Really confused - basic special relativity

[WannabeNewton]

The photon always travels along a null curve through spacetime. If we conveniently define the axes of spacetime, then we have c²dt² = dx². And so from a frame of reference where the photon is stationary, dt=0. This means that for a photon, it is absorbed at the same instant it is emitted (from its frame of reference). So the photons that reached our eyes from distant galaxies have left at the same time they reached us (from their frame of reference).

Of course, it doesn't matter how much time it takes from the photon's frame of reference, since a photon doesn't have a half-life, so it doesn't affect decay rate or anything like that.

A photon is never stationary. In order for there to be a frame in which the photon is stationary, your reference frame would have to be at rest with respect to the photon which is impossible. Again, you cannot lorentz boost to the frame of a photon so saying how time is in "its frame" is completely meaningless. Proper time intervals being zero on a null geodesic doesn't translate literally to "time does not pass for a photon".

 

[ghwellsjr]

Out of all the things I've said, this one was just a dumb metaphor. Of course I don't need an actual person made a photons to give meaning to "how does the photon experience time?". Particles experience time, for example, in that they decay. And clocks tick without having "emotions". It just seemed easier at the time of writing to give the photon human senses. I obviously wasn't aware I'm speaking with the most correct physicists out there :-) And I'm not dissing - it is clear to me one needs to be correct about it when actually wanting to understand, but I sort of assumed that what I meant is clear.

Some massive particles like neutrons and muons decay but photons don't, so how exactly do you think photons experience time?

 

[Dale]

I mean an inertial rest frame moving with the speed of light

Here is the self-contradiction in a nutshell. Part of the definition of an inertial frame is that light moves at c. So indeed, that speed is singled out in the definition of inertial frames. And that is precisely why you cannot have an inertial frame moving at c, because then it wouldn't be an inertial frame.

 

[BruceW]

Proper time intervals being zero on a null geodesic doesn't translate literally to "time does not pass for a photon".

Why not? Between two events on a null geodesic, the space separation equals the time separation (with c=1).

 

[Tomer]

Some massive particles like neutrons and muons decay but photons don't, so how exactly do you think photons experience time?

No idea. If I knew I wouldn't have asked.
I wonder though, what the difference between "not experiencing time" and "experiencing time not passing" is. I know, philosophy. But it physicist are allowed to, and should, be philosophers from time to time.

 

[Dale]

Not here.

 

[Tomer]

Not here.

Show me if you will where it's written, and try not to confuse your preferences with rules. There's nothing wrong in seeking the point of view of scientists on subjects related to philosophy and I can't think of a more appropriate place to have a discussion on it. I wasn't asking about gender studies or Karl Marx, I was asking about the possibility of the existence of a photon's rest frame. That it diverges one micro step from the almighty model doesn't mean it's meaningless/dumb/forbidden to discuss it, and if it doesn't belong to this forum, where does it belong to?
Especially if the question I posed could maybe be answered physically. The borders between the two are pretty shady as it is.

Please stop this witch hunting, it's ridiculous and insulting. I would much prefer it if you simply don't answer me if you find my questions so irrelevant. If moderators demand, I'll stop asking questions here, but I'll find it very sad.

Physics:
About the inertial frames - I was referring to the classic "not-accelerated" inertial frame. Of course the "inertial frame moving with velocity c" would have to have the exception that the speed of light there isn't c, otherwise I agree it's a conflict.

My next question is therefore: (Physics:)
Does it create some sort of paradoxes, assuming that the speed of light at a photon's "rest frame" isn't c?

 

[Dale]

Show me if you will where it's written

https://www.physicsforums.com/showthread.php?t=414380

 

[Dale]

I was asking about the possibility of the existence of a photon's rest frame.

And you were answered, very clearly.

About the inertial frames - I was referring to the classic "not-accelerated" inertial frame. Of course the "inertial frame moving with velocity c" would have to have the exception that the speed of light there isn't c

Then it isn't an inertial frame, by definition. That is the core of the self contradiction inherent in the question.

Repetition and becoming irritated with the responses is not going to change a self contradictory premise into a self consistent one.

 

[BruceW]

Physics:
About the inertial frames - I was referring to the classic "not-accelerated" inertial frame. Of course the "inertial frame moving with velocity c" would have to have the exception that the speed of light there isn't c, otherwise I agree it's a conflict.

My next question is therefore: (Physics:)
Does it create some sort of paradoxes, assuming that the speed of light at a photon's "rest frame" isn't c?

It doesn't really work like this. There is no way to measure the speed of light (in vacuum) to be anything other than c. That's one of the postulates of relativity.

There is no conflict. Imagine you see a photon and you want to try to measure its speed from its frame of reference. You would try to accelerate to catch up to it, but no matter how much you accelerated, it would keep moving away from you at c.

 

[Tomer]

*sigh*. Ok. :-)

 

[WannabeNewton]

Why not? Between two events on a null geodesic, the space separation equals the time separation (with c=1).

When you impose an indeterminate riemannian metric on a manifold, this makes it semi - riemannian in nature. This structure is what, to an extent, leads to null geodesics having zero length on intervals. This does not warrant you the ability to lorentz boost to the frame of a photon and justify that time is not passing at all for it. The statement has no meaning whatsoever.

 

[BruceW]

When you impose an indeterminate riemannian metric on a manifold, this makes it semi - riemannian in nature. This structure is what, to an extent, leads to null geodesics having zero length on intervals. This does not warrant you the ability to lorentz boost to the frame of a photon and justify that time is not passing at all for it. The statement has no meaning whatsoever.

I'm not trying to justify anything by using a Lorentz boost. I didn't mention Lorentz boosting.

Once we define a time axis, there are three types of null vector: the zero vector, the future directed null and the past directed null. So the type of null vector depends on the direction we specify for the time axis. So for the path of a photon, we could define the time axis such that its 4-velocity was given by the zero vector (0,0,0,0).

Is there anything wrong with doing this?

 

[WannabeNewton]

I'm not trying to justify anything by using a Lorentz boost. I didn't mention Lorentz boosting.

Once we define a time axis, there are three types of null vector: the zero vector, the future directed null and the past directed null. So the type of null vector depends on the direction we specify for the time axis. So for the path of a photon, we could define the time axis such that its 4-velocity was given by the zero vector (0,0,0,0).

Is there anything wrong with doing this?

Well a photon has no defined 4 - velocity. d\tau ^{2} = 0 so \frac{dx^{\mu }}{d\tau } is not defined. I assume you meant the photon's wave 4 - vector k^{\mu }. For either future null directed or past null directed, k^{\mu }k_{\mu } = 0 but this is, again, a consequence of the nature of a semi - riemannian manifold. It does not mean that time literally does not pass for a photon because you are not making a tangible physical statement about photons.

 

[Dale]

So for the path of a photon, we could define the time axis such that its 4-velocity was given by the zero vector (0,0,0,0).

Is there anything wrong with doing this?

I don't think that is true. I don't think that the choice of time axis can change two topologically distinct events into topologically indistinguishable events. The topology is more fundamental than the coordinate basis.