THE RETURN of two opposite travelling photons

[Blogical]

THE RETURN of two opposite traveling photons!

This is a question which has been intriguing me, if two photons A and B travel in opposite direction, what would be the relative speed between them?
If it is 'c', then after one year the distance between the two photons would be 2 light years, somehow it does not add up!
I know that considering a frame of reference in photon's view is wrong and hence this is a physically wrong question, but still i continue with asking this question, just too interested
hence forgetting the stupidity.

 

[harrylin]

This is a question which has been intriguing me, if two photons A and B travel in opposite direction, what would be the relative speed between them?
If it is 'c', then after one year the distance between the two photons would be 2 light years, somehow it does not add up!
I know that considering a frame of reference in photon's view is wrong and hence this is a physically wrong question, but still i continue with asking this question, just too interested
hence forgetting the stupidity.

There has been quite some confusion related to such questions because different people mean different things with "relative speed". So, I'll give you two answers, each corresponding to a different meaning of the term.

1. ("old-fashioned" meaning): The relative speed is the absolute value of the vector subtraction of velocities with respect to an implied or specified reference system.
In your case any inertial reference system may be implied, and the relative speed between your photons is 2c because the speed of each is c and they propagate towards each other. Now your calculation adds up.

Compare: many older textbooks as well as Einstein's 1905 paper, section 3 (search for "relatively") http://www.fourmilab.ch/etexts/einstein/specrel/www/

2. ("Newspeak"): The relative speed is the speed with respect to an inertial reference system in which one of the two objects is in rest.
a. The relative speed between your photons is undefined.
b. A new term has been invented: the "closing speed" between your photons is 2c.

Compare: many recent textbooks.

 

[ghwellsjr]

This is a question which has been intriguing me, if two photons A and B travel in opposite direction, what would be the relative speed between them?
If it is 'c', then after one year the distance between the two photons would be 2 light years, somehow it does not add up!
I know that considering a frame of reference in photon's view is wrong and hence this is a physically wrong question, but still i continue with asking this question, just too interested
hence forgetting the stupidity.

Your question is rather confusing because of the words "THE RETURN" in your title. I can only assume that you mean a return to the subject rather than that the two photons are returning to each other. So if that is what you meant, then in a frame in which two photons are emitted from a common source, it is true that after one year, the distance between them is 2 light years, and you are correct that we cannot consider a frame of reference in which a photon is at rest, but we can still use the formula that Einstein gave in section 5 of his 1905 paper for "The Composition of Velocities" and see what velocity we would get if we plugged in c for both v and w in his equation:

And we would get c as the resultant speed.

However, if we want to follow the rules, we could change your question slightly to what would be the relative speed between a photon emitted in one direction and an observer traveling at under c in the other direction?

In fact, Einstein even makes this point when he says:

It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain

If you don't see why this is true, just multiply the top and bottom of the fraction by c:

c(c+w)/c(1+w/c) = c(c+w)/(c+cw/c) = c(c+w)/(c+w) = c

He is making the point that no matter what the speed of the observer is, the photon is still traveling at c relative to him.

Does this all make perfect sense to you?

 

[A Dhingra]

If it is 'c', then after one year the distance between the two photons would be 2 light years, somehow it does not add up!
I know that considering a frame of reference in photon's view is wrong and hence this is a physically wrong question, but still i continue with asking this question, just too interested
hence forgetting the stupidity.

Just wondering, are we not suppose to find out the length or the distance between the two photons, considering say the frame moving with c and Length contraction!
But you said that the frame of photon's view point can't be considered, why so?
(Strange, i am asking a question to you in a thread started by you! I thought may be that has something to do here, not sure..)

 

[ghwellsjr]

Just wondering, are we not suppose to find out the length or the distance between the two photons, considering say the frame moving with c and Length contraction!
But you said that the frame of photon's view point can't be considered, why so?
(Strange, i am asking a question to you in a thread started by you! I thought may be that has something to do here, not sure..)

When we say the frame of an object, we mean the inertial frame in which the object is at rest. But photons are defined to travel at c in all inertial frames. So there is no frame in which a photon can both be at rest and traveling at c.

 

[A Dhingra]

But even a frame moving with uniform velocity with respect to a rest frame is inertial, can't we consider this??

 

[Jeronimus]

This is a question which has been intriguing me, if two photons A and B travel in opposite direction, what would be the relative speed between them?

The question itself is invalid or incomplete if you want when talking about relativity. The question implies a third observer. Only then you can talk about two photons "traveling" in opposite directions.
Even if it was rockets and not photons, the above statement applies. You can say two rockets travel at 0.5C RELATIVE to each other which is ok and does not imply a third reference frame necessarily.An observer in rocket A considers himself at rest and describes rocket B as traveling at 0.5C away of him OR towards him. An observer in rocket B considers himself at rest and describes rocket A as traveling at 0.5C away of him OR towards him. Again, this is what it means when you say they move at 0.5C relative to each other. The notion of an opposite direction cannot be applied here.The third observer, assuming a spacemonkey on a spaceplatform observing the two photons, sees photon A travel at C in one direction, closing distance to photon B and moving towards the other direction always at C as demanded by Einstein's postulates.
If we somehow managed to get a rocket up to 99.9999999..% speed of C to follow photon B for a long long time then an observer inside the rocket would see the photon A travel at C OF COURSE, as demanded by Einstein's postulates you can derive SR from.

