A Common Misunderstanding Of Special Relativity

AnssiH · July 23, 2014

Well, since a slightly different analysis implies different results, at least one or both must be overlooking something. Let's see if we can spot what it is.

The critical factor is the fact that what has just been described is identical to leaving the solar system as it is and instead adding 0.999999c in the opposite direction to every element in the rest of the universe. So let us look at the consequence of that act.

I would propose there must be a difference in whether or not we consider the trajectories of all the photons-in-flight to also change, or whether we just accelerate the sources. You suggested "every element in the rest of the universe" but I wasn't sure if you meant only massive elements or light also.

The visual distortion being discussed in earlier posts is related to the idea that apparent trajectories of light would change as a function of the velocity of the observer (similar to the apparent angle of rain falling down), so if in your thought experiment the trajectories of "photons" are not considered to change along with the velocity of the rest of the universe, it would take 4.2 years for anyone on earth to notice that Proxima Centauri is moving. If the trajectories are considered to change (simultaneusly in earth frame), the apparent angles would have to change the same way as they would if the observer or earth being accelerated in that "sea of photons".

Either way, let me develop a thought experiment that shows how I'd figure the above, and displays few other important details.

* Consider a lab frame, where there's a pipe placed orthogonal to the ground; pointing straight down.

* Consider a pulse of light moving through the pipe; the trajectory of the light pulse is plotted as moving orthogonal to the ground.

* Consider a ship in the lab frame moving parallel to the ground. Plotting the same situation from the frame of the ship, the pipe is moving parallel to the ground, and the light pulse moving down through the pipe is plotted as moving in a non-orthogonal angle to the ground.

The above reflects a change in the plotted angle of any light as a function of velocity of the observer (or symmetrically the velocity of the source if it was moving when the light was emitted; moving the source after the emission cannot affect the angle of the light anymore)

But this implies that also a natural observer must catch the light in varying angles as a function of his own velocity, as follows;

* We have the ship fly underneath the pipe so that the pulse of light is passed through a conveniently placed hole on top of the ship, and then passed again through a conveniently placed hole on the bottom of the ship.

* Plot this situation in the lab frame, and it is obvious the holes on the ship cannot be aligned so that they are together orthogonal to the ground (i.e directly above each others). The time it takes for the pulse of light to travel from the top of the ship to the bottom of the ship, the ship has moved forward. The appropriate location of the bottom hole is a function of the speed of the ship.

* Plot the same situation in the ship's frame. Now it is clear any observer standing where the bottom hole would be, must observe the pulse of light as a blink spotted through the hole on the ceiling (a hole which is not directly overhead of the observer, it is in an angle towards the front). At the moment of the blink, the observer must also see any other collection of photons that were emitted from the source at the same time; they see the device that was used as source passing past the hole at that moment, making it appear the device is in front angle from the ship. They must also see all of this "through the pipe"; meaning they see the pipe as non-orthogonal to the ground.

Note; The moment the beam of light reaches the observer, both of the frames plot the device as "actually" sitting directly overhead of the observer. In the lab frame the pulse of light can only be received directly underneath the pipe, in the ship the observer sees the source where it was in the ship frame when the light pulse was emitted; that would be somewhere towards the front, even though in Newtonian diagram the source is plotted as existing directly overhead of the observer (in both frames).

To get an idea of the magnitude of the distortion;

* Consider that the ship must be plotted as length contracted in the lab frame. Under this plotting, near the speed of light the length of the ship is approaching 0, and the light pulse entering the ship must be seen as hitting the back wall of the ship almost instantly after entering the ship.

* Plot the same situation from the point of view of the ship, and an observer at that back wall looking towards the front must see the blink of light through the hole on the ceiling (and the source device and the rest of the objects in the lab frame); meaning he must see everything approaching from an extreme front angle.

This idea is reflected in the equation;

http://en.wikipedia.org/wiki/Relativistic_aberration

Which first appeared in here;

http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

Page 16.

Now to clarify the difference between the source or the receiver starting to move in the middle of the experiment;

* Consider a version where the ship is at rest in the lab frame, waiting underneath the pipe. Clearly in this configuration, any observer inside the ship sees the light source as directly overhead through the pipe.

* We arrange a situation so that as soon as the pulse of light has passed through the hole on the top of the ship, the ship accelerates to a moving frame (instantly or otherwise). We do this so that once again the pulse will hit a particular point on the floor on the ship. Once again any observer by that location at the floor (which may well be the same observer as who was looking through the top hole, since the observer will not actually have to dmove in lab frame) will catch the light at an angle; as coming through the hole on the top. Whether we plot this in the ship or the lab frame makes no difference, the observation is the same; the blink of light, the source device, and the rest of the objects in the lab will visually distort towards the direction of acceleration, so that they appear to be visible through the hole.

This displays one thing that I think a lot of people do overlook; even in terms of relativity, the light beams must be seen as if they are "raining in" from their sources; an observer accelerating back and forth inside the rain will be catching the light in varying angles (much like what rain looks like from the window of a train). The speed of the photons is set to C as per convention, and the lengths of the paths are defined accordingly, but the angles must change.

And accelerating the sources around will not yield observable effects to the observer until the rain catches the observer, just like moving a showerhead will yield a delayed response, as follows;

* Do the reverse thing; start moving the light source, the pipe, and the rest of the lab, when the light has passed through the top of the ship (in terms of the simultaneity of the lab frame). The observer standing directly underneath the top hole will still receiving that pulse of light (the light-in-flight cannot change direction due to the source changing direction after it has been emitted). It is only little time later when the lab will start to look "distorted" as the light being emitted after the lab started moving finally catches the ship and the observer inside.

* To understand the symmetry here, consider a light beam that is emitted from the light source almost directly to the opposite direction of the motion of the lab right after the lab started moving. The beam is only slightly tilted downwards towards the ship. Consider that from the lab's (new) frame, the ship is now plotted as moving away almost at speed C racing right beneath the beam of light, and the beam of light is edging towards the ship slowly (the light is plotted as moving exactly at C, and the ship is moving slightly less). When the beam of light has moved down far enough, it will eventually pass through the hole on the top of the ship, and neatly land onto an observer standing almost directly beneath the hole. This observer will at that moment see the light source directly above; that shall be the last moment he sees the source through the hole. That light beam is plotted as almost completely orthogonal to the ground by the observer, but it must be plotted as almost completely parallel to the ground by the lab.

This displays the difference between starting to move the receiver (immediate aberration effect) or starting to move the source (delayed aberration effect), even though afterwards it cannot be defined which one is moving.

Now remove the idea of holes and such on the ship, and make the complete ship transparent, and all the results must be the same across the board; if the ship starts moving through the universe, the rest of the universe must immediately appear to distort towards the direction of motion, because the angle that the light is entering the ship must start changing.

See if you can spot anything I'm overlooking?

Applying the above to your thought experiment;

If you consider a situation where the solar system has always been moving in relation to Proxima Centauri (i.e. we are ignoring any idea of one or the other starting to move at some defined point in time, when Newton's coordinate system plots Proxima Centaury as being "directly overhead" so to speak (when the *plotted* direction towards Proxima Centauri is orthogonal to the direction of motion between earth and Proxima Centauri), we would have to see PC in the position it was 4.2 years ago (as measured in Earth frame). Which yields a very extreme angle towards the front.

If we include the idea that the solar system and the rest of the stars are initially in the same reference frame, and then one or the other start moving at some point in time, there must be a difference between the idea that the solar system starts to move close to C, or that the rest of the stars start to move. If the rest of the stars start to move, we will notice Proxima Centauri as starting to move 4.2 years later. If we start to move, the light having beamed from PC 4.2 years ago will instantly become catched in a new angle because we hit the trajectories in new angle, visually displacing PC almost directly towards the front angle; exactly the same location where it was 4.2 years ago in terms of this new reference frame. 4.2 years later, it would visually appear to be directly overhead again. If the solar system would now stop back to the original frame, the angle in which we catch the light from PC would again start to shift, making PC again visually shift towards the direction of acceleration, revealing that we have in fact moved very far away from it.

Note that this case becomes symmetrical if in the case of moving the rest of the universe, we would also change the direction of all the light already-in-flight accordingly to their sources changing direction after the light was emitted.

In a nutshell this distortion would be simply related to the changes in angle that the "same light" would have to be catched in different reference frames due to the fact that it is moving at finite speed.

Edited July 23, 2014 by AnssiH

Doctordick · July 24, 2014

Anssi, you are making exactly the error I was talking about when I made my comments about the standard relativistic transformation equations.

What is critically important is that these representations do not represent what will be seen but is rather a representation of the dynamic system as represented in the space time coordinate system Newton invented to display the dynamics of his physics. In order to make that issue clear, let me point out that when one looks at the stars, one does not see them as where they are now (what Newton's coordinate system is designed to display) but rather where they were some time ago.

