The Limitations of Applying Relativity to Photons

[NewLocality]

[SOLVED] Is relativity Uncertain

Simple question: -
If a photon does not experience time how does it know how many times to
wiggle as it crosses space.

 

[DB]

It oscillates (wiggles) because of its energy, its frequency can decrease or increase to keep propagating at c through any field. But a photon doesn't know to die or better yet decay. A photon emmited during the big bang is still here today at 0 years old. It is emmited at a certain frequency and wavelength and travels its path. A photon doesn't experience time because it travels at c. Anything traveling at c will have a zero variation world line. But of course nothing with mass can travel at c as Einstein prooved.

 

[NewLocality]

Hi DB

Your answer really does not address the problem. To say the photon oscillates at a given rate because of its energy is merely referring to Planck’s equation. Planck himself never attempted to explain the mechanism behind the formula.

It is true at a superficial level the photo-electric effect and the Compton effect seem to substantiate the idea of the photon but at a deeper level we find that it is inconsistent with the theory of relativity and also some aspects of quantum mechanics. Special relativity denies it the time for it to come into being and quantum mechanics demands that it be in all places at the same time. I admire your confidence in attempting to answer my question but I think a little more is required.

 

[Hans de Vries]

Simple question: -
If a photon does not experience time how does it know how many times to
wiggle as it crosses space.

A light wave doesn't wiggle in time. It's shape stays constant (in vacuum)
and just shifts along with the speed of c. At the most it can spread because
not all parts move exactly in the same direction. The EM components at the
head or the tail of the light pulse stay always directed in the same direction.

Regards, Hans

 

[NewLocality]

Hi Hans

Does this mean relative to its own frame of reference the light has an infinite wave length?

Regards, NL

 

[Hans de Vries]

Hi Hans

Does this mean relative to its own frame of reference the light has an infinite wave length?

Regards, NL

Yes, it's frequency is zero in it's own reference frame. It's maximally Red shifted:

$$f' \ \ = \ \ \sqrt{\frac{c-v}{c+v}} \ f$$

Regards, Hans

 

[Hans de Vries]

Yes, it's frequency is zero in it's own reference frame. It's maximally Red shifted:

$$f' \ \ = \ \ \sqrt{\frac{c-v}{c+v}} \ f$$

Regards, Hans

A few tips to understand what's going on:


You can draw a Minkovsky diagram. The light phases are bands at 45
degrees in parallel with the x' and t' axis.

You'll see that the frequency over the t' axis is zero.

And this one: The 'head' of a photon can never see it's own tail since
this would need propagation > c. The "head" can only see it's own phase,
(=> 0 Hz) The other parts of the photon are all mapped on it's light cone.

(What we call photon is spatially extended on the x-axis)


Regards, Hans

 

[NewLocality]

In that case when a photon interacts with an absorber system, the photon will see the absorber system as having infinite mass and a Debroglie frequency that is also infinite. The photon could never be fully absorbed because the universe would have grown old and died long before the process could be completed.

 

[JesseM]

A photon doesn't have a valid reference frame, because you get infinities if you try to apply the Lorentz transformations using v=c. You can consider the limit as v approaches c, but this gives you weird answers, like that a photon would experience its entire path throughout history as a single point in space which takes zero time to traverse.

 

[NewLocality]

A photon doesn't have a valid reference frame, because you get infinities if you try to apply the Lorentz transformations using v=c. You can consider the limit as v approaches c, but this gives you weird answers, like that a photon would experience its entire path throughout history as a single point in space which takes zero time to traverse.

So what's that telling us about the nature of the photon and its relationship with relativity?

Why is not possible to have theory which is free from these infinitives and we can apply mathematics to it in a self consistent way?