options. Hmm, what's wrong here?

Wait, maybe I made a mistake in the approximation. Let's do it more carefully. Let's let r1 = 1/(plx + d(plx)), r2 = 1/plx. Then N = 4π n (r2³ - r1³). The derivative dN/d(plx) is 4π n * d/d(plx) [ r³ ] evaluated at r = 1/plx. Using the chain rule, d(r³)/d(plx) = 3 r² * (-1/plx²) = -3 (1/plx²)³ * (1/plx²) )? Wait, no. Let me compute it step by step.

r = 1/plx. So dr/d(plx) = -1/plx². Then, d(r³)/d(plx) = 3 r² dr/d(plx) = 3 (1/plx²) * (-1/plx²) = -3 / plx^4. Therefore, dN/d(plx) = 4π n * (-3 / plx^4 ). But since we're looking at the absolute value (number per unit parallax), it's 12π n / plx^4. Wait, but the negative sign indicates that as plx increases, the number decreases, which makes sense. So the number per unit parallax is proportional to 1/plx^4. So the answer would be B. But earlier when I thought of the volume element, I also got 1/plx^4. So this seems consistent.

But wait, the problem statement says "stars are uniformly distributed in the sky." If they are uniformly distributed in the sky, that would mean their angular distribution is uniform, but their distances are not necessarily uniform. So the three-dimensional density might not be uniform. Wait, but earlier I assumed uniform three-dimensional density. Maybe the problem is not specifying whether the density in three-dimensional space is uniform or not. 

Alternatively, if the stars are uniformly distributed in the sky, perhaps they are uniformly distributed in the spherical coordinates (theta, phi), but their distances are random. Then, the number density in three-dimensional space would be n(r) = (number per unit area on the sphere) * (1/(4πr²)) * (number per unit depth dr). Wait, that might not be the right way. Let me think.

If the stars are uniformly distributed in the sky (angularly), then the number per unit solid angle is constant. But the solid angle dΩ = dθ dφ (for small angles). The total number of stars is the integral over all solid angles of the number per solid angle. However, their distances can vary. So the number of stars in a shell between r and r+dr is the number per solid angle times the solid angle of the shell. The solid angle of the shell is 4π (since it's all directions) multiplied by dr / (r² + ...). Wait, no, the solid angle of a shell at radius r is 4π (r² dr) / r² = 4π dr. Wait, that can't be. Wait, the solid angle subtended by a shell at radius r is the area of the shell divided by r². The area of the shell is 4πr² dr. So the solid angle is (4πr² dr) / r² = 4π dr. So the solid angle of a shell dr is 4π dr. Therefore, the number of stars in the shell dr is (number per solid angle) * 4π dr. If the number per solid angle is uniform, say k, then the number in shell dr is 4π k dr. Then the number density in three-dimensional space is 4π k, which is uniform. So if the stars are uniformly distributed in the sky (angularly) with uniform surface density, then the three-dimensional density is uniform. 

Wait, that's an important point. If the stars are uniformly distributed in the sky (angularly) with a uniform surface density k (stars per square radian), then the number of stars in a shell dr is 4π k dr. Therefore, the number density in three-dimensional space is 4π k, which is uniform. So in this case, the number density n = 4π k is constant. Then, the number of stars per unit parallax would be proportional to the volume element in parallax space times the density. 

Earlier, when assuming uniform three-dimensional density, we found that the number per unit parallax is proportional to 1/plx^4. But if the three-dimensional density is uniform, then the number per unit parallax is 12π n / plx^4, which is still proportional to 1/plx^4. So the answer remains B. 

But wait, the problem states "stars are uniformly distributed in the sky." Does that imply uniform surface density (angular), leading to uniform three-dimensional density? Because if the stars are uniformly distributed in the sky, their angular distribution is uniform, but their distances can vary. If their surface density is uniform, then the three-dimensional density is uniform. But if their distances are randomly distributed, then the three-dimensional density might not be uniform. 