Your comment --* Plot this situation in the lab frame, and it is obvious the holes on the ship cannot be aligned so that they are together orthogonal to the ground (i.e directly above each others). The time it takes for the pulse of light to travel from the top of the ship to the bottom of the ship, the ship has moved forward. The appropriate location of the bottom hole is a function of the speed of the ship.-- is incorrect.

When you perform the relativistic transformation per the theory of special relativity you will not obtain the result that the device that was used as a source was directly above the hole at that moment. However the observer on the ship will indeed see the device that was used as source passing past the hole at that moment.

In the rest frame the device will indeed appear to be directly above the rest observer as it is not moving; however it is not necessarily the position when the light began its trip. When you perform the relativistic transformation you cannot use that position. You must instead correct that apparent position to where it was when it began its trip. When we are looking at a star (or anything for that matter) whose velocity is relatively slow with respect to the speed of light, that correction is generally negligible so we don't generally worry about it.

However, when you perform the correct relativistic transformations the result will be totally different. The transformation will yield the position of that object where it was in the moving observers frame when the light began its trip as reckoned in the moving observers frame. If the source were at rest in the rest observers frame it will be moving in the moving observers frame. When the moving observer corrects for that apparent motion he will end up with the apparent position being directly above him at the moment he sights the flash.

There is another way of looking at the same issue. Consider a moving observer with a tube pointing straight up with a light shining through that tube. Consider an object a slight distance directly above him at some fixed position in the rest frame. As he travels along the path he will see nothing until that beam of light strikes that object. When that light strikes the object both he and any observer in the rest frame will see a flash. Let that flash define the origin in both frames of reference (makes the relativity transformations easy).

If that object is a large distance above him he will see the flash as going and coming in the same direction however the round trip will take more time. He will again perceive the flash as being directly above him.

What will the moving observer see? It should be clear that the moving observer will see that flash as being directly above him (that is how his tube is pointed).

What will the rest observers see? It should be clear that they will see a beam of light starting before the moving observer gets under the reflection point which reflects off the object at exactly the moment the moving observer is directly under the object. They will also uniformly agree that the moving observer will see the flash when he is exactly the same distance beyond the object as he was before the object when the beam of light left his tube.

This is where the relativistic change in time comes into play. The rest observers will see the diagonal light path to the object as considerably longer than the path seen by the moving observer. But they will both exactly agree as to the speed of light since the rest observers will see the moving observers clock running slower by exactly the factor required to make the speed of light a constant.

Now replace the example I just gave with one where the flash originated as an explosion on the object. Everyone will see exactly the same results. Then consider the angle to some other point on the moving frame of reference. When the rest observers correct for the measurements to maintain the correct value of c they will obtain the exact same angle. You ought to be able to figure that case out.

Does that clarify things a bit?

Have fun -- Dick

Edited July 24, 2014 by Doctordick

AnssiH · July 24, 2014

What is critically important is that these representations do not represent what will be seen but is rather a representation of the dynamic system as represented in the space time coordinate system Newton invented to display the dynamics of his physics.

Ah then we've fallen prey to the good old semantical ambiguity that is so common in all the presentations of relativity. I thought your comment "exactly how does the universe appear when observed from a interstellar ship moving at an extreme relativistic velocity" was referring to what things look like optically, not what things look like when plotted in space-time diagram.

You referred to some of the extreme distortions effects being mentioned in earlier posts (for instance the very last comment on my post #32), so you must have assumed those were claiming a distortion as things are plotted on a spacetime diagram. But I've been also referring to optical distortions, namely aberration of light.

I didn't go back to check my terminology on all the posts, but I know I constantly spot myself using ambiguous terms, and then I try to change them into something being either "plotted", or "seen"/"observed", the former meaning plotting into a spacetime diagram, and the latter being what things appear to a natural observer when he just looks.

I got a clue that you must be using the word "see" also to refer to something being plotted, when you said "...they will see a beam of light starting before the moving observer gets under the reflection point...". With that in mind, I see what you were trying to say.

This probably makes my earlier posts seem much more sensical. You will see I have commented essentially on exactly the issue you were trying to clarify; that any optical effect like this must be compensated for in their navigation by calculations. That means exactly figuring out where the objects are "now" in their coordinate system.

Doctordick · July 28, 2014

Anssi, I think that you have once more misinterpreted me.

In the paragraph beginning with "What is critically important…" I am specifically referring to the coordinate transforms calculated via the standard relativistic transformations. The rest of the presentation is indeed talking about how things appear optically to the two observers here. Let me attack the issue again being a little more careful as to what I am doing.

We have two frames of reference here: the rest observer's frame of reference and the moving observer's frame of reference. In order to examine any phenomena from these two frames of reference, we must use the standard relativistic transformations in order to transform the actual phenomena from one frame to the other (that is the correct transformations required to maintain the actual physical phenomena per the Newtonian perspective).

However, since we are discussing how things will actually appear to the observers, we must be very careful that the analysis performed is explicitly in one frame or the other and not confused by making incorrect transformations during our analysis. That is why I set the experiment up the way I did. Let me do it again (slightly differently) and try to follow very carefully how and why I set it up the way I do.

Again I will first examine the moving observer's examination via a vertical tube; in this case, the tube will have a mirror at the bottom. I will let the moving observer have a laser pointing horizontal in the y direction at the top of his tube and a point object in the rest observer's frame on the x axis exactly a negligible distance above the path of the top of that tube. In addition, I will add a strip of photographic paper in the rest frame directly above the x axis.

When the moving observer's laser lights up that point object, both observers will see a flash of light when the point object is exactly in the path of the laser. I will use that flash to designate the origin at t=0 of both coordinate systems. Just to keep calculations simple, let the tube be 186,000 miles long (since it is vertical, both observers will agree that its length is exactly 186,000 miles long as z'=z per the relativistic transformation required to yield the Newtonian perspective).

Now let us discuss what both observer's will see. Both will clearly see the opening flash as occurring at the origin a t=0 and both will perceive a photon pulse heading vertically downward: i.e., they will both see the flash as lying on their z axis together with a specific component of that flash heading downward on their z axis which, when reflected off a horizontal mirror 186,000 miles below that flash, will yield a vertical pulse returning exactly to the origin of their personal coordinate systems two seconds later. That is why I made the moving observer use that tube: i.e., to point out that the relevant photon pulses observed by the two observers are different pulses (I didn't put the rest frames pulse in a tube because I didn't want the two tubes to interfere with one another and since it is moving straight down in the rest frame it is easy to comprehend where it is going in the rest frame observer's analysis).

At any rate, we clearly know exactly what the rest observer will see happening in his frame and what the moving observer will see in his frame. The real question is what does the rest observer see happening in the moving frame? It should be clear that he will see a photon pulse heading down the tube; however, as that pulse proceeds downward, the moving observer will be proceeding to the right (the positive x direction) thus the rest observer will not see that pulse as going straight down as that pulse must go at an angle (it has to in order to stay inside the tube). When it reaches the bottom of the moving tube, it will reflect off the mirror and head back up the tube (still going at an angle in order to stay inside the tube). Eventually that photon pulse will reach the top of the tube where it will impact that photographic paper just above the x axis.

We know exactly what the moving observer will see. He will see a photon pulse go straight down that tube, reflect off the mirror at the bottom and return to the top exactly two seconds later. In his frame that tube is oriented straight up and down.

The rest observer will see the same thing occurring in his rest frame but he will see something quite different in the moving observers frame. He can go over and inspect that photographic paper at his leisure. For the fun of it, let us presume the moving observer is moving at 99.9999 percent of the speed of light to the right (that would be 185,999.81 miles per second). We already know that the pulse will reach the paper in the moving observers frame exactly two seconds after the original flash but (via the relativistic transformations required to maintain the correct Newtonian physics) the rest observer will see the moving observers clock as running slow. In fact, by the standard relativistic transformations the rest observer will conclude that path took 447.2272 seconds as measured on his clock: i.e., the rest observer will see the distance between the origin and the exposed point on the photographic paper as being roughly 83,184,174.2 miles.

Notice that the above has very little to do with the cause of the flash being observed or where it came from. The only point of interest was that it passed through the origin of the two frames at t=0. Suppose the actual source of the flash were some event somewhere else in the universe. If the rest observer saw that pulse as proceeding straight down his z axis, how would the moving observer perceive it? In this case, in order to convert between the two coordinate systems (to maintain the Newtonian perspective) we would have to work from the position of that source when the flash originated (where it was when that flash occurred).

