Two abreast
This is probably my last dabble with relativity (for a while).
Prove that if in a given inertial frame, two photons are travelling in parallel and abreast, at a distance r apart, then they do so in all inertial frames.
Hints: Consider the case when they are not travelling abreast and then when they are abreast. There is a nice way to do this.
Link to the reference book for this problem:
here
Prove that if in a given inertial frame, two photons are travelling in parallel and abreast, at a distance r apart, then they do so in all inertial frames.
Hints: Consider the case when they are not travelling abreast and then when they are abreast. There is a nice way to do this.
Link to the reference book for this problem:
here
Labels: funphysics
posted by Chris at
5:46 PM
47 Comments:
Hi Chris. If I want to consider photons with respect to SR, I should treate wave 4-vectors
k^i=(w,k),
where k is wave 3-vector, w=|k|. Then apply Lorentz to k^i. Because of two photons a travelling abreast,
their 4-vector differs only by total multiplier (if photons have different frequencies). Therefore Lorentz will lead to aligned 3-vectors k again. Maybe I should write how to perform arbitrary Lorentz. Introduce coordinate system with z-axis along v. Thus
x and y components of k will not change and for both photons z component change will give aligned propagation direction again.
By the way, distant r has no meaning here. Moreover, if we speak about single photon, not wave packet, i.e. we have only one frequency, we have no coordinates at all. It means that photon is not localized. Some finite range of coordinates has only wave packet. It's clear from a lot of considerations. For example, uncertainty principle. But if you perform a measurement of photon coordinate, your intervention will destroy the photon and create a wave packet.
thats what i was about to say *cough*
Hi quantense, I had in mind a semi-classical approach, whereby the two photons are treated as pointlike particles. I accept that measurements cannot be made without disturbing the photons.
The result I'm looking for presumably only applies to the average behaviour of a swarm of photons. i.e. the wavefront of a beam of light is perpendicular to the direction of travel (in vacuuo!) in all inertial frames.
Thnk you for your response, it was unexpected (that's good), and definitely has given me something to think about. I will try an LT out to see what happens. As with the "phasers on stun", I remember it from Wolfgang Rindler's book.
I decided to post this problem, rather than one about a supersnapshot, where different inertial observers take the same photograph of distant objects perpendicular to their direction of motion - the photos are the same shape, but presumably of different colours.
Consider two photons, one is ahead of the other by a distance l and their paths are a distance r apart and they're travelling in parallel. Consider two arbitrary events at the two photons. For convenience, let one photon happen to be at x=y=z=t=0 and also let that photon be travelling along the positive x-axis. Assume that z=0 throughout i.e. the photons are in the x-y plane.
At some arbirary time t, the other photon will be at x=l+ct, y=r. The spatial separation between the two events is sqrt(r²+(l+ct)²) = sqrt(r²+l²+2lct+(ct)²) and the interval s between the two events is then sqrt((ct)²-r²-l²-2lct-(ct)²)=sqrt(-r²-l²-2lct). So, if l<>0 then the interval s a function of t. However, if l=0 (=>photons abreast) then s = sqrt(-r²) and is independent of t. As s is independent of the inertial frame, we have the result required. i.e. spatial separation is r in all inertial frames, and the interval is independent of t in all inertial frames => unique "signature" of abreast photons.
Honestly, Chris, I can't agree with your solution. But as oft happens in physics, it leads to the correct final expression. The thing is that, as I've claimed from your post, you believe that there's a frame where first photon remains at (0000), i,e, you bound your frame to the first photon. This is illegal and leads to loads of paradoxes in all senses (one of them is v+c-c=c). But lets do this. If once the first photon is at (0000), the second photon event is (0,l,r,0). The trajectory is
for the first (t,ct,0,0)
for the second (t,l+ct,r,o).
Interval is constantly rt(l^2+r^2). And I get your
rt r^2=r in the case l=0. The same expression, but different way!
I didn't say the photon remains at rest, its worldline happened to include x=y=z=t=0 i.e. an event. The first photon's worldline is x=ct, y=z=0.
