Can a Rotating Sphere at Relativistic Speeds Create a 'Black Spot'?

[yuiop]

... the last one was informative. It looks like contraction is the only relevant Lorentz effect.

I agree.

 

[starthaus]

Thanks kev the last one was informative. It looks like contraction is the only relevant Lorentz effect.

False, you need the complete set of Lorentz transforms, length contraction alone is not sufficient for explaining the effect. You can buy the Penrose paper here

 

[Austin0]

False, you need the complete set of Lorentz transforms, length contraction alone is not sufficient for explaining the effect. You can buy the Penrose paper here

Thanks for the link but I can't really take advantage of it as I can't luxuries right now.

Once again ,you seem to be very familiar with the original paper so could you just quickly say what effects and where they are applied. No detailed explanations needed.
25 words or less

 

[Austin0]

Here is an excellent website that gives you all the mathematical details.

PS I checked out this site ALthough fasc8inating as I have gotten heavily into computer 3-d animation in the past it didnt seem to have any relevant math regarding the original derivation. It was all for writing functions and scripts for a 3-d program.

Thanks

 

[yuiop]

It looks like contraction is the only relevant Lorentz effect.

Yes. As far as I can tell, we only require the length contraction transformation of Special Relativity, to work out that an object that is physically a sphere in its rest frame S, is physicaly an oblate spheriod in frame S' when it has relative inertial motion. After that, all that is required is standard ray-tracing, taking into account the velocity of the oblate spheriod in S' and the finite speed of light, to work out that it can visually appear to be a sphere to observers at frame S'.

We can also note that this apparent visual unobservability of the length contraction of a sphere is only aproximately true, very close to the object and at greater distances from the sphere, the length contraction is increasingly visually observable. For non-spherical objects, the apparent inability to visually observe the length contraction is even less true. It is odd that the very special case of the inability to visually observe the length contraction of one specific shape of object at very limited distances, has led to the popular misconception / myth that length contraction of any object at any distance, is not visually observable.

 

[starthaus]

Yes. As far as I can tell, we only require the length contraction transformation of Special Relativity,

There is no such thing as "the length contraction transformation of Special Relativity".
The Lorentz transforms are space and time transforms and you need both of them to solve this problem:

$$x'=\gamma(x-vt)$$
$$t'=\gamma(t-\frac{vx}{c^2})$$
$$y'=y$$
$$z'=z$$ to work out that an object that is physically a sphere in its rest frame S, is physicaly an oblate spheriod in frame S' when it has relative inertial motion.[/quote]

You need to find all points at $$t'=k$$ in S' (line of simultaneity in S'). In order to do that you will need the second Lorentz transform:

$$k=\gamma(t-\frac{vx}{c^2})$$

The above, solved for $$t$$:

$$t=k/\gamma+vx/c^2$$

Substitute $$t$$ into the expression for $$x'$$ :

$$x'=x/\gamma-vk$$

or:

$$x=\gamma(x'+vk)$$

Subsitute the above into the equation of sphere in S:

$$R^2=x^2+y^2+z^2$$

$$R^2=\gamma^2(x'+vk)^2+y'^2+z'^2$$

meaning that the object is an ellipsoid in S'. The cross-section of the ellipsoid for a viewer situated on the common x-axis is a circular disc. Indeed

$$x'=a$$ means:$$y'^2+z'^2=R^2-\gamma^2(a+vk)^2$$

Note that :

$$y'^2+z'^2<R^2$$

For observers not situated along the common x axis, the situaton is more complicated, there is no obvious proof that such an observer obtains a circular disc as the photograph of the ellipsoid.

After that, all that is required is standard ray-tracing, taking into account the velocity of the oblate spheriod in S' and the finite speed of light, to work out that it can visually appear to be a sphere to observers at frame S'.

Can you prove the above? Mathematically, I mean.

We can also note that this apparent visual unobservability of the length contraction of a sphere is only aproximately true, very close to the object and at greater distances from the sphere, the length contraction is increasingly visually observable.

It is still unobservable if the sphere is not textured. Only if the sphere is textured, it is observable.

For non-spherical objects, the apparent inability to visually observe the length contraction is even less true.

Yet, no experimental proof exists (to date). No one has managed to photograph length contraction.

It is odd that the very special case of the inability to visually observe the length contraction of one specific shape of object at very limited distances, has led to the popular misconception / myth that length contraction of any object at any distance, is not visually observable.

There is no such "myth" amongst people who know physics.