For the rest observer, that fact makes little impact. The velocity of the stars (or any other objects of interest) is negligible relative to the speed of light therefor the rest observer would essentially see the source as directly above him (along the z axis). At least that is where it was when the pulse started towards him.

The moving observer is a completely different story. If that source were essentially at rest in the rest observers frame, the moving observer would take it to be moving to the left at 99.9999 percent the speed of light and, in his frame, he would be looking at it where it was earlier: i.e., at that earlier moment in time it must appear to be directly above him (for our analysis, it must duplicate the situation where the laser produced the flash) .

What we really need to know is where is the moving observer in the rest observer's Newtonian frame of reference? Remember, the lock on position must correspond exactly to the moment that the flash on that object slightly above the top of the tube was at exactly the same point in both reference frames. Since we are concerned with the correct relativistic transformation, life gets a bit complex. In order to simplify matters, I suggest we presume the distant object is perfectly at rest in the rest frame of the rest observer. That is certainly a possibility and it is one which is perfectly consistent with the rest observer's observations. This puts it exactly on the z axis of the rest frame.

Since we want the photon pulse to arrive at that point exactly equivalent to the point object the laser illuminated, we need to know the exact position of the moving observer in the rest observer's Newtonian space: i.e., we need to look at his frame relativistically transformed into the moving observers frame. That is not a trivial matter.

In order to make that transformation a little easier, I will make one simple change in our picture. This is not a change in the physics but is rather a simple change in the defined phenomena itself. The pulse of interest will occur in that distant star, proceed to the entrance of the moving observer's tube, reflect off the mirror at the other end of that tube and then continue out back to a horizontal line describing the path of the star which produced that pulse as seen in the moving observer's frame. What is important here is that the path the pulse takes in the rest frame of reference is perfectly symmetric around reflection point at the bottom of that tube.

The critical issue here is where is the moving observer in the rest observer's frame of reference when the flash occurs at that distant star? Since the rest observer sees the moving observer as moving to the right at 185,999.81 miles per second, he will perceive the position of the moving observer as many billions of miles to the left of the rest observers position (that would be the time it takes the light to come from that distant star times the velocity of the moving observer).

To correctly calculate that distance to the nearest quarter inch is essentially impossible; however, there is another critical factor we can use implicitly. We can consider where the moving observer will be when that reflected pulse gets back to that symmetric path introduced above. Since the circumstance is perfectly symmetric around the reflection point of the mirror in the moving tube, the apparent path within the tube is well defined and the rest observer must see the top the tube to be exactly at x=0 when the pulse enters the tube.

Just out of interest, note that the rest observer sees things "where they were when the light he sees arrives at his eyes" not at the physically correct position in his Newtonian frame of reference. It should be clear that the light entering the tube arrives at the rest observer's eyes 223.6136 seconds before the reflected light arrives and will furthermore thus see that reflection point as being directly below the moving observer's origin (note that the relativistic transformations do not depend upon y or z): i.e., the moving tube in the Newtonian transform is dynamically vertical to the path of the moving observer. However, when he calculates where that mirror actually was in his frame of reference, he will conclude the light must have taken 223.6136 seconds to reach him (again, one second in the moving frame) and thus must have already been 41,592,129.6 miles to the right of the z axis. Furthermore, the time between the two events, the pulse entering the tube and exiting the tube in the rest observer's frame must be 447.2272 seconds (two seconds in the moving frame). Thus it is that the observation (the exposed point on that photographic paper) must be at x=83,184,174.2 miles from the origin ( an observer at that point must be looking down the tube to see the light). Clearly he must be looking at an angle towards the reflection point. (I don't comment on the fact that he can only look at a very specific time, that tube he is looking down is moving very fast past him.) The moving observer would be looking straight down.

The moving observer, who would see that star as directly above him, would conclude the actual position of that star as being directly above him man many years ago (when the flash occurred) and, correcting for its apparent velocity, would presume that it was actually many billions of miles to the right and is now many billions of miles to the left.

I hope you can follow the logic of that analysis and comprehend that both observers would "see" the star as directly over them.

So let us look at another circumstance. Once again I will use a relative velocity of 185,999.81 miles per second sighting tubes 186,000 miles long (that lets me use the same factors I used above). Suppose that tube was tilted at some specific angle theta in both the rest frame and the moving frame. The important parameter here is [math]d=sin\; \theta[/math]. (This time I will use two tubes and presuming you can think of them as not interfering with one another.) The first thing we must do is to work out the Newtonian representation of these two tubes. Using the standard relativistic transformations, the apparent height of the two tubes would be identical. The parameter d (measured in both frames) is clearly d as seen in the rest frame and would convert to d' = 0.004472d (that is d in the moving frame from the Newtonian perspective as measured in the rest frame). That is the relativistic foreshortening in the x direction predicted by special relativity.

Once again, let us set t=0 (the origin of both frames) at exactly the same point and consider a flash on some star which passes a photon pulse through that point at t=0. Once again the moving observer's tube is moving to the right in the rest observer's frame of reference as the light pulse goes down to the bottom and we have to know exactly how that phenomena appears to the observer in the rest frame. Once again both observers will see the star as being in the direction the tube is pointed and will use the correct speed of light to calculate where that light pulse actually came from. In this case, the critical issue is, does the path of that light pulse hit the side of the moving tube? If it does, then the two observers will find that the star will appear to be in a different position at t=0: i.e., the moving observer will not see the star as being in the direction the tube is pointing: i.e., at that angle theta.

In the rest frame, the observer will see the light pulse proceeding to the other end of his tube in exactly one second, reflecting and returning to the opening of the tube after two seconds. Since the star is at rest with respect to him he will see the tube as pointing at the star.

We must now follow the supposed path of the light pulse down the moving tube and see if it hits the side of the tube. The real issue is, does it hit the mirror at the other end after exactly the correct time required by the movement of that tube.

Since the rest observer will "see" that reflection as if it were at the position of the moving mirror when the reflection occurred, it must appear to be at exactly the position of that mirror 223.6136 seconds after the pulse entered the tube (that is one second in the moving frame) the mirror would thus appear to be 41,592,129.6 -d' to the right of the x=0 plane.

Again, the actual time for the pulse to return to the top of the moving tube would be 447.2272 seconds and a rest frame observer looking down that tube at that moment would have to be x=83,184,174.2-d' miles from the origin (since the tube is moving so fast in his frame, the only reasonable way to perform that measurement would be with that photographic strip going all the way back to x=0. It should be clear that this is indeed exactly the path required by the analyzed circumstances: i.e., the light coming into the moving tube at t=0 would indeed exit at exactly the position of the top of that tube after reflection.

In this case, we need not worry about where that star actually is as it must "appear" to the moving observer to be in the direction his tube is pointed. Since in his analysis it is moving to the left at 185,999.81 miles per second its actual position is once again many billion miles to the left.

In essence what I have just shown is that that the "apparent" distribution of objects in the universe as seen at a specific moment from a specific point will be exactly the same to all observers no matter what their velocity through that universe happens to be.

Check out what I have done very carefully and see if you can find an error in my analysis. What is important here is that the relativistic transformations do not yield how things actually look but is rather a transformation into the proper dynamic Newtonian space time representation which conserves the supposed dynamic phenomena.

In essence, that transformation provides a universal picture where the speed of light is the same in all reference frames. The light going through those tubes described above moves at exactly c in everyone's frame of reference and that has some important consequences.

Have fun -- Dick

AnssiH · July 29, 2014

I hope you can follow the logic of that analysis and comprehend that both observers would "see" the star as directly over them.

Yeah I think I was able to follow that completely and I did not spot any errors in it (although I would really encourage you to avoid using the word "see" to describe how something is plotted. For instance, neither of the observers can "see" the light moving down the tube).

But you are not analyzing the same thing as I was originally describing, because you are setting t=0 to the emission at the point source above the tube. That way you are analyzing what the transformation looks like around that point, not what happens from the observer's perspective. In all of my examples the t=0 is implicitly the point where the two observers pass each others at the moment of observing the same pulse of light arriving at the bottom of the tube, because that's what I'm interested of; how it appears to the observers who are at the same place at the same time, but just in different inertial frames (i.e. catching the same light in different angles)

I can see exactly what is going on here; exactly as you correctly described; when you perform transformation at your reference point, what you get is that both observers are looking at two completely different pulses of light when they catch a pulse that arrives directly over their heads (respectively).

And as a flip-side of the same coin, when the observers are passing each others, and they catch the same pulse of light, the implied angle of that same light is different for different observers, of course, just like you describe.

Effectively, when you say both observers will see the star as straight above their heads, what you are expressing is what they will see when they are at completely different locations from each others (and also time-like separated btw).

