In the perspective of the photon

Discussion in 'Physics & Math' started by Cyperium, Nov 16, 2010.

1. Green DestinyBannedBanned

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I can play your twisted games as well. And what did you mean by a rest frame, you tit?

3. alephnullyou can count on meRegistered Senior Member

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Nonsense. Hilarious nonsense.
I can't believe you are trying to argue that what you said might be meaningless but it's not wrong!

5. AlphaNumericFully ionizedRegistered Senior Member

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Obviously you can't or else you'd have responded to the individual points I make. Instead you did precisely as you always do, quote a post in its entirety and then ignore pretty much everything in it. You didn't step up to anything. Do you seriously expect people to believe your claims you're right when you never defend against any criticisms, you just whine "Oh you're mean!"?

Seriously? Your retort, your attempt to play my 'twisted' game, is to ask me to define a rest frame? The rest frame for an object is that frame where the object, at a given moment in time, is stationary. More formally its velocity takes the form $v = v^{0}\frac{\partial}{\partial t}$. If the object is experiencing no external forces then it will remain at rest, in that frame. If its experience forces like gravity then its rest frame will actually be time dependent.

I actually gave that definition, in terms of components, previously. Didn't you understand it? I guess not, despite the fact you'd quoted the component expansion definition of a general vector. Obviously the place you copied and pasted from you didn't take the time to understand. If you're even capable of that, which I have ever increasing doubts about.

You're not going to be able to play my 'twisted game' back at me because I take the time to understand what it is I'm talking about, I don't need to mindlessly copy and paste things Google found for me.

7. RJBeeryNatural PhilosopherValued Senior Member

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GD and AN, you guys seem to have a strange symbiotic bully / victim dynamic going on...I just can't always discern the victim (besides of course the thread!

)

8. RJBeeryNatural PhilosopherValued Senior Member

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If you want to get philosophical, I'm not so sure the "photon exists". There are two relevant events to a photon's lifespan, which are emission and absorption, that from the photon's hypothetical perspective happen simultaneously. I'm not convinced we can say anything else about its existence "in between" those events. Any measurement made simply replaces the absorption event of the photon. To believe otherwise is to subscribe to counterfactual definiteness which, combined with locality, contradicts most QM interpretations.

We can slow the progression of light through a medium, which might appear to make the OP question more legitimate, except a quantum analysis of this system suggests a series of absorption/re-emission events by the atoms, while "a" photon travels with a velocity of c between them. One could argue that these are not the same photon.

9. ULTRARealistically SurrealRegistered Senior Member

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Hi Alpha,
GD is actually being quite useful..no seriously..A lot of the math, I don't get, so all your corrections and explanations are actually very helpful for someone who never even did calculus! You'd probably expect a Bio guy like me to run a mile from this stuff, but no, not me..I studied biology to find out how the world ticks..Now i want to know about the universe! it never ends!

10. arfa branecall me arfValued Senior Member

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Ok, that seems to correspond to what Hawking says in The Nature of Space and Time, that a Cauchy surface S has a future development which can be predicted from data on S.
Hawking says that the Cauchy horizon of S is a consequence of S being compact. I'm not so clear about what "compact" means, or what "geodesically complete" means.
Again, Hawking cites the compactness of the development of S, that implies the horizon is also compact. But this means, quote: "the null geodesic generators will wind around and around in a compact set. They will approach a limiting null geodesic ...that will have no past or future endpoints in the Cauchy horizon".

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True enough, but one can still consider an inertial reference frame moving relative to another at a speed arbitrarily close to c, and ask what sorts of things happen in this limit. And some interesting stuff pops out, like electricity and magnetism switching roles (or something along those lines - it's been many years since I attended the lecture that covered this in undergrad. Anybody want to expand/clarify?).

12. AlphaNumericFully ionizedRegistered Senior Member

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In that context 'compact' is the mathematical notion of 'compact', ie closed and bounded. A less formal (and not entirely accurate as a result) version would be its of finite size with a particular kind of clear edge.

A simple example is the interval [0,1]. Its compact, as its closed and bounded. Remove '1' to get the interval [0,1) and its no longer compact, as its not closed (consider the sequence $a_{n} = 1 - \frac{1}{n}$).

Geodesically complete, if memory serves, means that if you pick any two points in the space there is a geodesic between them. Flat space is an example, as the geodesic is just the straight line between the two points.

My comment on the realm of Cauchy surfaces and Hawking singularity theorems is 'there be dragons', ie the kind of eye wateringly complicated differential geometry and topology theorems that many people who spend a lifetime publishing work in GR don't ever go near.

Electric and magnetic fields are just different view points of the same underlying concept and applying a Lorentz transform to Maxwell's equations mixes them together. Its possible to write down a system which is entirely electric fields in a given frame and then boost to a new frame where its all magnetic.

