Dror Posted February 25, 2005 Posted February 25, 2005 If a photon has mass then it can't possibly reach the speed of light, but if a photon is massless then it can't be affected by gravitational fields, and surely in reality, it is affected. The way I see it, if a photon has 0 mass, then not even the mightiest of black holes can affect it's course.. One argument could be that the photon just transports mass from place to place, while not actually having the effects on itself. But, saying that theres no effect on the photon by the mass it carries, is like saying that the photon is not a player in the gravity game. Ofcourse this is a paradox.. And I'm probably missing something. Anyone?
blike Posted February 25, 2005 Posted February 25, 2005 If a photon has mass then it can't possibly reach the speed of light, but if a photon is massless then it can't be affected by gravitational fields, and surely in reality, it is affected.Photons are affected by gravitational fields. Think of Einstein's interpretation of gravity as bent space-time. A photon travelling in a "straight line" through spacetime will follow the curves therein.
swansont Posted February 25, 2005 Posted February 25, 2005 And the amount of curvature is related to the energy density, not just the mass.
Dror Posted February 25, 2005 Author Posted February 25, 2005 The curvuture is the product of gravity.. And gravity is the interaction between masses. If something has no mass why would it be affected by gravity - affected by the "curvuture"? I'm not trying to say einstein was wrong, I'm just trying to understand whats behind this.
Dror Posted February 25, 2005 Author Posted February 25, 2005 swansont now I get it... So gravity affects the 'forces' as well? Does it affect gluons for example, theoretically? From my point of view it does not matter how weak a gravitational influence is on something, as long as it isn't 0.
gamefreek_01 Posted February 26, 2005 Posted February 26, 2005 lol a photon is not mass less it has a mass although being very small
Dror Posted February 26, 2005 Author Posted February 26, 2005 lol a photon is not mass less it has a mass although being very small A direct implication of something having mass, is something not reaching the speed of light ever. I'm repeating my self here.. How do you settle this?
Dror Posted February 26, 2005 Author Posted February 26, 2005 This will probably not be accurate but forgive me as I'm not a physicist nor close to being one, yet. It is no secret that in order to accelerate a mass to c, you need infinite energy. By intuition I think that as infinity is involved, it doesn't really matter how small the mass is.. It could get closer to c at the same energy as oposed to a bigger mass, but can just keep trying.. I think this has something to do with the expodential nature of energy and velocity. As an example: If I drive my car at 20mph and use up 3 units of energy for that (imaginary numbers), then to reach 40mph i must use 9 units. This is probably not new to any of you. So by accelerating you just increase the demand expodentially. As or not as a result, by saying that mass X needs infinte energy to accelerate to the speed of light, X could be any positive number. It doesn't matter how small. Be gentle if I said something really stupid.
gamefreek_01 Posted February 26, 2005 Posted February 26, 2005 ok i dont understand it couldnt take infinte energy because then we wouldnt have any ergy left in the universe because light travel at c and it then must use up infinite energy thus we should run out which we havent so its impossible. But it doesnt matter its very easy photon is mass or else we could see because the light particles interact with the particles of our retina and actual our etire body, but if it had no mass then we couldnt see which we do so it does have a mass
Dror Posted February 26, 2005 Author Posted February 26, 2005 You better do some reading.. In takes infinite energy for a mass to accelerate to light speed, and thats why that wont happen.. Saying that it would take all the energy in the universe is quite silly, as there isn't enough throughout the whole universe. I wouldn't argue with the infinite energy idea as its not my idea.. Furthermore, saying that a massless particle is not entitled to be called a particle, as your saying is ludicrous.. A photon conveys energy, that affects mass. The fact that it conveys extra mass to matter as it hits it does not mean it has mass in itself. The fact that a photon travels at light speed ensures that it's mass is 0. The wierd thing for me WAS that as a massless particle, its still evidantly affected by gravitational fields. But as stated earlier by swansont, gravity affects energy as well.
Sayonara Posted February 26, 2005 Posted February 26, 2005 If you don't understand, stop trying to tell people how it works.
