I want to understand Einstein's relativity. Below I demonstrate my ignorance in the hopes of expelling it.

Question

I am observing a rigid rod moving at half the speed of light, parallel to me. On the rod is a light source and a mirror. An observer is also on the rod. To the observer on the rod, light moves at the speed of light, both approaching the mirror and returning from the mirror. But to me as the "stationary" observer, light should also appear to be moving at a constant speed in both directions. Assuming the rod is approaching me at half the speed of light, how can it be that I observe light to be traveling at a constant velocity both toward the mirror and from the mirror while at the same time the observer on the rod sees this.

Grant me your wisdom oh theoretical physicist gurus...

Okay. What I am looking for is someone to correct my scenario so that relativity is illustrated correctly. Someone? Buehler?

First Answerer, note third answerer's response. This is basically the set up I am considering. However, third answerer only considers when the light on the ship is traveling away from me. When the light on the ship is traveling toward me, shouldn't time appear to contract rather than expand, and isn't this an inconsistency? In other words, doesn't the usual explanation for relativity as given by answerer 3 contain a logical flaw? And if not, then what am I not seeing?

Answer 1

Your question doesn't make any sense for two reasons.

1) How can a rod be "moving at half the speed of light, parallel to me" and at the same time "approaching me at half the speed of light"? Either it is moving parallel or it is approaching me. It can't be doing both.

2) If light is moving from the passenger on the rod to the mirror and being reflected back to the passenger then you can not observe that light at all. It never reaches you. That light has a round trip that remains within the length of the rod. Your FOR is outside the length of the rod. You can't possibly observe that light.

>>>EDIT

So you want someone to ask your question for you, and then answer it for you? Is that it?

Answer 2

The following is my own crackpot notion and not to be confused with accepted scientific theory.

But I like this visualization for the apparent contradiction of the "I am standing on a train moving very fast, and I shine a flashlight in the forward direction of the train, why doesn't the light go at the sum of the speeds of the train and 'c'"

And my visualization sort of implies an 'ether' which I don't mean to imply, but then again, why not? Here goes:

Imagine you are in a rowboat on a still lake and you have been rowing vigorously and worked up quite a velocity. You decide to rest and lift your oars. They are still wet and dripping drops onto the lake surface below them. As each drop hits the lake, it creates a circular ripple that expands outward from the point of contact.

In this model, the ripple is expanding at the speed of light (speed of propagation of the wave front in the medium). The boat is also moving through the same medium. But the ripple is expanding from the point where the drop hit the lake. That energy release happened at the moment of contact and did not inherit the boat's velocity at all. Hence the boat is now simply 'racing the ripple' and there is no additive connection between them.

Hence getting back to my flashlight, the physics which causes photons to be created by the flashlight does not bind the forward velocity of the flashlight itself to the photons created by it. In fact, since each photon can be thought of as an ever expanding sphere of energy, the center of that sphere is actually being left behind as the train moves forward. Hence we are racing the wavefront and the train's velocity, by some measure, now makes that wavefront seem just a little slow to us...

..which violates all we hold scientifically holy, so hold on a second.. umm...

so I am hold a great stopwatch and start it when the photon is born and stop it when a photodetector (moving with the train, and held in front of the flashlight) sees the wavefront hit it... and we try to ignore the propagation time of the 'information' about when we sense the hit... The time I measure is ultimately the time it took the wavefront to move the distance from the flashlight to the detector PLUS the distance the center of the sphere has fallen behind us due to the train speed. Which again makes the light seem a bit slow to me.

OK, so the stopwatch and my consciousness exist at the surface of the flashlight. when the wavefront hits the detector, the information about the hit travels back to me at the speed of light, so I am slow to stop the stop watch and the time appears even longer.

Welll. I never liked relativity anyway ;p Time is the thing which should be constant, and let everything else adjust as needed to fit.

And I LIKE the oar metaphor, so I am going to stick with it. Maybe someday I will be proven right.

Answer 3

Your question is not trivial and it is true that no matter what the velocity of the observer, the speed of light will be the same. This strangeness is resolved by the Special Theory of Relativity, and the result is that time and space are not the same for the two observers. I will provide a simple exampe (similar to yours but simpler):

You are on a spaceship travelling away from earth at close to the speed of light. You shine a light toward the front of the ship, and note that it takes d/c seconds to get to the front, where d is the distance to the front and c is the speed of light. Now I observe you doing this from earth. I also see the light travel from you to the front of the ship. I also observe the speed of light as c. However, I see the front of the ship moving away from the light; the space ship speed is close to that of the light; as far as I'm concerned, it will take a long time (maybe days) for the beam to reach the front of the ship. Yet you, the traveler on the ship, sees the light reach the front in a very short time! Therefore, time is not the same for me and the traveler.

The full theory also predicts dimension and mass changes as well.

Answer 4

The answer to your question is yes both you and the observer on the rod will see the light beams moving at the same constant speed of C. However, you and the observer will not agree as to when the beam of light hit the mirror and then return back to the observer.

For the observer on the rod, if the rod has length L, then he sees the beam of light traveling at speed C going from him to the mirror at the other end of the rod. The time it takes for the light beam to move from one end of the rod to the other is t=L/C. Therefore it will take twice that or 2t for the light beam to return back to the observer from the mirror.

Now you the stationary observer on the ground see the rod as L'= L/gamma, gamma = 1/sqrt(1-v^2/C^2) = 1.155 for v=0.5C
or L'=0.866L This is called Lorentz Contraction.

Because of Lorentz Contraction, you will see the light beam hit the mirror at time t'=L'/C=0.866L/C=0.866t. Hence, you will see the light beam hit the mirror before the observer on the rod. Then the light beam will return back to the observer at 2t'=1.732t.

(Please note: the above answer assumes that the rod is traveling towards you for the entire duration of the light beam moving from one end of the rod to the mirror and returning.)

So there is no contradiction or violation of Special Relativity here.

Answer 5

But to some it up, it basically comes from the fact that to you, the light is not only moving up and down, but sideways along with the rod, causing it's path from your point of view to appear diagonal.

This means on it's way from one mirror to the other, the light has a longer path to take from your point of view. In order for you, and the person on the rod to see the light stike the same mirror at the same time, from your perspective, since the light has to take the longer path, it must speed up, BUT it CAN'T speed up because the speed of light is constant. As a result, you have to wait a lot longer to see the light strike the mirror. In fact, the same is true for EVERYTHING you observe on the rod. Meanwhile, the person on the rod observes the same phenomena when looking at you. Thus, you both see time running slower when you look at eachother.

Now, the faster the rod goes, the longer the path you see the light taking becomes but the up down distance remains the same. Eventually, as the rod approaches the speed of light, this diagnal path will become so long that it will appear almost as a straight line to you. This means you NEVER see the light strike the mirror. Thus, to you, time on the rod appears to have stopped.

Answer 6

The entire scenario demonstrates why it is called relativity. Each element of your question is relative to the physical position you put yourself in at the time and place you pose the question. Depending on where you are in the moment your perception of time and space is relevant to your present reality.