Re: gravitational bending of light, surprising?

On May 9, 9:32 am, mluttg...@xxxxxxxxx wrote:
On 9 mai, 16:01, PD <TheDraperFam...@xxxxxxxxx> wrote:



On May 9, 4:43 am, mluttg...@xxxxxxxxx wrote:

On 9 mai, 00:43, PD <TheDraperFam...@xxxxxxxxx> wrote:

On May 8, 5:33 pm, mluttg...@xxxxxxxxx wrote:

On 8 mai, 23:33, PD <TheDraperFam...@xxxxxxxxx> wrote:

On May 8, 4:23 pm, mluttg...@xxxxxxxxx wrote:

On 8 mai, 18:31, PD <TheDraperFam...@xxxxxxxxx> wrote:

On May 8, 11:17 am, mluttg...@xxxxxxxxx wrote:

On 8 mai, 16:29, Dono <sa...@xxxxxxxxxxx> wrote:

On May 8, 6:53 am, mluttg...@xxxxxxxxx wrote:

On 8 mai, 13:07, "OG" <o...@xxxxxxxxxxxxxxxxxxx> wrote:

"david" <david_lawrence_pe...@xxxxxxxxx> wrote in message

news:a7b0e74b-9bf4-4efb-8fce-eeb63f7421eb@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

It would seem that even Newtonian mechanics predicts that light will
follow a curved path in the presence of a gravitational field.  If
that's not obvious, think about the following thought experiment.  If
we shoot a bullet perfectly horizontally from, say, six feet above the
ground, and simultaneously drop a stone from six feet above the
ground, we know that they will hit the ground at the same time. So
just consider what would happen if we were to shoot the bullet at the
speed of light;  it would still hit the ground at the same time as the
stone, according to Newtonian mechanics. So a particle of light should
do the same.

Your logic is flawed.

Of course it wouldn't be obvious how a relativistic bullet would
behave in a gravitational field when we switch over from Newtonian
mechanics to special relativity (pretending for the moment that we
don't know anything about general relativity), but a very reasonable
speculation would be that it would still hit the ground at the same
time as the stone. And from that assumption, we should be able to
derive a differential equation that predicts the path of light in a
gravitational field.

You make a false assumption then use this to speculate about a possible
derivation of an equation.

So this leads to a couple of possibly troubling questions.  First of
all, why were physicists surprised by Einstein's prediction that the
path of light would bend in a gravitational field (after all, even
Newtonian mechanics predicts that it would)?   But more importantly,
is there a difference between what would be predicted by the
differential equation as described in the above paragraph and what is
predicted by general relativity, and are experimental measurements of
the bending of light by the sun accurate enough to distinguish between
the two predictions?

If you are interested, why not work it out yourself - it'll be challenging,
but it will be all your own work

What's troubling is that we (i.e. non-experts) are led to believe that
the bending of starlight (as in gravitational lenses, etc.) should be
accepted as a confirmation of general relativity, but it's far from
obvious that it is.  Certainly that first measurements of the bending
of light by Eddington that catapulted Einstein into fame weren't
accurate enough distinguish general relativity from other possible
theories.

Calculating the 'Newtonian' bending of light gives a value that is wrong by
a factor of two to that calculated by General Relativity. GR gives the right
answer.

According tohttp://www.theory.caltech.edu/people/patricia/lclens.html

"If we want to fake the propagation of light in Newtonian gravity, we
can set
the energy E = m v^2/2 = m c^2/2 so that (2 E/m) = c2. The angular
momentum per
unit incoming mass (L/m) becomes L/m = Ro c."

The author ignores that for light, E = mc^2, hence that (E/m) = c^2.
Using this last value, Newton gets the same result as Einstein.

Marcel Luttgens

Lattkes

E=mc^2 does NOT apply to photons. You didn't know that , old fart?

Tell that to Steven Weinberg, stupid!

See for instance "Gravitation and Cosmology", 1972, p.84 :

"When a photon is produced [...], its inertial mass changes by an
amount related to the photon frequency Nu1 he observes, that is,
by delta(m1) = -hNu1, etc..."

Marcel, how dishonest! What does the word "its" refer to in the quoted
sentence?
Hint: it is NOT the photon!
For shame!

I agree that 'its' could be misleading.
But read a little further:
"an observer in a freely falling system will see the apparatus
change in inertial mass by an amount related to the photon
frequency Nu2 he observes, that is, by delta(m2) = hNu2."

Clearly, noting that Weinberg, like many GR specialists, uses
c = 1, the mass change of the apparatus is corresponds to the
mass m2 = hNu2/c^2 of the photon of frequency Nu2, whose energy
E is of course equal to hNu2.
Thus, for Weinberg, m2 = E/c^2, or E = m2 * c^2.

No. The mass of the apparatus changes. This mass is converted to the
*energy* of the photon, not a *mass* of a photon. E=mc^2 does not
apply to photons, and you CANNOT say, "Well, if I can translate an
energy emission by the apparatus to a loss of mass of the system, then
I can translate the energy carried by the photon to a mass of the
photon." That is incorrect.

Delta(m2), that is the mass change of the apparatus, corresponds
to hNu2/c^2, which is the mass m of the photon of frequency Nu2.

No. The mass of the photon is zero. The Energy of the photon is given
as you indicate. The mass of the photon is sqrt[E^2 - (pc)^2]/c^2,
which you will find invariably to be zero for an observed photon.

http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence:

"Even a single photon traveling in empty space has a relativistic
mass, which is its energy divided by c2."

"Relativistic mass" is nothing but energy, scaled by 1/c^2. It has no
bearing on the mass as that term is modernly used: m^2 = E^2 - p^2.

Iow, you agree that the relativistic mass of the photon is
m = E/c^2 = hNu/c^2.
So, it is clear that to calculate the deflection of light
by a massive object, for instance the Sun, E = mc^2 should
have been used instead of E = (1/2) mc^2 in the 'Newtonian'
calculation.

No, for the reasons mentioned. Just taking the energy and turning into
something that looks like mass and even has a historical label that
involves the letters m-a-s-s does NOT mean that this is the physical
mass, the gravitational mass, or any other equivocation you want to
insist on making.


Marcel Luttgens



One can quibble about such claim, but the relativistic mass
m = E/c^2 = hNu/c^2 of the photon can be used to interpret
results of a series of experiments, for instance those obtained
by Pound & Rebka.

So, I maintain my opinion:

As the energy E of the photon is hNu, its mass m can be expressed
as E/c^2, hence E = mc^2, which is the value that should have
been used to calculate the deflection of light, instead of
E = (1/2)mc^2.

Then, Newton would have gotten the same result as Einstein.

Marcel Luttgens

.

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