You are effectively pointing out that the reference for which pulse of light is moving "straight up" will change upon the relativistic transformation (which is true), but when you hinge the transformation around an origo outside from the observers, you are not analyzing the observer perspective at all.

So setting t=0 to the observer to get a result of what would happen from the point of view of the observer as they are changing their frame via accelerating to some direction. To make matters simpler, I'm assuming instant acceleration to some new frame, and then just figuring out the respective angles of some light-ray of interest.

Which is the same thing as imagining two observers, one moving, one at rest, superimposed over each others so that they will receive some pulse of light just when they pass each others.

So the moment of passing is set to be the same moment as when the pulse of light is received (the same pulse of light), and that is set as t=0.

If you do that analysis, the result you get is actually consistent with everything you said, and the implied angle of that same ray of light is a function of the speed of the observer. And also consistent to your statements, that angle rotates so that the light is seen as coming more from the direction of acceleration (i.e. in any new frame, you get the angle where the light source was in terms of that frame when the light was emitted)

Note also that if you take your exact thought experiment to this point;

The moving observer, who would see that star as directly above him, would conclude the actual position of that star as being directly above him man many years ago (when the flash occurred) and, correcting for its apparent velocity, would presume that it was actually many billions of miles to the right and is now many billions of miles to the left.

And now add an observer who is first tacking along with the rest observer, and at the moment of passing tacks along to the moving observer.

That third observer first must see the star as directly above (the star is stationary above him), and after tacking along to the moving observer, only sometime later he will see the light *again* as directly above him (he must see what the moving observer sees). So what must happen during the change of frame is the apparent position of the star moves somewhere far forwards; at relativistic speeds almost directly forwards. This is completely due to the same light getting catched in different angle, and that is exactly the distortion referred to earlier.

So in a nutshell, all of my thought experiments are concerned with representing what the *same* pulse of light looks from different frames, in order to figure out how things shift for an observer who is accelerating through the universe), not which pulse of light would the reference for "straight up" in each frame.

Edited July 29, 2014 by AnssiH

Doctordick · July 30, 2014

Anssi, I still get the feeling that you are not following my analysis carefully. Instead of referring to "catching a pulse that arrives directly over their heads", I am referring to the apparent direction of a pulse which enters the top of the two tubes at exactly the same time and place. In your mind, replace the top of the tube with a small camera whose shutter is fast enough to open and close essentially instantly and has a film fast enough to record that instant exposure. Let the distance between the film and the pinhole to be relatively small. The only reason I made the tube 186,000 miles long was to clearly demonstrate that the observed angles ended up being exactly the same in both frames.

Certainly we are concerned with the positions of the two observers. You would rather set the origin at the bottom end of the tube. Ok, let us do that and examine the differences. Once again I will point out that the relativistic correct transformations yield the distortions required to obtain the correct Newtonian referance frames necessary to figure out the correct dynamics in the two different frames. In particular, I want to point out that there are no distortions in the y or z direction. Thus the z axis in both frames (the center line of the tubes) points in exactly the same direction in both frames. No one has any argument with that conclusion.

Now, you want the observers to set their origins (the point where x'=x=0 and t'=t=0) at their eyeballs at the bottom of the tubes. Then, if their observations are to be made at exactly the same time when they are in the same place (the origin). I am afraid that once again you must conclude that they are looking at different photons. Again, let the source be a flash on some point; say the closest star to us (that would be roughly 4.3 light years by the way). Just to keep the calculations easy let us say that the flash occurred on a planet which, at the instant the flash occurred, was perfectly at rest with respect to our rest observer.

When that flash reaches the rest observer, it essentially constitutes a spherical surface 4.3 light years in radius (that is 135,693,964.8 miles). Notice that the rest observer will presume the photon the moving observer will see will enter the moving observers tube one second (that is in the moving observer's frame) before the moving observer detects it. The rest observer will conclude that this specific event occurred 223.6136 seconds prior to the critical detection at his origin (that is the one second in the moving observers frame converted to the rest observers time measurement). On the other hand, the rest observer will presume the photon he detects at the origin entered his tube only one second prior to his detection. The two observers will certainly agree that they are not detecting the same photon.

The rest observer will conclude that the moving observer was 41,592,129.6 miles short of the origin when the photon he will detect entered his tube (that is 223.6136 seconds times 185,999.81 miles per second). Notice that this exactly places the top of the moving observer's tube on the rest observer's z axis at that specific time: it follows that the event occurs exactly on the z axis of both frames of reference and the spherical nature of the wave front is of no significance. On the other hand, it also points out that the event being observed can not be a flash. From the rest observer's frame, the moving observer would see nothing as no photon from that flash could have yet reached that point.

That was the problem I avoided by having the origin at the top of the tubes. Notice that, in this case, this specific difficulty vanishes when the flash actually reaches both observers at the defined origin. What it really says is that we cannot use a flash. The light the two observers detect must be on for at least something like four minutes. Actually that is relatively unimportant; the original question was where would they see the object and I introduced the flash to avoid certain calculation problems. I began with a flash as seen by both observers in order to make the calculations convenient but it also required that reflection off the bottom to complete the construct.

Notice further that the frequency of that light is quite different in the two frames. The moving observer will also find the clocks in the star system to be running slow. The apparent frequency of any activity will appear time dilated. What is central here is that the actual observation can be difficult to unravel and you must be very careful so as to take exact account of what relativistic factors are critical and take them into account properly.

There is one other interesting observation I could make here. From the moving observer's perspective, the star he is observing is moving to the left at 185,999.81 miles per second. It follows that every second he watches it, it moves 185,999.81 miles. However, being 4.3 light years away, that is an angle of only 0.0785 degrees. It would take about 12.73 seconds for it to appear to move one degree. What I am pointing out here is that most of the stars out there are considerably further away and their apparent movement would likewise be slower.

I also noted that you brought up acceleration. That brings in general relativity and the consequences of general relativity is much more complex and very difficult to do correctly. I can also prove (quite easily as a matter of fact) that Einstein's theory of general relativity is erroneous. Thus it is important that you should make every effort to avoid including any acceleration at all.

Have fun -- Dick

AnssiH · July 31, 2014

Anssi, I still get the feeling that you are not following my analysis carefully. Instead of referring to "catching a pulse that arrives directly over their heads", I am referring to the apparent direction of a pulse which enters the top of the two tubes at exactly the same time and place.

Yes, of course if you have two tubes in there it's not possible to arrange a single pulse of light pass through them.

Let me comment step by step on few points so you can establish better what I have going on in my mind;

In your mind, replace the top of the tube with a small camera whose shutter is fast enough to open and close essentially instantly and has a film fast enough to record that instant exposure. Let the distance between the film and the pinhole to be relatively small. The only reason I made the tube 186,000 miles long was to clearly demonstrate that the observed angles ended up being exactly the same in both frames.

Certainly we are concerned with the positions of the two observers. You would rather set the origin at the bottom end of the tube. Ok, let us do that and examine the differences. Once again I will point out that the relativistic correct transformations yield the distortions required to obtain the correct Newtonian referance frames necessary to figure out the correct dynamics in the two different frames. In particular, I want to point out that there are no distortions in the y or z direction. Thus the z axis in both frames (the center line of the tubes) points in exactly the same direction in both frames. No one has any argument with that conclusion.

I think here our thoughts are exactly aligned.

Now, you want the observers to set their origins (the point where x'=x=0 and t'=t=0) at their eyeballs at the bottom of the tubes. Then, if their observations are to be made at exactly the same time when they are in the same place (the origin). I am afraid that once again you must conclude that they are looking at different photons.

But here we aren't, because I specifically want them to catch the same photon. That is to say, just have one tube, place it in the rest frame; in rest with the light source, and in rest with the rest observer.

Then adding a moving observer to pass the rest observer so that he will catch the same exact pulse of light when they pass (which they must be capable of), I want to figure out where does the moving observer see that light as coming from, i.e. what is the apparent angle of the pulse of light that the rest observer sees as directly over head.

What is the result do you get if you do that setup?

Then the next point of discussion would be, is this setup meaningful in order to analyze how the visual view would change when a natural observer changes his inertial frame? To me it seems to be, but of course I may be overlooking something.

And yeah, I wouldn't want to bring up acceleration, which is exactly why the original version was a setup where two inertial observers are passing each others, and just when they overlap, we see how do they see the universe as from that exact location (which can now be identified as "the same location" because they are there at the same time)

The only reason I brought up the possibility that a third observer may catch onto a hook to switch his inertial frame is that, once his acceleration ceases, he must be seeing the universe just like the inertial observer sees it.

