blackholes vs. massless photons

[gib65]

I'm confused about something. I've been hearing that photons are massless, yet it is predicted that blackholes exist. The appearant contradiction here is that blackholes are predicted to exist because light is subject to gravity in the same way matter is. That is, light curves toward the source of gravity. They say that blackholes are objects with so much gravity that they do not let light escape (i.e. Light curves so much that it always ends up travelling directly towards the object's center of gravity). But how can light be influenced by gravity in this way if light, being composed of photons, is massless? Doesn't it require mass to be "heavy"?

[5614]

you know what.... that is a bloody good question and i dont have a clue!

 

(i hope someone else does though!)

[Ophiolite]

I have a fantasy in which, following a question such as yours, forty eight thousand physicists around the world raise their hands to their chins and say "Shit! I never thought of that."

 

Is the explanation not that space is curved in the vicinity of the black hole. The light still moves in straight lines - its just the straight lines are curved! The same is true around any mass, but especially so for black holes, where space curves in on itself?

 

But what do I know. Simple minded geologist. Still, if dropped from a tall building I would fall at the same speed as Newton or Einstein, so in that sense I am their equal.

[Gilded]

Hmmmh... Interesting topic, indeed. Was there space-curving in the graviton exchange theory, or was it a totally different thing?

[5614]

http://www.wonderquest.com/thin-air-black-holes-smallest-egg.htm

Much of what you were taught is valid even under the terrible constraints of a black hole.

We first tested Einstein’s General Theory of Relativity with this effect. His theory predicted that a massive object' date=' like the Sun, would bend light. Sure enough, during the solar eclipse of 1919, we observed the light from a star behind our Sun only because the Sun’s gravity field bent the star’s light. Otherwise, the Sun would have blocked the starlight. So, we have observed gravity influencing light. [/quote']

[5614]

http://www.physlink.com/Education/AskExperts/ae180.cfm

(on the non-mass of photons)

[timo]

I've been hearing that photons are massless, yet it is predicted that blackholes exist.

Yes, that´s both correct. Note, however, that "photons are massless" refers to their rest-mass (a term which is a bit misleading because photons cannot be at rest). The fact that the term "mass" can either mean the rest-mass or the total energy (relativisic mass) can be a bit confusing if you´re not used to keep that in mind.

But that´s not really important for your question, just wanted to say it.

 

But how can light be influenced by gravity in this way if light, being composed of photons, is massless?

Simple answer: The movement of a particle (due to gravity) is independant of it´s mass. Example: A feather will fall as fast as a cannon-ball if there´s no air-resistance. This is even partly true in Newtonian Mechanics where the "heavy mass" and the "inert mass" are experimentally verified to be proportional and hence cancel each other out in the equations.

[timo]

I have a fantasy in which' date=' following a question such as yours, forty eight thousand physicists around the world raise their hands to their chins and say "Shit! I never thought of that."

[/quote'] You wish .

 

Is the explanation not that space is curved in the vicinity of the black hole. The light still moves in straight lines - its just the straight lines are curved!

That´s absolutely correct. Those "straight lines" are called Geodesics which should sound familiar to you. In fact, not only light but all particles move on Geodesics.

[Pete]

I'm confused about something. I've been hearing that photons are massless, yet it is predicted that blackholes exist. The appearant contradiction here is that blackholes are predicted to exist because light is subject to gravity in the same way matter is. That is, light curves toward the source of gravity. They say that blackholes are objects with so much gravity that they do not let light escape (i.e. Light curves so much that it always ends up travelling directly towards the object's center of gravity). But how can light be influenced by gravity in this way if light, being composed of photons, is massless? Doesn't it require mass to be "heavy"?

 

Light does have mass (aka inertial mass aka relativistic mass). It has zero proper mass (aka rest mass) and that's what you were thinking about. As such it is effected by gravity. One simply has to turn to The Feynman Lectures Vol -I page 7-11. Section entitled Gravitation and Relativity

One feature of this new law is quite easy to understand is this: In Einstein relativity theory, anything which has energy has mass -- mass in the sense that it is attracted gravitationaly. Even light, which has energy, has a "mass". When a light beam, which has energy in it, comes past the sun there is attraction on it by the sun.

That´s absolutely correct. Those "straight lines" are called Geodesics which should sound familiar to you. In fact, not only light but all particles move on Geodesics.