(To make this even worse. EVERY observer within an ARBITRARY inertial frame of reference traveling at an arbitrary speed in an arbitrary direction seen from the spacemonkey's reference frame will see that photon and EVERY other ARBITRARY photon travel at C. This is the very basic demand of Einstein's postulates which build up the formulas of SR)

edit: In particular one of the two postulates required to build the formulas of SR states.

"The speed of light in free space has the same value C in all inertial frames of reference."If you do not like a third observer, then an observer inside the rocket following photon B considers himself at rest like every observer within his inertial reference frame does. Motion is relative. You are at rest within your frame, and describe all motion relative to you.
This is why the notion of opposite directions is nonsensical when only two reference frames are in question.

If it is 'c', then after one year the distance between the two photons would be 2 light years, somehow it does not add up!

This is exactly the distance between the two photons they would have, SEEN from the spacemonkey's reference frame.

I know that considering a frame of reference in photon's view is wrong and hence this is a physically wrong question, but still i continue with asking this question, just too interested
hence forgetting the stupidity.

I am not sure it is completely wrong but my it delves into the unthinkable or away of the humanly understandable. This is why you can cheat and use two rockets traveling at close to C at 99.99999999999999999999999999999999999999999999...% of C.
So what distance would those two rockets see after two years WITHIN the spacemonkey's reference frame?
The spacemonkey we already said would see ~2 lightyears of distance between them. Because of the length contraction however, those 2 lightyears would look like L = L0 * sqrt(1-(v^2/c^2)) which for 99.9% of C would already be in the range of 0.09 lightyears compared to 2 lightyears the spacemonkey sees it after 2 years.
An observer in any of the two rockets would see the other rocket at 0.09 lightyears away given the spacemonkey sends both rockets a laser pulse which reaches both rockets after 2 years SEEN from the spacemonkey's inertial frame of reference and they both look out the windows to check the distance at exactly that event.

This is important to state, because of time dilation the time is not 2 years measured from observers within the two rockets.

As you can see, if we somehow managed by a miracle to speed up those two rockets to C so they ride along with the photons, the distance would shrink to zero. The spacemonkey could not reach the rockets with the laser pulses. All kinds of spooky stuff we really don't want to go into.

 

[ghwellsjr]

But even a frame moving with uniform velocity with respect to a rest frame is inertial, can't we consider this??

Only if the second frame is moving at less than c with respect to the first. The Lorentz Transformation process is the way you convert coordinates between two frames and it won't allow them to have a relative motion of c or greater.

 

[harrylin]

Just wondering, are we not suppose to find out the length or the distance between the two photons, considering say the frame moving with c and Length contraction!
But you said that the frame of photon's view point can't be considered, why so?[..]

Well, a frame moving with c would have infinite inertia (a physical impossibility in this universe); moreover its rulers would have zero length and its clocks would be stopped. No "Einstein synchronization" between distant clocks can be made, not even in theory. What would you want to with such a theoretically impossible and defect system?

And even if hypothetically something could be done with it, an impossible system is allowed to give wrong answers.

 

[ghwellsjr]

1. ("old-fashioned" meaning): The relative speed is the absolute value of the vector subtraction of velocities with respect to an implied or specified reference system.
In your case any inertial reference system may be implied, and the relative speed between your photons is 2c because the speed of each is c and they propagate towards each other. Now your calculation adds up.

Compare: many older textbooks as well as Einstein's 1905 paper, section 3 (search for "relatively") http://www.fourmilab.ch/etexts/einstein/specrel/www/

I did this search and of the 23 hits, I could not find one that corresponds to the current definition of "closing speed". As far as I can tell, Einstein always said "relatively to a system [meaning frame]" or "relatively to another object". Where did you find it?

 

[ghwellsjr]

Well, a frame moving with c would have infinite inertia (a physical impossibility in this universe); moreover its rulers would have zero length and its clocks would be stopped. No "Einstein synchronization" between distant clocks can be made, not even in theory. What would you want to with such a theoretically impossible and defect system?

And even if hypothetically something could be done with it, an impossible system is allowed to give wrong answers.

I think it is wrong to discuss what a frame moving at c with respect to another frame would be like. We are talking about Einstein's theory of Special Relativity. That theory has a method to define what an inertial frame is including synchronization of distant clocks. The point is not that the synchronization breaks for clocks moving at c, it's that you can't build a clock out of just photons. Clocks require massive particles in their construction. So saying that its clocks would be stopped is misleading. You can't get any clocks to travel that fast.

It is far better to point out that even in a frame that is traveling at just a hair under c, everything appears normal, just like the original frame, it's a perfectly legitimate frame, and in this incredibly fast moving frame, the speed of light is c, you're no closer to the speed of light in this fast moving frame than you were in the original frame.