When that flash reaches the rest observer, it essentially constitutes a spherical surface 4.3 light years in radius (that is 135,693,964.8 miles). Notice that the rest observer will presume the photon the moving observer will see will enter the moving observers tube one second (that is in the moving observer's frame) before the moving observer detects it. The rest observer will conclude that this specific event occurred 223.6136 seconds prior to the critical detection at his origin (that is the one second in the moving observers frame converted to the rest observers time measurement). On the other hand, the rest observer will presume the photon he detects at the origin entered his tube only one second prior to his detection. The two observers will certainly agree that they are not detecting the same photon.

Also in addition to that, in your setup the moving observer cannot even see the star through his own tube when he is passing the rest observer; there is no such path of light that could have passed through his tube without hitting its walls. He can only see the star much later; as directly overhead. Implying it is at the moment of passing visually somewhere up ahead.

I.e, if he just plain catches the same photon that is coming down the rest observer's tube (which I presume you agree is wholly possible setup to do?), he must catch it from an apparent forward angle.

If this is valid, then any other photons coming off from the interior walls of the tube (if the inside of the tube is illuminated) and hitting the eyes of the moving observer, will be catched similarly in forward angle; the entire tube (and everything else in the rest frame) must appear visually distorted.

Note that even if it is visually tilted, the moving observer is obviously still plotting the tube as pointing straight up.

I find your reasoning to be effectively correct in your setup, and furthermore what you find is exactly that the moving observer will see the star as directly overhead; only he will see it so as much later than the rest observer (and they are time-wise separated, so we can define the order of events here).

The fact that he sees the star as overhead much later than the rest observer implies exactly the result that the visual view must distort forwards if you change inertial frame as a natural observer.

So my problem is that I don't think it is valid to say "there is no distortion since both observers see the light as overhead" when they do so at different locations. Valid thing to say would be to say the apparent angle of light is distorted so that the moving observer will see the light as overhead far off-set from the location where the rest observer sees it as overhead; that's exactly the distortion I am referring to.

See if you can find errors from the above.

Edited July 31, 2014 by AnssiH

Doctordick · August 2, 2014

Anssi, I apologize sincerely for the misunderstanding behind my presentation. I was clearly performing a subtle rearrangement of the circumstances behind the scene which you were totally unaware of. In essence you and I were talking about two very different events and it was entirely my fault for not making my picture clear. I was comparing two very different events. I was certainly not looking at the moment they would catch the same exact pulse of light. I was correcting for the fact that they both see those distant stars where they were years ago and correcting for that factor behind the scene; something I should not have been doing.

Again, you are absolutely correct. At the exact moment the two observers occupy the same location (and catch the same exact pulse of light) they will see that pulse as coming in at a different angle. The problem is that the specific moment is not the moment we should really be interested in.

Back on January 5th (post number three of this thread) you said,"stars directly in front of the ship would appear to be further away for the moving ship, while the stars towards the rear would appear to be closer. Only when calculating away the optical effect of aberration, the stars would appear to be in identical distances (seems to me)", I inferred that you were referring to the idea that the moving observer would see a distorted image of the universe. In essence I have no disagreement with that assertion at all. My position is that the specific moment referred to is the wrong moment to consider. Since what he is actually seeing is the universe as it existed in the past, and his time references are considerably different from rest observers we should not expect them to be seeing the same thing. On the other hand, a signal between them would travel no distance at all as they passed so they would both consider themselves to be at the origin of the coordinate system at that moment. What is important here is that the "apparent" y,z plane will not constitute the same events for both observers. The moving observer's clock will appear (to the rest observer) to be running quite a bit slower and thus in the rest observers universe, the stars the moving observer is looking at were actually far in the past (not in the rest observers y,z plane). This is the correction I was implicitly handling without being clear.

If the rest observer is looking at a star he sees as directly overhead and we want to know what the moving observer sees, we really don't want to compare what he sees as he passes through the origin of the rest observer. As the moving observer sits in his ship and watches the universe go by, he will eventually see that very star as being directly overhead (it once appeared to be ahead of him and will, sometime in the future, appear to be behind him). If we are concerned with exactly how the universe appears to him, we need to examine what he sees at that moment (at the moment he perceives he is passing through the y,z plane of the rest observer. It actually makes little difference that he sees these stars as they were years ago, the rest observer also saw the universe as it was years ago. What is important is where he sees the other stars (the ones not directly overhead) at that very same specific moment. I was explicitly converting to that moment without actually specifying the fact. Something I should not have been doing without explanation.

Go back to my post of July 28th (post #38) and see if it makes a little more sense now. In particular, the issue of that factor d which describes the alternate angle to the star not directly overhead. In essence since their clocks run so differently, the moving observer will see the universe as it looked long before he and the rest observer occupied the same point. On the other hand, he will see the rest observer's clock running slow thus there will be very little change in the rest observers clock (and in the configuration of the star systems) during the time required for him to reach the point where he sees the universe as the rest observer saw it.

Now I am going admit that it is possible that I have made some serious errors here (when you are doing relativistic changes, it is awfully easy to make an error and I do make errors). In particular, I am a little worried about relativistic foreshortening of the x measurements of the "rest" universe. I don't think I have taken that into account. Note that the foreshortening would be in exactly the opposite direction of the aberration you refer to.

Sorry to put forth such a confusing presentation. Relativity is not an easy subject at least not from Einstein's perspective.

Have fun -- Dick

LaurieAG · August 3, 2014

Hello Doctordick,

What is critically important is that these representations do not represent what will be seen but is rather a representation of the dynamic system as represented in the space time coordinate system Newton invented to display the dynamics of his physics. In order to make that issue clear, let me point out that when one looks at the stars, one does not see them as where they are now (what Newton's coordinate system is designed to display) but rather where they were some time ago.

Here's a few things to think about.

Øyvind Grøn in http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf looks at the various ways to represent what is seen when viewing a relativistic rolling wheel and provides a solution that uses a wheel frame and a road frame that both exist in the same 2D plane i.e. x, y (and t) dimensions with z=0. Basically the problem is to provide the timings and order, for 16 equidistant points around the rim, so that the photons emitted from each point reaches an observer on the road at the exact same time (as point 1 on the rolling wheel reaches the road).

And as everything in our universe is rotating you can go one step further and plot the location of the light paths produced by a dynamic Newtonian system over one cycle of rotation. In the following image the two sources at 1,0 and 3,0 have undergone one complete rotation and have returned to their start positions just as the light, that was first emitted when they were previously at the start location, arrives at the observer during the observation period (at t0).

Now Grøn's SR solution is different from the light path image in that the observer is not in the plane of rotation of the wheel but is effectively stationary at a fixed distance wrt the wheels hub. The wheel is shown at a 45 degree angle as there would be no apparent shift if the plane of the wheel was 90 degrees to the observer. The red and blue colors of the photon paths designate the motion of the source wrt the observer at the point of emission.

The photons emitted from the rotating sources at 1,0 and 3,0 travel 2 * Pi * r to the observer in one complete rotation where r equals the radius of rotation and the sources rotation speed equals c. Photons from sources rotating < c will travel 2 * Pi * r * (c/v) during one complete rotation. (i.e. at c is 2 * Pi * r, at 0.5 c is 2 * Pi * r * 2, at 0.1 c is 2 * Pi * r * 10, at 0.001 c is 2 * Pi * r * 1000)

If you make the center of mass of the sources move towards or away from the observer (at 0.25 c) you stretch or compress the apparent photon paths between them.

So you can think about what you see when you travel faster if you have a good idea about what you are seeing when you aren't travelling at all.

Bombadil · August 4, 2014

Again, you are absolutely correct. At the exact moment the two observers occupy the same location (and catch the same exact pulse of light) they will see that pulse as coming in at a different angle. The problem is that the specific moment is not the moment we should really be interested in.

Doctor Dick I think that you are going to have to convince me of this. From what I have read about Terrel rotation the result is that they will agree on the angle to the star, what they wont agree on is the orientation of the objects that they see.

As a example lets return to your thought experiment where there is a pulse that is traveling in the direction of the rest observer and the moving ship timed so that at some point the light pulse will pass though the end of the pipe. Lets say that it is a laser and we can in fact see the laser inside of the pipe by filling the pipe with smoke.

The question is what is the course of the laser as seen by an observer that is at the far end of the pipe. Well firstly we have set it up so that the pulse will pass though the far end of the pipe, what happens now, will it hit the ship, if so then it must of course pass though the pipe and the observer on the ship will say that it is directly over head, well when the light pulse enters the end of the pipe the observer will see the ship to be where it was in fact one second ago for the moving observer or 223.6136 seconds ago for the rest observer, we know this since this is how long it takes the light to travel down the pipe and so the observer can just look down the pipe to see the ship.

But the rest observer will then have to say that the length of the pipe is over ((223.6136*185,999.81)^2+c^2)^1/2 in length, in other words the rest observer will see the tube to have a long bend in it that will end at the other ship. This really makes a lot of sense if we stop and think about it.

