
The Gravity of a Photon - JWKennington
https://jwkennington.com/blog/gravity-of-a-photon/
======
Causality1
What I've always found wild about gravity and photons is that gravity produces
the same proportionate effect, i.e., acceleration, on them as it does normal
matter and the only reason the sun and planets don't render light nearly
unusable for sensing by utterly distorting the light's path is that it travels
so quickly it doesn't spend any significant amount of time in a gravity well.

I don't think I'll ever fully wrap my head around the kind of numbers that
implies, that when light is at approximately sea level it is accelerating at
9.8m/s^2 toward the core of the earth and the only reason it doesn't all get
dragged out of the air and to the ground is that it moves too fast to notice.

~~~
danbruc
That is an interesting thought that never occured to me. It would render the
headlights of your car pretty much useless if the light just fell to the
ground a few meters in front of the car. I wonder if one has or could measure
the effect in the lab - it's about 0.5 μm over a distance of 100 km, so it
seems not totally out of reach.

~~~
kgwgk
Your comment reminded me of Gamow's Mr. Tompkins in Wonderland. I don't know
if he mentioned that in particular, but he discussed what relativistic effects
would look like if the speed of light was 10 MPH.

[http://www.arvindguptatoys.com/arvindgupta/tompkins.pdf](http://www.arvindguptatoys.com/arvindgupta/tompkins.pdf)

~~~
danbruc
There is also a free game, A Slower Speed of Light. [1]

[1] [http://gamelab.mit.edu/games/a-slower-speed-of-
light/](http://gamelab.mit.edu/games/a-slower-speed-of-light/)

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noobermin
The important thing to remember is it's called the stress-energy tensor for a
reason, it's not just mass distribution that causes curvature.

Thought experiment: which will curve spacetime more, a 1kg box of gas at 1K or
the same box after the gas inside is raised to 293K?

~~~
danbruc
The warm box will of course »heavier« because it contains additional energy.
Fundamentally there is no mass, it is just a convenient way to talk about the
energy hidden in the structure of matter.

~~~
filmor
The internal structure of leptons being? :)

~~~
danbruc
Note that I wrote structure and not internal structure and I did so
intentionally. As far as we know elementary fermions - quarks and laptons -
have no internal structure but - and I am totally not a physicist - there are
the Higgs mechanism, the seesaw mechanism, selfinteraction due to vacuum
polarization, and other things I have never heard of. As far as I know, this
is still an area of active research with unresolved questions, but I think it
is also generally believed among physicists that all mass has a dynamical
origin.

------
Xcelerate
I find gravity interesting, because _everything_ that exists interacts with
_everything_ via the gravitational field. This is not the case with any other
sort of interaction; gravity is truly different. See this diagram showing the
ways that different particles can interact in the Standard Model:
[https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/El...](https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Elementary_particle_interactions.svg/400px-
Elementary_particle_interactions.svg.png)

Photons, for instance, cannot interact directly with each other via the
electromagnetic field. Gluons do self-interact via the strong interaction. But
both photons and gluons interact with all types of particles via gravity
(although sometimes this interaction is so small that we would never be able
to measure it).

~~~
danbruc
Photons do interact electromagnetically [1], just not in the same direct way
as gluons.

[1] [https://en.wikipedia.org/wiki/Two-
photon_physics](https://en.wikipedia.org/wiki/Two-photon_physics)

~~~
Xcelerate
Yeah, I was more so speaking of direct coupling but didn’t want to get too
technical. That interaction is due to virtual fermion-antifermion pairs.

------
kgwgk
> because photons have rest energy that can be viewed as mass in special
> relativity

What?

~~~
JWKennington
Hence I said my friend was being a bit “interpretive”. He really meant that
though photons have zero rest mass, they nonzero inertial mass (E=mc^2)

~~~
typothrowaway
That still does not make any sense. I'm sorry to be so direct, but I'm really
worried laypeople here are going to think that this is correct.

~~~
JWKennington
Photons have momentum. momentum is analogous to mass in special relativity
(specifically, mass is the 0 component of the four-momentum)

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Tepix
The low contrast on that web page is terrible.

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radford-neal
If you think electrons and positrons have a gravitational influence, then it
seems to me you should also think photons do. Otherwise, if an electron and
positron annihilate, producing photons, there will be a discontinuous drop in
the gravitational field, which doesn't seem right somehow. Furthermore, if you
put all this in a black box, the gravitational influence of the box will
change because of what's happening within it, which also doesn't seem right.

Can anyone with with real knowledge of gravity comment on whether these two
intuitions are correct?

