Gravitation and light

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Given that light can be pulled in by a large gravitational force, what would stop light being pulled by the gravitational force of a galaxy or group of galaxies or another gravitational body and being pulled into an orbit???

Raavin :wink:
 

climbhi

Originally posted by Raavin
Given that light can be pulled in by a large gravitational force, what would stop light being pulled by the gravitational force of a galaxy or group of galaxies or another gravitational body and being pulled into an orbit???

Raavin :wink:
Nothing! Though I've never heard of light being bent into an orbit, does anyone know if this can happen, I imagine it would take a tremendous mass (supermassive blackhole maybe?). If it can that's kind of neat, I wonder if Keplers laws would apply to the light in the orbit...
 

mathman

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Near a black hole, it is possible for photons to be in orbit. In general, photons go too fast (much greater than escape velocity) to be trapped into orbit.
 

pmb

Re: Re: gravitation and light

Mathman wrote:
Near a black hole, it is possible for photons to be in orbit. In general, photons go too fast (much greater than escape velocity) to be trapped into orbit.
Yup. The photons can orbit at a distance of r = 3M (in geometrican units). The sphere r = 3M = constant is called the "Photon Sphere". This orbit is unstable however and the light, if not exactly at r=3m, will either spiral in or will not remain in orbit.

Pete
 
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Photons do orbit the central galactic engine (galactic black hole) of ~108 MSun. A general rule is that photons are bent into orbit only around a black hole. Black hole mass and radius, in cgs units, are related by MBlack Hole=1028 RBlack Hole

I wonder if gravitons likewise orbit a black hole. The Universe is nearly a black hole, and I also imagine that photons circulate near its horizon.
 

steppenwolf

maybe i should start a new thread but it seems such a waste of space...

how are photons affected by gravity when the value of the force of attraction is dependant on the mass of each object, the mass of a photon being 0 you would end up with 0 wouldn't you? my physics teacher tried to explain this by saying that as a photon has energy this equates to it having mass, but then why would we say that a photon has no mass? help!
 

Janus

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Originally posted by steppenwolf
maybe i should start a new thread but it seems such a waste of space...

how are photons affected by gravity when the value of the force of attraction is dependant on the mass of each object, the mass of a photon being 0 you would end up with 0 wouldn't you? my physics teacher tried to explain this by saying that as a photon has energy this equates to it having mass, but then why would we say that a photon has no mass? help!
When one says a particle is "massless" it is meant that it has zero "rest" mass. This would be the mass left after subtracting that due to the particle's kinetic energy. This mass due to energy is sometimes called "Relativistic" mass. This term isn't used in modern scientific circles anymore however. Nowadays, when a physicist mentions "mass" he is talking about rest mass alone. They just call relativistic mass "energy".

Thus a photon has energy, and no mass, but couples with gravity through its energy.
 

pmb

re - "Nowadays, when a physicist mentions "mass" he is talking about rest mass alone. They just call relativistic mass "energy". "

That's not true in general. Depends on the physicist. A rather new relativity text by one of the most fanmous relativists (i.e. Wolfgang Rindler) is not that way. There are several more like that.

Pete
 

russ_watters

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Though I've never heard of light being bent into an orbit, does anyone know if this can happen, I imagine it would take a tremendous mass (supermassive blackhole maybe?). If it can that's kind of neat, I wonder if Keplers laws would apply to the light in the orbit...
Thats pretty much the definition of a black hole. Since light has a constant velocity, there is a single orbital distance that it can have - called the event horizon. Further away and it escapes, closer in and it is captured. This is the event horizon:
Yup. The photons can orbit at a distance of r = 3M (in geometrican units). The sphere r = 3M = constant is called the "Photon Sphere". This orbit is unstable however and the light, if not exactly at r=3m, will either spiral in or will not remain in orbit.
how are photons affected by gravity when the value of the force of attraction is dependant on the mass of each object, the mass of a photon being 0 you would end up with 0 wouldn't you? my physics teacher tried to explain this by saying that as a photon has energy this equates to it having mass, but then why would we say that a photon has no mass? help!
Relativity. Einstein re-defined gravity to not be an simply interaction but to be a warping of space-time by massive objects. So two objects with mass will still interact the way Newton says (almost) but light still curves around the bend in space-time around a massive object.
 

pmb

It should also be pointed out that a body need not be a black hole in order for light to orbit the body. So long as "r", the "reduced circumferance" of the body is in the range 2M < r < 3M.

