So you're saying that not even a black hole, essentially the most powerful thing in the universe, can affect a photon's speed.....
(brain...........hurting............ow)
So you're saying that not even a black hole, essentially the most powerful thing in the universe, can affect a photon's speed.....
(brain...........hurting............ow)
The speed of light C can depend upon the geometry of the local space, thoughOriginally Posted by Wina
Well actually Photons travel at C regardless, they appear to travel slower in some mediums because they get absorbed and retransmitted, so you can't really slow one down. They either are travelling at C, or they don't exist. Pick one and only one state at any one point in time.
edited to add: Just noticed the second page. Photons always travel at C, even in a black hole. The high gravity inside of the black hole distorts space and time, meaning that time slows down. As far as the photon is concerned it's travelling at the speed of light, but to an observer outside of the BH, it would appear to slow down and stop. It's all relative.
By definition, you can't observe a photon from outside a black hole though....... lol gotcha!
Well in reality you can't observe a photon in -or- out of a Black Hole, so......Originally Posted by Wina
If energy and mass are equivalent. Then the photon does have mass. The energy of a single photon is 6.62x 10(-34) Joules or watt/seconds.
I hope I have posted in the correct forum.
Sorry if I haven't.
howard2
Light doesnot have mass and therefore we say that to achieve the speed of light we have to make mass of an object approach zero.
Energy and mass are equivalent, but they are not the same.Originally Posted by howard2
Planck's constant gives you the conversion from frequency to energy, but is not the energy of a single photon.
energy doesn't have to equal mass. Use the full equation for e=mc^2.
E^2=sqrt(p^2*c^2+m^2*c^4).
Energy is rest mass and momentum. notice, for a stationary object (p=0) it reduces to e=mc^2. For a massless particle (m=0), all the energy is in the momentum (p), unless p=0...in which case there is no energy, so why are you using the equation at all?
Light falls into the second category, stationary masses into the first, moving masses uses the whole thing.
Light dont have mass. Light is composed of photons right? So we could ask if the photon has mass. Well, to my understanding the answer is definitely no. The photon is a massless particle and according to the theory it has energy and momentum but no mass. And this has been confirmed by experiment to within strict limits.
Titana.
Uh,.... shouldn't either the ^2 on the left or the sqrt on the right be left off?Originally Posted by Ricimer
Originally Posted by Tensor
E=pc+mc^2
E^2=(pc)^2+(mc^2)^2
E^2=sqrt(p^2*c^2+m^2*c^4)
No, that's right.
Wanna try again, a bit more slowly?Originally Posted by G O R T
E = pc + mc^2....Are you kidding me?
sqrt(a^2 + b^2) IS NOT EQUAL TO (a+b)
Let us begin with...
E^2 = p^2*c^2 + m^2*c^4 (eq. 1)
Done, now go get yourselves a coffee.
edit: I can't stop chuckling.
Established Member
Dang it Alan, he was supposed to work it out.Originally Posted by alainprice
Thought experiment:
Take two perfectly straight, non-diverging lasers and fire them into the flattest available space in perfectly parllel beams.
If light has no mass, the beams would remain parallel. If light has mass, the beams would be drawn together.
The equivalence principle says that everything is equally affected by gravity. But does light's momentum count for anything? Would it curve space, causing the two beams to converge?
oops, my bad. I often use one form or the other, and I guess I use both this time.
it is either E^2 or sqrt.
The whole issue here is, does one mean rest mass, or energy-equivalent mass (i.e., E/c^2). Note that gravity comes from energy (and pressure), not just rest mass, so it is the latter meaning that is relevant to gravity. So yes, the light beams converge to the extent that you crank up the energy of the beams. If you treat the light as a test particle with essentially zero energy just to probe the background spacetime, then the convergence would be neglected.Originally Posted by mugaliens
Isn't the most powerful thing a first generation star going supernova? Either that or a supermassive black hole feeding.Originally Posted by Wina
Helloo... have we forgotten the cosmos that soon? as far as I am concerened the cosmos is made up of two parts. Yep.. just two part. the first part of the cosmos is "MASS", the second part of the cosmos is "ENERGY & MOMENTUM". Energy needs a medium to be propagated. That means that Light is the mass through which "LIGHT ENERGY" is propagated. Energy cant travel in a vacuum; so what it does is that it generates light, and through light energy can be successfully transfered in a vacuum. That is why the suns energy reach us.