This can be seen formally from the fact that $E_{i} = F_{0i}$ and $B_{i} = \frac{1}{2}\epsilon_{ijk}F_{jk}$ in a given frame (modulo minus signs I've probably forgotten) and under a Lorentz transformation you have $F_{ab} \to \tilde{F}_{ab} = \Lambda^{c}_{a}\Lambda^{d}_{b}F_{cd}$. If you work out how this alters the E and B fields you'll find that's precisely how Maxwell's equations transform too, which is to be expected since $F_{ab}$ is a slick way of doing electromagnetism (compared to how Maxwell did it).

13. chaos1956BannedBanned

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Does the statement "the photon" reflect only a singular photon? How do we know what we believe the quanta of light to currently be in all these calculations is the singular perspective of a single infinitly small particle? The second you get to its size to view its perspective it just gets smaller from your perspective.

14. arfa branecall me arfValued Senior Member

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Oh snap. There goes my show and tell for next Monday.

15. CyperiumI'm always meValued Senior Member

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Ok, so you say that light travels its course because the universe travels even faster in infinity? But that doesn't apply to the light we see, does it?

Let me explain again, the true nature of light would be in the perspective of light itself, as this is what light is. How come that we perceive a different truth? How can the photon be both at every point in every time and at a specific observable point at the same time? Where the photon has never been specific about that point.

It means that the observable universe is different from its true nature...that none of the parts has actually been as we see them.

So you say that the photon behaves in a undeterministic way and because of that it can exist in a specific place when in actuality it exists at all places and at all times, what determines where it should be then?

I mean, what determines the specific point where it is to observers when it actually is at all points? I get that we can only see one time, so that is what determines what time it should be in, but we do perceive many points.

How does it relate to the OP?

Ok, so you've taken on the string perspective of things, but does it solve the problem?

You have a different agenda, the radiation you refer to is Hawking Radiation and is because of quantum fluctuations where a pair of particles starts to exist near a black hole's horizon and the particle with negative mass gets drawn in and the other escapes as radiation leading to the decrease of mass of the black hole (because the particle that is sucked in has negative mass).

What do you mean?

Ok, I think so too.

Last edited: Nov 19, 2010
16. Farsight

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That's how it might seem to the photon, Cyperium, but it isn't actually how it is. You can jump out in front of that photon and absorb it at a given time and place. It is subject to events. To really appreciate this, imagine you 're travelling at c like a photon. We all know you can't actually travel at c, but just imagine you can. Then you don't actually exist at all points, and distances don't actually shrink to zero. You might think that they do from your perspective, but actually, you don't have any perspective, none at all. That's why you don't notice when I put an asteroid in your way. BLAM! And your "everlasting moment" is over.

17. CyperiumI'm always meValued Senior Member

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But that would require altering special relativity though. Sure I can do that in my imagination, but it wouldn't make it real. Special relativity says that all distances would shrink to zero when travelling that fast, and time would stop. If I imagine that, then yes I would be at all points as there isn't a preffered reference frame (the one that makes most sense). To me I wouldn't just hit your asteroid but all other points as well in the universe.

Even though the asteroid is closer to me than all other points, the distance to all other points have shrunk to zero and therefor the asteroid is no special case from those other points anymore. Only to a outside observer does the asteroid seem closer to my starting point.

The first photon that ever existed would have seen the universe pass away infinitely fast as time has stopped for it, also it would be at all points in the universe for all that time.

18. AlphaNumericFully ionizedRegistered Senior Member

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The length contraction only occurs in the direction of your motion, not in all directions. The 2 directions orthogonal to your motion do not experience length contraction.

19. przyksquishyValued Senior Member

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Just to pick a nit, this isn't entirely true. The reason is that the quantity E[sup]2[/sup] - B[sup]2[/sup] is invariant. If the field is all electric in some frame then this quantity is positive. If it's all magnetic then it's negative.

20. AlphaNumericFully ionizedRegistered Senior Member

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Fair point. On further reflection I think what I was half remembering was that given a combined electric and magnetic field there's a frame such that it 'rotates' into purely one of the two, the other field is turned off. Since $K^{2} = E^{2} - B^{2}$ is invariant you can determine which field can be turned off before hand and then all you're doing is rotating into the other field such that the new field has modulus K (that's not standard notation, I just had to pick a letter).

21. M00se1989BannedBanned

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Though it could be If you were to make $K$ the tripple point.

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23. Farsight

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It doesn't mean altering special relativity at all, Cyperium. All the maths still holds good. What it probably does mean is altering your understanding of what that maths means. For example, if you're in a spaceship and you hit the gas and move fast towards a star, the star doesn't "really" change from a sphere to a discoid. Your measurement and perception of the star change, but not the star itself. You know this because I'm in another spaceship, and our measurements don't tally, so we know they're not "really" real. Have a read of The Other Meaning of Special Relativity by Robert Close. All the maths is exactly what you know, but the interpretation is totally different. Good paper.

That's the distance according to your measurements, but the thing is, if you're travelling at c, you can't make any measurements at all.

But think on this: that first photon hasn't finished seeing the universe pass yet. And it never ever will.