Wait, the problem says "uniformly distributed in the sky," which typically refers to the angular distribution. So the stars are spread out uniformly over the celestial sphere. But their distances from Earth are not specified. However, the question is about how the number of stars per unit parallax changes with parallax. 

Assuming that the stars are uniformly distributed in the sky (angularly) but their distances are random. Then the number density in three-dimensional space isn't uniform. The number density in three-dimensional space would be n(r) = (number per solid angle) * (1/(4πr²)) * (number per unit depth dr). Wait, no. Let's clarify. 

The number of stars in a small solid angle dΩ is proportional to dΩ. The solid angle dΩ is 2π sinθ dθ dφ (for a small area). The volume element in three-dimensional space is r² sinθ dr dθ dφ. So the number density in three-dimensional space is the number per solid angle divided by the volume per solid angle. The volume per solid angle is r² dr dΩ. Therefore, the number density n(r) = (number per solid angle) / (r² dr). But if the number per solid angle is constant (uniform distribution), then n(r) = constant / r². So the number density decreases with r². 

Wait, that's different from the previous assumption. So if stars are uniformly distributed in the sky (angularly), the number density in three-dimensional space is n(r) = k / r², where k is a constant. Then, the number of stars in a shell dr is n(r) * 4π r² dr = (k / r²) * 4π r² dr = 4π k dr. So the number density in three-dimensional space is uniform? Wait, no. Wait, n(r) dr is the number per unit length along dr. But integrating over dr, you get 4π k dr, which is uniform in dr. So the number of stars per unit length along the line of sight (depth) is uniform. 

But in terms of parallax, which is inversely proportional to distance, we need to relate dr to d(plx). Since plx = 1/r, dr = -1/plx² d(plx). The number of stars per unit parallax is the number per unit dr multiplied by dr/d(plx). 

The number per unit dr is 4π k. But dr = -1/plx² d(plx), so the number per unit d(plx) is 4π k * |dr/d(plx)| = 4π k * (1/plx²). Therefore, the number per unit parallax is proportional to 1/plx². But that contradicts the earlier result. Wait, now I'm confused. 

Wait, perhaps I need to be careful with the definitions. Let's start over. 

If the stars are uniformly distributed in the sky (angularly), the number per solid angle is constant. Let's say the number per steradian is N. Then, the number of stars in a shell of radius r and thickness dr is N * 4π r² dr * (dr / r²)?? Wait, no. The volume element is r² sinθ dr dθ dφ. The solid angle is dΩ = 2π sinθ dθ (for a given φ). The number of stars in the shell is N * dΩ. So the number of stars in the shell dr is N * 2π sinθ dθ * dφ. But integrating over φ from 0 to 2π and θ from 0 to π, the total number of stars in the shell dr is N * 4π r² dr. Wait, that can't be. Wait, no. The total number of stars in the shell dr is N * dΩ * r² dr? No, I think I'm mixing up terms. 

Wait, the number of stars per steradian is N. The solid angle of the shell is 4π (since it's all directions) multiplied by dr / r². Wait, no. The solid angle of a thin shell at radius r is 4π * (dr / r²). Wait, how? The solid angle Ω is the area on the sphere divided by r². The area of the shell is the surface area of a sphere of radius r plus dr minus the surface area up to r. The differential area is approximately 4πr² dr. Therefore, the solid angle Ω of the shell is (4πr² dr) / r² = 4π dr. So the solid angle is 4π dr. Therefore, the number of stars in the shell dr is N * Ω = N * 4π dr. 

Therefore, the number density in three-dimensional space is dN/dr = 4π N. Wait, that's a constant. So the number density in three-dimensional space is uniform, n = 4π N. 

But then, when we relate this to parallax, the number per unit parallax would be the number of stars in a shell corresponding to a change in parallax. Since parallax plx = 1/r, dr = -1/plx² d(plx). The number of stars in a shell of parallax d(plx) is dN = 4π N * |dr| = 4π N * (1/plx²) d(plx). Therefore, the number per unit parallax is 4π N / plx², which is proportional to 1/plx². So the answer would be A. But this contradicts the earlier conclusion. 

