Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jeffreyH on 05/01/2020 01:03:47

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

That is, does a photon's momentum affect its path through the gravitational field in the same way a massive object's velocity would?
The path of a massive object is not a function of its momentum. A geodesic is a geodesic.
Technically, the more massive photon would mass more and thus pull the larger object a little closer as it passed, thus bending the path a bit more. This is negligible in the same why that there is a negligible difference in the time it takes to drop a feather or a brick in a vacuum. The brick wins by an immeasurable amount because it sucks Earth up to it more than does the feather.

Surely, if the velocity of an object is equal to the escape velocity of the larger mass, then the momentum must affect the path. Momentum is not a constant.

In my first post the velocity of the massive object can be any legitimate value. The higher the velocity the nearer it gets to matching or exceeding the escape velocity. In the same way that the photon can be assumed to have any energy value.
I hope this helps to clarify.

That is, does a photon's momentum affect its path through the gravitational field in the same way a massive object's velocity would?
The momentum of a photon does not affect it's path through a gravitational field. Or another way to say this is that photons of different energies will follow the same geodesic.
Additionally the mass of an object does not affect it's path through a gravitational field. If you shoot a feather and a cannon ball in a vacuum at the same velocity and angle, they will both follow the same parabolic path. Mass will not change the path.

Surely, if the velocity of an object is equal to the escape velocity of the larger mass, then the momentum must affect the path.
As Origin points out, two objects (feather and brick) both with identical velocity (escape velocity or otherwise), will follow (nearly, to at least 10 digits) identical paths from identical starting points. These objects have very different momentum, so momentum plays no role in the path followed.
Momentum is not a constant.
Nobody suggested otherwise. I said the path taken was not a function of the momentum.

That is, does a photon's momentum affect its path through the gravitational field in the same way a massive object's velocity would?
The momentum of a photon does not affect it's path through a gravitational field. Or another way to say this is that photons of different energies will follow the same geodesic.
Additionally the mass of an object does not affect it's path through a gravitational field. If you shoot a feather and a cannon ball in a vacuum at the same velocity and angle, they will both follow the same parabolic path. Mass will not change the path.
In other words a gamma ray photon will follow the same path as a radio wave photon. Is there any concrete observational evidence for this assumption?

In other words a gamma ray photon will follow the same path as a radio wave photon. Is there any concrete observational evidence for this assumption?
it is not an assumption, it is a direct outcome of GR. I recall an experiment that showed radio waves followed the same geodesic as visible light around the sun. I can look it up if you feel it is necessary.

Momentum, velocity, energy aside, one must remember that photons are quantum things andthere is relativistic bending of light in addition to the spatial curvature. A large mass acts as a refracting lens due to the time dilation at different depths in the gravity well. This causes the photon to interfere with itself and bend its path beyond that of its geodesic path. It would mean that light from say a distant white source would appear as a spreadout spectrum if it passes by a gravity source with a steep gradient such as a small black hole. I wonder if this has ever been observed.

In other words a gamma ray photon will follow the same path as a radio wave photon. Is there any concrete observational evidence for this assumption?
it is not an assumption, it is a direct outcome of GR. I recall an experiment that showed radio waves followed the same geodesic as visible light around the sun. I can look it up if you feel it is necessary.
If you can supply the reference it would be helpful.

in the same way a massive object's velocity would?
In my simple understanding: I think the path of an object through a gravitational field is determined by its velocity
And so the path of a photon through a gravitational field it is like the limit of a massive object's path as the massive object's velocity approaches c.
 ie the path is determined by velocity rather than momentum
 and the velocity of all photons is c (whether gamma rays or radio waves)
 Disclaimer: when the speed of a photon is measured in a vacuum, in the local frame of reference (ie not as seen by an observer far outside the gravitational field)...
The difference is that a massive object cannot ever quite reach c
 So a massive object can orbit around a star
 But light will not orbit a star unless the escape velocity exceeds c (ie a black hole)
 A more speedy massive object will follow a hyperbolic path when travelling past a star (like Comet Borisov, to pick a current example)
 Light always follows a hyperbolic orbit when travelling past a star
 If you accelerated a proton up to nearly c, it would nearly follow the same path as a photon, because its velocity is nearly the same

I'd still like a reference to experimental verification.

