michel123456

That's what i asked in another thread. If there is only one photon for each star in the sky, that makes billions of billions at any place. I mean, if I look at a star and change position taking care to keep my eye on the star, there is no gap, there is no place where there is no photon coming from this star. Photons from any faint star are literally everywhere. The vacuum must be full like an egg.

There are still windows open to research into "varying constants". That changes a lot in physics but the basic laws remain the same. Like VSL for example.

O.K. thanks. So it is 2.77 x 10^18 photons/cm^2 sec 10^9 looks too few. That is 10^3 on each edge. (correction below) see below

Only for the French-speaking (...) un nouveau Pape est appelé à régner. Araignée ! quel drôle de nom, pourquoi pas libellule ou papillon ? Jacques Prévert

cross posting But there are a lot of other photons coming from all over the universe passing by this sphere, going in all directions that do not cross our planet.

Yes. But the result over volume should be more than the result over surface. No?

From ramblings over the net there may be a 10^9 missing. ?? Found this:

I disagree with Savain. quoted from the link: is incorrect the correct is Δ = dt/dx where Δ is the inverse of velocity expressed in second/meter. It is the amount of meter (the distance) you need to "travel" one second in time.

10^9 only? Is that compatible with a flux of 2.77 x 10^18 photons/cm^2?

Right.

I believe you but I don't understand your calculation. The sun radiates in all directions. What the Earth receives is a tiny part of it. Where do the 500s come from? ------- mm. the 500s is the orbital radius. So you are estimating the volume of the cylinder from Earth to the Sun. ------ Isn't there something missing?

How many photons do exist by units of volume of "empty space"? Of course there is no empty space if photons are going through it: there are photons. So, when I look at the sun (don't do that) my eye catches a bunch of photons coming from the star. "catch" meaning that the photons hit my eye. When I look anywhere from any direction, I can catch some photons coming from some source. If I change my eye with a telescope, turning to Alpha Centauri, I catch some photons from there. I can theoretically catch photons coming from the whole observable universe. So, at the same instant, in outer space, how many photons are coming from all sources in a single cm3? If the stars are billions of billions, one photon from each one makes billions of billions photons not in a cm3, but in space small as 1 photon. Logically, I guess. That makes a lot in 1 cm3. Or am I wrong somewhere?

This thread is derailing IMHO To the OP: Of course NO: mathematics alone is not a safe medium. Mathematics is a tool, you can use it wisely or not. There are examples where mathematics give results that are considered unphysical and are dismissed at first sight. Like results giving negatives where evidence shows only positives. Observation & hypothesis, yes, but not only. I am convinced that many scientists use also ideas (inspirations) out of the strict scientific process. After all, if you don't know what you are looking for you will never find it. Expecting to find everything from observation is like the scientist wanting to dig the entire Earth to find a lost civilization, sorry no, a bone of australopithecus, oops no an unknown bacteria, no an unknown [put in here what is unknown].

Thanks for clarification, I thought I got crazy. That's not the first time. ------------- @Przemyslaw Unless the amount of observed photons is frame dependent, I don't see how your theory can stand.

?? If total energy E is frame dependent, then E divided by a constant (c^2) should also be frame dependent.

That nobody cared making an evaluation of the gravitational effect of light.

Nobody cared making an evaluation? That is 13 billion years of burning stars radiation + the radiation of the BB that did not change in massive particles. I am amazed.

So what's the question? i would be very amazed if astronomers did not count for those gravitational effects in their calculations.

Correcting post #54 I don't understand (still). A photon has no invariant mass, O.K. But the photons that we observe (and the others that we don't) are not in state of rest. They have energy and momentum. Those photons that travel at C in all directions exert and are sensible to gravity, as much as I know. Correct?

Relativistic mass is frame dependent, no? Like momentum. How can that be "real mass"?

Oh sorry, no, I ment "observe". The Universe is full of photons that travel in all directions and do not reach the Earth & our eyes. A photon has not even relativistic mass?

I don't understand (once again). a photon has no rest mass, O.K. But the photons that we catch (and the others that we don't catch) are not in state of rest, thus they have mass, be it relativistic. Those photons that travel at C exert and are sensible to gravity, as much as I know. Correct?

Wow. Where were you all this time?

I don't think that a fish evolved into a whale. Moontanman where are you? Please explain the difference between a fish and a marine mammal. --------------- For the OP: a sceptic.

O.K. How then do you change the energy of a particular photon? Do you mean that the doppler-shift formula matches the diagram (it is also a hyperbola) ?