Mathematical model of photons that can accelerate

[Johnny5]

I am interested in modeling a photon as a particle that can accelerate, but I am not sure about where to begin.

 

What I have been doing is this.

 

I first define the position of the photon's center of inertia in a frame as follows:

 

[math] \vec r = x \hat i + y \hat j + z \hat k [/math]

 

Then velocity as follows:

 

[math] \vec v = \frac{d\vec r}{dt} [/math]

 

Now the magnitude of the photon's velocity is its speed in the frame.

 

So this is a theory of a variable speed of light.

 

I use the following formula for the speed of a photon in a frame:

 

[math] |\vec v | = f \lambda [/math]

 

I am not sure how to interpret photon frequency just yet, and photon wavelength just yet, but I am treating the relation as valid in the frame.

 

Now, to get to the magnitude of the acceleration i have:

 

[math] |\vec a| = \frac{d}{dt}(|\vec v |) = f \frac{d\lambda}{dt}+ \lambda \frac{df}{dt} [/math]

 

Has anyone seen this approach to variable speed of light theories? What is the next step? I want to relate the general photon speed v, to the speed c, relative to the emitter.

 

Note:

 

[math] \frac{dc}{dt} = 0 [/math]

 

Regards

[5614]

ummm, right, i cannot realy go much further than this post in this thread because I do not know enough physics mathematics to do so, but photons always travels at c in a vacum... they do not accelerate and decelerate, they maintain a constant speed of c.

[Johnny5]

ummm, right, i cannot realy go much further than this post in this thread because I do not know enough physics mathematics to do so, but photons always travels at c in a vacum... they do not accelerate and decelerate, they maintain a constant speed of c.

 

You have the speed of a photon being independent of the motion of the source, does that not bother you?

 

Regards

[5614]

it depends on your frame of reference i suppose... but how does that effect the fact that photons travel at a constant and do not accelerate/decelerate?

[Johnny5]

it depends on your frame of reference i suppose... but how does that effect the fact that photons travel at a constant and do not accelerate/decelerate?

 

Why do you keep insisting that the speed of a photon is a constant?

[5614]

here:

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

is a long, good and thorough article on it!

 

(I'm going to bed now, 10:10pm in london and I've got exams tomorrow and the day after (maths then physics), so I'll continue our conversation in a few threads tomorrow.)

[swansont]

Why do you keep insisting that the speed of a photon is a constant?

 

Because when I drive in my car I can still see and my radio works.

[The Rebel]

Would be interesting to see a mathematical side to it though, thus proving dc/dt = 0.

 

Although I can't help thinking this is already hidden in Maxwells' equations somewhere.

[ydoaPs]

...photons always travels at c in a vacum... they do not accelerate and decelerate, they maintain a constant speed of c.

 

afiak, photons ALWAYS travel at c. isn't n(index of refraction) because photons asre absorbed and reemitted by matter?

[Johnny5]

I don't want this post to end here.

 

Suppose that before a photon is emitted, it is whirling in a circle in the rest frame of the emitter. Just like Newton's pail. Then we let it go, and it flies off in a straight line.

 

There were parameters to the motion of the photon before emission. The photon had a period T, and a circumference [math] 2 \pi R [/math]

 

Its tangential speed would be:

 

[math] v_t = \frac{2 \pi R}{T} [/math]

 

f = frequency = 1/T, hence

 

[math] v_t = 2 \pi f R [/math]

 

and

 

[math] \omega = 2 \pi f [/math]

 

Which gets us here:

 

[math] v_t = \omega R [/math]

 

Which is correct for circular motion. Setting the tangential speed vt equal to the speed of light relative to the emitter gives:

 

[math] c = \omega R [/math]

 

Now I guess use the galilean transformations from here.

 

Has anyone seen this before?

 

Regards

[ydoaPs]

if it is whirling in a circle, it is acccelerating.

 

i think of it kinda this way: photons are kinda like dorps of water and an atom is like a soaked sponge. if a water droplet hits the sponge, it is absorbed another expelled somewhere else.

[5614]

yub yub yourdadonapogos, my mistake, the photons do travel at c but are absorbed and reemitted by matter, I did it in my coursework just a few days ago and I still made a mistake!

 

I suppose the water droplet is an ok method of visualising it, the energy of the photon is absorbed straight away and then re-emitted, there is no stage of (de/ac)celeration.

[Johnny5]

I don't want this post to end here.

 

Suppose that before a photon is emitted' date=' it is whirling in a circle in the rest frame of the emitter. Just like Newton's pail. Then we let it go, and it flies off in a straight line.

