[PmbPhy]

Yes, there is a centre of mass frame of the universe //

Wrong. No such frame exists. When physicists use that term it really means zero momentum frame.

[yor_on]

And it's ok being slow, I think?

For 50 years I took that description as a qualified compliment but now I am starting to see it as "just slow"
 

Nah, being slow is cool. If people use it as a negative then that says more about them than about you. Take your time, it's yours.

[yor_on]

Photons/light are defined to never be 'at rest', but the way I naively look at it myself I allow for the possibility of two 'photons' propagating aside each other, and then define them as being 'at rest' with each other

Wouldn’t this be defining the photon frame as an inertial frame?
My understanding is that it is not, so velocities with respect to it cannot be measured.

True, but I said naively. There is no rest frame as there can't be a observer. And that makes the other proposition with light not 'propagating' a better one to me. You see, as long as we talk about a propagation, and 'particles' you naively can imagine such a circumstance. Doesn't make it fit though, does it :)

[yor_on]

Actually, talking about 'propagation'. Imagine two lasers (set together) pulsing at a same rhythm towards some measuring device. According to that device the two lasers 'photons' hit it simultaneously. Are those 'singular corpuscles/photons', or not, 'at rest' with each other under their 'propagation'?

If not, what makes it 'simultaneous' at the measuring device?

And of course we will have to assume the 'system'  of lasers/measuring device as well as however you will define a observer, to be 'at rest' with each other too
=

Remember that this is a 'ideal system', so we will presume that everything is ideally as well as perfectly set up at a same distance etc etc.

[Jaaanosik]

This states that L = V + U.

Correction. L = T - V, where T = Kinetic energy and V = Potential energy. If the potential is a velocity dependent one then the letter U is used. This is particularly important in electrodynamics.

There is an issue with the Lagrangian that I brought up in the pendulum thread:
https://www.thenakedscientists.com/forum/index.php?topic=73606.0
The understanding/description of physics is in trouble here.

[Bill S]

If not, what makes it 'simultaneous' at the measuring device?

My initial thought is that the photons arrive simultaneously relative to the measuring device.  It says nothing about simultaneity in the RFs of the photons, since we cannot define such RFs.

[yor_on]

That's one answer. Anyone else prepared to define why you can't talk about them 'being at rest with each other' in whatever frame now 'propagating photons' might be said to exist :)
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Do notice that I'm not using 'rest frame' above.  From a 'frame' of a photon time is nonexistent, and as a distance/speed is described by a ruler and a clock, that distance/speed disappear from a photon's view. Expressed otherwise you need a arrow of time to use that ruler to measure a distance.

It becomes a contradiction in terms as from the observers point of view you emit a 'photon' from the laser, or in our case one 'photon' each from our co-joined lasers, that then slightly later registers at the measuring device. That's how we get to 'c', well, we use the 'two mirror experiment' but still, you want that clock and ruler and then measure how long it took that photon to 'propagate'.

Never the less, all distances must be non existent from that photons point of view, and so emitting and absorbing should be one and a same event. And if we then look at frames in where we find 'things' being at rest with each other, as being in a same frame of reference, then all photons must share that same frame of reference. And actually, there is no 'motion' in that frame, is there :)

Only relative proper mass aka matter

===

And yes, photons are indeed 'timeless', to the best of our knowledge. We can see them from the Big Bang, as good as new :)  13. 7 ~ billion years new that is. So asking oneself whether a 'photon' can decay is a very meaningless question. Another very interesting point is then if you can attach some sort of mass to a photon,  http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

You might want to argue that as we find a defined speed then it has a 'mass' of sorts, relative us. But if we by it would mean 'rest mass' we no longer would be able to argue a 'time less ness', and yes, now photons should 'decay' as I think about it. Which then would question everything we know from studying early light.

Arguing that it has a energy isn't that expressive actually. Everything has, or at least is expected to have, even a  'perfect vacuum'. And thinking of that one might notice that 'time' has a lot to do with the definition of 'virtual particles', too transient to make a difference

[Bill S]

Never the less, all distances must be non existent from that photons point of view, and so emitting and absorbing should be one and a same event.

That's where my "logic" took me before I realised that I was extrapolating SR to include a mass-less particle, travelling at c.  I think that's outside the scope of SR.

[yor_on]

Then your logic did you a good turn Bill. Because it fits, a 'photon' is time less, we can actually confirm that by being able to 'see' 13.7~ly. While we can't be 'at rest' with it other 'photons' definitely should be able to be, presuming everything mainstream being correct, about it being a 'particle' 'propagating' at 'c'
=

Also, presuming that 'c' is correct, there can only be one 'frame' for those photons, unless someone wants to argue that the energy content differs and so then should the 'frames'? But being 'at rest' or sharing a same 'frame of reference' doesn't discuss the 'energy content', as far as I understand.
=

Let me widen that definition slightly, Presuming that 'c' is 'observer dependent, and from that finding that depending on somethings 'uniform motion' relative some other objects proving' that 'c' differs, that as both come to 'c' locally which 'can't be correct'. That as you can prove that you and the others objects 'uniform motion' definitely differ from each other, but it will still be a fallacy. The main reason for it being so is that you then must presume there to be a 'golden standard'. A 'absolute frame' from where we can define all other motion.

There is none.

[yor_on]

In fact, to get it right we need to stop looking at 'c' as a speed defined. It's a lot more than just a 'speed'

[Toffo]

Energy is not observer dependent. (Except kinetic energy)

If many types of energy are observer independent and one type of energy is observer dependent, is energy then observer dependent? I would say no.

If we manage to understand why the one type of energy is observer dependent, then we understand all of energy's observer dependence. And that seems quite easy to understand.

[yor_on]

oh yes, energy is definitely observer dependent Toffo :)

Presuming that a photon hits your retina its energy will differ depending on direction of your motion, relative the photons 'propagation'

[Toffo]

Presuming that a photon hits your retina its energy will differ depending on direction of your motion, relative the photons 'propagation'

Well that tells us that energy of a photon is kinetic energy.

[yor_on]

You need to define your terms there. What kind of 'energy' do you consider to be observer independent?

[Toffo]

You need to define your terms there. What kind of 'energy' do you consider to be observer independent?

Do you know what energy is? Do you know what kinetic energy is?

Observer independent energy is energy, except kinetic energy.

Things with momentum p and speed v have kinetic energy p*v.

Things with energy E and kinetic energy 0 have observer independent energy E.

[yor_on]

Hmm :)

Would that make your definition a absolute frame then
=

Maybe you're thinking of the energy one define while being 'at rest' or in 'same frame of reference' with whatever you measure. And as we can't be at rest with light it has no other definition than 'observer dependent' from our point of view.  But when we come to relative motion all uniform motions becomes the same and there is no golden standard defining what is a 'null motion' in this universe.

[yor_on]

Relativity has two definitions, accelerations/decelerations and uniform motion. Of those two only the first definition is (locally) provable. You can prove to your satisfaction that there exist different uniform motions relative yourself, but all uniform motions express themselves equivalently, locally defined.
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Actually it's not that I think you wrong, I have the same sort of feeling that there should be a 'golden standard', and the way you go about defining it is similar to mine, because it's 'local'. Locally defined, as in being at rest with something, or sharing the same frame of reference. But doing it that way turns it up side down, suddenly we lost the 'whole universe' and the 'objective 'global' perspective', instead going for local definitions. Which I actually (again) think makes sense :)

Although doing it that way introduce a new way of looking, and should lead to new definitions too.
We'll see

[yor_on]

Actually. if 'energy' is a coin of exchange it has to have a consistent value. Other wise that 'coin' will just give you a headache