Inertial Ref: Can Light Photons Be a Privileged Frame?

[Opressor]

Can light photons be considered as a privileged inertial frame?

 

[Ibix]

No. A photon is not a frame.

Assuming that you actually mean to talk about an inertial reference frame moving at the speed of light, the answer is still no. One of the postulates of relativity is that the speed of light is the same in all inertial reference frames. A reference frame traveling at the speed of light would describe some light as "at rest". So a reference frame traveling at the speed of light is a contradiction in terms - light would have to be both stationary and moving at 3×10⁸m/s at the same time.

 

[Opressor]

so a photon of light can not be an inertial frame?

 

[Ibix]

As far as I can tell your last post simply restates your first post, but in Portuguese (Edit - I see it's now been translated). The answer is the same.

Please note that you are expected to post in English in this forum. While many of us speak more than one language, English is the one we have in common.

 

[bhobba]

As far as I can tell, your last post simply restates your first post, but in Portuguese. The answer is the same. Please note that you are expected to post in English in this forum. While many of us speak more than one language, English is the one we have in common.

I just want to reiterate, with my mentor hat on, we post in English here, and we certainly try to add value, not simply repeat a previous answer. It translates as 'so a photon of light can not be an inertial frame?'. The answer of course is - it can be in an inertial frame - it just can't be at rest in one.

Thanks
Bill

 

[kent davidge]

Portuguese:
Suponha que você ficou sabendo, de algum jeito, que um referencial $A$ é inercial. Se outro referencial $B$ for inercial, então os dois referenciais podem ser conectados por uma transformada de Lorentz. Mas um raio de luz viaja sempre com a mesma velocidade (c) com respeito a um referencial inercial, desde que não haja matéria interagindo com ele. O fator de Lorentz, $\gamma = 1 / \sqrt{1 - v^2 / c^2}$, tende para o infinito quando $v \longrightarrow c$. Logo, não é possível fazer a transformação. Então um referencial viajando a velocidade da Luz não é inercial.

English:
Suppose we learned by whatever means that a reference frame $A$ is inertial. If another reference frame $B$ is inertial, then the two frames can be related through a Lorentz transformation. But a light ray travels at speed $c$ w.r.t. the frame $A$ as long as there's no matter interacting with it. The Lorentz Gamma factor assumes the behaviour $\gamma \longrightarrow \infty$ when $v \longrightarrow c$. Therefore, it's not possible to connect frames $A$ and $B$ through a Lorentz transformation. We conclude that a frame attached to light is not inertial.

 

[stevendaryl]

Portuguese:
Suponha que você ficou sabendo, de algum jeito, que um referencial $A$ é inercial. Se outro referencial $B$ for inercial, então os dois referenciais podem ser conectados por uma transformada de Lorentz. Mas um raio de luz viaja sempre com a mesma velocidade (c) com respeito a um referencial inercial, desde que não haja matéria interagindo com ele. O fator de Lorentz, $\gamma = 1 / \sqrt{1 - v^2 / c^2}$, tende para o infinito quando $v \longrightarrow c$. Logo, não é possível fazer a transformação. Então um referencial viajando a velocidade da Luz não é inercial.

English:
Suppose we learned by whatever means that a reference frame $A$ is inertial. If another reference frame $B$ is inertial, then the two frames can be related through a Lorentz transformation. But a light ray travels at speed $c$ w.r.t. the frame $A$ as long as there's no matter interacting with it. The Lorentz Gamma factor assumes the behaviour $\gamma \longrightarrow \infty$ when $v \longrightarrow c$. Therefore, it's not possible to connect frames $A$ and $B$ through a Lorentz transformation. We conclude that a frame attached to light is not inertial.

It's interesting (I'm not sure if there is any significance to it) that in Rindler coordinates (the coordinate system that would be used by people on board a rocket that was undergoing constant acceleration), there is a height-dependent apparent gravitational field. This gravitational field tells you the proper acceleration necessary to remain at "rest" at that height. This apparent gravitational field is singular (goes to infinity) at height 0. The path of an object that is at constant height of zero is the path of a photon. So in terms of Rindler coordinates, the path of a photon is not inertial, but is, on the contrary, a path with infinite proper acceleration.

 

[SiennaTheGr8]

It's interesting (I'm not sure if there is any significance to it) that in Rindler coordinates (the coordinate system that would be used by people on board a rocket that was undergoing constant acceleration), there is a height-dependent apparent gravitational field. This gravitational field tells you the proper acceleration necessary to remain at "rest" at that height. This apparent gravitational field is singular (goes to infinity) at height 0. The path of an object that is at constant height of zero is the path of a photon. So in terms of Rindler coordinates, the path of a photon is not inertial, but is, on the contrary, a path with infinite proper acceleration.

 

[PeterDonis]

in terms of Rindler coordinates, the path of a photon is not inertial, but is, on the contrary, a path with infinite proper acceleration.

This is not correct. First, since Rindler coordinates are singular at $x = 0$, you can't draw any conclusions from Rindler coordinates about any "path of a photon" at $x = 0$. Second, if we ignore the coordinate singularity and take limits as $x \rightarrow 0$ (and ignore another fact about the surfaces of constant $t$ in Rindler coordinates--see below), the locus $x = 0$ in Rindler coordinates is not a single lightlike curve: it's a "V" shape made of two of them (the future and past Rindler horizons), and the "infinite proper acceleration" is just an artifact of the "kink" in the V at the origin (meaning the point where the two horizons cross). Third, the surfaces of constant $t$ in Rindler coordinates all meet at the origin, so in fact the "locus" $x = 0$ is not actually the future and past horizons, but just the single point at the origin, so that locus doesn't even describe a curve at all, just a single point.

But fourth, and most importantly, there are lots of lightlike curves in Rindler coordinates that pass through positive $x$ values, and it's simple to show that they are geodesics, so the paths of photons in the non-singular region of Rindler coordinates are inertial.

 

[stevendaryl]

This is not correct. First, since Rindler coordinates are singular at $x = 0$, you can't draw any conclusions from Rindler coordinates about any "path of a photon" at $x = 0$. Second, if we ignore the coordinate singularity and take limits as $x \rightarrow 0$ (and ignore another fact about the surfaces of constant $t$ in Rindler coordinates--see below), the locus $x = 0$ in Rindler coordinates is not a single lightlike curve: it's a "V" shape made of two of them (the future and past Rindler horizons), and the "infinite proper acceleration" is just an artifact of the "kink" in the V at the origin (meaning the point where the two horizons cross). Third, the surfaces of constant $t$ in Rindler coordinates all meet at the origin, so in fact the "locus" $x = 0$ is not actually the future and past horizons, but just the single point at the origin, so that locus doesn't even describe a curve at all, just a single point.

But fourth, and most importantly, there are lots of lightlike curves in Rindler coordinates that pass through positive $x$ values, and it's simple to show that they are geodesics, so the paths of photons in the non-singular region of Rindler coordinates are inertial.

Yes, you're right. I goofed.