I've assumed exact positions of the photons at the expense of not being able to know their momenta. I've also assumed exact times for the photons at the expense of not being able to know their energies. Is that unreasonable?
Using the 4-vector notaion (t,x,y,z) order. The first photon has worldline (t,ct,0,0) and the second (t, l+ct, 0, 0) just as you say. I chose an event for the first photon at t = 0 and for the second photon at t=T (I'm using T to avoid ambiguity). So the interval between these two events is sqrt(c²T²-(l+cT)²). It may seem a bit strange, but I was trying to avoid using Δx, Δy,Δz,Δt as it's a pain to type. Greetz.
Hi Chris, when you want to determine spatial distance between events in the flat space-time, you should take interval between this events at the same coordinate time. You couldn't use one time t=0 for one photon and another t=T for the second one.
I didn't mean uncertainties any more in my previous post. I can admit such classical approach here. But it doesn't mean I can admit photon to stay in definite coordinate state as long as I want. Such narrow wave packet will spread almost simultaneously and after wery short time will have almost infinite coordinate range.
Hi quantense. That you must choose Δt=0 or t=T, is simply not true. If I was measuring a length of a moving object, then of course I'd probably choose Δt=0 as that is the eaiest way. If I knew the speed of the object then I could measure the positions of the end points at at different times, but I'd have to take into account that the positions will have changed by vΔt to find the length - but that isn't a fundamental problem - it's just engineering.
I have simply looked at two events, each associated with a different photon. Those can have an LT applied, and the associated intervals will be the same in all inertial frames - this is very basic.
If I said there was a car that was in London at 12am and there was another car in Manchester at 5pm you'd probably accept that the distance between those two points was about 200 miles. That's all I'm doing.
If I was trying to determine a speed of a particle, say, then I'd have Δx, Δy, Δz and Δt to play with - you wouldn't insist that Δt=0 any more than you'd insist that Δy=0.
As I said a while ago, I was only intending to do "off-beat" physics. So that may be why my problems are proving difficult for you to understand - I don't mean that you don't understand physics, I mean that the problems are not quite the standard stuff that you will be familiar with.
Chris, please read any SR textbook to figure out that spacelike length is interval at the same time moments in flat spacetime. Or opposite thing, if so, give me a citation.
Chris, this is my last post untill you give me any textbook referense for verification of your words or accept you are wrong. I completely shure that you are wrong, so I feel myself very good. But I wanna make an accent on that I have some amount of life experience, not as much, as yours, of course, so I'm sure that your last post was consequence of the fact you can't convince me in physical sense and just try to say: "when you grow up you will understand I was right". Chris, I'm not a kid. Yes, I didn't understand Phasers on Stunfirst time because of misty formulation or my English knowledge or both. Of course I make mistakes sometimes, even Einstein was wrong in some questions, not deep, just ordinary physics. But it is a consequense of my approach to physics: I try to find some beauty in any problem, to feel it physically, and sometimes it closes my eyes on ordinary stuff. But this approach almost everytime helped me to solve really off-beat problems on exams in MIPT. That problems were really harder I find there. I've already said you, that I like conversation and practising English, that's why I visit this cite.
Hi Chris, the thing I really mean was that physical distanse between two spacelike separated events is interval between them. This means we should choose the frame where there is the same time coordinates of two events, as you understand. Physical distanse is invariant for spacelike separated points. That's all (as self time for timelike). My 8.23 post was about this. But of course the coordinate length is relative value and it changes, and this is connected with relativity of simultaneity.
Chris, you've removed other your posts ten minutes after I've read them.
Hi quantense. I promise that I'll leave you alone after this post.
The posed problem came directly from such a textbook (Wolfgang Rindler- "Relativity: Special, General and Cosmological"). It was a set book at Imperial College, London, where my wife was studying theoretical physics. I lent her my copy. I've made a link directly to the page in that book at the top of this page. The link takes you directly to the relevant page. Unfortunately, Rindler doesn't break the problem down into the steps that I used, so you may not accept that he's done exactly the same physics as I have done.
You asked me to cite any text book, that says you can determine a distance at different times. My response is, every text book that includes (Δt,Δx,Δy,Δz) or Δs²=c²Δt²-(Δx²+Δy²+Δz²) will do, because Δx, Δy and Δz are spatial separations (distances) determined at different times, the time difference being Δt.