That's not quite right. While it is true that light moves on geodescics is not quite right to say that a geodesic is a "straight line". Its more appropriate to say that its the straightest possible line. In this case it is a geodesic, not in space, but in spacetime. Light does not travel on a straight line is space. It travels on a curved line in space.

 

Pete

[timo]

That ["straight lines"=geodesics] is not quite right. While it is true that light moves on geodescics is not quite right to say that a geodesic is a "straight line". Its more appropriate to say that its the straightest possible[/i'] line.

Partly agreed. I rather see a straight line as a special case of a geodesic for a flat space. Well, I can live with your statement but I think it cries for new confusion (keep in mind that most people here don´t know GR).

 

But the thing that raised my attention in your post was the following statement:

In this case it is a geodesic, not in space, but in spacetime. Light does not travel on a straight line is space. It travels on a curved line in space.

Is it really possible to tell space from time in spacetime in general? Or do you speak about the coordinates of your coordinate system (which are rather meaningless due to your freedom in chosing a coordinate system)?

Or in other words: Is it possible to tell if a trajectory is curved or non-curved without using a coordinate system other than saying "they are all non-curved because they are geodesics"?

[Molotov]

Photons will travel the curvatures created by gravity. In the case of a black hole space curves so much that light is directed into the center. Or so I think...

[1veedo]

Yes, you're right Molotov.

 

The history of a photon (or any particle for that manner) is a path in spacetime with ever increasing entropy (conservation of information). If the future spacetime path of a particle curves to a point then these spacetime paths will converge and intersect at that point. If this were to happen in the past, such that the spacetime paths emerge from the point of intersection and gradually curve away from each other, the point of intersection would appear to constitute the beginning of spacetime. If it were to happen in the future, such that the spacetime paths emerge to a point and thus gradually curve toward eachother, the point of intersecatin would appear to be the end of spacetime.

[Pete]

Partly agreed. I rather see a straight line as a special case of a geodesic for a flat space. Well, I can live with your statement but I think it cries for new confusion (keep in mind that most people here don´t know GR).

People should keep examples of geodesics in mind so as not to be confused. A few examples go a long way. Here are a few which are illuminating

 

(1) geodesics on an infinite flat plane

(2) geodesics on a sphere

(3) geodesics on a cylinder

(4) geodesics on a cone

 

For #1 the geodesics are straight lines.

For #2 the geodesics are great circles.

For #3 the geodesics are spirals. See Figure 4 at

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

For #4 the geodesics are hard to picture. See figure 2 at

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

 

Is it really possible to tell space from time in spacetime in general?

Absolutely. Its quite important to distinguish the two. You measure "space" with a ruler and "time" with a clock. As Einstein himself said in Science in Feb 17, 1921 issue page 783

..it follows that, in respect to its role in the equations of physics, though not with regard to its physical significance, time is equivalent to space coordinates.

Consider the inside of an elevator in flat spacetime in an inertial frame of reference. Let there be a rectangular set of spatial axes, i.e. an xy coordinate system whose axes are mutually perpendicular. A beam of light will move on a spatial curve described by x = ay + b. This is a straightline in space and is thus also a geodesic in space. The worldline is a line in spacetime for which x = ct (when x and "ct" axes are perpendicular). This too is a straight line, a straightline in spacetime - a geodesic. However if the elevator is uniformly accelerating then the curve the light will move on a spatial curve which is not a straight line. The curve in spacetime is also not a staight line. However it is a trajectory in spacetime is still a geodesic.

 

Pete

[timo]

Thx for the answer Pmb.

 

However, I either didn´t understand your answer or there has been a misunderstanding because your answer actually backed up what I said: That it´s not possible to tell space from time in spacetime.

I´ll also assume a flat spacetime:

 

- >>"You measure "space" with a ruler and "time" with a clock."

Yes. And the time I measure is the length of my path through spacetime. So assuming we both start from the same point and since we are obviously heading in different directions the direction in spacetime you call time is a different one. In fact, I would say that what you call a time-interval is indeed a mixture of time and space. There is no objective separation of time and space in spacetime.

Guess you were speaking of the coordinates (of our both frames of rest in this case), then.

 

- Your elevator example is a good example for my claim that you cannot tell if a trajectory is curved or not:

I hope we can agree that the path of the light is completely independent of what the elevator does. As you said the beam will be a line for an observer in the elevator if it isn´t accellerating but will appear curved if it accellerates. So since both reference systems are equivalent there´s no way telling if the trajectory is curved or not.

