How the surface area of a cone changes; special relativity

In summary, the surface area of a cone is S=πh2tanθ0 if it is at rest, and S=πh2tanθ1 if it is moving. If the cone has maximum radius R in the rest frame, it should remain R in the frame moving along the symmetry axis.
  • #1
Hiero
322
68

Homework Statement


A cone has half angle θ0 and lateral surface area S0 in the frame in which the cone is at rest. If someone moves at relative speed β=v/c along the cones symmetry axis, what surface area will they see for the cone?

Homework Equations


I believe the lateral surface area of a cone is S=πh2tanθ, where θ is the half angle and h is the height of the cone.
Length contraction is also relevant; h1=h0√(1-β2)

The Attempt at a Solution


S0 = πh02tanθ0
S1 = πh12tanθ1

If the cone had maximum radius R in the rest frame, it should remain R in the frame moving along the symmetry axis (because R is measured normal to this axis).
So we have tanθ0=R/h0, and also tanθ1=R/h1, so we see tanθ1=(h0/h1)tanθ0

Putting all this together (with the length contraction relation) it seems to be that S1=S0√(1-β2)

However my book claims S1=S0√(1-(βcosθ)2); so where is my mistake?
 
Physics news on Phys.org
  • #3
Ahh. For some reason, when I thought over that formula, I had in my mind that "R=h*sinθ" instead of "R=h*tanθ"
(I even remember thinking this formula over a couple times, and each time I kept incorrectly saying this to myself! I am tired.)

So then the correct formula is actually S=πh2tanθ/cosθ.

So then my answer is off by (needs to be multiplied by) a factor of cosθ0/cosθ1

If you take the ratio of these two equations
cosθ0=1/√(1+(R/h0)2)
cosθ1=1/√(1+(R/h1)2)=1/√(1+(R/h0)2/(1-β2))
and simplify a fair amount (then use R/h0=tanθ as well as 1/(1+tan2θ)=cos2θ) then multiply it by my incorrect answer then it indeed becomes the correct answer, so that was the only mistake.

Thank you Orodruin
 

FAQ: How the surface area of a cone changes; special relativity

1. How does the surface area of a cone change as the height increases?

The surface area of a cone increases as the height increases. This is because the surface area of a cone is calculated by adding the area of the base (a circle) to the lateral surface area (a curved surface). As the height increases, the curved surface becomes longer, resulting in a larger surface area.

2. Does the surface area of a cone change if the radius is changed?

Yes, the surface area of a cone changes if the radius is changed. The surface area of a cone is directly proportional to the radius, so as the radius increases, the surface area increases as well. Conversely, if the radius decreases, the surface area will also decrease.

3. How does special relativity affect the surface area of a cone?

Special relativity does not have a direct effect on the surface area of a cone. Special relativity deals with the relationship between time and space, and how they are affected by the speed of an object. The surface area of a cone is a geometric property and is not affected by the speed or motion of an object.

4. Is the surface area of a cone affected by the shape of the base?

Yes, the surface area of a cone is affected by the shape of the base. A cone with a circular base will have a different surface area than a cone with a triangular base, even if they have the same height and radius. This is because the shape of the base affects the lateral surface area of the cone.

5. Can the surface area of a cone be negative?

No, the surface area of a cone cannot be negative. Surface area is a physical property that represents the total area of an object's surface, and it cannot have a negative value. Even if the cone is inverted, the surface area will still be positive.

Similar threads

Back
Top