This is why Einstein said in his paper in the middle of section 4, "the velocity of light in our theory plays the part, physically, of an infinitely great velocity" because no matter how close you are to it, it remains just as far away.

 

[harrylin]

I did this search and of the 23 hits, I could not find one that corresponds to the current definition of "closing speed". As far as I can tell, Einstein always said "relatively to a system [meaning frame]" or "relatively to another object". Where did you find it?

Yes that's quite right. Thus, "the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v". Perhaps you would say: the closing speed between the ray and the initial point of k, when measured in the stationary system, is c-v.

 

[ghwellsjr]

Yes that's quite right. Thus, "the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v". Perhaps you would say: the closing speed between the ray and the initial point of k, when measured in the stationary system, is c-v.

Ok, so you're saying that even though Einstein talked about what is now called "closing speed", he did so without applying any terminology to it at all, except the algebraic notation, correct?

 

[harrylin]

Ok, so you're saying that even though Einstein talked about what is now called "closing speed", he did so without applying any terminology to it at all, except the algebraic notation, correct?

No, that is not correct. Instead, I cited how he used the terminology of relative motion.

 

[arindamsinha]

I would really appreciate some opinions from forum members who have a solid, in-depth understanding of Relativity (no disrespect meant to earlier responders)...

This discussion has gone on in two separate threads recently, including this one.

The question is:
- Is it, or is it not possible for a third party inertial observer to observe a closing speed/relative speed of nearly 2c between two high-velocity particles approaching each other (e.g. a particle physicist watching colliding high-energy particles in a particle accelerator)?

One answer seems to be - YES. The observed closing speed/relative speed can be near 2c, as observed in the inertial frame of the third party observer (not the inertial frames of the high-velocity particles).

The other answer seems to be - NO. No observer in any inertial frame can ever observe a closing speed/relative speed (or any other speed for that matter) reaching c between any two particles, under any circumstances.

My understanding is that the first answer is the correct one, and that it does not violate SR principle. The reference Wikipedia article is http://en.wikipedia.org/wiki/Faster-than-light#Closing_speeds.

What is the actual answer, in the opinion of the more knowledgeable members in the forum?

 

[harrylin]

I would really appreciate some opinions from forum members who have a solid, in-depth understanding of Relativity (no disrespect meant to earlier responders)... [..] The question is:
- Is it, or is it not possible for a third party inertial observer to observe a closing speed/relative speed of nearly 2c between two high-velocity particles approaching each other (e.g. a particle physicist watching colliding high-energy particles in a particle accelerator)?

One answer seems to be - YES. The observed closing speed/relative speed can be near 2c, as observed in the inertial frame of the third party observer (not the inertial frames of the high-velocity particles).
[..]
My understanding is that the first answer is the correct one, and that it does not violate SR principle. The reference Wikipedia article is http://en.wikipedia.org/wiki/Faster-than-light#Closing_speeds.

What is the actual answer, in the opinion of the more knowledgeable members in the forum?

Your understanding is correct. That is the actual answer, and it did not change; it's even a mathematical necessity. This has been explained by knowledgeable members here (in multiple threads), and so did Einstein already in his first paper on this topic.

Note: Contrary to what you seem to think, there is no disagreement about the physics. I tried in vain to prevent possible misunderstanding about words with my clarifications in post #2 of this thread.

 

[Jeronimus]

I would really appreciate some opinions from forum members who have a solid, in-depth understanding of Relativity (no disrespect meant to earlier responders)...

This discussion has gone on in two separate threads recently, including this one.

The question is:
- Is it, or is it not possible for a third party inertial observer to observe a closing speed/relative speed of nearly 2c between two high-velocity particles approaching each other (e.g. a particle physicist watching colliding high-energy particles in a particle accelerator)?

One answer seems to be - YES. The observed closing speed/relative speed can be near 2c, as observed in the inertial frame of the third party observer (not the inertial frames of the high-velocity particles).

The other answer seems to be - NO. No observer in any inertial frame can ever observe a closing speed/relative speed (or any other speed for that matter) reaching c between any two particles, under any circumstances.

My understanding is that the first answer is the correct one, and that it does not violate SR principle. The reference Wikipedia article is http://en.wikipedia.org/wiki/Faster-than-light#Closing_speeds.

What is the actual answer, in the opinion of the more knowledgeable members in the forum?

Wikipedia does not really make it much easier to understand when you read things like

"The rate at which two objects in motion in a single frame of reference get closer together is called the mutual or closing speed."What we have here are two objects observed by an observer inside a specific inertial frame of reference.
The two objects themselves ARE NOT "in" this single frame. If they were, they would be at rest relative to every other observer in the frame.
Each moving object observed by an observer in that "single" frame of reference is in it's own inertial frame of reference considering "itself" at rest.
Cannot blame the author much, as it is quite difficult to stay precise in the wording when we talk about relativity.Also closing speed and relative speed are NOT the same as you put them together separated by a line would possibly imply.When talking relativity, the relative speed is a COMMON FACTOR between two objects/observers/reference frames.
When someone tells you that an observer A moves at v relative to observer B then you should imagine two situations in your mind.
One is where observer A is at rest, describing B's motion inside his inertial frame of reference and the other is observer B describing A inside his inertial frame of reference.