Dror Posted February 26, 2005 Author Posted February 26, 2005 sayonara, the only thing I didn't understand was how come a massless particle is affected by gravity. swanston cleared it out by saying that gravity affects both matter and energy. End of story.
Dror Posted February 26, 2005 Author Posted February 26, 2005 Oh, hehe.. No harm done then. Good thing I didn't flame you..
Deified Posted February 26, 2005 Posted February 26, 2005 Maybe I'm misunderstanding, but it seems that you are saying that photons convey energy. Photons have no rest mass, but they are NEVER actually massless. I hope you are familiar with E^2=(c^2)(p^2)+(m^2)(c^4). (when p=0 it reduces to E=mc^2). This equation works to illustrate both of my points, firstly, photons NEVER slow down. They always go at c. So if p=c in the above equation, then we have a substantial amount of energy that is ALWAYS present in the photon. The second point is, Einstein showed that mass can be converted into energy and the other way round too. So we often refer to this as mass-energy since they are basically the same thing. So photons can interact with our eyes and they are affected by gravity because they have mass-energy, ALWAYS. If I made any inaccurate statements PLEASE correct me, I'm learning this stuff too.
Dror Posted February 26, 2005 Author Posted February 26, 2005 I fully agree with your first point. I'm having a bit ouf trouble with the second point though. Yes, matter can be converted to energy at extreme conditions, and the same holds for the other way around. Furthermore, even at "non extreme" conditions matter and energy can interact, but with gravity being the middle man for the APPEARENTLY mass related consequences. But after the fact, at our universe as it is, matter isn't energy and energy isn't matter. The confusion can arise from the fact that both matter and energy convey gravitational fields. However this does not mean that energy has mass, it only means that as energy interacts with matter, theres a slight increase to the matters gravitational field. This is firmly backed up by the undeniable fact that no mass whatsoever regardless of how tiny can accelerate to c in this universe's life time. Infinite energy is required for that to happen. And as a result of this fact and the fact that light travels at c - photons are massless. And by massless I mean absolute 0.
timo Posted February 26, 2005 Posted February 26, 2005 The path of a (sufficiently small) particle that is affected by a gravitational field is independent of it´s mass. That´s true in both Relativity and Newtonian gravity. In Newtonian gravity: [math] m \vec a = \vec F = m \vec g \, \Rightarrow \, \vec a = \vec g[/math] g of course describes the gravitational field. Whether this equation is still true for m=0 is not completely certain from this (there are arguments for both views) but it´s not important as systems with a relevant light deviation are relativistic anyways. In General Relativity: [math] \forall i : \, a^i = \sum_{jk} -\Gamma^i_{jk} v^j v^k [/math] Even without fully understanding the relativistic movement equation my point should become clear: The Gamma ([math] \Gamma [/math]) is the analogy to g in the Newtonian equation. It describes the gravitational field. The mass of the particle affected does not appear in the equation. In short: Movement in a gravitational field is independent of mass.
swansont Posted February 26, 2005 Posted February 26, 2005 Maybe I'm misunderstanding' date=' but it seems that you are saying that photons convey energy. Photons have no rest mass, but they are NEVER actually massless. I hope you are familiar with E^2=(c^2)(p^2)+(m^2)(c^4). (when p=0 it reduces to E=mc^2). This equation works to illustrate both of my points, firstly, photons NEVER slow down. They always go at c. So if [b']p=c[/b] in the above equation, then we have a substantial amount of energy that is ALWAYS present in the photon. The second point is, Einstein showed that mass can be converted into energy and the other way round too. So we often refer to this as mass-energy since they are basically the same thing. So photons can interact with our eyes and they are affected by gravity because they have mass-energy, ALWAYS. If I made any inaccurate statements PLEASE correct me, I'm learning this stuff too. emphasis added - was this a typo? p = c has the wrong units. "mass-energy" is usually referring to the amount mc2, where m is the rest mass. The mass-energy of a photon is zero. Photons are affected by the curvature because they follow the path of least time - in curved space, the curved path is the shortest distance between the two points, so it "seems straight" in the context of that geometry. Think about lines of longitude - they are the shortest path from north pole to south on the curved surface of the earth, but they are not straight lines.