At this point lets move to the observer on board the ship, and admit that there must be some point when he can look down the pipe and see the rest observer at the other end of the pipe. Further more this must correspond to when he could see the pulse as the observer is so close to the pipe that there is no other time when he can be seen to be at the other end of the pipe, this is also where the pulse is entering the end of the tube, notice that I have said nothing about at what angle the pulse is entering the tube. But if it travels all the way down the tube the moving observer will say that it is over head.

Think about this, the moving observer will say that the pulse will travel for one second down the pipe but he will also say that any rest observer will experience 223.6136 seconds during this time. He also thinks that his pipe is perfectly straight. My conclusion is that he will see that pulse as directly over head just like the rest observer right next to him.

Can any one see an error with this analysis?

AnssiH · August 4, 2014

Anssi, I apologize sincerely for the misunderstanding behind my presentation. I was clearly performing a subtle rearrangement of the circumstances behind the scene which you were totally unaware of. In essence you and I were talking about two very different events and it was entirely my fault for not making my picture clear.

I was comparing two very different events. I was certainly not looking at the moment they would catch the same exact pulse of light. I was correcting for the fact that they both see those distant stars where they were years ago and correcting for that factor behind the scene; something I should not have been doing.

Don't worry about it. I figured it was probably something like that. So then we should discuss why do we think it is meaningful to analyze one or the other version.

I think I see what you are getting at, and I think I can kind of figure out why you are analyzing the particular event of emission that you are analyzing, but here are some of my thoughts about the meaning of the particular analysis you are performing.

So you take the moment the observers are passing each others, and then you take a simultaneous event in the light source (simultaneous to the observers passing each others).

And since the light source is directly overhead of the observers at that moment, both of them plot the event as simultaneous, and occurring directly overhead, in their frames (Lorentz transformation yields no changes to Z, Y plane).

And then you figure out what is the direction the observers will see the light from that event as coming from.

Like you presented, it is true that both of them will see that particular event as coming directly from over their own head at the moment when they finally receive the light coming from the event. There are few quite trivial ways to see this, and one is to consider that the moment both of them are directly underneath the pipe (passing each others), both of them actually do plot the light source as reciding directly over their heads at that moment. Thus, both of them must also plot any light coming off from that source as coming directly down towards them (i.e. the speed of the source is not taken as affecting the velocity of the photons) in their own frames. And like you said, they are in this case considering different photons, of course.

But I don't think the moving observer can actually consider the moment he receives the light to be identified with the Z,Y plane of the rest observer. I.e,

If we are concerned with exactly how the universe appears to him, we need to examine what he sees at that moment (at the moment he perceives he is passing through the y,z plane of the rest observer.

appears problematic. Consider that the distance where the moving observer will receive the light from that event, depends entirely on the distance between the light source and the rest observer. (Since you have already analyzed this in terms of the light path to the moving observer forming a triangle, you can see a light source higher above the rest observer will cast the light path further away to catch the moving observer; and that is exactly where the moving observer would see the light source as over his head)

In other words, if we add a series of light sources on top of the rest observer so they all line up (i.e. the rest observer only sees one of them; the bottom one), then the moving observer will see those light sources directly above him one by one after passing the rest observer. Thus he can't meaningfully identify any particular light source being overhead as the Z Y plane of the rest observer. He would have to arbitrarily choose a light source to decide.

On the other hand, the moment he sees them lining up is exactly when he is passing the rest observer (Only, he sees them lining up somewhere in front of him, forming apparent diagonal line). Of course both observers are seeing past events at this point, but that's not an issue as that's always the case.

So to me it seems arbitrary to consider where do the observers see an event that is simultaneous to their passing of each others; each star is at different distance from them and they will receive the information from each star at different times. The rest observer will always line up with any event if he is not moving in relation to the stars at all. But if he is also moving in relation to the stars, the picture gets even more arbitrary, as he will also have displaced himself in the frame of each star by the time he receives the light from the event-of-interest.

I don't think the omissions related to x-axis shortening are relevant to the results as long as we are not interested of the exact magnitudes of the effects, since the shortening does not cancel out any of the effects we are talking about (I commented about this in a post to Bombadil earlier in this thread).

And yeah, relativity is sometimes a *****... I'm not sure if it's common or not, but I can perform a Lorentz transformation for a 2D diagram in my head quite easily, just because I first happened to figure out how it works that way. That also allowed me to see easily how it's really just a convention of representation. When you have bunch of variables that define each others entirely (simultaneity, C, lengths, time), it's possible to change anything as long as you make the rest follow suit coherently. If you want to set C to be isotropic, ain't no one going to stop you as long as you change the other variables accordingly. That, to me, is relativity in a nutshell.

I remember you have commented something similar about the way you came to view relativistic relationships. Isn't it just so true, that different representations of the same thing really make different issues easier to think about. And that's what human semantics is all about... But that's another story.

Edited August 4, 2014 by AnssiH

AnssiH · August 4, 2014

Can any one see an error with this analysis?

Yeah there are some errors;

From what I have read about Terrel rotation the result is that they will agree on the angle to the star, what they wont agree on is the orientation of the objects that they see.

That's incorrect; actually the apparent curving and the disagreement on the angles are resulting from the same mechanism.

Think about a box-like building, and a box-like car moving along a street parallel to the building.

If someone is hiding behind a corner, and will shoot at the car parallel to the wall, can he hit the front of the car?

If the speed of the bullet was infinite, he could not. If the speed of the bullet is finite, why yes he can.

And depending on the speed of the car, the more likely it is the bullet will hit the front of the car.

Consider photons being emitted from that wall towards the street. The faster the car is moving, the more likely it is to catch those photons with its front; some even in considerable forward angles. The faster the car is moving, the higher the maximum angle he will still catch the photons at.

If you can plot the situation in the rest frame of the building, and figure out what angles will hit the car at the front, then from the perspective of the car, those photons are coming in somewhere from the diagonal front direction. In fact they define his front direction (if he is using relativistic conventions). When the car catches those photons, that's where you can take him as forming his picture at.

Consider the fact that photons emitted from the far end of the wall will be catched later, and correspondingly in more forward angle (exactly analogous to what I said in my previous post about a series of light sources). That is why the far end of the wall seems to reside orthogonally on the side from the direction of motion into further distance than the closer edge of the wall. The end result is apparent "curving" of a continuous wall.

So really, it's just a trivial consequence of the aberration of light.

Consider now that each part of that wall is seen further ahead, than where a rest observer (in the same location) would see it at. This is simply because the apparent arriving angle of the light is exactly where we see the object at (and we plot it by extrapolating its velocity to the future).

Or a different way to put it, if we think we are at rest and the building is moving, then we see the building where it was in our coordinate system when the light that we see was emitted.

That is to say, we plot the building as passing us in perfectly orthogonal angle, and if we consider the moment we would plot the wall to just barely become visible, then from that instant, we will have the light arriving to our eyes from the far end of the wall sometime later than the light arriving from the front end of the wall; thus it would appear the front end of the wall is moving ahead of the far end of the wall (but the wall would look straight)

This is simply another way of saying that the pipe (in rest) in my version of the thought experiment is seen in an angle from the point of view of the moving observer.

Bombadil · September 9, 2014

If someone is hiding behind a corner, and will shoot at the car parallel to the wall, can he hit the front of the car?

If the speed of the bullet was infinite, he could not. If the speed of the bullet is finite, why yes he can.

OK I can see how this works, one thing that was throwing me though at first was that you mean literately the front of the car that is the side of the car orthogonal to the wall of the building or parallel to the one that some one is standing behind, there is something else here that seems worth noting and that is that for the person hiding behind the wall the front of the car is never truly visible, for some reason I was thinking originally that a situation could not be created that resulted in only one direction that light could make the trip in, or in this case that a bullet could be fired from the front of the car and hit the person standing behind the wall, but this is clearly not the case.

The front of the car can be hit by the person behind the corner but someone at the front of the car can't hit the person standing behind the corner without shooting though the side of the car.

Consider the fact that photons emitted from the far end of the wall will be catched later, and correspondingly in more forward angle (exactly analogous to what I said in my previous post about a series of light sources). That is why the far end of the wall seems to reside orthogonally on the side from the direction of motion into further distance than the closer edge of the wall. The end result is apparent "curving" of a continuous wall.

The other thing that was throwing me is that what we see is not where something is but rather, what we see is where we intercept the light reflected off of something. The part that was throwing me though is that this may be long after something has happened even if something happened relatively close, if we are traveling at relativistic speeds.

