If a sphere is moving perpendicular to an observer and at relativistic speed will the sphere appear to be an ellipsoid or will it still look like a sphere?
To an outside observer, the sphere will appear shorter in the axis along which it is travelling.
What if it’s spinning?
Say, arbitrarily, it’s ROLLING left-to-right.
Will the top (moving faster) seem shorter than the bottom (moving slower)? So it’d look like its sagging?
That is also what I thought and I have even seen the correction to the retardation of an electric field explained this way. But if length contraction can be explained via a spacetime rotation then it would seem a sphere would appear as a sphere.
I also found this from Classical Dynamics page 544 by Marion and Thornton:
“Similarly the customary statement that a moving sphere appears as an ellipsoid is incorrect; it appears still as a sphere.” Supposedly this is due to what is called Terrell rotation. This is very confusing.
Sorry NardoPolo you snuck your post in on me.
A bump to see if any weekday participants might know the answer to this question.
In my Modern Physics class, the teacher told us that what you measure is not the same as what you see. That is, sticking a ruler next to the object and making a comparison you’ll measure it as being an oblate spheroid. But if you observe it via the photons hitting your eye, you’ll see something else. He said that a cube would look rotated in a sense, but we didn’t get any mathematical explanation of it.
Well, I’ll give a shot at explaining this. Let’s start with a cube.
There will be two vertical edges in the rear, one closer to you than the other; call this closer one A, and the other B. The cube is moving, and if it wasn’t clear, by “rear” I meant “in the back, given the direction of motion.” Clear so far?
Now, suppose at some time t photons from edge A and edge B both reach you. Hence, you see the two edges. Good. But edge A is closer, so its photon left later, right? And if it left later, it left from further forward, right? Thus, you see A as being further forward than B, and you can therefore see part of the back. Naturally, it also looks to be foreshortened.
So you’ve got two things going on: the object is foreshortened, and you can see part of the back. What’s special about a sphere is that the amount of the back you can see precisely cancels the amount of foreshortening, so that the entire thing looks like a sphere still. This means that if I painted the front half of a sphere black and the back half white, and sent it whizzing by you at 0.9c, you would see a sphere with less than half black and more than half white. You would, however, still see a sphere.
Obligatory helpful link.
Thanks that answers the question.
With a grid of clocks and measuring rods attached to your frame you’ll measure the sphere to be an oblate spheroid but you’ll still see it as a sphere.
Great link, I just wish I was a good enough searcher to have found it.
The reason that appearances and measurements are different in relativity is that it takes light some time to reach your eye. A relativistic measurement takes this time lag into account, whereas a mere observation does not. For the rest of this post, I’ll be talking in terms of measurements, not appearances.
Quoth NardoPolo:
A simple thought experiment will show that it doesn’t “sag” in shape. Suppose that we have our rolling ball contained in a non-rotating spherical shell. When we’re moving along with the ball (such that its center is at rest relative to us), the ball clearly must fit inside the shell. But the fact that the ball fits in the shell can’t depend on our reference frame. So the ball rolling past us at relativistic speed must have the same overall shape as the same ball moving without rotating.
Wy doesn’t the sagging occur? I’m pretty sure it’s because you’ve got to consider themotion of each part of the ball. A point on the front of the ball, for instance, isn’t just moving forwards. It’s moving down at an angle, so that’s the direction of its contraction. Do this for every point on the ball, and the effect on the overall shape should be the same as for no rotation.
However, there will be distortion. If we have a pattern printed on the surface of the ball (say, a globe of the Earth), then the continents will be moved around. Without running the numbers, I think that the result is that they’d be squished up together at the top of the ball (the part that’s moving fastest, at any given time).
I think it may also have something to do with the fact that the equations that hold for Special Relativity only work with inertial reference frames, ie, ones that aren’t accelerating. Since rotation implies an acceleration, SR can’t account for it. What the effects of General Relativity would be on the sphere, I can’t say with any degree of certainty.
Liked your thought experiment, BTW. Very clever.
Jeff