Same as above: A judgement like "this path is curved" can only be made within a coordinate system and is dependant on it. Frame-dependant statements, however, are quite useless. That´s the point in the whole tensor stuff (so you can tell if a trajectory is a geodesic or not because that´s a tensorial statement).

[Gilded]

Ahh yes, I forgot that only the rest mass is 0. Well, that explains a few things.

[Pete]

Thx for the answer Pmb.

 

However' date=' I either didn´t understand your answer or there has been a misunderstanding because your answer actually backed up what I said: That it´s not possible to tell space from time in spacetime.

I´ll also assume a flat spacetime:...

[/quote']Perhaps I was unclear as to why you can tell the difference between space and spacetime but its quite true that you can. You do it all day long. I can move my car from x = 0 to x = 1 and I can then move my car from x = 1 back to x = 0. However, while its true that I can move my car from t = 0 to t = 1 it is not true that I can move my car from t = 1 back to t = 0. (x,y,z) detnotes a point in "space" while "t" denotes a momentum in "time." Spacetime is the collection of points whose coordinates are (t,x,y,z). It is true that mathematically space and time can in some ways be treated on the same footing mathematically they cannot be treated on the same footing physically.

 

- >>"You measure "space" with a ruler and "time" with a clock."

Yes. And the time I measure is the length of my path through spacetime.

 

Why do you think that's true and what is it you mean by "length"?

 

Normally in GR the term "length" as it pertains to spacetime is not defined the same way as it is in Euclidean geometry. typically the term is used to denote the "interval" between two points. For instance, the spacetime interval between two points on a worldline of a photon in flat spacetime in an inertial frame is zero, even if the actual photon has traveled light years. The spatial displacement equals t = x/c and the temporal displacement equals x = ct, the spacetime interval [itex]\Delta s[/itex] is zero. Althought the spatial and temporal coordinates are proportional in this case they are not identical.

 

- Your elevator example is a good example for my claim that you cannot tell if a trajectory is curved or not:

I hope we can agree that the path of the light is completely independent of what the elevator does. As you said the beam will be a line for an observer in the elevator if it isn´t accellerating but will appear curved if it accellerates. So since both reference systems are equivalent there´s no way telling if the trajectory is curved or not.

Why do you think that there are "special" coordinate systems in which light is "really" moving on a straight spatial trajectory? Why true in Newtonian mechanics it is false in GR.

 

Same as above: A judgement like "this path is curved" can only be made within a coordinate system and is dependant on it.

That's correct. And that's why I was explaining to you. I.e. that whether the spatial trajectory of a beam of light is curved or not depends on the frame of reference.

 

 

Frame-dependant statements, however, are quite useless.

Then the question "Is the spatial trajectory is curved" is quite useless. But I can still speak of it. Its not true that frame-dependant statements are useless. The observer lives in a particular frame and the observer himself is a geometric object. Therefore statements like "spatial path is curved" has a geometric meaning independant of a coordinate system and depends only on observer and phenomena.

 

That´s the point in the whole tensor stuff (so you can tell if a trajectory is a geodesic or not because that´s a tensorial statement).

The whole point of tensor stuff is that relativity demands that the laws of physics not depend on a particular coordinate system. It does not mean that observations are coordinate independant.

 

Pmb

[swansont]

The history of a photon (or any particle for that manner) is a path in spacetime with ever increasing entropy (conservation of information).

 

entropy <> information

 

It's awkward, at best, to discuss the entropy of a single photon, as it's a macroscopic phenomenon.

 

AFAIK there is no "second law" of information theory

[5614]

can someone explain the exact difference between:

 

inertial mass,

relativistic mass &

rest mass.

[YT2095]

rest mass is the mass of a "body" that`s stationary.

 

if the "body" is moving its mass increases proportional to its velocity.

 

other than that, Hell if I know

[Pete]

can someone explain the exact difference between:

 

inertial mass' date='

relativistic mass &

rest mass.[/quote']The inertial mass of a body is defined as the quantity m such that the quantity mv is conserved in particle collisions as observed from an inertial frame of reference. If you're discussing relativity then some people like to use the term relativistic mass for this same term. I don't like it since it makes people think that its somehow different than inertial mass. If the body can travel at speeds less than the speed of light then it turns out that the mass is a function of speed, i.e. m = m(v). The quantity m₀ = m(0) is called "rest mass" since v = 0, i.e. the body is at rest.