So am I correct in saying that if in your thought experiment of driving down the road, if we were to be traveling at a relativistic speed we would see that the side of the wall that is facing the street would appear to warp around towards the front of the car, this would result from the light reflecting off of the wall in front of us coming in at a steeper angle then it normally would and so it will appear to warp towards our direction of motion, meanwhile the wall behind us would appear to warp away from the road and the light coming from it would be coming in at a shallower angle.

On a related note the effect that the Lorenz contraction has on what is seen I am not so sure about, but while the adoration of light must take place it seems that the Lorenz contraction must also be considered.

Now the Lorenz transformation would seem to have the effect of making the wall that is facing the street appear shorter, however there are a few things to consider, for one everything that has been said so far and the whole point of this thread seem to be that the rest observer and a moving observer will agree on the distance between the edges of the wall.

Now this is where I am still having a hard time understanding what is going on. From the beginning of this thread, the whole point seems to be that the length of the building will be seen to be the same length for both a moving and a stationary observer. That is the distance between the two points where there is no longer a path for the light to travel from the wall to the street must be the same for a rest and a moving observer. But if we want to know what the moving observer sees don't we have to apply the Lorenz transformation and the appearance of the wall warping in towards the street, so how is it that both observers can agree on this distance?

At first I thought that this is where the Terrell rotation actually comes in and results in the apparent length of the building never changing independent of speed, however on closer inspection this appears to only be the result in some sort of local space where the angle between the front and back of an object is small and so while this may be the result for the apparent distance between two points directly ahead or behind I don't see how this can hold in a situation like that suggested in the opening post?

What I found rather interesting was the fact that my Euclidean mental construct seemed to always give exactly the same answers as his theory. Actual calculations were sometimes easier in his picture and sometimes easier in my picture so, whenever I took a test, I used the mental construct which was more convenient to the specific question being asked. Not once in my entire life did my picture ever give me an incorrect answer. Nevertheless, since Einstein was the authority, I presumed my picture was erroneous. Clearly, Einstein's theory included nothing analogous to that projection I had dreamed up and I could conceive of no mechanism to justify such a projection.

So this means that we can use a euclidean metric in place of the Mankowski metric, and the Lorenz transformation will then be derived by instead of assuming a constant speed of light. As has been assumed, we assume a constant distance traveled in the 4 dimensional euclidean space? I say assumed here only because I see no justification for the constant speed in the 4 dimensional space in this context.

In my shadow picture, every point in the universe follows some path through this four dimensional space. Movement along this path defines time (not what clocks read). Light is clearly moving perpendicular to that “time” axis (a clock moving with a photon shows no change in time along the path). Any object moving parallel to the “time” axis will appear to be at rest in the shadow universe no matter what its velocity might be. In fact its velocity is not a measurable thing. The simplest view is that the velocity of a point has nothing to do with which way it is going; it is no more than a measure of its movement along its path over time and setting it to “c” is no more than setting the measurements of that time axis to the same units used on the other three axes.

Isn't this kind of like saying that it just so happens to be c or is there some fundamental assumption that we must use to measure the speed of light that is also defining its speed?

Unless you are saying that the distance that is moved along that 4th axis is measured by our comparing it to our state of movement and agreeing on the same units of distance, and then the definition of c is nothing more then a defining of a measure of that forth axis. But this still brings me back to the same question of why should everything travel though that 4 dimensional space with the same speed.

HydrogenBond · September 10, 2014

At first I thought that this is where the Terrell rotation actually comes in and results in the apparent length of the building never changing independent of speed, however on closer inspection this appears to only be the result in some sort of local space where the angle between the front and back of an object is small and so while this may be the result for the apparent distance between two points directly ahead or behind I don't see how this can hold in a situation like that suggested in the opening post?

According to Einstein's theory of special relativity, velocity impacts distance, time and mass. In the twin experiment, the clocks move at different rates with the moving twin aging slower. This twin does not suddenly get older, by returning to the earth reference. The time dilation sticks with him staying younger but now aging faster like his twin.

Distance appears to work differently, because he does not remain contracted in distance; skinny, but loses whatever skinny it appeared to have. As far as his mass, his relativistic mass returns to normal for the earth, but energy conservation requires the potential energy of the moving reference relativistic mass be conserved as something else.

It is not clear where the energy conservation goes and whether the time conserved is connected to energy conservation. In the case of the twin, he has extra time on earth, relative to his twin; implies energy conservation into time potential. But the energy does not conserve distances that had been contracted in the moving reference.

Edited September 10, 2014 by HydrogenBond

AnssiH · September 21, 2014

The other thing that was throwing me is that what we see is not where something is but rather, what we see is where we intercept the light reflected off of something.

In other words, we see the things where they were, when the light that we receive was emitted. Of course; that's the information that is reaching us.

The part that was throwing me though is that this may be long after something has happened even if something happened relatively close, if we are traveling at relativistic speeds.

Yeah, because if you are moving close to the speed of light, some of the light off of the wall that will eventually hit you, is moving almost parallel to your direction of motion. In the frame of the "street" that is.

So am I correct in saying that if in your thought experiment of driving down the road, if we were to be traveling at a relativistic speed we would see that the side of the wall that is facing the street would appear to warp around towards the front of the car, this would result from the light reflecting off of the wall in front of us coming in at a steeper angle then it normally would and so it will appear to warp towards our direction of motion, meanwhile the wall behind us would appear to warp away from the road and the light coming from it would be coming in at a shallower angle.

Hmm, yes, maybe... It's a bit hard to think about this because your description only makes sense for a very brief moment when you are passing the wall.

On a related note the effect that the Lorenz contraction has on what is seen I am not so sure about, but while the adoration of light must take place it seems that the Lorenz contraction must also be considered.

Now the Lorenz transformation would seem to have the effect of making the wall that is facing the street appear shorter, however there are a few things to consider, for one everything that has been said so far and the whole point of this thread seem to be that the rest observer and a moving observer will agree on the distance between the edges of the wall.

...

At first I thought that this is where the Terrell rotation actually comes in and results in the apparent length of the building never changing independent of speed, however on closer inspection this appears to only be the result in some sort of local space where the angle between the front and back of an object is small and so while this may be the result for the apparent distance between two points directly ahead or behind I don't see how this can hold in a situation like that suggested in the opening post?

Well the opening post situation is based on the idea that if some photons enter the two holes of the rest ship in particular angle, then those two photons would also enter the two identical holes of the moving ship - and that angle would be the same if you correct for the aberration. Typically this would mean correcting to the rest frame of the source. If you don't do the correction, you get where the star was in your frame when the light was emitted. If you do the correction, you get where the star is "now", i.e. where it is if you stop into the frame of the star.

So far it's easy to prove, but the case of photographing pictures of fast moving objects makes me wonder...

So there are these websites that are saying that Penrose demonstrated that you can't photograph length contraction, but thinking about it a bit I would like to see what type of argumentation he used. I just can't find it from anywhere.

Think about it this way. Let's first of all idealize a camera, let's say the shutter is just a pinhole, and when it opens for a moment, it lets some photons through just as they are crossing each others (flipping the picture). What those photons are will define the picture it takes. The point here is that the picture is effectively approximated as what a point entity receives, ignoring the area of the lense, the motion of the shutter, and the geometry of the plate. I'm not entirely sure if this approximation is a critical error or not.

Now let's say we are photographing a semi-transparent glass tube, and a metal rod passing straight through the tube. These objects have the same length at rest.

Now, if you trace what is the picture taken by one camera, then you can add another camera superimposed into the same location at same time, but moving at some speed to some direction.

Which is significant because any camera you add, must catch the same photons; it must take the same picture. Only askew to some direction as per aberration. But the events displayed in the pictures must all be the same. If in some picture the glass tube is seen as shorter or longer than the metal rod, it must also be so in the pictures of all the cameras in the same spot at the same time.

Actually first let's just refresh on the role of simultaneity here;

Pay no attention to the ridiculous narrator. Everybody always talk about these things as if observations can be used to draw conclusions about reality, when in actual fact this is all about self-consistency of our definitions, and working conventions.

So, if we use the Einstein convention of synchronizing clocks, based on the convention of isotropic C, the passenger on the train must assume the flash from the front happened before the flash at the back. Since he knows he is sitting in the center seat.

Now imagine the lightning bolts leave marks onto the ground right where they hit the train (or imagine ground poles here if you want). Since the observer on the platform is using the convention of isotropic C, following that convention he thinks the front end and the back end of the train were at those spots simultaneously. Thus the distance between those spots is exactly the same as the length of the train in terms of his frame.

For the passenger in the train, if he is using his own notion of simultaneity as per Einstein convention, he will say that when the front end got hit with a lightning, the back end had not yet been hit; and thus the back end had not yet reached the position where the ground would also get hit.