Pmb
 
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Originally posted by Janus
When one says a particle is "massless" it is meant that it has zero "rest" mass. This would be the mass left after subtracting that due to the particle's kinetic energy. This mass due to energy is sometimes called "Relativistic" mass. This term isn't used in modern scientific circles anymore however. Nowadays, when a physicist mentions "mass" he is talking about rest mass alone. They just call relativistic mass "energy".

Thus a photon has energy, and no mass, but couples with gravity through its energy.
Is it not that when we say photon's rest mass is null actually in a cool kind a way we state that photon never
"rests" ? It just keeps going with its c speed. :frown:
 
This all means that the gravity of a light emmitting star will decrease as 1/r you know.

:E
 

russ_watters

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It should also be pointed out that a body need not be a black hole in order for light to orbit the body. So long as "r", the "reduced circumferance" of the body is in the range 2M < r < 3M.
If there exists such an "r" for a body, then doesn't that mean that light within that "r" can't escape to infinity? Thats the definition of a black hole.
 

marcus

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Originally posted by pmb
re - "Nowadays, when a physicist mentions "mass" he is talking about rest mass alone. They just call relativistic mass "energy". "

That's not true in general. Depends on the physicist. A rather new relativity text by one of the most fanmous relativists (i.e. Wolfgang Rindler) is not that way. There are several more like that.

Pete
Rindler, who got his PhD back in 1956, is in a tiny minority of hold-outs.
His text "Essential Relativity" was first published in 1977. A revised edition "Essential Rel. Special, General, Cosmological" came out, if my information is correct, in 2001.

Rindler is an eminent and admirable guy. I have seen his clinging to the "relativistic mass" phrase described by colleagues as attachment to a "now old-fashioned heresy"----but he is otherwise very respectable. Must be nearing 70 now, or already in his 70s.

Is he still teaching in Texas or is he emeritus?

It is a really confusing phrase in general discussion---at least as you use it Pete

Everytime on PF I have pointed out that the righthand side of the Einstein eqn is an *energy* density----which is the normal view----you automatically answer back "oh no! you misunderstand! it is a "relativistic mass" term!!

(as if "relativistic mass" was your codeword or jargon for energy)

In other words, the equation relates curvature to energy density

Gmu,nu = 8pi Tmu,nu

with Gmu,nu having units of curvature, and
Tmu,nu having units of energy-per-volume (formally the same as pressure).

But no one can point out the obvious fact that the units on the righthand side are energy-per-volume without your piping up that the energy is *really* not energy but something else which modern physicists dont use as a concept (except a few holdouts like Rindler now I guess 70-years old)

that said, Pete,
congratulations for knowing about the PHOTON SPHERE of a black hole! I gather that it is not familiar to a lot of posters at PF.
And it should be-----it's introduced in introductory general astronomy courses right along with the Schwartzschild radius.

In brief, light cant escape from radius 2M
and if projected horizontally at radius 3M it goes into circular orbit.

But, to answer Russ Watters, it could escape from a point at distance 3M if, for instance, projected straight upwards.
 
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pmb

marcus wrote
Rindler, who got his PhD back in 1956, is in a tiny minority of hold-outs.
What evidence do you have to support that conclusion? I don't see it that way in the literature. One notable example is Alan Guth at MIT. In his class notes he writes

http://www.geocities.com/physics_world/guth-mass.htm

(http://web.mit.edu/8.286/www/)
======================================
We are perhaps not used to thinking of electromagnetic radiation as having mass, but it is well-known that radiation has an energy density. If the energy density is denoted by u, the special relativity implies that the electromagnetic radiation has a mass density

(7.3) rho = u/c^2

To my knowledge nobody has ever "weighed" electromagnetic radiation in any way, but the theoretical evidence in favor of Eq. 7.3 is overwhelming - light has mass. (Nonetheless, the photon has zero rest mass, meaning that it cannot be brought to rest). The general relation for the square of four-momentum reads p^2 = -(mc)^2, and for the photon this becomes p^2 = 0. Writing out the square of the four-momentum leads to the following relation for photons:

p^2- E^2 = 0, or E = cp.