 

Pete

[swansont]

can someone explain the exact difference between:

 

inertial mass' date='

relativistic mass &

rest mass.[/quote']

 

Inertial mass is the mass in F=ma. As far as anybody can measure, this is the same as the gravitational mass in F=GMm/r²

 

relativistic mass is the mass from E = mc² if you use total energy. This is frame dependent.

 

rest mass is the frame independent mass in E²=p²c² +m²c⁴

[5614]

so what is a photon's relativistic or intertial mass?

[swansont]

so what is a photon's relativistic or intertial mass?

 

In a sense they are undefined - science doesn't use the terms anymore.

 

mass is the invariant mass (aka rest mass) - that's the only term you can really discuss, because it's reference-frame independent.

 

for the people that still refer to relativistic mass, it's E/c², but it's not a useful quantity since it's not invariant. So it's not generally used.

 

inertial mass is even more problematic, because if you look at it from point of view of resistance to acceleration (i.e. m=F/a), you have to see that for a massive object moving at a reasonable fraction of c, it's easier to accelerate it perpendicular to the direction of motion than in the direction of motion. Which means inertial mass isn't a scalar, and that's a problem. It's a purely classical phenomenon that loses meaning in QM and relativity.

 

more on the topic.

[Pete]

Inertial mass is the mass in F=ma.

That is incorrect. That is not the correct definition of inertial mass. That is a relationship between force and inertial mass in those cases when the mass is constant. Force is not defined as F = ma. Its defined as F = dp/dt.

 

As far as anybody can measure, this is the same as the gravitational mass in F=GMm/r²

The qauntity F/a does not have the same value as the quantity that might be considered to be "gravitational charge" as is the quantity m in in F=GMm/r². The expression for gravitational force in relativity is not F=GMm/r². Its the m in

 

[tex]G_k = m\Gamma^{\alpha}_{k\beta} v_{\alpha}v^{\beta}[/tex]

 

..relativistic mass is the mass from E = mc² if you use total energy.

That is incorrect. The m in that expression is an equality under certain circumstances. Not all circumstances however. There are cases when E = mc² is invalid. Note: E = mc² is only total energy when the potential energy V is zero.

so what is a photon's relativistic or intertial mass?

For a photon the inertial mass can be expressed in various ways. If the magnitude of the photon's momentum is p then its inertial mass is given by m = p/c. If the photon's frequency is f then since E = hf the inertial mass is m = E/c² = hf/c².

 

In a sense they are undefined ...

That is incorrect. The inertial mass of a photon is a very well defined quantity.

..- science doesn't use the terms anymore.

That is also incorrect. The concept is widely used. For example;

 

Textual Examples ---

Relativity: Special' date=' General and Cosmological,[/i'] Rindler, Oxford Univ., Press, (2001)

Cosmological Physics, John A. Peacock, Cambridge University Press, (1999)

Understanding Relativity: A Simplified Approach to Einstein's Theories, Leo Sartori, University of California Press, (1996)

Basic Relativity, Richard A. Mould, Springer Verlag, (1994)

Introducing Einstein's Relativity, Ray D'Inverno, Oxford Univ. Press, (1992)

Gravitation from the Ground Up, Bernard F. Schutz, Cambridge University Press, (2003)

 

Journal Examples

Apparatus to measure relativistic mass increase, John W. Luetzelschwab, Am. J. Phys. 71(9), 878, Sept. (2003).

Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995).

In defense of relativistic mass, T. R. Sandin, Am. J. Phys. 59(11) 1032 (1991).

A simple relativistic paradox about electrostatic energy, Wolfgang Rindler and Jack Denur, Am. J. Phys. 56(9), Sept. (1988).

An elementary development of mass-energy equivalence, Daniel J. Steck, Frank Rioux, Am. J. Phys. 51(5), May (1983).

A Short Course in General Relativity, Foster & Nightingale, Springer Verlag, (1994).

 

Observed Relativistic Mass Increase for 0.3 eV Electron, G. Gabrielse and H. Dehmelt, Bull., Am. Phys. Soc. 25, 1149 (1980).