That's not even roughly right, Ruffly.
Everyone is entitled to his own opinion, but not his own facts.
The cosmos does consist of two parts (at least you got that right). The rest isn't even close.
Space-time and mass-energy. Everything else is details.
There is a trick to this “always c” business.Originally Posted by Wina
As the Einstein theory goes, an atomic clock slows down its tick rate inside a gravity field and light slows down its speed in the same gravity field. When both happen, the atomic clock in the gravity field still “measures” the speed of light to be “c”, even though the light speed is slower in the gravity field than outside it. It is “measured” to be “c” because the atomic clock inside the gravity field has slowed down it’s tick rate.
Of course light slows down in a gravity field, but you’ll need to judge that slowdown by an atomic clock that is located outside of a gravity field.
Light is massless.
"We believe so. The special theory of relatvity says that the energy of an object with mass traveling at the speed of light would be infinite." - Fortunate
For something to travel at the speed of light it must be massless, and since light travels at the speed of light it has no mass (as far as we know).
Note that this is an unusual interpretation of special relativity, because SR is basically a set of perfectly natural measurement conventions and their observed ramifications. One of those conventions is that if you want to measure time, you should be able to use your own clock, not refer back to someone else's clock at some distant location. Thus it is with this interpretation of the meaning of time that we say light does not slow down in a gravity field, but from someone else's perspective who is not in the gravity field, it might be said to, although I think if you also use that distant person's sense of distance, you will again reach the same c. The bottom line is, one must avoid absolute statements of reality, and instead be clear on one's measurement conventions, or just use the agreed-on conventions of SR.Originally Posted by Sam5
General relativity?Originally Posted by Ken GHow do you do that, and still get results like gravitational lensing?I think if you also use that distant person's sense of distance, you will again reach the same c.
Yeah, that's what I meant, you're right.Originally Posted by hhEb09'1
Gravitational lensing need not have a change in lightspeed if it monkeys with the geometry of space itself. I'm no expert, but I think you can always take two different perspectives, which differ in how you are coordinatizing things (after all, coordinates are arbitrary, and so are the words you use to describe what is happening). One is, you treat space as normal but alter the speed of light. That is not the standard coordinatization, however, because it does not correspond to the way space would be locally measured, and as I said, relativity is the ramifications of natural measurement conventions. I think the standard approach actually bends straight lines, without altering the speed of light at all. Note the difference to what is happening to light in a glass lens-- there the geometry of a straight line is not altered, but the speed of light is. So we see that the phrase "gravitational lensing" is probably a misnomer that may cause more confusion than it solves.Originally Posted by hhEb09'1
Alright, I'm going to try to get back on topic and actually give answers to this guy's questions.Originally Posted by TwAgIssmuDe
The short answer is that photons have a rest mass of zero, but they also carry energy, and, as a consquence, momentum. The force of the black hole is able to change the energy and momentum of the photon, so that a photon trying to escape a black hole, while actually travelling away from it at the speed of light, gets redshifted so much by the time it escapes that it has zero energy, which means there is no photon left. The speed of the photon is never changed, but its energy is, because it takes more energy to escape the black hole's gravitational pull than the photon has.
Here's the long answer: As I just mentioned, the two properties we are most interested in of a photon are its energy and its momentum. Both of these are directly proportional to mass, so you might think that since a photon has zero mass that it has zero momentum and energy too. That isn't the case, because it only has zero rest mass. The mass used to calculate momentum and energy is equal to m=γ*m0 where γ(gamma) is the Lorentz factor, or 1/SQRT(1-v^2/c^2) and m0 is the rest mass. A photon always travels at the velocity of light, c, and has a rest mass of zero so that means we get m = 0 * infinity. 0*infinity could be anything though, so that isn't very useful to us, and it turns out that not only do we not need the mass to find the energy and momentum, but actually that you need the energy or momentum to find the mass.
