i'm just wondering, does light have weight and mass? if so, what is the weight and mass of light?
photons dont have a rest mass, they do however have momentum, [tex]E=hf=pc[/tex]
where f is the freq of the light, h is plank's constant, p is the momentum and c is the speed of light.
i dont know GR so well, but there is an effect of space bending near mass
which makes it look as if light is attracted by gravity... someone else here might know more and elaborate on this...
i also know of an experiment which showd that if you emmit light upwards from earth, it undergoes red-shift as if it had to invest energy for the potential energy of graviation it gained.
Light doesn't have rest mass (it can only travel at c) but it has energy and GR says that energy and mass curves and warps spacetime; although a photon would only curve it very very slightly i guess you could say that it does have some kind of mass ONLY in the sense that it curves the fabric of spacetime other than that it has no rest mass
i dunno GR super well either, but one of the "equivalence" thought experiments of Einstein was that of a person standing in a spaceship that is accelerating in the direction that his/her head is pointing by 9.8 m/s2 (the acceleration of gravity of the earth), then, according to the equivalence hypothesis, that person cannot tell the difference between that and standing on the earth's surface. now if a beam of light came in through a window of the accelerating space ship, even though that beam is really travelling a straight line, it would appear to the occupant standing in the space ship to be curving "downward" (toward his/her feet) because he/she is accelerating "upwardly". now if the equivalence principle is true, than the corresponding observer standing on the earth's surface would see the same downward curvature of a corresponding beam of light implying that gravity is acting on it as if it has weight.
I considered this question quite a bit earlier. I could probably dig up my math somewhere. My idea was to (theoretically) get a perfectly reflecting box and trap the light in there so I could try to weigh it. I worked out the math based on the momentum of the photons and found that it had extra inertia and weight exactly as though it were filled with a gas of the mass you would expect from E = mc^2.
Ok except for one minor detail. Use two beams coming in at different altitudes in the space ship. Since gravity changes with distance from the center of mass, if's it gravity affecting the beams, then one of them bends more than the other (may nead a really "tall" space ship to dectect this, but I'm talking hypothetical with infinitely accurate measurements). If it's acceleration, then both beams bend at the same rate.
I'm not so convinced about equivalency either. Light may bend around a massive object simply because space is warped. It's also possible that a strong gravtiational field attracts some form of dark matter or energy (stuff we can't detect) that simply refracts the light.
The other issue with photons, is that for them, the speed of light is not constant from their frame of reference. Each photon "sees" it's own speed of light as zero, but "sees" other photons from other beams as non-zero.
this accelerating space ship is out in the middle of nowhere and we're ignoring the graviational attraction of the space ship itself because its mass so much less than the mass of a planet.
yes, both beams bend the same amount.
that's true . that is what GR says. in both cases the beam of light is in reality perfectly straight but appears to the observers be slightly bent because, in the spaceship case, the observer is in an accelerated frame of reference and in the earth's surface case, there is warping of spacetime because of the great mass the observer is in the vicinity of.
that's non sequitur, as far as i can tell.
but it's not, I'm referring to the case of a very tall space ship resting on a large planet. The beams won't bend by the same amount in this case.
You wouldn't even need light. I could just weigh a known weight and see if the weight decreased with altitude within the spaceship.
this is literally what we mean by the term "non sequitur". i was the one bringing up a concept of an observer in a space ship and i was talking about the thought experiment, i believe originally attributed to Einstein, that led him to the "Equivalence Principle". your counter example is non sequitur. it is not what the conversation is about.
Yeah, but my point is that given accurate enough instruments, you can tell the difference between gravity and acceleration. I don't like the space ship analogy because there is a way to detect the difference in this case, even with single beam. Just coat the interior with mirros and the rate of bend changes as the beam move towards or away from a gravity source.
Equivalency should be explained as follows: that an infinitely thin beam of light bends the same amount in a gravity field as it would if observed from an accelerating reference point where the acceleration would generate the same force on objects as the gravitaional field. If the beam isn't inifitely thin, then the difference in strength of the gravitational field would spread the beam a bit.
This also ignores the fact that during acceleration, time and distance dialation occurs, and this would affect the results as well.
Probably just easier to state that light is affected by gravity by the same amount that mass would be if the mass could travel at the speed of light.
then your point is wrong. it is contrary to one of the basis principles of the Theory of General Relativity.
nowhere in this EP is the assumption of a varying graviational field (such as inverse-square). it's a perfectly uniform graviational field.
time and distance dialation occur because of a velocity difference between two different observers. what you're saying here is still non sequitur. it is not what we are talking about. or at least what i'm talking about and the subject i am talking about is the Equivalence Principle that GR is built on.
yes, but not a zero mass. if light had zero mass, there would be no bending in the presence of a gravitational field.