In this set of notes we will examine the role which the mass of
electromagnetic radiation plays in the early stage of the universe.
======================================

I'd hardly say that Guth was out of touch. And this is what he's teaching students at MIT.



>Is he still teaching in Texas or is he emeritus?

I think he's emeritus.

>It is a really confusing phrase in general
> discussion---at least as you use it Pete

When you say "confusing" do you find it confusing or is it you think that others are confused? If you find it confusing then in what sense are you confused?

>Everytime on PF I have pointed out that the righthand side of
>the Einstein eqn is an *energy* density----which is the normal
> view----you automatically answer back "oh no! you misunderstand!
> it is a "relativistic mass" term!!

They differ by a constant. So you can write that equation either way. But if you follow the actual usage in, say, the American Journal of Physics, or Peebles text on cosmology then they tend to be more precise when they use terms like "mass" - it can mean one of 3 things

(1) proper mass
(2) inertial mass (i.e. relativistic mass = This is how Schutz uses it)
(3) active gravitational mass
(4) passive gravitational mass

Einstein's eqution is the general relativity equation that replavced Poisson's equation - and in that equation it's mass. and yes mass density =- but I thought it was obvious that it was density.

And I realize that you think it's out of use by all "up to date" folks. Fine - believe that. I'm not trying to force you to believe anything. But I will state my opinion and what I see relativists doing.

By the way. Rindler is just *one* example. There are others. E.g.

"Basic Relativity," Richard A. Mould
"Introducing Einstein's Relativity," Ray D'Inverno
"Understanding Relativity: A Simplified Approach to Einstein's Theories," Leo Sartori

And then there is my astronomy text 'Perspectives in Astronomy' I think it's callled (It's somewhere around here. Ref furnished upon request)

Then there are the articles from the literature, i.e. physics jorunals written in support of relativistic mass and others who disagree with the con-relativistic mass people.

However I question the reason that its becoming porpular. I don't think its based on pure logic and for purely physics oriented reasons. I think some authors wrote in their text that relativistic mass is "old fashioned" when it really wasn't. Students eager to learn relativity and who want to impress their friends with it surely want to learn the "up to date" physics. Reading that relativistic mass is "old fashioned" then leads them to think "Gee! I certainly don't want to be old fashioned and the guy who wrote my, very expensive, text must surely know what he's talking about. They must have done an experiment that proved it wrong. etc."

When the fact is that some of the articles in the physics literature had some highly misleading arguements as well as some which were just plain wrong! In actuality there is nothing wrong with the notion. In fact it best fits the notion of what mass is.

After all. The student is in no position to fully understand such broad sweeping statements as are made in their texts. And the physics I, II, III course barely scratch the surface of relatvity.

Physicists for the last (almost) 100 years have been studying and testing these ideas. Some have gone through great pains to cover all the intricate ideas and notions that go into such statments such as m = E/c^2 with all of its implications.

To dimsiss the notion as "old fashioned" is extremely misleading. Some of these claims of old fashioned were made when relativistic mass was IN fashion. Something of this magnitude of an idea doesn't become olf fashioned over night. It'd take many decades of complete non-use. And that can hardly be said for current physics today. Intro Physics texts (For physics I, II and III) are not written by cosmologists and GR-experts in general. And these are the greatest users of the idea.

marcus - Pick up the Peebles' cosmology text and look under (1)inertial mass (2) active gravitational mass and (3) passive gravitational mass

Also look under cosmic strings and vacuum domain walls. The former has zero active g-mass and the later negative active g-mass



Pete
 
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marcus

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Originally posted by russ_watters
If there exists such an "r" for a body, then doesn't that mean that light within that "r" can't escape to infinity? Thats the definition of a black hole.
Hello Russ,

Pete said:
"It should also be pointed out that a body need not be a black hole in order for light to orbit the body. So long as "r", the "reduced circumferance" of the body is in the range 2M < r < 3M."

Then you replied with a question, which was never answered, and the thread was dropped.

Instead of saying "reduced circumference" which means C/2pi---circumference divided by 2pi----I will simply say "radius".

If the body Pete mentioned has radius less than or equal to 2M (in his notation) then it is a black hole.

