Light beam path from a moving light source

One argument.

If case B was true, how would a distant observer ever form an image of the source as it moved?
 
Ned Bate - I apologize for missing your response, I was working in another thread. I agree with you - "Yes, you have it exactly right. The path of the photon will also determine where the source appears to be for anyone who happens to perceive the light." - At the time the light was emitted...tell me, when you look at the sun (not wise) are you seeing the sun where and as it is exactly at that moment? or are you seeing it where and as it was 8 minutes ago? - KitNyx
 
Or is this the case:

photons2.gif
 
superluminal said:
I think the drawings are misleading as they present light "rays" appearing instantaneously.

What do you think?
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Well, I think that the rays should propagate at the finite rate of c, but I see your point. Give me time to learn how to make an animation
 
Thought experiment:

A tube, one light MINUTE long and one light SECOND wide, is moving uniformly at 0.5c similar to the animation arrangement above. A pulse of light 1 second long is emitted. What happens?

Use SR to show what must happen.

(you may have guessed that I think I know the answer to this whole thing... I'm very interested in everyones answers.)
 
superluminal said:
Thought experiment:

A tube, one light MINUTE long and one light SECOND wide, is moving uniformly at 0.5c similar to the animation arrangement above. A pulse of light 1 second long is emitted. What happens?

Use SR to show what must happen.

(you may have guessed that I think I know the answer to this whole thing... I'm very interested in everyones answers.)
Click to expand...
Who's second are we using for the pulse emission time? The formulation suggests the tube frame second, so I'll assume that, below.

We'll assume time 0 for the time of emmision for both the tube and embankment frame.

From the tube frame: The pulse definitely hits the detector, because the emitter is pointed straight at it, neither are moving, and light propagation is isotropic. The pulse is emitted for one second in the tube frame, so it is detected as being 1 second long. It hits the detector at tube time 60 sec. to 61 sec.

In the embankment frame: The tube and embankment frame agree on events happening, so the pulse definitely hits the detector. That means that in the enbankment frame the pulse is travelling at an angle so as to hit the detector.

Why is this so? Doesn't the isotropy of light propagation imply that photons move in the manner of your (i.e. superluminal's) first picture? Well, no. Both observers must agree on the events that happen, so if light strikes the sides in one frame, the it must also strike the sides in another frame. But light striking the sides is inconsistent with the assumption that light propagation is isotropic in all frames, which implies that the light must hit the detector in the tube frame. And if it didn't, it would imply an absolute reference frame.

For fun, let's calculate some times.

The tube is seen as length contracted in the direction of travel, but the tube's long side is perpendicular to its direction of travel, and thus still 60 lightseconds long. Hence, length contraction is not a factor in this thought experiment.

Let the time for the light to get from the emitter to the detector be T. The path of the pulse is the hypotenuse in a right angled triangle with sides 60 lightseconds and ½c * T long, the hypotenuse itself being of length cT. Setting c = 1 (which factors out, anyway), we have (Pythagoras)

T^2 = (½T)^2 + 60^2

which solves to approx. T = 69.28 sec. Note that this gives us gamma = 1.15 (ca.) directly because both observers agree the light hits the detector at tube time 60 to 61 sec. (And this gamma factor is (whew) the same as if we calculate it directly by the standard formula)

Now, the pulse is emitted for exactly 1 second on the tube clock, which is time dilated by the factor gamma in the enbankment frame, which we've measured directly. So the pulse is emitted for 1.1547 enbankment seconds, hence the detector is hit by the pulse at enbankment time 69.28 sec. to 70.43 sec. (ca.)

Yes, I know that's probably not quite what you had in mind, but it was fun, nonetheless.


I'd say it is case B, and the isotropy of light propagation proves it.
 
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KitNyx said:
Is there someone here who does not subscribe to case A? Just curious...sometimes it is hard to tell exactly what people are trying to say (including myself I am sure).

- KitNyx
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Yes, me, by the reasoning above (in case people find that too long-winded.)
 
Both the cases are wrong. JamesR was partly right, and I too was inclined to agree with him at first. I'm sure what he meant was what I'm going to say, from his ball-throwing analogy. The 'B' figure is partly correct, in that the ends of the rays will be as shown. But the actual paths will be as in the figure below.