You would draw two minkowski diagrams, one for each observer representing their inertial frame of reference and draw every event you know of happening in observer's A reference frame into observer's B reference frame as well.
Because an arbitrary photon(light) travels at C always, observed from within any arbitrary inertial frame of reference, the only way to map events happening in observer's A reference frame into observer's B reference frame, is to shift the time and space coordinates of any given event. This is why time dilation and space contraction occur.

 

[altergnostic]

I think the most fundamental issue here is that this question is impossible regardless of what theory or interpretation you choose. You can't see two photons traveling perpendicular to you, as someone said here, this implies a third observer. You can only see a photon, or calculate its velocity, when it reaches you directly. It's impossible to be even aware of a photon moving in any direction except when it hits you right in the eye. Its like asking what is the force you feel when a rock hits you when its moving away from you, it doesn't make any sense.

 

[Blogical]

I have read the arguments put forth by all the members here and i come to the conclusion that we can define relative velocity which can be greater than c for two photons, but will it be valid in context of particles traveling at relativistic speeds but less than c, in this scenario we can define an inertial reference frame of rest for one of the particles and calculate the velocity of the other, which will be greater than c??
In case of photons we can't define a rest frame hence it would be valid!

 

[harrylin]

[..] What we have here are two objects observed by an observer inside a specific inertial frame of reference.
The two objects themselves ARE NOT "in" this single frame. [..]

All objects are in all inertial frames and SR explains how to map them from one inertial frame to another one. That would be impossible if they were not in both "frames" (in fact: reference systems).

Also closing speed and relative speed are NOT the same as you put them together separated by a line would possibly imply. [..]

See my post #2: it's all explained there, although it was to no avail to post it only once or even twice as the misunderstandings that it prevents still occurred afterwards in this same thread...

 

[harrylin]

I have read the arguments put forth by all the members here and i come to the conclusion that we can define relative velocity which can be greater than c for two photons, but will it be valid in context of particles traveling at relativistic speeds but less than c, in this scenario we can define an inertial reference frame of rest for one of the particles and calculate the velocity of the other, which will be greater than c??
In case of photons we can't define a rest frame hence it would be valid!

No speed of an object as determined with a standard inertial reference frame can be greater than c according to SR; the speed of light is the maximum object speed. See again posts #2 and #3. If, somehow, it's still unclear to you due to the different way of presentation, I can merge them for you into a single summary post (if you like).

 

[Jeronimus]

All objects are in all inertial frames and SR explains how to map them from one inertial frame to another one. That would be impossible if they were not in both "frames" (in fact: reference systems).

Unfortunately you are right. My brain somehow mixed up being in an inertial reference frame, also implying that inertial reference frame being the object's rest frame.

Unfortunately, because i would now have to correct every of my texts. This is a disaster.

 

[harrylin]

Unfortunately you are right. My brain somehow mixed up being in an inertial reference frame, also implying that inertial reference frame being the object's rest frame. [..]

I can offer consolation for the "somehow": probably sloppy language (jargon) of others "got" you. See the discussion in parallel here: https://www.physicsforums.com/showthread.php?p=4095748

 

[altergnostic]

Unfortunately you are right. My brain somehow mixed up being in an inertial reference frame, also implying that inertial reference frame being the object's rest frame.

Unfortunately, because i would now have to correct every of my texts. This is a disaster.

I've just experienced the same problem in another thread. I think its no big issue if understand the concepts beneath the terminology. Even though all observers are in all frames, once you pick a frame as your reference frame in SR, all observations are done from there and nowhere else (meaning you can't exchange frames with other observers, you can only know what someone else sees through secondary means, like he telling you what he saw or taking a picture). If you understand that and WHY it has to be so, i think you are good to go. Only one frame can be your reference frame, your rest frame, but you can draw it in as many different configurarions you like. Frames and coordinates may coincide between observers, since you can draw the axis' in any direcions you like, but you can only see things with light coming directly into your eyes, and it is with this light that you directly relate the observed coordinates of objects to your frame, and no other frame (light that reaches you doesn't reach anyone else, the photons - or beam or patch of emitted light - that enters your eyes does not enter any other eyes, two people can't observe the same photons). In this sense, only you can directly draw your frame and be in the origin of your own frame - two bodies can't occupy the same place at the same time. You just can't claim that you are in your own frame and in no one else's, and that events are only in your frame or anything in that line of reasoning.

 

[harrylin]

[..] I think its no big issue if understand the concepts beneath the terminology.[...] you can only see things with light coming directly into your eyes, and it is with this light that you directly relate the observed coordinates of objects to your frame, and no other frame (light that reaches you doesn't reach anyone else, the photons - or beam or patch of emitted light - that enters your eyes does not enter any other eyes, two people can't observe the same photons). [..]