If it has radius greater than 2M but less than 3M then light can escape from it (and it is not a black hole) but only if the light is pointed enough upwards so that it doesnt curve around and fall back.

In the simplest model of black hole, un-charged and non-rotating, with the simplest formulas. The radius of the event horizon (from which there is no escape) is 2M and the radius of the *photon sphere* (where light can orbit if you give it a horizontal start) is 3M.

Pete is saying there could be a body which is NOT a black hole, which light could escape from, but which was dense enough that light could go in circular orbit around it----but only if you started the light off horizontally.
The light would escape from radius 3M if you tilted your flashlight upwards.

this is an informal discussion and there is doubtless lots to quibble about--but anyway thats the general idea.

PS if you dont know what Pete meant by M, you should ask.
It is obviously a length benchmark associated with the mass of the body---because he is comparing radiuses to it. It is obviously an important length scale belonging to whatever body is being discussed so you should ask if you didnt know what he meant by it and are at all curious. Otherwise what he said wouldnt mean anything-----the bit about 2M < r < 3M.
 
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pmb

marcus wrote
I have seen his clinging to the "relativistic mass" phrase described by colleagues as attachment to a "now old-fashioned heresy"----but he is otherwise very respectable. Must be nearing 70 now, or already in his 70s.
That's not quite accurate. Some physicsts say that. Others say differently. And I don't see what age has to do with it. Take Wheeler as an example. Wheeler is about 90 or so and he disagrees with Rinlder on this mass thing. Okun who is around the same age as Rindler disagrees with him too. I can give you a list of people from each side of this arguement as well as a healthy list of physicists on both sides. But to what end?

This is a rather complex issue. To fully address this in all its detail would require about 40 pages of text. But is that the purpose of this thread?


Pmb
 

Alexander

Any energy has both inertia and gravity. Mass (being just a form of energy) is not an exeption.
 

pmb

Originally posted by Alexander
Any energy has both inertia and gravity. Mass (being just a form of energy) is not an exeption.
I have one exception to that rule. And that is simply a side note as to what energy one is speaking of. If a particle is moving in a conservative field then the energy W defined by

W = "Kinetic Energy"(T) + "Rest Energy"(E_o) + "Potential Energy"(U)

W = T + E_o + U = E + U

where as usual E = T + E_o

The potential energy U does not contribute to the inertia. However that is that the energy of the particle in the field. The energy of "System + Particle" will have an increase in inertia and gravitational mass.


Pmb
 

pmb

I wanted to addrress these comments more but wanted to think about it for a while first. So I think I'm ready to respond.


marcus wrote
Everytime on PF I have pointed out that the righthand side of the Einstein eqn is an *energy* density----which is the normal view----you automatically answer back "oh no! you misunderstand! it is a "relativistic mass" term!!
This happens because this is a discussion forum. When one person has one view and uses it to explain something to someone then it's expected that a person with another view will post his view as well to the person asking the question.

Now regarding my comment regarding relativistic mass. First off I don't recall stating it exactly as you say. And in the second place I do chime in because it's my opinion that you're wrong. In particular with respect to your claim "which is the normal view." This I don't agree with. To see this consider where Einstein's field equation comes from. To illustrated this I'll quote Ohanian. I would guess that we can all at least agree that he's an expert in GR right?

From "Gravitation and Spacetime - 2nd Ed.," Ohanian and Ruffini, page 140
The first thing we must do is to decide what the source of gravity is going to be; that is, we must ask what quantity plays a role analogous to that of the electromagnetic source j^u. The answer is given byu Newton's principle of equivalence, according to which (total) inertial mass of a system gravitates, that is the energy gravitates. The source of gravitation must therefore be the energy density. However it is impossible to construct a Lorentz-invariant theory of gravity in which the energy is the only source of gravitation [...] Before we proceed, a remark on a possible alternative. We could take the trace of the energy-momentum tensor. [ ...] The trouble with [] is that for the electromagnetic field T^u_u = 0 () Hence electromagnetic energy would not gravitate, in contradiction with the principle of equivalence.
Hence, in deriving Einstein's equations, Ohanian postulates that the total (inertial) mass of a system gravitates. Ohanian also states that this is exactly energy (or energy density)! It's for arguements such as this one of Ohanian's that I object to your claim that it's energy and not mass that gravitates since they are one in the say and differ only by a constant c^2. But that is exactly what relativistic mass is. And that is also why it's called mass-energy!