I don't even need accurate instruments. I can see if I'm standing on the Earth or out in a spaceship accelerating. It's obvious you can tell the difference. I don't think that's the point here. The point is that light is affected by gravitational fields.
Now if you're referring to the general concept of equivalency, which doesn't involve light at all. Then the effects
of on a mass from a gravitation source is the equivalent of acceleration. But still there's the issue that gravitation
fields vary in strenth depending on distance from the center of the source (and assuming your outside the surface
of the source).
Ok I sit corrected, but such fixed gravitation fields don't exist in real life.
Would the gravity field from an ifinitely large flat plane do the job?
if the acceleration in space was the same as the graviational force divided by the mass of the "stationary" planet-bound observer (which has the same units/dimension of acceleration), you may think it's obvious you can tell the difference, but, if you cannot see the stars moving outside your spacecraft, there is no way you can measure the difference. no matter how sensitive your instruments are or how obvious you think it would be.
no, that's wrong.
what if your source of gravity is an infinite plane of mass? would the strength of gravity be different for different distances from that infinite plane?
Yes, from a point source, it's 1/r^2, from an infinitely long line it's 1/r, and from an infinitely wide plane it's constant.
Even so, I can see if I have motion relative to other objects, like the stars (except you took away my windows).
but that's the point of the thought experiment that leads to the EP. the other half of that thought experiement is comparing the observer in a spaceship out in the middle of nowhere and not accelerating at all to the observer in an elevator on earth that is accelerating downward at a rate of g (or in the so-called "vomit comet"). both observers think they are weightless and observe a beam of light to be perfectly straight.
Ok, I get your point, but to the Apollo 10 crew, approaching the earth at 24,861 mph, it was pretty clear to them that they were rapidly approaching the earth. They stated that they could see the rate of apparent increase in size of the earth as they got close. Unlike the shuttle, this is pretty close to escape velocity and if they messed up on aiming for re-entry, it was going to be a very long and lonely ride.
Since, by definition, mass is defined as the ratio of momentum to speed it follows that since light has momentum and speed it therefore must have a non-zero mass. However the proper mass (intrinsic mass) is zero. The force of gravity acts of light if that's what you were wondering. To measure the weight of something you must be able to support the object in a gravitational field. The force required to support the object is called the
"weight" of the object. You can't support a beam of light, only a gas of photons in a box may be weighed. The result is the weight of the box + the weight of the gas.
Jeff, your ability to jump from one topic to another one completely unrelated is remarkable. this is exactly what is meant by the term "non sequitur". i have no idea what you mean by the use of the word "but".
Back to the original question. Light has no rest mass, hence no weight. Light has both wave and a particle character. As as wave it's hard to imagine any rest mass, and as a photon phenomena rest mass also appears to be zero. E=hf=pc is the only quantification I am aware of, but I'm an amateur. Space curving near a mass and light following the usual shortest distance as it travels (the curve here) is Einstein's alternate explanation to light having momentum (p= mv) and being bent by gravity. We do know that when light passes a mass ( like the sun) it does bend. Of course nobody knows exactly what gravity is nor negative gravity (Einstein's cosmological constant) and that makes a true explanation and understanding murkier. So we have to make do with some math to make predictions.
how did you draw that conclusion? is light "at rest" in any reference frame?
Well a photon is at rest relative to itself.
Photons only exist at the speed of light.
One thing I've been curious about, Assuming that process of electrons capturing or releasing photons isn't instaneous, what happens during the transition? Assuming that the photon emerges at the speed of light, is it emerging at the speed of light from the electron's frame of reference? I ask this because of the way lasers work. Photons passing through high energy molecules cause the eletrons to release photons with the same phase and direction, and I'm wondering how this release happens relative to the orbital path of the electron at the during the release period. The photons direction is always the same, but what about the eletron's path and speed during the moment of release?
well, i'm not sure that this would be a legitimate frame of reference. why? well you have answered that:
and so has Einstein:
(or google "riding a beam of light" along with "Einstein" and pick the web story of your preference.)
A good answer is http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html" [Broken].
Just to provoke:
what is the meaning attatched to the quantity
m = \hbar \nu / (c^2)
Separate names with a comma.