See in contrast:

physicsforums.com/showthread.php?p=4097067

https://www.physicsforums.com/showthread.php?p=4099114

As well as: http://www.bartleby.com/173/9.html (two oppositely traveling light flashes with many photons):
" the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of [..]the embankment. [..]. Now [..] the observer [in the train] will see the beam of light emitted from B earlier than he will see that emitted from A."

The fact that some light reaching only you has nothing to do with SR suggests to me that there is, in fact, a misunderstanding; it's just unclear what that misunderstanding is...

 

[altergnostic]

See in contrast:

physicsforums.com/showthread.php?p=4097067

https://www.physicsforums.com/showthread.php?p=4099114

As well as: http://www.bartleby.com/173/9.html (two oppositely traveling light flashes with many photons):
" the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of [..]the embankment. [..]. Now [..] the observer [in the train] will see the beam of light emitted from B earlier than he will see that emitted from A."

The fact that some light reaching only you has nothing to do with SR suggests to me that there is, in fact, a misunderstanding; it's just unclear what that misunderstanding is...

I'm not sure I completely follow your point, but I think this is important in SR for two reasons:

1 - The very fact that two observes observe the same event and assign different space and time coordinates is only possible because light that reaches one observer does not reach the other. All sets of photons (or beams or patches of light) carry the same visual information (the emission event). The relativistic effects arise from the distance light has to travel to reach each observer and the time it takes to cover those distances. This is clear that it is not the same light / photons / patch that reach each observer, if it was, both observes would have to be at the same location and see the same thing, and assign the same coordinates. In this sense it is only helpful to remember that one observer sees a differente set of photons than another, so you don't think that both of them see "the same thing from the same perspective at the same time".

2 (and this is more important) - Some diagrams and thought problems completely ignore that you can only see light locally, and light at a distance is not only invisible, but impossible to diagram properly. Some famous diagrams of the light clock problem contain this very error:

http://home.comcast.net/~peter.m.brown/sr/image_gif/sr05-im-01.gif

At first it seems ok, but this diagram is a mess. It tries to diagram light at a distance. Think about it, the local observer wouldn't draw that triangle, obviously. He would see the beams from the stationary frame, as in the left diagram. The only observer that could conceive those triangular light paths is an observer that sees the light clock moving relative to him. But those beams of light are NEVER observable to him. He couldn't possibly draw this diagram. They can't be part of his data. Even though it seems logical to do it (especially because it is the accepted way to do it and I can't discuss too much about this here or I'll get banned), it is a conceptual mess. It can't be done. Those beams are reflecting from mirrors back and forth, the never reach any distant observer. The only way such an observer can know about these light beams is if a light signal is sent from each mirror towards this distant observer at every moment light reflects from each mirror, but then he would do the transforms on those light signals. And those would be coming at him in a straight line from the point of emission. He would not see those paths of light moving at an angle like in the diagram, he would have transformed the light signals and, as a result, the light beams of the light clock would be diagramed just like the stationary observer would diagram them, locally. When you try to diagram light at a distance like that, you are trying to diagram an event in another system without transforms. That event hasn't been observed by you yet. You are basically trying to diagram a distant event happening "now", in clear contradiction with SR postulates. There's no consideration of how would you see or know about those diagonal paths, because you wouldn't.

You see, the fact that light can only be seen (or properly diagramed) locally – only when it enters your eyes or receptors (and logically nowhere else) – is a fundamental issue, not just a bald assertion. If people keep in mind that light at a distance is unknowable, undetectable and invisible to any observer, diagramming distant light or light seen by someone else would be disallowed, unless you apply transforms on the light that reaches you directly – provided it exists in the thought problem. By doing that, you are transforming the coordinates of an event in a moving system (light signals sent to you) into the coordinates of the event as seen by an observer at rest relative to the event. And then light would never propagate at angles like above. This is a key to understanding how light can be constant for every observer - because it is always measured / observed locally, from a stationary frame.

 

[harrylin]

[..] I think this is important in SR for two reasons:

1 - The very fact that two observes observe the same event with different space and time coordinates is only possible because light that reaches one observer does not reach the other. [..] the relativistic effects that arise from the distance light has to travel to reach each observer and the time it takes light to cover those distances are clearly due to the fact that it is not the same light / photons / patch that reach each observer. If it was, it would mean that both are at the same location. [..]

SR asserts in essence the contrary of what you here assert, just as phrased by Einstein in the citation in my last post: it is assumed that effectively the same "patch" of light reaches both observers. That according to SR this is in practice not exactly true (and according to QM not at all), is totally irrelevant. Einstein assumes that both observers see the same light flash, and this is not a problem because according to SR the fact of observation cannot affect the speed of light before the observation. It's a mix-up between QM and SR to think that in SR the fact of observation changes what is observed in an essential way.