Lucky for us though, there are other ways to find the energy and momentum of a photon. We know, for example, that the energy of a photon is E = hf = hc/λ (where h is planck's constant, f is the frequency of the photon and λ is its wavelength). We also know, from einsteins famous equation: E=mc^2=Sqrt(m0^2*c^4+p^2*c^2) (which, when v<<c, making the second term negligible, reduces to E=m0c^2) since m0=0, the first term is 0, leaving us with E=pc=hc/λ, or E/c=p=h/λ
So for energy, we have E=hc/λ
And for momentum, we have p=h/λ
And the photon's mass, you could find it from either of those: E=mc^2=hc/λ, m=h/λc; p=mc=h/λ, m=h/λc. From now on, I'll be stating everything in terms of E, the energy of the photon though, so simply:
p=E/c
m=E/c^2
But we also know that c is constant, and that all photons travel at c, and can't be slowed down*. So how does a black hole trap a photon? Well, since the photon has a mass while it is moving, which we have just calculated, the black hole also exhibits a force on that photon: F=GMbMγ/r^2 (Mb is the mass of the black hole, Mγ is the mass of the photon, G is the universal gravitational constant and r is the distance from the center of the black hole to the photon). So if the photon is moving away from the black hole, then the hole does a negative amount of work on it equal to W=-FΔr, but since F changes with r, we have to do this in infinitesimal steps and use calculus, so the real equation is dW=-Fdr=-GMbMγdr/r^2. So, say we want the photon to move from a distance r0 to a point where the black hole has no gravitational influence on it at all (infinity). We just integrate this equation from r=r0 to infinity: W=GMbMγ/(infinity)-GMbMγ/r0=0-GMbMγ/r0=-GMbMγ/r0. Now, if we set that value equal to the energy of the photon, then it will take ALL of the energy of the photon to escape from the black hole from that radius r0. If we solve for that radius r0 where W=E, then that will be the radius of the event horizon of the black hole: E=GMbMγ/r0=GMbE/r0c^2, 1=GMb/r0c^2; r0=GMb/c^2. So at any value of r0 smaller than GMb/c^2, it will take more energy than a photon has to escape that black hole, which means it simply can't. And that radius, r0=GMb/c^2 is the radius of the event horizon.
Also keep in mind that a black hole is simply any object with a density large enough that its event horizon has a larger radius than the object itself. At small distances, the density required for this would be enormous (An event horizon of 1Å would require a mass of 1.35*10^17kg, the equivalent of an asteroid 50km across) But since r0 scales with Mb, which itself scales with Rb^3, doubling the radius of an object while maintaining the same density effectively octuples the radius of its event horizon.
Still, I disagree with your comment that Sam5's comment is an unusual interpretation. Notice that you even use the word "normal" in describing his point of view!Originally Posted by Ken G
Also, as you point out, the term "gravitational lensing" is slightly at odds with what you say is the usual interpretation--and it corresponds to Sam5's notion. Gravitational lensing is a very commonly used term, so I doubt you can justify saying that it is an unusual interpretation. In fact, it is probably the common one outside of the physics community, and not uncommon within.
I should also point out, as I meant to in my first post, that the photons of light itself do not actually slow down in media other than a vacuum, they simply encounter so many obstacles and bounce around so much that they appear to be travelling more slowly. Light always travels at c.
What occurs with gravitational lensing is sort of the same thing as what I just described with a photon trying to escape an object, except that this time, the light is actually passing by the object, which exerts not only a force parallel to the light as we discussed before, but also a perpendicular one. This perpendicular force is able to change the direction of the momentum of the photon (Fdt=dp, remembering that F and dp are both vectors). This requires absolutely no modification of the shape of spacetime to accomplish, and is simply light being bent by its gravitational attraction to ab object. It also happens that, for black holes, the radius at which the force is big enough to trap the photon in a circular orbit is the same as the event horizon radius we calculated earlier.