Conversation with Tom Bailey about how a photon can have momentum even though it has zero mass 9 September 2012 at 17:57

Transcription

1 Conversation with Tom Bailey about how a photon can have momentum even though it has zero mass 9 September 2012 at 17:57 Only me Tom Bailey Could Planck constant be seen as a minimum possible mass if E=MC2? 5 hours agochristian Salas Hi. The minimum possible mass is zero...e.g. photons have zero mass. Some theoretical people suggest that there might be things with 'negative mass' in the universe (though this is not proved yet) in which case there would be no minimum mass as such. So my answer to you would incline towards 'no'... 2 hours ago Tom Bailey It just confuses me how something can have momentum but no mass 2 hours ago Tom Bailey A photon has no rest mass. But a photon is never at rest. E=mc2 is for particles E=hv is for waves So E=mc2=hv No rest mass, but an effective mass on account of its energy. 2 hours ago Tom Bailey And in 2003 Luo et al gave the upper limit for the photon mass of 1.2x10-54 kg through a rotating torsion balance experiment 59 minutes agochristian Salas Hi again. You are saying lots of confused things above. You really need to study special relativity properly in order to fully understand this. First, the equation E = mc^2 is the 'rest energy' or 'mass energy' of an object. It is the energy that an object has due to having a mass m. 57 minutes ago Tom Bailey

2 People have tried to explain special relativity to me. Waves don't have mass, so couldn't every subatomic particle be seen as having no mass when we aren't observing them 56 minutes agochristian Salas Conversely, the mass m of an object is defined as E/c^2 which is the energy the object has in a frame of reference in which the object is at rest. Just listen to what I am saying for a minute...try to grasp what I'm saying first...then reply. I will try to explain... So the equation E = mc^2 only applies to a particle at rest in a particular frame of reference. The m in this equation is the only 'mass' that there is. 51 minutes agochristian Salas So my first point is that you cannot write E = mc^2 = hv as you did above. This equation doesn't make sense because the mc^2 bit is the 'rest energy' of a particle. 48 minutes agochristian Salas I'll explain to you now how a photon can have momentum even though it has mass m = 0. You will hopefully understand it if you read this next bit minutes agochristian Salas In Newtonian mechanics, momentum is p = mv But when you are talking about things at speeds close to the speed of light, the equation p = mv gives incorrect results (it violates the 'law of conservation of momentum'). 42 minutes agochristian Salas The correct equation for momentum in all cases (whether close to the speed of light or not) is the relativistic equation p = γmv where γ is called the 'Lorentz factor' and is given by the equation γ = 1/sqrt(1-(v/c)^2) 39 minutes agochristian Salas

3 Similarly, the correct relativistic equation for the TOTAL energy of an object is E = γmc^2 This total energy consists of the 'rest energy' mc^2 which I mentioned above, plus the 'kinetic energy' which is the energy the object has due to its motion. 38 minutes agochristian Salas Now, here is the crucial bit. The Lorentz factor in the above relativistic equations gets bigger and bigger as the speed v gets closer and closer to the speed of light c. And when the speed v actually equals the speed of light c, the Lorentz factor γ becomes 'infinite'. 35 minutes agochristian Salas This is why it is impossible for any object with a non-zero mass m to ever reach the speed of light. The equation E = γmc^2 shows that the energy of such an object would have to be infinite (which is impossible) since the Lorentz factor on the right is infinite when the speed v equals the speed of light. 32 minutes agochristian Salas Similarly, the momentum p = γmv would become infinite if an object of non-zero mass m ever reached the speed of light. Again, this is impossible. 31 minutes agochristian Salas But these two relativsitc equations DO NOT go to infinity if we reduce m towards zero 'fast enough' as we let the speed v get closer and closer to the speed of light c. 28 minutes agochristian Salas So in the equation for total energy E = γmc^2 what happens as the speed v gets closer and closer to the speed of light c is that γ is heading to infinity, but this is being COUNTERBALANCED by making the mass m smaller and smaller. 25 minutes agochristian Salas Similarly, in the equation for relativistic momentum p = γmv as the speed v gets closer and closer to the speed of light c, the Lorentz factor γ is getting bigger and bigger towards infinity, but this is being COUNTERBALANCED by making the mass m smaller and smaller towards zero. 23 minutes agochristian Salas

4 In the limit, when v actually reaches the speed of light c, and at the SAME TIME the mass m has fallen to zero, the expression for the total energy becomes E = pc where p is the momentum given by p = hf/c These are the energy and the momentum of a photon with frequency f. 21 minutes agochristian Salas So the momentum of a photon is what you get when you let the speed v get closer and closer to the speed of light c while AT THE SAME TIME making the mass m get closer and closer to zero. In the 'limit' of this process, you end up with the momentum p = hf/c of a photon, and with the photon energy given by E = pc. 20 minutes agochristian Salas For any other object with NON-ZERO mass m, both its energy E and its momentum p would go to infinity as the object's speed v got closer and closer to the speed of light c. So this is why a photon's mass MUST be zero (at least in traditional relativistic physics), and the above explains how the photon has momentum even though its mass m is zero. END OF MESSAGE... :) 17 minutes ago Tom Bailey Holy crap That was amazing 15 minutes agochristian Salas But do you get it? 14 minutes ago Tom Bailey I think so 14 minutes agochristian Salas The momentum of a photon is a 'limiting momentum' which you get when you let the mass m get closer and closer to zero AT THE SAME TIME that you are letting the speed v get closer and closer to c.

5 13 minutes ago Tom Bailey M must be zero to stop the Lorentz factor from rising (rising proportionally into infinity as E rises) into infinity 13 minutes agochristian Salas In the limit, which is the situation for a photon, you have mass m = 0, speed v = c, and then the limiting momentum of this object MUST be p = hf/c 13 minutes agochristian Salas Yep...I think you are getting it. 12 minutes agochristian Salas For any other object which has NON-ZERO mass m, the momentum MUST be p = γmv and this will ALWAYS go to infinity as v gets closer to c. 11 minutes ago Tom Bailey I'm pretty sure I get it Because of the lorentz factor rising infinitely 11 minutes agochristian Salas It is only for the photon and any other massless particles, which have ZERO mass m, that we can get a non-infinite momentum (and a non-infinite energy) when speed v equals c. Yep minutes ago Tom Bailey If it was infinity then it would be a nonsense equation since you can never have an infinite amount of energy? So must therefore either be slower than the speed of light, or have no mass 5 minutes agochristian Salas exactly...

6 3 minutes ago Tom Bailey I managed to scribble down everything you said Will read it until I memorise it Anyway, thanks a bunch Really excellent explanation Supplement 1 to the above conversation. I will show here the derivation of the equations E = mc^2 for the rest energy of an object, and E = pc for the energy of a photon, from the relativistic momentum and energy equations p = γmv and E = γmc^2. First, starting with p = γmv we can solve for v^2 as follows: Now we use this to eliminate v^2 from the equation for E^2 as follows:

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8 For an object at rest we have zero momentum, and setting p = 0 in the above we get E = mc^2.

9 Supplement 2 to the above conversation. Here is a simple derivation of the relativistic momentum p = γmv. The time dilation formula referred to is The simple derivation is then as follows:

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