Wait, but earlier when I assumed uniform three-dimensional density, I got 1/plx^4. But now, assuming uniform angular distribution leads to uniform three-dimensional density, which gives 1/plx². Which one is correct? 

I think the confusion arises from the definition of "uniformly distributed in the sky." If it means uniform in solid angle (angular distribution), then the number density in three-dimensional space is uniform (n = 4π N). But if "uniformly distributed in the sky" is interpreted as uniform in three-dimensional space (i.e., uniform density in volume), then the answer would be different. 

The problem states "stars are uniformly distributed in the sky." The term "sky" typically refers to the angular distribution. So it's more likely that the stars are uniformly distributed over the celestial sphere in terms of angle, but their distances are such that the three-dimensional density is uniform. 

Wait, but if the three-dimensional density is uniform, then the number of stars per unit parallax would be proportional to 1/plx². But earlier calculations under different assumptions gave conflicting results. 

Alternatively, perhaps the correct approach is to consider the number of stars in a given parallax range. Let's model the parallax distribution. 

Parallax plx is 1/r. The number of stars with parallax between plx and plx + d(plx) is equal to the number of stars in the distance range r to r + dr, where dr = -1/plx² d(plx). 

If the stars are uniformly distributed in three-dimensional space (i.e., uniform density n), then the number of stars in dr is n * 4π r² dr. Substituting dr = -1/plx² d(plx), the number is n * 4π (1/plx²)² (1/plx²) d(plx) = 4π n / plx^6 d(plx). So the number per unit parallax is proportional to 1/plx^6. But this contradicts previous results. 

Wait, maybe that's not the right way. Let's think again. 

If the stars are uniformly distributed in three-dimensional space (density n = constant), then the number of stars in a shell between r and r + dr is 4π r² n dr. Since r = 1/plx, dr = -1/plx² d(plx), the number per d(plx) is 4π (1/plx²)² n (1/plx²) d(plx) = 4π n / plx^6 d(plx). So dN/d(plx) = 4π n / plx^6. Therefore, the number per unit parallax is proportional to 1/plx^6. But this is not one of the options. So perhaps my initial assumption is wrong.

Alternatively, if the stars are uniformly distributed in the sky (angularly), leading to uniform three-dimensional density, then the number per unit parallax is proportional to 1/plx². But the problem states "stars are uniformly distributed in the sky," which might mean that their angular distribution is uniform, but their distances are random. 

But the problem is asking about the number of stars per unit parallax. The parallax depends on the distance. If the three-dimensional density is uniform, then the number of stars in a shell dr is 4π n r² dr. Substituting r = 1/plx and dr = -1/plx² d(plx), the number per d(plx) is 4π n (1/plx²)² (1/plx²) d(plx) = 4π n / plx^6 d(plx). So again, 1/plx^6. But the options don't include this. 

Alternatively, perhaps the problem is simpler. Let's think in terms of probability. If stars are uniformly distributed in the sky (angularly), then the number of stars per unit parallax is determined by the area on the celestial sphere. 

The parallax is inversely proportional to distance. For a given parallax, the stars lie on a sphere of radius 1/plx. The area of this sphere is 4π (1/plx)^2. The number of stars per unit area on the sphere is uniform. Therefore, the number per unit parallax is proportional to the area of the sphere at that parallax. 

Wait, but parallax is a measure of distance, so for a given parallax, the stars are at a distance r = 1/plx. The number of stars in a shell of radius r and thickness dr is proportional to the area of the shell (4π r²) times the thickness dr. But dr is related to d(plx). 

But if the stars are uniformly distributed in the sky (angularly), the number per unit solid angle is constant. Let's say the number per steradian is k. Then the number of stars in a shell of radius r and thickness dr is k * 4π r² dr. Since r = 1/plx, dr = -1/plx² d(plx), substituting gives k * 4π (1/plx²)² (1/plx²) d(plx) = k * 4π / plx^6 d(plx). So the number per unit parallax is proportional to 1/plx^6. But again, this isn't one of the options. 

I must be making a mistake here. Let me think differently. 

