[Halc]

The path of a massive object is not a function of its momentum.

Not true. objects with differing momentum move on different geodesics. This holds for all tardyons but not luxons.

Unclear to what 'this' refers.  My statement or your statement only holds for tardyons?

As for luxons (and quantum scale tardyons for that matter), per the principle of locality, any quantum scale objects do not have a position until said position is measured, and hence it is meaningless to speak of its path.

Also, per my reply 8 above, quantum scale objects are subject to gravitational refraction which affects their motion in addition to inertial effects, so I would be open to things like photons taking non-geodesic paths.

What does "mass more" mean??

Has more (frame dependent) mass, as opposed to having more frame independent proper mass.

and thus pull the larger object a little closer as it passed, thus bending the path a bit more.

That would violate the principle that the rate of fall of an object does not depends on its mass.

First of all, I didn't say 'fall rate', and the principle should be worded that the acceleration of an object does not depend on its mass.  That much I would not deny.  I said a brick will hit the ground before a feather, not because it accelerates more, but rather because the ground does.

This is not obvious when playing with bricks and feathers, but it does very much become obvious with larger things.  Drop a brick from 385,000 km up (where the moon is) and it will take significantly more time (about 7 more minutes) to fall to Earth than stopping (relative to Earth) the moon in its tracks and letting it fall from that altitude.  This is true even if both are point masses.  This illustrates that your principle above is wrong as worded.
My comment you quote above refers to that difference.

[PmbPhy]

Unclear to what 'this' refers.

The statement that precedes it.

As for luxons (and quantum scale tardyons for that matter), per the principle of locality, any quantum scale objects do not have a position until said position is measured, and hence it is meaningless to speak of its path.

Same holds for geodesics.

Its unwise to group quantum mechanics into a thread abut generality since there's no theory of quantum gr. Nothing in your posts speaks to QM in this threasd.

This is not obvious when playing with bricks and feathers, but it does very much become obvious with larger things.

A feather will only fall faster than a brick if there's a non-gravitational force acting on it like air pressure. Put the two in a vacuum and they will fall at the same rate. It's an odd thing to watch.

A perfect example of a geodesics being a function of momentum is a tardyon moving at constant speed in flat spacetime in an inertial frame. The more momentum the faster the particle will move. Draw the geodesic in the spacetime diagram and it will be a strainght liine whose slope depends on the particles momentum and hence its velocity.

[PmbPhy]

If quantum mechanics is taken into account then as you said, photon's won't move on a classical trajectory. In fact only position measurements can be taken making moving on a worldline of any kind not meaningful, never mind a non-geodesics world line.

[Halc]

Its unwise to group quantum mechanics into a thread abut generality since there's no theory of quantum gr. Nothing in your posts speaks to QM in this threasd.

If there's no quantum GR, then you can't speak of how photons (quantum things) behave under GR.  So there very much is quantum effects in GR theory, and in particular: refraction.  Light refracts if its speed is medium dependent, and under GR gravitational wells, it very much is.

A feather will only fall faster than a brick if there's a non-gravitational force acting on it like air pressure.

I never suggested a feather falling faster than a brick, nor any suggestion of drag.
To be specific, I described how a brick dropped from a height will hit the ground in less time than a feather independently dropped from the same height.

Put the two in a vacuum and they will fall at the same rate.

Then stop repeating your assertions and tell me why my explanation of the brick getting there first is wrong.  I computed the difference for the moon falling if that helps.  The example is pretty irrelevant for the photon thing, but you're the one that brought up the principle.

Draw the geodesic in the spacetime diagram and it will be a strainght liine whose slope depends on the particles momentum and hence its velocity.

It's a function of its velocity in this case and nothing else.  If the particle was twice the mass and thus twice the momentum, it would take the same path.  If you knew only its position and velocity, you'd know its path.  If you knew only its position and momentum, you'd not know its path.  That's what I mean when I say its not a function of momentum.  That momentum doesn't define the line slope.

How long does it take for something to travel 1 km at 100 kg-meters/second of momentum?  Can't answer?  Then it's not a function of momentum.

