usenet weirdness

That's what most sources say: m means mass as used by Newton and Einstein, and is what's in gravitational equations, and photons don't have any.

So when two gammas collide and create a particle pair, mass appears where there was none.

Right so far?

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

No. Both gammas have zero rest mass but do have mass due to the energy they contain. Photons always travel at the speed of light, so never at rest, but they always have mass.

Joe

john larkin snipped-for-privacy@glen--canyon.com wrote or quoted:

While a single photon does not have mass, a pair of photons can have mass. When two photons collide with opposite momenta, all their energy is mass, and this mass is conserved during the process. So it is /not/ correct to say, "mass appears where there was none".

Possible causes of why many people do not seem to get this (I already proved it before [1] in a recent post of mine):

Some may confuse mass with /matter/. Matter usually is taken to be fermions. In the process, non-matter (bosons) is in fact converted into matter (fermions). So you /can/ say, "matter appears where there was none".

While energy and momentum is additive, mass is /not additive/. You can add two photons, both of which have /no/ mass, and get a compound system /with/ mass. However, in the everyday world, mass /is/ additive because we do not deal with relativistic systems there. So, some people might transfer their everday concept of mass to relativistic situations where it does not apply.

|Let's call the momenta of the two photons p0 and p1. | | We may assume p1 = -p0 as the two photons are moving towards | each other from opposite directions. Let's call the momentum | of the system of these two photons "p", then we have: | |p = p0 + p1 = p0 +( -p0 )= p0 - p0 = 0 | | . Let's call the energy of this pair "E" and its mass "m". From | |E^2 = m^2 + p^2 | | (in units with c=1) and | |p = 0 | | , we get, | |E^2 = m^2 | | for the pair. I.e., all its energy is mass. And this is the | mass the particle pair has after the collision.

If that's enough from me about physics, here's for example a sort of discussion about medicine, starting with a bit about physics and later about some philosophy.

The, "Analysis and Methods" bits there, get into some of the considerations of the algebraic or properly the derivation.

"Retrospectivus"

Yes. (there is no such thing as 'mass-equivalence, there is only mass-energy) For example, all those photons inside the sun, on their way out, do contribute to the Sun's gravitational mass.

No. 'There are no stupid questions.' But you will be in trouble if you try asking undergraduate level questions in a high level research gathering. (because such a gathering doesn't have the purpose of educating you)

OTOH, is you ask the right level questions in such a gathering you will be fine, undergraduate or not. It's a matter of the right question in the right place,

Jan

I doubt that you would be happy having for example a young Richard Feynman in the audience shouting about how silly and impossible your idea is.

Have you ever read things about Los Alamos in the good old times?

Jan

If e and m are the same thing, why do people use two symbols?

Do photons attract one another? Do they bounce off one another?

If you apply Newton's law of gravitation to photons, the force will be enormous when they get in a close intersection. Wouldn't that fuzz up images at cosmological distances?

Just asking.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Yes, lots of books, including The Los Alamos Primer and the Rhodes books. And cool pics and movies of bomb tests.

My wife had an uncle who photographed the h-bomb shots in the Pacific. He died young of cancer.

We instrumented a bunch of stuff for the big linear induction xray machine (1/4 mile on a side) at Los Alamos.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

john larkin snipped-for-privacy@glen--canyon.com wrote or quoted:

Energy and mass are not the same thing. For one example, a photon has energy, but not mass. (I already gave this example before.)

The capital "E" is mass, the small "e" is an electron.

There is photon-photon scattering in QED (virtual charged particle loops), but it is a very weak effect that rarely matters in practice.

In general relativity, photons should attract each other, but this effect should be very weak and should hardly be observable.

Both effects are so small they have never been measured directly.

Photons are usually not that localized like tiny balls that have a sharp distance. They are being described by quantum electrodynamics, which does not take gravity into account. On the other hand, general relativity does not take quantum effects into account. We do not yet have a theory of everything.

They are not the 'samething', they are two different things, E and M.

Before the Big Bang only E existed.

The Big Bang was the creation of M.

fire and earth are two elements.

snipped-for-privacy@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

The capital "E" is energy, the small "e" is an electron.