I now ask you to in return to cite any textbook (or decent website) that says you have to measure spatial separations at the same time. I mean as a broadly general statement i.e. don't tell me about measuring the length of a moving object (such as a ruler) - even if you do, I'll refute that allegation in less than a minute. Please confine the citations to SR books only. If possible, please include at least one that can be checked on the internet.
---
In case you doubt Rindler, here's some background on him:
http://en.wikipedia.org/wiki/Wolfgang_Rindler
I can't find much on James Terrell either, but you should be able to convince yourself he's not a nut by Googling "james terrell physicist".
This post has been removed by the author.
Here we got abreast photons, and they stay abreast at the same distanse because alteration of their position is therefore lightlike, which doesn't change spacelike initial separation.
Hi quantense. I'd been tidying up. I realised that most of the posts I made would probably not achieve a useful purpose. As you were becoming irritable I decided to combine my most important ideas into a single post.
I'll put them back if you want.
I'll publish this, but will add another in a few minutes.
Ther's something wrong with the order of postings, Chris, read 10.40, please
Hi, I see you've deleted one as well. My response was: If you doubt that you made such statements, then check your posts: Sep 19, 8:23 AM, first sentence; Sep 19, 10:18 AM, only sentence.
I'll watch for your response now.
You shouldn't repeate. I think we can't agree if make a point on what each of us was intending to say.
It was the 10:40 post that I thought you deleted. At lest your talking to me again. That's such a relief after your earlier posts. Appreciate that I'm also totally sure that I have only done good physics. You make statements that are untrue. You have several times said that I have somehow stuck a photon for eternity at single point in time - that statement doesn't even make sense as (0,0,0,0) is a single point in space-time. You've seemed to imply that I've also stuck that photon at (t,0,0,0) - I haven't - you have ignored that I have said the worldline is (t,cv,0,0) and that I only have considered an instantaneous event at t=0.
I'm publishing before you run away.
I had a misprint. I wrote timelkeinstead of lightlike. I noted you've removed your post just for a joke. Just to show you I may be late to read them.
Chris, don't hurry. I promise stay here at least an hour :)
Not sure which "repeat" you are referring to, I only recall repeating that the photon wasn't stationary and that you can measure distances at different times because you kept on ignoring my responses to that error.
In my 10.40 post I tried to clarify what I was talking about. Have you read it?
Ok I'll slow down. I just noticed a typo in my 10:54 post, I meant that the worldline was (t,ct,0,0).
Burning question, do you yet accept that I might not have made a mistake? and Greetz.
Yes, Chris, you were right. But read my 10.40 to understand I was talking about different thing.
What I was talking about is physical distanse. What were you is coordinate. Physical is invariant, coordinate changes (relative).
And physical is interval in the frame where two events are simultaneous.
The bunch of posts I deleted were OK, but I really was afraid of trying your patience. I didn't want to make an enemy by trying to force my opinion onto you. None of them were silly or rude, they just didn't serve a useful enough purpose to be left up. I had been deleting them before I realised you were online, not BECAUSE you were online.
Don't worry about time-like space-like typos (too much), I have to think carefully to decide what that means. I have to say to myself, timelike is when there is an inertial fram where the events are at the same place but at a different time, and space like is when their is a frame where the events happen simultaneously at different places.
are we friends again?
I wonder if the misunderstandings may have been avoided if I hadn't tried to keep the original explanation short by cunningly arrranging for one photon to pass through (0000).
The saddest thing about the misunderstandings is that I considered the de Broglie wave of simultaneity and this problem as being very nice - almost beautiful.
Yes Chris, we were friends even when we were arguing. When to men can't agree with each other and both sure they are right, there's 80 percent probability they are talking about different things.
Hi. You had started to seal down the hatches - I was upset when you did that - but I'm glad you're back again.
I don't know if I dare do the one with a man with a long pole successfully running into a small room according an observer in the room. Do you like the sound of that?
