[SOLVED] Reflection-speed of light

olga11

1. The problem statement, all variables and given/known data

You hold a mirror in your hands and you look yourself on the mirror while you are moving with almost the speed of light. What is the difference with your reflection on the mirror when you are not moving?

2. Relevant equations



3. The attempt at a solution
I believe that when I move with the speed of light I will see myself bigger because of the length increase. My reflection's dimensions will increase and I will be twisted.

rl.bhat

You can see your face when light from your face falls on the mirror and reflects back to your eye.And it requires cirtain finite time interval. To complete this process, when you are moving with the speed of light, light must emitt from your face with a velocity more than the velocity of light which is impossible. So you won't see your face when your are moving with speed of light.

Dick

The problem said "almost the speed of light". The problem doesn't make sense "at the speed of light". You and the mirror are travelling in the same inertial frame. What makes you think you will see anything unusual at all?

olga11

In the beginning I thought that something "extraordinary" would happen at nearly speed light. Now I have studied more and I believe the answer is the following:

Special Theory of Relativity by Einstein

1. The speed of light in a vacuum is a constant.

2. The rules of physics have to work the same in all frames of references — moving or at rest — as long as the moving frames are not accelerating.

So if I am traveling at a relativistic speed I can see myself in the mirror in the same way I see myself when I am moving.The rules of nature are the same for both situations.

Light from my face bounces off my mirror and reflects back at light speed, even though I am traveling at nearly the same speed. Since I am not accelerating I can see myself in the same way.

Is this answer correct and satisfactory enough? What do you think?

ANOTHER QUESTION

An observer is standing still on the surface of the Earth. A spherical spaceship moves above him at a relativistic speed. What shape does the observer on Earth think that the spaceship have?

Can we say again that since the spaceship is not accelerating the shape remains the same?

Thank you both for your interest.

learningphysics

Your answer to the first question looks good to me.

But your answer to the second question doesn't. In the first question... you were together with the mirror... whether you were still or moving... you and the mirror were in the same inertial frame of reference. Like you said, the laws of physics are the same... the situations were identical.

In this second question, you need to compare what an observer in the same inertial frame as the spaceship will see (ie the moving frame)... with what an observer in a different inertial frame of reference will see (earth frame). The situations are not identical. An observer in the same frame of reference as the spaceship will see a sphere... but not the person in the earth frame.

olga11

Let us suppose that the spaceship moves with the x-axis parallel to the earth. The observer on Earth sees this dimension bigger (length contraction in the moving frame), so the shape is now an oblate spheroid. Or is it not?

Shooting Star

"Seeing" and observing are two different things. The latter pertains to measurements, which will show the spaceship to have contracted in the direction of motion.

Strangely enough, If a photograph is taken of a sphere moving at a relativistic speed, the image on the film will be a sphere, but rotated. So, the answer to your second question is that the person standing on the earth will see a sphere.

Shooting Star

It's called the Penrose-Terrell Rotation. Read the basics here:

http://math.ucr.edu/home/baez/physic...R/penrose.html

olga11

It is a little difficult, I have to admit, and English is not my mother-tongue.

So we can say that the spaceship will look flattened, but it will be a sphere on a film.

Shooting Star

No, the spaceship is flattened, according to any possible measurement you can make. But light takes different times to come to your eyes or the camera from different parts of the sphere, and the result is that it looks spherical, but a bit rotated.

olga11

http://www.aps.org/meetings/unit/dpp...upload/vay.pdf

I read this article some time ago and I had the impression that the spaceship would look flattened.

My answer would be:

Length in the direction of velocity is contracted from the point of view of the observer on Earth, so the spacecraft would be a flattened sphere.


But with your explanations I think I am able to understand my mistake. I considered time to be universal, not dependent on the reference frame.

Thank you, Shooting Star.

olga11

http://www.spacetimetravel.org/fussball/fussball1.html

"A rigid body which, measured in a state of rest, has the form of a sphere, therefore has in a state of motion - viewed from the stationary system - the form of an ellipsoid of revolution [...]. While the Y- and Z-dimensions of the sphere are not modified by its motion, the X-dimension is shortened in the ratio 1:[1-(v/V)2]1/2, i. e. the more the larger v ist." (Albert Einstein, 1095 in [1]. Here, X is the direction of motion of the sphere and v is its velocity; V is the speed of light.)

Is this picture wrong?

Attached Thumbnails

 

Shooting Star

(Are you trying to KILL me or something? That was written by you know who...) [BTW, it should be 1905.] But, yes, the picture is wrong.

The word "viewed" used above means "measured". The Penrose-Terrel rotation pertains to the situation if we could actually see the moving sphere with our eyes. A better option would to photograph it with a high speed or whatever camera. Then the image would not seem flattened, if the sphere was passing you just in front of you. The sphere will seem elongated if it's approaching you.

More here: http://en.wikipedia.org/wiki/Terrell_rotation

olga11

you mean 1095? I just did "copy" and "paste". What is BTW?

olga11

FINAL ANSWER

"Seeing" and observing are two different things. The latter pertains to measurements, which will show the spaceship to have contracted in the direction of motion.

The contraction can be measured, but the measurement is frame-dependent.

According to Terrel-Penrose rotation,thanks to the differential timelag effects in signals reaching the Earth observer from the spherical spaceship's different parts, a receding spaceship would appear contracted, an approaching spaceship would appear elongated and the geometry of a spherical spaceship passing you just in front of you would appear skewed.

Strangely enough and in contradiction with a previously-popular description of special relativity's predictions, in which an observer sees a passing object to be contracted (for instance, from a sphere to a flattened ellipsoid), if a photograph is taken of a sphere moving at a relativistic speed, the image on the film will be a sphere, but rotated.

And since I am asked to present a simulation of it in 2-d in excel, I will answer that I cannot do better from

http://www.spacetimetravel.org/fussball/fussball1.html

References: Shooting Star.

How does it look to you? Excellent????????????

Thank you very much.

Shooting Star

It should be 1905 because that was the year Einstein published "On The Electrodynamics Of Moving Bodies", which we know now as the basis of special relativity. BTW means "by the way".

I have a feeling you have understood the concept "observing" and "seeing" quite well. Excellent, I do say!

BOL. (Best of luck.)