 

Online Examples

 

http://web.uniud.it/cird/girepseminar2003/abstracts/pdf/mulaj.pdf

http://physics.syr.edu/courses/PHY106/Slides/PPT/Lec16-Special-Relativity_2.pdf

http://physics.syr.edu/courses/PHY106/Slides/PPT/Lec17-Special-Relativity_2.pdf

http://www.physics.mq.edu.au/~jcresser/Phys378/LectureNotes/SpecialRelativityNotes.pdf

http://astro.wsu.edu/allen/courses/astr150/Einstein.pdf

http://www.phys.virginia.edu/classes/109N/lectures/mass_increase.html

http://www.eas.asu.edu/~holbert/eee460/Relativity.pdf

http://www.ucls.uchicago.edu/Academics/depts/science/physics/Relativity.pdf

http://www.dur.ac.uk/Physics/students/physics_specialrelativity.html

http://www.ph.rhul.ac.uk/course_materials/PH154/Relativistic%2520mass%2520and%2520dynamics.pdf

http://www.lima.ohio-state.edu/physics/113sp03/113Lectures/113rela2.pdf

http://www.lima.ohio-state.edu/physics/113sp03/113Lectures/113rela2.pdf

http://www.physics.fsu.edu/users/ng/courses/phy2054c/hw/Ch26/ch26.pdf

 

http://www.astro.washington.edu/tmurphy/phys110/faqs/AB05.05.html

But the most honest answer to your question is yes--light has mass.

 

Particle Accelerator Labs

 

http://humanresources.web.cern.ch/humanresources/external/training/tech/special/AXEL2003/AXEL-2003_L02_24Feb03pm.pdf

http://www.neutron.anl.gov/hyper-physics/inertia.html

http://aether.lbl.gov/www/classes/p139/animation/sr.html

http://www.fpm.wisc.edu/safety/Radiation/2000%2520Manual/chapter12.pdf

 

 

mass is the invariant mass (aka rest mass)..

That assertion is dependant on how one defines "mass" in the first place.

 

.. - that's the only term you can really discuss, because it's reference-frame independent.

That is incorrect. One can always speak of frame dependant quantities in relativity because those are the only quantities which are measureable. Its not quite accurate to say that relativistic mass is not invariant unless you're quite specific about what you mean by that statement. It can be very tricky. For example: If the 4-momentum of a particle is P and the 4-velocity of an observer is U then the relativistic mass as measured by that observer is proportional to the scalar product of P and U[/b]_obs, i.e.

 

m_obs = P*U_obs

 

In case you think this sort of thing doesn't appear in the relativity literature then I recommend reading

 

Energy Conservation as the Basis of Relativistic Mechanics II, J. Ehlers, W. Rindler, R. Penrose, Am. J. Phys. 33, 995-997 (1965). From page 996

For convenience, we introduce instead of u a new variable, namely the Lorentz factor

 

[tex]\gamma = \gamma(u) = (1-u^2/c^2)^{-1/2}[/tex]

 

and we temporarily write E(u,S) = E[[itex]\gamma[/itex]] when only one scalar state is under discussion. We recall that the 4-velocity associated with the 3-velocity u is given by U = [itex]\gamma[/itex](u)(u,c); hence the 4-velocity of an inertial observer relative to his own rest frame is V = (0,c). Thus

 

[itex]\gamma[/itex] = V*U/c²

 

i.e., the Lorentz factor of a particle relative to any inertial observer is given by the scalar product of the two corresponding 4-velocities divided by c². And this product, being invariant, can be calculated in any inertial frame.

 

It is good to keep in mind that the components of a vector or tensor are defined in terms of the scalar product of the tensor or vector with a basis vector and as such is a scalar.

 

..for the people that still refer to relativistic mass, it's E/c2,

For people who do that they run the risk of making an error in those cases when p/v does not equal E/c². There are cases where the relationship p/v = E/c² does not hold and therefore E = mc² does not hold. A simple example would be the mass of a gas in a box.

 

...but it's not a useful quantity since it's not invariant. So it's not generally used.

Its both useful and as such still used.

 

inertial mass is even more problematic, because if you look at it from point of view of resistance to acceleration (i.e. m=F/a), ..

 

Since that expression is invalid this arguement is flawed. Mass is the proportionality quantity between velocity and momentum. It therefore represents the resistance to changes in momentum, not velocity.

 

 

More on this subject can be found listed in the references here

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

 

and in the journal articles listed here

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

 

Pete

[JaKiri]

Whilst swan's explanation may be technically incorrect, it still illustrates the correct definitions of the terms; I don't think the answers are geared towards physicists.