If the photons are travelling at the speed of light (duh) wouldn't time dilation make any effect "over time" go to zero anyway? I.e. it doesn't matter how much photons attract each other if there's no time in which to be affected by it.

(I am *not* an expert here)

Please trim your quotes to the relevant minimum:

Please fix the address header field of your postings so that it conforms to network standards:

Please stop amok-crossposting (crossposting without Followup-To for no reason):

Further ignorance of these Usenet *basics* will cause your post> [...] Thomas 'Po>>> and it would create a symmetric e-field pulse that

I posted Unicode characters, not this garbage. This is 2026, not 1986. Unicode has been around since 1991, and in Usenet since 2016 at the latest. Please use up-to-date, sufficiently bug-free, conforming software, e.g.

See also: <news:news.newusers.infos>.

I think we do not know that yet because we do not have a quantum field theory of gravitation ("quantum gravity") yet. General relativity, with which we describe this, is a classical field theory. You can do quantum field theory in a curved spacetime, but ISTM that will not provide the full picture.

But assuming that photons as the quanta of the quantum-electromagnetic field are excitations of that field in the same way as electromagnetic waves are excitations of the classical electromagnetic field, then they contribute to the energy--momentum density in a region of spacetime, and therefore their existence corresponds to the curvature of spacetime.

The *equivalent* (NOT: actual) mass of a photon of human-visible light, say with the peak wavelength of sunlight of lambda = 550 nm, is

m_eq = E_γ/c^2 = ℎ f/c^2 = ℎ c/(λ c^2) = ℎ/(λ c) ≈ 4.019 × 10^-36 kg.

IOW, the gravitational influence of a single photon is very small. It does NOT explain the deflection angle of light near Sol or black holes; that is correctly explained by the spacetime curvature that the existence of objects with (equivalent) mass (like Sol) correspond to instead, which light simply follows (much the same as other objects follow curved spacetime geodesics).

However, it has been hypothesized that if enough electromagnetic energy is contained in a small enough volume of space, that could produce a black hole, called "kugelblitz" (from the German word for a rather legendary atmospheric phenomenon of the same name; literally: "ball lightning"):

No, but:

F'up2 <news:sci.physics.relativity>

Then you must have an idea of the sometimes rough interactions between the various physicists assembled there.

Like many of the Los Alamos crowd. Feynman even managed to die of two independently acquired rare forms of cancer,

Jan

Guess you mean E. And the answer is backward compatibility. E = mc^2 has been reduced to merely a trivial conversion between units. (even more trivial because c = 1 in any half-way decent unit system)

So E and m have been given different meanings, making the new general connection E^2 = m^2 + p^2.

Certainly. See under photon-photon scattering. (too weak to observe under lab conditions, except indirectly)

And no, you really shouldn't try to visualise photons as tiny billiard balls.

See above.

Wait for answers from a theory of Quantum Gravity, or better yet, the Theory of Everything,

Jan

When a photon is emitted by a lamp at time t0 and later detected by a camera at time t1, we assume that for times t between t0 and t1 the photon's distance from the lamp is ct (the product of c and t), where c is the speed of the photon (the speed of light, assuming vacuum).

This assumes that the times are measured in a coordinate system where the observer is at rest, so the whole process and the movement of the photon can be described without the need to take time dilation into account.

Observations confirm photon deflection by the Sun (Eddington 1919) and gravitational lensing, measured in Earth-frame time.

So, there /is/ time, viz. the duration t1-t0, in which the photons can be affected by gravity.

There is no need to transform the whole process into the "point of view" of the photon itself. There probably is no such point of view, although a very young Einstein is reported to have wondered what the world would look like to a photon.

Photons lack a valid reference frame since Lorentz transformations are undefined at v = c.

As if you would have any clue what you are talking about.

So, finally you saw the light? ;-)

It _does_ have mass. That has to do with the fact that the mass of a

See also:

PBS SpaceTime: The Real Meaning of E=m c^2

"Realistic"? Otherwise this is correct.

Misleading. The energy density of electromagnetic radiation was larger than the energy density of matter then. See also:

Photons have all the time of the world. (they travel an infinite distance in zero proper time)

If you want to be naive about it, that is,

Jan

Who appointed you Obergruppenfuehrer here?

Jan