Suppose the stars are uniformly distributed in the sky (angularly), and their distances are random. Then the number density in three-dimensional space is not uniform. The number of stars in a shell of radius r and thickness dr is proportional to the solid angle times the number per steradian times the volume. 

Wait, the number per steradian is k. The solid angle of the shell is 4π (dr / r²). So the number of stars is k * 4π (dr / r²) * r² dr? Wait, that can't be right. Wait, the volume of the shell is 4π r² dr. The solid angle subtended by the shell is (4π r² dr) / r² = 4π dr. Therefore, the number of stars in the shell is k * 4π dr. So the number density is k * 4π. 

But then, the number of stars per unit distance (dr) is constant. So the number per unit parallax (d(plx)) would be proportional to dr / (plx²), since dr = -1/plx² d(plx). Therefore, the number per unit parallax is proportional to 1/plx². Which would be option A. 

But this contradicts the earlier result. I must be missing something. 

Alternatively, let's use a different approach. The parallax is plx = 1/r. The number of stars in a shell between r and r+dr is n(r) * 4π r² dr, where n(r) is the number density at radius r. If stars are uniformly distributed in the sky (angularly), then n(r) is constant because the number per unit solid angle is constant. So n(r) = k, a constant. Then the number of stars in the shell is k * 4π r² dr. Substituting r = 1/plx and dr = -1/plx² d(plx), the number per d(plx) is k * 4π (1/plx²)² (1/plx²) d(plx) = k * 4π / plx^6 d(plx). So again, 1/plx^6. 

But the problem states that stars are uniformly distributed in the sky. If this means uniform in three-dimensional space, then the number per unit parallax is 1/plx^4. If it means uniform in angular distribution (solid angle), then the number per unit parallax is 1/plx². But the options don't have 1/plx^6. 

Wait, maybe the problem is simpler. Let's think in terms of volume. The number of stars per unit parallax is the number of stars in a shell where parallax is between plx and plx + d(plx). The parallax is inversely proportional to distance. So the volume element in parallax space is related to the distance cube. 

But I'm getting stuck here. Let me look for another approach. 

The number of stars per unit parallax is the number of stars in a shell of radius r = 1/plx and thickness dr, divided by dr. But since dr is related to d(plx), we need to relate the volume element in terms of parallax. 

The volume element in terms of parallax is dV = 4π r² dr. Since r = 1/plx, dr = -1/plx² d(plx). Therefore, dV = 4π (1/plx²)² (-1/plx²) d(plx) = -4π / plx^6 d(plx). The negative sign indicates direction, but since we're considering absolute values, the number of stars per unit parallax is proportional to 1/plx^6. 

But again, this isn't one of the options. The options are A: 1/plx^2, B: 1/plx^4, C: 1/plx^3, D: 1/plx^1. So perhaps the correct answer is not among the options, but that's unlikely. 

Alternatively, perhaps the problem is considering that the number of stars per unit parallax is proportional to the solid angle. Since solid angle is 4π, which is constant, but that would imply a constant number per parallax, which isn't the case. 

Wait, maybe I should think in terms of the number per parallax bin. For a given parallax, the number of stars is proportional to the surface area of the sphere at that distance. The surface area is 4π r². The number per unit parallax is the surface area divided by the parallax range. But parallax is 1/r, so for a small change in parallax d(plx), the corresponding change in distance dr is dr = -1/(plx)^2 d(plx). The number of stars is proportional to 4π r² dr. Substituting r = 1/plx and dr = -1/(plx)^2 d(plx), we get 4π (1/plx²) * (1/plx²) d(plx) = 4π / plx^4 d(plx). So the number per unit parallax is proportional to 1/plx^4. Which is option B. 

Ah, so this seems to align with option B. Therefore, the correct answer is B. 

But earlier when I considered uniform three-dimensional density, I also got 1/plx^4. So if the stars are uniformly distributed in three-dimensional space (constant density), then the number per unit parallax is 1/plx^4. If they are uniformly distributed in the sky (angularly), leading to uniform three-dimensional density, the same result holds. 

Therefore, the answer is B. ~ 1/plx^4.
</think>