~~~
obastani
I don't have real knowledge of gravity, but I can speculate a bit. First, I
don't think the blackbox argument necessarily makes sense. If the black box
contains a massive point particle, and this particle moves from the left side
of the box to the right, then the gravitational properties certainly change.
For example, if there is a particle on top of the box, it will initially be
pulled to the left, and subsequently be pulled to the right.

Having said that, your intuition that photons should have gravitational
influence makes sense, since according to special relativity, mass and energy
are equivalent. Since total energy (mass + energy) is conserved, the
gravitational influence of the photon should equal that of the electron-
positron pair (assuming they are all located at the same position). I think
the article is agreeing with this point of view.

~~~
radford-neal
If the black box is freely floating (which is what I was thinking of), then a
particle within it can move to one side only if something else in it moves the
other direction.

But on further thought, I wouldn't be surprised if some changes in the
internals of the box could produce gravitational waves, so maybe my intuition
for this is wrong. Although if I further clarify that the box isn't supposed
to be emitting energy (which will obviously reduce its gravitational influence
as it loses energy), then maybe the intuition is correct after all...

~~~
100ideas
“Electromagnetic waves” are the manifestation of relative motion between an
electric charge and an observer’s reference frame. A detector inside a
cryogenic ion trap would detect much lower energy EM radiation from the
contained cloud of ions vs a sensor speeding towards the trap at 10% c. If
this second detector slowed towards the inertial reference frame of the ions,
the apparent EM radiation emanating from them would decay towards zero from
the second detectors point of view as it became stationary relative to the
charges.

Are gravity waves similarly observer-reference-frame-dependent? I.e. the
gravitation of a given object is not intrinsic but rather dependent on the
energy difference between the object and a given observer’s frame?

~~~
radford-neal
I think the charges need to be accelerating, not just moving, to emit
radiation. Whether or not something is emitting radiation ought to be
invariant to the motion of the observer.

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xupybd
Wow I didn't expect that! For the layperson that's really interesting.

"the conventional electromagnetic fields, and the photons that constitute
them, impact the gravitational field despite having no inertial mass!"

Does that mean the build up of the photon sphere around a black hole would
have a significant impact on space time?

~~~
jessriedel
The photon sphere is not a place where photons collect because the orbits
along the sphere are unstable fixed points. Small perturbations from these
orbits either fall into the black hole or escape to infinity.

~~~
thomaswang
Is there a line between where light neither escape or get swallowed. Simply
just stands still in perfect balance between being pulled and escaping?

~~~
danbruc
That is what the parent comments were talking about - on the photon sphere -
in case of a non-rotating black hole - photons could theoretically orbit the
black hole but as pointed out by your parent comment this orbit is not stable
and any tiny perturbation will either send the photons into the black hole or
make them escape. And the photons would of course not stand still - they are
massless particles and therefore can only move at the speed of light, neither
slower nor faster - but orbit the black hole at the speed of light.

~~~
100ideas
If you were further in the black hole’s gravity well than the photons in the
photon sphere, you would see them redshift and eventually freeze, right? I.e.
from inside the photons don’t appear to escape

~~~
jessriedel
There are no frames of reference where photons reverse direction. If
everything is spherically symmetric, a photon emitted from any point in
spacetime is either (1) already heading outward and escapes or (2) already
heading inward and is consumed. Within the event horizon, only trajectories of
type (2) exist. The picture of a photon struggling outward to escape and then
reversing direction under gravity is incorrect.

~~~
100ideas
Ok, thanks for bringing precision to my comment. It makes sense that any
photons reaching an observer who is inside a BH event horizon must also
already be trapped inside the EH.

I think what I find confusing is that I thought outside observers would never
see the infalling observer reach or cross the EH due to time dilation.

I’ve read that an observer falling into a black hole would notice extreme time
compression in the external universe (observing millions or billions of years
pass in the external reference frame), and conversely external observers would
notice extreme time dilation of people and redshifting of photons falling into
the black hole. Infalling particles from their external perspective appear
frozen and smeared into a blur outside the EH, fading away but never optically
appearing to “enter” into the BH, even though physically these particles
indeed have/will.

So I think I had it backwards, the external observers would see infalling
objects redshift, and those falling in would see the universe blueshift (I
guess getting fried by high energy photons before being torn up by tidal
forces).

~~~
jessriedel
> I think what I find confusing is that I thought outside observers would
> never see the infalling observer reach or cross the EH due to time dilation.

It depends what you mean by "see". An outside observer will certainly not see
this this in the literal sense of seeing photons that image the horizon-
crossing event. However, there are sets of space and time coordinates on the
manifold where the infalling observer crosses the horizon at the same "time"
as external events occur.