2 (and this is more important) - There are some visualizations of beams of light that completely ignore that you can only see light locally, and light at a distance is not only invisible, but impossible to diagram properly. Some diagrams of the light clock problem contain this very error:
[moving light clock llustration]
At first it seems ok, but this diagram is a big mistake exactly because it tries to diagram light at a distance. Think about it, the local observer can't draw that triangle, he would see the beams as the stationary diagram to the left. The only observer that could draw those triangular light paths is an observer that sees the light clock moving relative to him. But those beams of light are NEVER observable to him. They can't be part of his data. [..] distant observer [..]

Strictly speaking one cannot observe a beam through vacuum; in such a case one can only observe the detections, for example with sensors at the mirror and the light source. But one can also have stationary detectors on both sides with which one registers the "ticks" immediately when they arrive, either by partly transparent mirrors or by means of relaying the detection. Such things are merely technological issues; for SR those are valid detections of the bouncing light in the "stationary" system.
Moreover, one can do the same in a cloud chamber, and then detect the scattered light simultaneously in both systems, by means of very nearby detectors. For SR this isn't an issue, and to think that it could matter reveals a misunderstanding of SR.

Note also that in the train example, both reference systems are "close-up" to both detections.

[..] If people keep in mind that light at a distance is unknowable, undetectable and invisible to any observer, directly diagramming distant light or light seen by someone else would be disallowed, unless you apply the transforms on the light that does reach you before you diagram the light seen by a distant observer. By doing that, you are transforming a distant moving observer's measured coordinates into locally measured coordinates, and light would never propagate at angles like above, and people would have an easier time understanding how light can be constant for everyone - because is always measured / observed locally, from a stationary frame.

I was afraid of something like that.

- The "fundamental issue" that you may be thinking of, is that distant simultaneity cannot be detected except as a function of definition. However:
- There is nothing that binds "stationary system" to "nearby detection", or "moving system" to "far away detection".

Such arguments are therefore only helping a misunderstanding of how the speed of light can be constant for everyone.

 

[altergnostic]

SR asserts in essence the contrary of what you here assert, just as phrased by Einstein in the citation in my last post: it is assumed that effectively the same "patch" of light reaches both observers. That according to SR this is in practice not exactly true (and according to QM not at all), is totally irrelevant. Einstein assumes that both observers see the same light flash, and this is not a problem because according to SR the fact of observation cannot affect the speed of light before the observation. It's a mix-up between QM and SR to think that in SR the fact of observation changes what is observed in an essential way.

SR states the opposite? All data in a real SR problem has to be either given information or data brought by light, reaching one observer. You can't directly use data that another observer is receiving at the same time he is receiving it. That's why you need transforms in the first place, because two observers have different perspectives, they are at different positions, times and velocities relative to the point of emission/reflection (i.e.: the back of the train). Hence, they may describe the coordinates in space and time of the same event in different ways.

We may assume they see the same light only for simplicity, as Einstein often tries to do. It is obvious that this assumption has no grounds on any real situation, or are you saying that if light is emitted at the midpoint between us, when we see it we see with the same patch of light/photons? This is exactly the same situation as if two laser beams are sent in opposite directions, one towards you, another towards me. I see one, you see the other. I don't see the one who reaches your eyes, and vice-versa. If you saw the same beam as I do, we are either at the same place or the beam has been reflected from me back to you, and then we wouldn't see the beam at the same time.

And where did I say that the act of observing changes what is observed? Nowhere i had this in mind. That's not SR.

Strictly speaking one cannot observe a beam through vacuum; in such a case one can only observe the detections, for example with sensors at the mirror and the light source. But one can also have stationary detectors on both sides with which one registers the "ticks" immediately when they arrive, either by partly transparent mirrors or by means of relaying the detection. Such things are merely technological issues; for SR those are valid detections of the bouncing light in the "stationary" system.
Moreover, one can do the same in a cloud chamber, and then detect the scattered light simultaneously in both systems, by means of very nearby detectors. For SR this isn't an issue, and to think that it could matter reveals a misunderstanding of SR.

All these examples show no contradiction to what I have been saying. In a cloud chamber, you need to scatter light so that two detectors can detect anything, meaning, you need light to move towards them. All detections are valid, of course, but they occur locally, and that's why they are described from a stationary system, as you yourself pointed out, just like in the left diagram above, but never like the one on the right, that's supposedly a diagram made by a distant observer in relative motion, but how can he possibly have seen those beams? And if he uses the detections, which are done in a "stationary" system, why on Earth would he try to diagram them like "seen" from a moving system? There's no possible way that you can diagram that triangle with real data. The detections are made locally, in a stationary system, they can only be diagrammed from a stationary system. There's no light detection at a distance.

If you don't see the relevance of this to SR, just look at the diagram I posted. That diagram is a mistake. It comes from wrong assumptions and poor visualization, as most SR paradoxes and misunderstandings come from. People have a hard time visualizing light, and they try to do it in ways they shouldn't. Strictly speaking, you don't observe light, you observe with light. Einstein was very careful to tell us that light is a special case. Diagraming unseen light moving at an angle due to relative motion is forbidden. First because light would have to go faster than c, since it has to cover a larger distance in the same time than in the stationary system. Second, because it is unrealistic, you have no way whatsoever to see that light, you can only guess that it traveled at those angles, and it is a wrong guess, because you didn't do the SR transforms on any light that was reaching you. Again, if you were sent light signals at each reflection, you'd need to apply the transforms on those light signals, and you would end up with a diagram just like if you diagrammed it locally, in the stationary system.