<P>
<Img src= "http://www.geocities.com/virusmakermad/laser.PNG"></Img>

<P>
I'll attempt to make this clear, and perhaps, also open a venue for arguments from MacM. Imagine that laser device also has a cap, as show in the section in the bottom half of the diagram. Section A shows what is seen by an observer who is moving along with the laser. In this frame, the beam travels directly upwards and reaches the cap. Section B shows our frame of reference (in which the laser is moving). In this frame also, the beam has to reach the cap. But by the time the beam (or the tip of it, actually) reaches the top, the laser, and the cap, will have moved away. So the beam actually has to take a diagonal path in order to reach the cap. Simple reasoning shows that the beam will take this path even if there is no cap. Thus solved.

<P>
Now, MacM, I suppose will want to know why the beam <I>has to</I> reach the cap in both frames of reference. Apart from the argument that this is what is happens according to experience, there are also reasons which occur to one's common sense. I'm in no mood to elaborate them right now.
 
No...Figure B shows a "beam" of light that MOVES with the emitter. I agree that light contains information about its source, but that is the EXTEND of their conection once the photons are emitted. The photons will continue to travel in a straight line until the energy is absorbed/ relected/ refracted etc. Anyways, this is why we see objects not where they are at this moment, but as they were at the moment of releasing the observed/translated photons. If the answer were ANYTHING other than Figure A then we would have a violation of causality and relativity.

- KitNyx
 
Rosnet, I agree with you that the paths of the individual photons will look like your diagram makes out. But how does the beam (consisting of many sequentially outputted photons) look? Your path analysis would lead to the beam looking like case B, wouldn't it?

KitNyx, consider superluminal's thought experiment. Do you honestly think the pulse doesn't hit the detector, as case A suggests? Wouldn't that be a blatant contradiction of isotropic light propagation?
 
Ok. From the rest frame of the moving tube, in which we sit sipping tea at one end, the light "beam" must appear to be completely normal, straight and true. SR tells us that in this inertial frame, all of physics behaves normally. The detector goes of and all is well.

From the rest frame of the embankment the detector still goes off (it is an event that happened after all). however, the tube has moved 1/2 light minute to the right (wrt us) since the first photons were emitted.

So from the tube the beam must look like A and from the embankment it must look like B:

beam1.gif

Notice how cleverly I even drew the emitter and detector in figure B as length-contracted in the direction of motion. Spacetime itself is skewed wrt the embankment due to relative speed alone.
 
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funkstar said:
Rosnet, I agree with you that the paths of the individual photons will look like your diagram makes out. But how does the beam (consisting of many sequentially outputted photons) look? Your path analysis would lead to the beam looking like case B, wouldn't it?
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Funkstar makes an important distiction between "path" and "beam". It is my fault for using both terms in the title of the thread.

Both Rosnet and Superluminal have drawn diagrams that show the light path on an angle in the same direction as the motion of the laser. However, I do not think that they really intended to claim an angle with respect to the laser. I believe these diagrams are just another way of choosing Case B.

My original question was whether there was an angle or not, and by drawing the diagrams this way, (although they are technically more correct when considering our rest frame where we are not allowed to even turn our heads)it becomes unclear as to whether the angle is being claimed to exist.
 
Neddy,

From the embankment the angle must exist. The photons travel in the skewed spacetime of the tube as seen from the embankment and must follow the path of the beam (any "Gunslinger" fans out there?) as drawn. It's ka.
 
superluminal said:
Neddy,

From the embankment the angle must exist. The photons travel in the skewed spacetime of the tube as seen from the embankment and must follow the path of the beam (any "Gunslinger" fans out there?) as drawn. It's ka.
Click to expand...
I agree that we see a forward angle from our rest frame, but unfortunately my original question (and drawings) were whether or not we would see the beam as being on some angle relative to the laser (or tube in your case).

I may have accidentally put us into the laser's frame, but all I meant us to do was to follow the moving laser with our eyes. We were supposed to detect whether there was an angle (backward) or whether the beam stays vertical with respect to the moving laser. I think that Case B is the same as your and Roswell's drawings, if you consider the location of the laser in each of my original drawings.

Consider my claim that Case A represents acceleration. In that case, drawing it as you and Rosnet have, you could have a vertical line that actually represents the beam hitting the sides of your tube. Technically, it is probably more correct to fix our rest frame as rigidly as you have, but it complicates the matter, in my opinion. I find it easier to just allow my eyes to follow the moving laser.
 
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