I'd still like a reference to experimental verification.
If you see a result of gravitational lensing as white curve, then photon energy doesn't affect its path. Otherwise, it does when you see a rainbow spectrum.

photon energy doesn't affect its path. Otherwise, it does when you see a rainbow spectrum.
When light enters a denser medium (like raindrops), it slows down.
 The amount of slowing is dependent on the photon energy
 This produces a rainbow
This mechanism doesn't operate in the vacuum of space.

photon energy doesn't affect its path. Otherwise, it does when you see a rainbow spectrum.
When light enters a denser medium (like raindrops), it slows down.
 The amount of slowing is dependent on the photon energy
 This produces a rainbow
This mechanism doesn't operate in the vacuum of space.
It doesn't matter if the photon slows down or not. What matters is if it bends or not.
If it slows down but not bends, you don't get the rainbow. On the other hand, if different photons bend by different amount according to the frequency, you get the rainbow, regardless they slow down or not. Think of diffraction gratings. No slowing down required.

If it slows down but not bends
If you trace the wavefronts of light striking glass (or any transparent substance) at an angle, you will see that:
 the wavefronts are closer together inside the substance
 the frequency is the same; the wavelength is shorter because the velocity is reduced
 the wavefronts within the substance are continuous with the wavefronts outside the substance
 that automatically means that a change of speed means a change in angle
 the angle changes towards 90°
 the only exception is if the light strikes the substance at a 90° angle  the light continues through at the same 90° angle, since it can't get any closer towards 90°.
See the graphic at: https://en.wikipedia.org/wiki/Refraction#General_explanation
Think of diffraction gratings. No slowing down required.
I agree that a diffraction grating does change the path of light in a wavelengthdependent way, without slowing the light.
 Diffraction is a different mechanism than refraction.
But I did hear of an infrared spectrometer for a spacecraft which used the "solid" side of the diffraction grating.
 This used the fact that the wavelength of infrared light is much shorter in silicon to reduce the size of the spectrometer by a factor of 4 in length, and an order of magnitude in volume.

A gravitational potential may red or blueshift light, but I'm not sure if it affect a geodesic depending on 'energy'. But then you have the idea of two photons affecting each others geodesics Jeffrey (in a 'empty' space). So in the end, yes, I think you're correct although I doubt it to be measurable.
=
What makes it even stranger is if you think of it as a emanation in a field. Either they exist or they don't, the 'path' I mean. If I define a geodesic as a path taken by light, then it somehow needs to 'interact'. And if you then apply quantum mechanics on it? Any interaction of it is that 'light', or 'photon', becoming 'real', as in having a defined spin.
=
Thinking that way would lock its spin at its 'emission'. That also means me thinking of 'gravity' as a force able to interact with lights propagation. To get away from it maybe I need to reinterpret gravity, and/or a 'propagation'
The point there being that if I think of gravity as a force interacting with the photons propagation, light can't be 'ageless' anymore, something ('energy'/'speed') needs to be 'lost' for that photon getting 'forced' into this geodesic. And what that 'energy' would be transformed into? (This as we never found light speed in a vacuum to differ.)
I think you can exchange it to 'momentum' too if you like. (that is if defining it as a momentum is a combination of a speed and a 'energy')
One more thing, if you think of it as entering a beam splitter then it means that one of the entanglement already was locked before entering it, not sure what difference that would make though? ahh damn, why not argue that the beam splitter break up the lock :) then we get back to the original in where it sets a opposite spin, us only knowing that they will be opposite.

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.
Technically, the more massive photon would mass more ...
What does "mass more" mean??
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.

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.

The path of a massive object is not a function of its momentum. A geodesic is a geodesic.
That is incorrect. If an object is massive (i.e. nonzero proper mass) then the path is a function of its momentum. If I through a stone with zero momentum then its path is a dot in that space. The more momentum I give it the faster it moves and the path will differ. This is also true for geodesics. I suggest that you don't use path and geodesics interchangeably since they don't mean the same thing. A path is an objects trajectory in space while a geodesics is an objects trajectory in spacetime having extremal proper time.
What do you mean by a geodesic is a geodesic? That's true only as a tautology. Geodesics can differ with their path through spacetime being different.

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 nongeodesic 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.

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 nongravitational 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.

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 nongeodesics world line.

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 nongravitational 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 kgmeters/second of momentum? Can't answer? Then it's not a function of momentum.

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 CenterofGravity of Earth +Brick+Feather.
Halc was calculating in the frame of reference of the CenterofGravity 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.

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

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.

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 lightspeed particle regardless of energy, but I don't necessarily agree that the photon follows such a path.

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 relmass). 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

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

I would appreciate the link to Feynman. Thanks for the other info.
Sorry. I just recallke4d that I lost that volume. It's in VII

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