Note also that in the train example, both reference systems are "close-up" to both detections.

It doesn't matter how close they are, they still see different patches / beams / photons. This is irrelevant.

I was afraid of something like that.

- The "fundamental issue" that you may be thinking of, is that distant simultaneity cannot be detected except as a function of definition. However:
- There is nothing that binds "stationary system" to "nearby detection", or "moving system" to "far away detection".

Such arguments are therefore only helping a misunderstanding of how the speed of light can be constant for everyone.

Let me clarify what i mean by distant observer. I am not claiming that distance alone causes relativistic effects (although sometimes it does) and I am not relating these distant observers to any specific type of frame, stationary or otherwise. I use distant in a non-relativistic sense. A distant observer is just another observer, distant is just to emphasize that it doesn't receive the same light as you do. Very fundamental. It is the fact that you can't directly see the same light as someone else, even if they are "close-up", that allows for SR operations in the first place. You have to transform your observed coordinates into the other system's (locally) observed coordinates.

I am not talking about simultaneity, although it is directly related to what I'm saying. I am talking about a more correct understanding of light and how to diagram it, or visualize it, properly. You don't even need equations to understand thought problems like the embankment/train. You need them to get the numbers, but the equations don't explain anything. That's why Einstein was so lengthy in logical explanations, he tried to explain SR in as many different ways as he could, without math, so that anyone could understand why you needed the Lorentz transforms in the first place. I am presenting no contradiction to Einstein, all I am saying is that you have to understand exactly how we interact with light before you go deep into SR, otherwise you end up with bad diagrams and visualizations like above.

 

[harrylin]

As this is really not on-topic, a last clarification from me in this thread:

SR states the opposite? [..] That's why you need transforms in the first place, because two observers have different perspectives, they are at different positions, times and velocities relative to the point of emission/reflection (i.e.: the back of the train). [..]

Yes, SR states that Maxwell's theory is valid for a single inertial frame; that theory applies to all kinds of emitters and receivers, no matter if they are moving or in rest. In other words, you do not really need transforms; they can just be extremely handy. And that is essentially the contrary of what you claim.

We may assume they see the same light only for simplicity, as Einstein often tries to do.

[..]
In a cloud chamber, you need to scatter light so that two detectors can detect anything, meaning, you need light to move towards them. All detections are valid, of course, but they occur locally, and that's why they are described from a stationary system, as you yourself pointed out, just like in the left diagram above, but never like the one on the right, that's supposedly a diagram made by a distant observer in relative motion, but how can he possibly have seen those beams? [..] There's no possible way that you can diagram that triangle with real data. [..] That diagram is a mistake. It comes from wrong assumptions and poor visualization, as most SR paradoxes and misunderstandings come from.

[bold face mine]
[rearrange]:

Let me clarify what i mean by distant observer. [..] I use distant in a non-relativistic sense. A distant observer is just another observer, distant is just to emphasize that it doesn't receive the same light as you do. Very fundamental. It is the fact that you can't directly see the same light as someone else, even if they are "close-up", that allows for SR operations in the first place. You have to transform your observed coordinates into the other system's (locally) observed coordinates.

That is exactly the error that I perceived; see next!

Once more: a cloud chamber scatters light over the whole trajectory. It is technically feasible to observe the diagram on the right with an array of close-up lateral detectors that are in rest in the "stationary" frame, and the same can also be captured far away with a CMOS camera that is mounted "in rest" in the "stationary" frame. Both diagrams are equally observable, with real data.

As a matter of fact, SR was first of all concerned with comparing real data from real measurements, and that diagram illustrates what according to SR really can be measured. I agree that most SR paradoxes and misunderstandings come from wrong assumptions, but your solution is erroneous.

[..] Diagraming unseen light moving at an angle due to relative motion is forbidden. [..] First because light would have to go faster than c, since it has to cover a larger distance in the same time than in the stationary system. [..] and it is a wrong guess, because you didn't do the SR transforms on any light that was reaching you. [..]

To the contrary, it covers a larger distance in the stationary system. SR uses Maxwell's model of light propagation, which has that light propagates at c relative to a system that is thought to be "in rest". According to SR, light takes more time (by a factor γ) to cover the distance depicted on the right.
Something similar is done in Einstein's first paper on this topic, in section 3 (however without suggesting that such light cannot be detected). Citing from section 3:

"it being borne in mind that light is always propagated [along the axes of Y and Z of the "moving" system], when viewed from the stationary system, with the velocity √(c²-v²)".

In other words: the kind of thing that you claim to be "forbidden" in SR, is just matter-of-fact in SR. The very transforms that you say are needed, are based on what you claim to be "forbidden".

If anything of this is still not clear after thinking it through, please start a topic about the remaining issue in the form of a question based on the literature; for it appears that more people are confused about these matters. If it deviates from the topic of this thread, it must be a new thread.