Dror Posted February 26, 2005 Author Posted February 26, 2005 Surely i will not argue with this. You obviously know what your saying. But it still strikes me as odd.. Can you maybe try and explain it non mathematically? This message was meant for atheist.
Dror Posted February 26, 2005 Author Posted February 26, 2005 Photons are affected by the curvature because they follow the path of least time - in curved space, the curved path is the shortest distance between the two points, so it "seems straight" in the context of that geometry. Think about lines of longitude - they are the shortest path from north pole to south on the curved surface of the earth, but they are not straight lines. So theres virtually no difference between why matter is affected by curvature and why energy is affected by curavture. Both convey gravitational fields themselfs. What I don't understand as of your remarks is if theoretically there was a particle that conveys 0 gravity, why would it be affected by other gravity? The way I understand it curved space is gravitatinal attraction. And a gravitational field/curved space regardless of how huge, cannot affect an object with 0 gravity.. (Not talking about photons). So curved space is just an abstraction?
5614 Posted February 26, 2005 Posted February 26, 2005 two things: 1) photons can have mass, the term 'mass' is not specific, there are different types of mass, we know that photons have energy, and that e=mc^2 now if an electron had 0 mass it'd have no energy, so in this sense photons do have mass, that is the relativistic mass... photons have 0 rest mass. 2) photons dont travel in straight lines, they take the shortest distance between two points, so if space-time is curved by gravitational fields then photons will travel that 'curved' path.
Deified Posted February 26, 2005 Posted February 26, 2005 Swansont, yes "p=c" was a typo. I understand mass-energy (as in the law of conservation of mass-energy) as being energy or mass, because they are different forms of the same thing. So if I say that a photon has mass-energy greater than 0, I don't see why this is inaccurate. It has 0 mass + a substantial amount of kinetic energy. The simplification E=mc^2 doesn't apply to photons. Here is another question, when a particle is accelerated in a particle accelerator it is said that it "gains mass" as it gains speed. Do they mean that literally ( it actually increases in size ) or in terms of mass-energy (it gains kinetic energy, and therefore indirectly gains mass). Thanks for helping me out, I'm pretty new to this stuff.
swansont Posted February 26, 2005 Posted February 26, 2005 Swansont' date=' yes "p=c" was a typo. I understand mass-energy (as in the law of conservation of mass-energy) as being energy or mass, because they are different forms of the same thing. So if I say that a photon has mass-energy greater than 0, I don't see why this is inaccurate. It has 0 mass + a substantial amount of kinetic energy. The simplification E=mc^2 doesn't apply to photons. Here is another question, when a particle is accelerated in a particle accelerator it is said that it "gains mass" as it gains speed. Do they mean that literally ( it actually increases in size ) or in terms of mass-energy (it gains kinetic energy, and therefore indirectly gains mass). Thanks for helping me out, I'm pretty new to this stuff.[/quote'] It's inaccurate in that it is a non-standard application of the term. Those that talk about a particle increasing its mass are getting into a hazy area and are using relativistic mass, from E=mc2, which can be confusing if you aren't careful about defining your terms. The momentum of the object is p = [math]\gamma[/math]mv. You have to want to use p=mv, and assign the Lorentz term specifically to the mass, to make the concept work. You'll get consistent results if you do it, so it's not wrong, per se - the math works. But the (relativistic) mass now depends on the frame in which you measure it - and that can be a problem. If you define C-12 to have a mass of 12 amu, you now have to define that to be at rest with respect with your frame of reference. Since having an invariant term like rest mass is useful, that's what is usually used. Note that in the objects own frame, it never sees the mass increase, since it always sees itself at rest. This is not related to an object changing size, though length contraction is a relativistic effect.
Deified Posted February 26, 2005 Posted February 26, 2005 It's inaccurate in that it is a non-standard application of the term. Fair enough. Those that talk about a particle increasing its mass are getting into a hazy area and are using relativistic mass' date=' from E=mc[sup']2[/sup], which can be confusing if you aren't careful about defining your terms. So what exactly is relativistic mass?
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