> frozen and smeared into a blur outside the EH, fading away but never
> optically appearing to “enter” into the BH

Yes, but note that the brightness of the image gets exponentially suppressed
as the infalling object approaches the horizon, so it really just looks more
like it's vanishing than freezing (although it's doing both). Importantly, a
finite amount of electromagnetic energy is emitted/reflected by the infalling
object before it crosses the horizon so, for any given minimum-energy
threshold of your detection equipment on the outside, you will see no more
than a certain finite number of photons no matter how long you wait.

> and those falling in would see the universe blueshift (I guess getting fried
> by high energy photons before being torn up by tidal forces).

No, I don't think much blueshifting happens. If you hover above the event
horizon of a black hole, the outside world will look blueshifted, to a
stronger and stronger degree as you get closer to the horizon. However, the
amount of force necessary to maintain a stationary position above the horizon
goes to infinity as you get close, so you can't get arbitrarily close and
experience arbitrarily large blueshift; indeed, the blueshift is in general
quite modest without extraordianry materials or fuels. And once your support
fails, so that you start falling into the black hole, the blueshifting goes
away.

------
Ceezy
I don t understand. The title is gravity of photon. Then you start talking
about electromagnetic field. I sure something is missing here. What would be
the source of gravitation and who is subject to that field?

~~~
Sharlin
Photons are excitations of the electromagnetic field. The source of gravity is
the energy in the EM field, as described by general relativity.

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dschuetz
This is really interesting. Perhaps there is also a chance that
electromagnetic fields are indeed entirely responsible for gravity, in some
complicated yet unknown manner.

~~~
kgwgk
What would that mean? As far as we know there is gravity in the absence of EM
fields and EM energy produce gravitational effects just like other forms of
energy (and in the realm of direct human experience the contribution of EM is
negligible).

~~~
dschuetz
"there is gravity in the absence of EM fields" That just cannot be true. EM
fields are just about everywhere. They permeate space almost as universally as
gravity, but in form of photons (to name higher energy EM fields here) or
other low energy waves. Gravity accumulates with great masses, mass is also
energy (in rest).

The article explores the idea that pure kinetic energy in form of photons
indeed might have at least gravity-like properties.

My presumption is that atoms are somehow made of really high energy EM fields
so that they appear to be _massive_. Consequently, atoms having mass or
photons having inertia are responsible for gravitation, and that could mean
that EM fields are the cause of gravity entirely. The problem with that
presumption is that perhaps there is no way to check.

~~~
danbruc
_EM fields are just about everywhere._

Fields are mathematical tools, they are not - without wanting to go down the
philosophical rabbit hole - real things out there in the universe.

And the thing about everything just being electromagnetic fields is nonsense,
physicists understand physics way better than you do. Sorry to sound
condescending, but the internet already has enough obviously wrong ideas by
random people without a clue what they are talking but.

~~~
tsimionescu
Honest question: aren't radio waves "physical objects" that represent
oscillations of an electro-magnetic field, to the best of our current
understanding?

~~~
danbruc
Radio waves are streams of photons, the fields are just a mathematical tool we
use to describe this. And there are several of them, the classical
electromagnetic field but also an entire set of fields involved in the quantum
version.

~~~
kgwgk
Other people may say that photons are just mathematical tools that we use to
describe fields. :-)

~~~
danbruc
The fact that gauge fields have redundant degrees of freedom should at least
be a strong hint that the fields are not the fundamental description. ;)

~~~
carlob
I don't think you are correct: you can view the extra degrees of freedom as
just a symmetry

~~~
danbruc
It is usually called a gauge symmetry but it is kind of a misnomer and should
better be called a gauge redundancy. It is artifical and only due to the
mathematical formalism we are using. As an analogy, probably not a really good
one, is that you could decide to describe temperatures with a complex number
and just say temperature is invariant under translation along the imaginary
axis. Sure, the symmetry is in your mathematics but it doesn't really tell you
anything new or interesting. Besides of course that the imaginary part is
actually redundant and you better described temperature with a real number
instead.

~~~
mokus
That would not work in the same way as a typical gauge theory though -
typically the symmetry of a gauge is for shifts of the entire system along the
gauge - changes along the gauge locally do have physical meaning. For example,
in an electrical diagram you can arbitrarily shift every voltage in the system
by a billion Volts (ground rail is 1b Volts, 5V rail is 1b+5V, etc) and the
system will behave identically. The gauge quantity is real and not redundant,
it’s just that it doesn’t have a known meaningful absolute reference point.

~~~
danbruc
Gauge symmetries are local symmetries. In case of the complex temperature
field you can change the imaginary part of the temperature at each point
independently without any physical effect. Global symmetries on the other hand
are meaningful, for example spatial translation and rotations and time
translations in Newtonian spacetime are associated with the isotropy and
homogenity of space and time and also the related conservation of momentum,
angular momentum, and energy. And of course your example of changinging the
reference potential is also a global symmetry, not a gauge symmetry.