 

[altergnostic]

For those interested in this discussion, i have started a thread as suggested by harrylin here:
https://www.physicsforums.com/showthread.php?t=643225

 

[dubiousraves]

Your question is rather confusing because of the words "THE RETURN" in your title. I can only assume that you mean a return to the subject rather than that the two photons are returning to each other. So if that is what you meant, then in a frame in which two photons are emitted from a common source, it is true that after one year, the distance between them is 2 light years, and you are correct that we cannot consider a frame of reference in which a photon is at rest, but we can still use the formula that Einstein gave in section 5 of his 1905 paper for "The Composition of Velocities" and see what velocity we would get if we plugged in c for both v and w in his equation:



And we would get c as the resultant speed.

However, if we want to follow the rules, we could change your question slightly to what would be the relative speed between a photon emitted in one direction and an observer traveling at under c in the other direction?

In fact, Einstein even makes this point when he says:

It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain


If you don't see why this is true, just multiply the top and bottom of the fraction by c:

c(c+w)/c(1+w/c) = c(c+w)/(c+cw/c) = c(c+w)/(c+w) = c

He is making the point that no matter what the speed of the observer is, the photon is still traveling at c relative to him.

Does this all make perfect sense to you?

Are the time dilation and length contraction attributes of SR specifically addressed in this equation, or elsewhere in SR? thanks (For some reason the quote did not pick up the equation, but it's the one in post #3.)

 

[ghwellsjr]

Are the time dilation and length contraction attributes of SR specifically addressed in this equation, or elsewhere in SR? thanks (For some reason the quote did not pick up the equation, but it's the one in post #3.)

The equations would have come through in the quote if you had just clicked on the QUOTE button in the lower right corner of the post you wanted to quote. It would also have provided a link back to the original post so that you wouldn't have to bother saying it was post #3. Here, like this:

Your question is rather confusing because of the words "THE RETURN" in your title. I can only assume that you mean a return to the subject rather than that the two photons are returning to each other. So if that is what you meant, then in a frame in which two photons are emitted from a common source, it is true that after one year, the distance between them is 2 light years, and you are correct that we cannot consider a frame of reference in which a photon is at rest, but we can still use the formula that Einstein gave in section 5 of his 1905 paper for "The Composition of Velocities" and see what velocity we would get if we plugged in c for both v and w in his equation:

And we would get c as the resultant speed.

However, if we want to follow the rules, we could change your question slightly to what would be the relative speed between a photon emitted in one direction and an observer traveling at under c in the other direction?

In fact, Einstein even makes this point when he says:

It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain

If you don't see why this is true, just multiply the top and bottom of the fraction by c:

c(c+w)/c(1+w/c) = c(c+w)/(c+cw/c) = c(c+w)/(c+w) = c

He is making the point that no matter what the speed of the observer is, the photon is still traveling at c relative to him.

Does this all make perfect sense to you?

Now to answer your question, no, this post is not about time dilation or length contraction. The purpose of the post was to show that the closest possible calculation of the relative speed between two photons going in the opposite direction is c and not 2c.

 

[harrylin]

Are the time dilation and length contraction attributes of SR specifically addressed in this equation, or elsewhere in SR? thanks (For some reason the quote did not pick up the equation, but it's the one in post #3.)

It's unclear to me what you mean with "specifically addressed". Time dilation and length contraction as well as clock synchronization are inherent in the Lorentz transformation, from which that equation is derived. So, without time dilation and length contraction that equation would be different, if that is what you mean; they are needed to find c for the one-way speed of light after clock synchronization.

 

[dubiousraves]

Sorry, I don't see any quote button. Perhaps it doesn't exist in Google Chrome?

Anyway, I posted my question on this thread because harrylin, the esteemed commentator here, suggested the answer to my question could be found in the post I referenced. It does seem my query would fit more comfortably in the time dilation thread, but I'll state my question here anyway and kick it over to wherever you experts think it belongs.

To wit: What are the time-dilation and space-contraction reasons for c being constant for an observer not only traveling in the same direction of a light beam (say at .75c, where I understand the time-contraction reasons for the traveler seeing the light beam at c), but for an observer heading directly toward a light beam coming at him?

Thanks.

 

[harrylin]

Sorry, I don't see any quote button. Perhaps it doesn't exist in Google Chrome?

Anyway, I posted my question on this thread because harrylin, the esteemed commentator here, suggested the answer to my question could be found in the post I referenced. It does seem my query would fit more comfortably in the time dilation thread, but I'll state my question here anyway and kick it over to wherever you experts think it belongs.

To wit: What are the time-dilation and space-contraction reasons for c being constant for an observer not only traveling in the same direction of a light beam (say at .75c, where I understand the time-contraction reasons for the traveler seeing the light beam at c), but for an observer heading directly toward a light beam coming at him?

Thanks.

Yes, you asked: "what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this? "

That is closer to the topic of this thread. See also post #7 as well as my earlier reply which was just before yours here.