My first aim was to provide consistent way to treate photons, and show, that it's impossible to treat them as points, even if they are in point coordinate states some time. Then I was astonished about physical length, as I would be if someone said me self time depends on coordinate distanse. We would save a lot of time if formulaation were "at a Coordinate distanse r apart". The thing is that almost every time during conversation I try to detalize problem formulation immediately and I don't like to solve the problem, formulation of which I can't understand. When verbally speaking, you can do it all very fast and don't pay attention on it. Here I've learned I should formulate my thoughts more clear.
I like your 80% probability quote. On quotes, I did try a blog that used about 10 quotes from Einstein as a kick start. Rajesh wisely removed it; I suspect that he thought it might not have gone down well with some people.
I completely assume that (Δt,Δx,Δy,Δz) are co-ordinate units already. I had never intended to make an LT. I was initially only really interested in Δs and r in one frame, but with the intention of appealing to the invariance of Δs and the time (in)dependence of it.
Just looked at your last post: I imagined that text books often treated photons as almost classical point like particles as a matter of course. It's only because of you first post that I seriously considered the QM aspects. That's when I realised I couldn't know anything about E or p (or ω or k). I don't definitely recall any text book mentioning that "small detail".
Of course, Chris, when people learn some stuff, they read textbooks which present problems with pointlike photons. Bit this is textbook - just for learn some aspects on some in-mind model. This is not real physics. But I suppose that here if we do smth, we do it as right as possible. We don't solve standard problems. which you were talking about when blamed me by saing I solved easiest thing all my life (please, don't pay attention on this. I know, you didn't mean it).
Actually I've learned something during posting here. In my everydaylife I discuss physics with my fellowd and lectures, who almost everytime understand me from the first word. Here another case.
It might be a language problem, I didn't intend to mean that you do simpler physics than in these problems. I'm sure you do very much more advanced stuff. Now that I'm consciously thinking about it, perhaps the comment was unwise, you probably see a lot of elegant proofs and so will be used to it. I also know that a lot of text book stuff can be have very elegant proofs. The posted problem is an example of that.
Some of the blogs on this site do have very nice logic in the answers, that's what makes them special - it's that beauty thing.
I know what you mean about not normally having to justify yourself, it's the same for me. Even on this site, most people don't challenge my reasoning. I think it is a good thing, because it forces you to really clarify your thoughts - nobody really likes losing an argument, especially when they've been quite certain they're right, it can be bad for the ego if you're not made of the right stuff. I don't mind losing an argument or making mistakes (very much), and usually laugh at myself when I do get it wrong. See my dreadful performance in Ragknot's "Return to WowDVDFilm" for evidence of what a nit I can be if I try. It was reletively easy to laugh at myself there as I was the one who was pointing out my own errors, with some gentle nudges from Ragknot.
Another thing or two is that I have a lot of patience, it's almost always the other person who gets angry before I do. I'll just go on and on trying to think of how to make my viewpoint obvious, unfortunately for my victim.
I like doing it, but I run the risk that people might think I'm bullying them.
OK, Chris, I'll take into account:)
Hi again, I've been elsewhere. You'll notice that I do post after post after post even when no-one is making an argument, I just like to be thorough. Also I'm stuck at home with a broken leg (although I am becoming quite mobile now) so have bags of time on my hands. This site keeps me entertained. But I'll be getting a life (as some of the trolls say) soon.
I've just re-read my last post: I only really had one run-in with someone ("Mixing it" a couple of weeks back). I learnt (more or less) to not feed the true trolls. The Anonymous there took all my posts the wrong way, but he had made statements that were factually incorrect, and I was just trying to help him see that. He then went into silly idiot mode as he couldn't make reasoned arguments. Bye for now.
For those of you who don't know relativity Δs²=c²Δt²-(Δx²+Δy²+Δz²) takes the role of the more familiar Δr²=Δx²+Δy²+Δz² (Δr being length or distance) that you are used to. Δs is called the "interval" in SR. Just as length is considered independent of the frame you measure it in (in Newtonian mechanics), the interval is independent of the inertial frame you measure it in, in SR.
It is also quite common to see, Δs²=c²Δt²-Δr² for obvious reasons.
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