As for Einstein's equations - You've chosen units that don't reveal much. Revert back to MKS

G^ab = -(8*pi*G/c^4)T^ab

This equation must be dimensionally consistent with Poison's equation. There is a c^2 for canceling the c^2 in "E = mc^2." There is a c^2 for the otherside of the equation. By that I mean that

[G^ab] = 1/m^2

To make this Einstein's equations dimensionaly consistent we multiply that by c^2 to get c^2/m^2 since Phi = gravitational potential has units which are the same as c^2. Einstein's equations then become dimentionally consistent with

del^2 Phi = 4*pi*G*rho

Since the quantity which is *really* the source of gravity is mass density then we could define M^ab = (T^ab)/c^2 and write


G^ab = -(8*pi*G/c^2)M^ab

Which is closer to reality. For if you look at Einstein's 1916 paper on GR, Einstein writes

"The special theory of relativity has led to the conclusion that inert mass is nothing more or less than energy, which finds its complete mathematical expression in a symetrical tensor of second rank, the energy tensor."

Thus it is seen that the source of gravity is mass - and since mass and energy are the same we use the stress-energy-momentum tensor - but we have to convert that to units of mass before we can get the units to be dimensionally consisent.





In other words, the equation relates curvature to energy density

Gmu,nu = 8pi Tmu,nu

with Gmu,nu having units of curvature, and
Tmu,nu having units of energy-per-volume (formally the same as pressure).
Yes. I agree. The units of G^ab has the units of curvature i.e. 1/m^2 (i.e. 1/length^2). The physical reason being that G^ab becomes, after we multiply by c^2, in the Newtonian limit del^2 Phi. This is the relativistic version of

del^2 Phi = 4*pi*G*rho

In Newtonian Physics del^2 Phi is simply the contraction of the Cartesian tensor known as the (Newtonian) tidal force tensor t_ij

See -- www.geocities.com/physics_world/tidal_force_tensor.htm

so where there is no tidal force there is no mass. However this does not mean that where there is no mass there is no gravitational field. It also does not mean that there is no gravitational field where there are no tidal forces - the uniform gravitational field being an example: see

http://www.geocities.com/physics_world/grav_cavity.htm

for an example of a uniform g-field.


But no one can point out the obvious fact that the units on the righthand side are energy-per-volume without your piping up that the energy is *really* not energy but something else which modern physicists dont use as a concept (except a few holdouts like Rindler now I guess 70-years old)
See above remarks as to why the coefficient of T^ab, i.e. "-8*pi*G/c^4", contains c^4. I also hold that what you claim "which modern physicists don't use as a concept" is not true as well. MTW being a good example as well as all the others I've mentioned. E.g. page 404 "Mass is the source of gravity." On page 141 "... They must be equal because energy = mass ("E = Mc^2 = M")". Take Schutz as another example: Page 98

"Thus, n is a scalar. In the same way that 'rest mass' is a scalar, even though energy and "inertial mass" are frame dependant, here we have that n is a scalar, the 'rest density,' even though number density is frame dependant."

Which clearly indicates that Schutz does not mean "inertial mass = rest mass" as one might be led to believe in other conversations regarding mass.

Again I note that age has nothing to do with this. This debate is as old as relativity itself. French even notes this in his "Special Relativity text." And he still thinks velocity dependant mass is a good idea. There is nothing new about people arguing over it. It didn't just begine in the 80's at all. And recall my remarks that Wheeler is older that Rindler and Wheeler is almost a mass=rest mass person. "Almost" because in MTW he clearly states that mass=energy just as Ohanian and Ruffini do.


Pmb
 
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Originally posted by Raavin
Given that light can be pulled in by a large gravitational force, what would stop light being pulled by the gravitational force of a galaxy or group of galaxies or another gravitational body and being pulled into an orbit???

Raavin :wink:

nothing, i suppose. the thing is a galaxy doesn't have enough localized gravity to do more than tweak light's path a bit. only black holes have enough energy to prevent light from leaving its event horizon. hence the name "black" holes!
 
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