THE RETURN of two opposite travelling photons!!

Blogical

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

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

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

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

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

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.



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

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

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

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

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

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

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

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

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-...Closing_speeds.

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

harrylin

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

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.