[evan_au]

tell me why my explanation of the brick getting there first is wrong

I think PmbPhy was measuring in the frame of reference of the  the Center-of-Gravity of Earth +Brick+Feather.
Halc was calculating in the frame of reference of the Center-of-Gravity of Earth+Brick (or Earth + Feather).

Because (Earth+Brick) has slightly higher total mass than (Earth + Feather), the brick will arrive quicker - by an immeasurably small time.
But not quite as quick as (Earth+Brick+Feather) - also by an immeasurably small time.

[PmbPhy]

Its common place to speak of photons in GR. When this is done they're treated like luxons. A pulse of radiation which is highly located has energy/momentum related by E = pc, just like a photon does. In Taylor and Wheeler's text "Exploring Black Holes" the authors were warned about using photons in GR by Philip Morrison. Its just easier to speak of photons than luxons because most people don't know what a luxon (or a tardyon for that matter) is.

The question as phrased assumed that he was asking about classical GR, not quantum GR. Its best to answer what the person is seeking to learn rather than what a literal interpretation might yield. If unclear then its best to ask. If you feel like giving the QM its best to state it as such. Since he was speaking about curves followed by photons he had luxons in mind.

Evan et al, see - http://www.newenglandphysics.org/physics_world/cm/two_accel.htm

[jeffreyH (OP)]

That is, does a photon's momentum affect its path through the gravitational field in the same way a massive object's velocity would?

These are two different questions. I'll address the one in the subject line.

The answer is no. Think of it like this - Imagine a particular photon (a given amount of energy) passing by the sun. The amount of deflection depends on the initial conditions of the photon (where it was, the direction it was going). Now imagine the same geodesics but this time another photon  moving side by side with the initial one. The geodesic is still a null geodesic as all are which photons more on. But now the amount of energy is twice as much regardless of how close the particles are moving next to each other, even so close as to be one entity for all practice purposes.

The geodesic a massive object (aka tardyon) moves on depends on the objects initial velocity which can vary.

This is all I wanted. A definitive answer. Thanks Pete. The reason I asked was because F = GMm/r^2 cannot work with a photon since it has no rest mass (m). I was wondering if, since energy has an equivalent mass, the photon's energy could be used to derive this mass term. However, it is moot point. The equivalence principle and the photon's constant speed in vacuum rule out any deviation in the geodesic. Thanks for clearing it up.

[Halc]

F = GMm/r^2 cannot work with a photon since it has no rest mass (m).

m is mass in that equation, not proper mass.  A photon has mass per what you said:

the photon's energy could be used to derive this mass term.

Just so.  If not, light could not move objects.  A lit torch (flashlight) in space will accelerate due to the reaction force against the mass of the photons emitted.

However, it is moot point. The equivalence principle and the photon's constant speed in vacuum rule out any deviation in the geodesic.

I had questioned that in post 8 since the photon going around a strong gravity well can refract, bending its path from the geodesic.  I agree that the geodesic path is the same for any light-speed particle regardless of energy, but I don't necessarily agree that the photon follows such a path.

[PmbPhy]

This is all I wanted. A definitive answer. Thanks Pete. The reason I asked was because F = GMm/r^2 cannot work with a photon since it has no rest mass (m). I was wondering if, since energy has an equivalent mass, the photon's energy could be used to derive this mass term. However, it is moot point. The equivalence principle and the photon's constant speed in vacuum rule out any deviation in the geodesic. Thanks for clearing it up.

You're most welcome my friend. A photon is deflected by a gravitating body because it has inertial mass (aka rel-mass). This is how Feynman explains it in his Lectures. Ever read that part? Want the reference?

A photon does not have constant speed in a gravitational field. Einstein proved that in the early history of GR. See

The Principle of Relativity, Lorentz, Einstein, Minkowski, Weyl, Dover Pub. See article which starts on page 99,

Here's a derivation which you may find easy to follow. http://www.newenglandphysics.org/physics_world/gr/c_in_gfield.htm

[jeffreyH (OP)]

I would appreciate the link to Feynman. Thanks for the other info.

[PmbPhy]

I would appreciate the link to Feynman. Thanks for the other info.

Sorry. I just recallke4d that I lost that volume. It's in V-II

[PmbPhy]

Evan - If Hacl was referring to the center of mass of a rock and feather then the feather will fall faster than the rock.