So in terms of his simultaneity notion, he must say the train is longer than the distance between the lightning marks on the embankment.

Add another identical train moving at the same speed but to the opposite direction on the next lane. And make the trains line up as seen from the platform, just as the lightning hits. The observer on the platform thinks the trains are the same length. The passenger in the newly added train thinks his train is longer than the markings, and furthermore both passengers symmetrically think their train is longer than the other train, as long as they both think their one-way C is the same as two-way C.

That is the connection between simultaneity notion, and length contraction. You can't actually measure the length of a moving object if you don't know the correct time delays for your observations (if you don't know where the front end and the back end of the object were simultaneously). And since you can't know one-way speed of light... If you choose to change the one-way speed as a function of the reference frame, you also change the implied lengths of things.

Now let's switch the two trains for the glass tube and metal rod passing through each others.

Let's look at the pictures that any observers standing on the platform would be taking of the situation. All the observers anywhere on the platform would make the same conclusions about the simultaneity, and thus about the plotted lengths of things. But the pictures they are taking would be quite different.

An observer in the middle of the platform would receive information about both lightning strikes simultaneously, and that is what would display on his picture; the ends of the rod and tube neatly matching.

An observer at one end of the platform could take a picture where the lightning strike at that end is visible, but the strike at the other end is not. The further away something is from the observer, the older states he is photographing.

Thus, if you take a photograph while standing at one end of the platform, with such timing that the end of the tube and the rod near you neatly overlap in your photograph, the far ends are still approaching each others in your picture (you photograph older states).

The end that is moving towards you looks elongated, and the end moving away from you looks shrunk in your picture. Symmetrically, this is exactly what you expect via aberration analysis as well (think of the rod or the tube as at rest, and the platform as moving).

So it's clear that any photograph is distorted by the fact that the events appearing on the picture are not simultaneous. However, this time the distortion is a function of your location, not a function of your speed!

If you add a cameras to this setup such that they are at rest with the tube or with the rod, and they will neatly co-incide with a camera on the platform just as that camera takes a picture, all the cameras co-inciding must display the same events in their pictures (only aberration distorted).

Thus, the observer in the middle of the platform taking a picture of neatly matching tube and rod, will match the pictures of the cameras moving along with the tube and with the rod, as long as those cameras just happen to be in the middle of the platform when the picture is taken.

This might be the kind of argument Penrose made to conclude length contraction can't be photographed, but it is actually an erroneous conclusion. The above doesn't mean that length contraction cannot be photographed, it just means length contraction of objects moving in opposite directions is the same.

Note that the cameras we added as moving with the tube and rod are not orthogonally aligned to the object they are at rest with. They are following their objects somewhat behind, and thus they do not think they are photographing simultaneous events. They think their picture is distorted by the fact that the events in the picture are at different distances to themselves, and that distortion is exactly counteracting their length contraction.

Let's look at a camera that is orthogonally aligned towards the metal rod. and situated in the mid-point of the rod, ready to take a picture as the glass tube is passing.

He must plot the glass tube as length contracted, and it seems trivial to setup a situation where you start tracing photons from the far ends of the metal rod at the moment the glass tube is passing somewhere in the middle of the rod. Trace those photons to the camera and take that picture, and it would be a picture of the tube appearing shorter than the rod.

Now add a camera B that is at rest with the glass tube, so that it co-incides with the picture we just took; it must take the same picture, and also see the glass tube as shorter than the metal rod.

The reason why the camera that is at rest with the glass tube also sees the glass tube as length contracted is that that particular camera must be trailing the tube somewhere behind it, and thus photographs older states of the further end. In its frame the metal rod is approaching, and in the picture it has taken the far end of the metal rod has not yet crossed the entry of the glass tube, when the front end is already passed through the exit.

And this is again symmetrical whichever way you look at it.

Sooo, right now I would say it is possible to take a picture of length contraction, contrary to what Penrose concluded. But it is also possible to interpret the reality of that same picture in all kinds of different ways. In relativisticy convention, your interpretation of the same picture is literally a function of what velocity you assign to the camera that took the picture.

And you get the same results of all of the above with other formulations of relativity. In Lorentz aether theory the pictures taken are all exactly the same, for instance. And also in DD's notation.

(ps, here's another video with nice visualization of Lorentz transformation. But again the comments in the video are quite naive)

So this means that we can use a euclidean metric in place of the Mankowski metric, and the Lorenz transformation will then be derived by instead of assuming a constant speed of light. As has been assumed, we assume a constant distance traveled in the 4 dimensional euclidean space? I say assumed here only because I see no justification for the constant speed in the 4 dimensional space in this context.

The constant speed requirement comes from epistemological considerations. The generality of the notation, if it is truly general as it appears to be, is to say it is always a valid choice to use this notation. But it is not to say it is the only valid way to represent epistemological constraints.

What it boils down to is that if you can generate rules where the entirely of the rest of the universe can truly be ignored when generating expectations for some subset of the universe, then those same epistemological constraints apply for any such sub-set of elements.

Meaning, it is possible to define laws of physics that apply in the same form to different objects independently.

But the associated expectations are expressed with a wave function of fixed velocity. If this is true for two different sub-sets of the universe (that can be seen as having their own coordinate systems), both of those objects can arrive at valid expectations by assuming fixed velocity isotropically in their own perspective. As long as that is true, then a valid transformation for any validly defined laws of physics can be accomplished with Lorentz transformation.

Did you see this;

http://foundationsofphysics.blogspot.fi/2014/09/epistemological-derivation-of-special.html

And also you should read this;

http://foundationsofphysics.blogspot.fi/2014/09/some-background-on-special-relativity.html

Isn't this kind of like saying that it just so happens to be c or is there some fundamental assumption that we must use to measure the speed of light that is also defining its speed?

Unless you are saying that the distance that is moved along that 4th axis is measured by our comparing it to our state of movement and agreeing on the same units of distance, and then the definition of c is nothing more then a defining of a measure of that forth axis. But this still brings me back to the same question of why should everything travel though that 4 dimensional space with the same speed.

C just arises as a part of self-consistent set of definitions. Just backtrack to Schrödinger's equation to see where it was defined (the book is availble through the blog in the links above)

Doctordick · September 21, 2014

Hi Anssi,

Diantha and I have been gone for over a month (She got new eyes in Denver this year -- cataract problems solved). At any rate, when we got back,I tried to look at forum to see what has happened and the whole thing has been changed. I haven't the slightest idea how to navigate the thing; how to find out what is going on. I would have just written you an e-mail note but thought maybe posting my disappointment with the new structure would excite some further complaints by others. This new structure is anything but convenient. You know, when one gets old, change is not an exciting thing!

Regarding your post (and some of the others I have managed to read) my central argument with modern physics is the fact that they use two different definitions of time which they assume are interchangeable. Time is "what clocks measure" and time is a coordinate which indicates when things can interact: i.e., things which exist at the same time and place can interact.) I have seen people complaining about my use of a constant speed of light as a hypothesis and not a factual measurement. They completely miss one of the central issues of my presentation.

Clearly if one uses time is what clocks measure, the twin paradox is a serious problem. The standard solution is to say the clock must be at rest in your frame of reference but that leads to serious complications the scientific community tends to avoid thinking about. In my "shadow" representation of reality (that fourth axis being projected out) actual positions in that fourth coordinate are not knowable (see the second paragraph on page 24). What most everyone seems to miss is the fact that this new four dimensional Euclidean geometry specifies interaction points under the definition: things at the same place and time can interact. Thus interactions add an additional parameter called "time" By this mechanism we are led to a picture of specific interactions "happening". Actual measurement of this parameter changing is totally undefined; however, momentum in the direction of that projection is later associated with mass. Since photons possess no mass, it follows that they must be traveling perpendicular to the projection dimension.

Since the parameter associated with these interactions is totally undefined, it is easy to associate it directly with the distance between interactions along its path. This is all a the idea of a constant speed of light accomplishes. Now, absolutely any motion in the projection direction is projected out. It follows that the "time rate of change" in that direction can not be defined. Yet we also have collections of interactions along the paths of those objects with non zero mass. Since the distances along the projection path are unknowable, we are free to define that anyway we wish. Setting the speed of those entities along their physical paths to be the same as the speed of light accomplishes a very nice constraint and makes everything rather simple (notice that we are free to set the time parameter any way we wish). No massive object can exceed the speed of light (that would be its apparent speed when it is moving orthogonal to the projection direction.

Have fun -- Dick

Edited September 21, 2014 by Doctordick

Doctordick · September 24, 2014

Well, I apologize for my previous post. My access to Hypography was really strange on the 21st and for some strange reason I could not even access the site on the 22nd or 23rd. But today everything is back to what it was before we left to Denver. I have no idea what is going on.

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