 

There were parameters to the motion of the photon before emission. The photon had a period T, and a circumference [math'] 2 \pi R [/math]

 

Its tangential speed would be:

 

[math] v_t = \frac{2 \pi R}{T} [/math]

 

f = frequency = 1/T, hence

 

[math] v_t = 2 \pi f R [/math]

 

and

 

[math] \omega = 2 \pi f [/math]

 

Which gets us here:

 

[math] v_t = \omega R [/math]

 

Which is correct for circular motion. Setting the tangential speed vt equal to the speed of light relative to the emitter gives:

 

[math] c = \omega R [/math]

 

Now I guess use the galilean transformations from here.

 

Has anyone seen this before?

 

Regards

 

To continue a bit, we have:

 

[math] \frac{dc}{dt} = \frac{d}{dt}(\omega R) = R\frac{d\omega}{dt} + \omega \frac{dR}{dt} [/math]

 

Now, invoke the postulate that the speed of light relative to the emitter is a fundamental constant of nature. Hence dc/dt=0 so that we have:

 

[math] 0 = R\frac{d\omega}{dt} + \omega \frac{dR}{dt} [/math]

 

From which it follows that:

 

[math] R\frac{d\omega}{dt} = - \omega \frac{dR}{dt} [/math]

 

Has anyone seen this before?

[lumi]

photon is a particle of light, it is sometimes called light itself. if the speed of light is constant then the speed of photon is constant because photon is a function of light

[□h=-16πT]

Light has no inertial mass. Light has no inertial frame. Light does not accelerate.

 

Why the dickens are you using the galilean transformation?

[Johnny5]

Light has no inertial mass. Light has no inertial frame.

 

 

Light which is emitted by the sun, doesn't get gravitationally pulled back to the sun. So it makes more sense to say that light has no gravitational mass. The rest frame of a photon is an inertial frame, but that's for another time.

 

Do you happen to have a proof that light has no inertial mass?

 

 

PS: Another way to see my point, is that photons communicate gravity, and so are not effected by it.

[□h=-16πT]

Yes I have a proof for that. I also have a proof that light has no inertial frame as well. But I can't be bothered to show them right now as I'm rather busy. The sun doesn't have an ergosphere, that is why it doesn't stop light back, or indeed a mass big enough to have a huge surface escape velocity. Light is lensed by gravity, black holes do not allow light to be emitted and, in addition, Kerr black holes can stop light emitted equatorilly.

[Johnny5]

Yes I have a proof for that. I also have a proof that light has no inertial frame as well. But I can't be bothered to show them right now as I'm rather busy. The sun doesn't have an ergosphere, that is why it doesn't stop light back, or indeed a mass big enough to have a huge surface escape velocity. Light is lensed by gravity, black holes do not allow light to be emitted and, in addition, Kerr black holes can stop light emitted equatorilly.

 

Well there is no rush to see your proofs, but I would be interested no matter whether they be right or wrong. Obviously, you make a certain number of assumptions, for a starting point, and you say that if your assumptions are true then such and such, so that your reasoning will be correct. So it is the assumptions I am interested in. But that is for some other time.

 

But... you did say something interesting, which is that black holes do not emit light.

 

I was talking to a fellow here, by the name of 'island', several weeks ago, and he got me thinking about that.

 

Obviously, black holes have an immense gravitational pull.

 

Obviously, they emit something, and that something curves the path of something approaching near. Island's point was simply, that the something emitted (if anything) could not be 'light' since light cannot escape a black hole. A conundrum.

 

Regards

[□h=-16πT]

I'm not going to even bother with your theory.

 

The processes of radiation emission from a blackhole are the Penrose effect and Hawking effect. The two aren't too disimilar. That something you're talking about would probably be a graviton.

 

If you want to know about the maths involved in showing that light is not emitted from a blackhole's horizon look up Schwarzshild geometry for the case of a stationary star or the Kerr metric for a static star.

 

I'm not going to discuss this here, because I'm already talking about GR with you in the relativity forum, so if you want to know anything ask me there.

[Johnny5]

I'm not going to even bother.

 

Why not?

 

Not bother with what?

 

That black holes cannot emit photons, something about GR, ???

[reverse]

Why not?

 

Not bother with what?

 

That black holes cannot emit photons' date=' something about GR, ???[/quote']

 

From what I can tell, he seems pretty keen to keep GR and SR arguments within their own niche.

Yes, it appears everything is relative.

Even relativity.