Tres Juicy Posted January 23, 2012 Share Posted January 23, 2012 Hi all, Just contemplating the speed of light... If a photon travels in direction A and is forced to emit another photon also in direction A. Then from the point of veiw of the first photon, the second photon would accelerate away from it at c. Does that mean the second photon is now travelling at 2c? Link to comment Share on other sites More sharing options...
swansont Posted January 23, 2012 Share Posted January 23, 2012 A photon is not in an inertial frame. We can't describe what is happening from a photon's point of view. In any event, photons don't emit photons. The two ideas here may or may not be related. Link to comment Share on other sites More sharing options...
Tres Juicy Posted January 23, 2012 Author Share Posted January 23, 2012 A photon is not in an inertial frame. We can't describe what is happening from a photon's point of view. In any event, photons don't emit photons. The two ideas here may or may not be related. Sorry, the emition of another photon was just to describe the scenario. Let me phrase it differently: If I was driving at the speed of light with my headlights on, from my point of veiw the light travels away from me at c but in reallity in order for that to happen it must travel at 2c Link to comment Share on other sites More sharing options...
D H Posted January 23, 2012 Share Posted January 23, 2012 Let me phrase it differently: If I was driving at the speed of light with my headlights on, from my point of veiw the light travels away from me at c but in reallity in order for that to happen it must travel at 2c The rest frame of a photon doesn't make sense. This is being discussed in another active thread, http://www.scienceforums.net/topic/63263-why-is-the-speed-of-light/ (post 18 and on). You are assuming velocities add as 1+1=2. Suppose you're driving along at 60 miles per hour and a motorcycle zooms past you, going 20 mph faster than you. How fast is the motorcycle going? Simple: 60+20=80. That isn't how velocities add in relativity. Suppose you are in a spaceship going at 3/4 the speed of light relative to some observer. Another spaceship going in the same direction zooms past you at 3/4 of the speed of light relative to your spaceship. Naively adding these velocities suggests that observer who sees you moving at 3/4 c will see that other spaceship as moving at 1.5 times the speed of light. That's not how it works. That observer will see that other spaceship as moving at 24/25 c. Link to comment Share on other sites More sharing options...
Tres Juicy Posted January 23, 2012 Author Share Posted January 23, 2012 The rest frame of a photon doesn't make sense. This is being discussed in another active thread, http://www.sciencefo...speed-of-light/ (post 18 and on). You are assuming velocities add as 1+1=2. Suppose you're driving along at 60 miles per hour and a motorcycle zooms past you, going 20 mph faster than you. How fast is the motorcycle going? Simple: 60+20=80. That isn't how velocities add in relativity. Suppose you are in a spaceship going at 3/4 the speed of light relative to some observer. Another spaceship going in the same direction zooms past you at 3/4 of the speed of light relative to your spaceship. Naively adding these velocities suggests that observer who sees you moving at 3/4 c will see that other spaceship as moving at 1.5 times the speed of light. That's not how it works. That observer will see that other spaceship as moving at 24/25 c. Doesn't that make the speed of light breakable just by changing the reference frame? Link to comment Share on other sites More sharing options...
swansont Posted January 23, 2012 Share Posted January 23, 2012 Sorry, the emition of another photon was just to describe the scenario. Let me phrase it differently: If I was driving at the speed of light with my headlights on, from my point of veiw the light travels away from me at c but in reallity in order for that to happen it must travel at 2c You can't move at the speed of light if you have mass. Doesn't that make the speed of light breakable just by changing the reference frame? On the contrary, it's what is necessary to preserve it. Link to comment Share on other sites More sharing options...
D H Posted January 23, 2012 Share Posted January 23, 2012 Doesn't that make the speed of light breakable just by changing the reference frame? It's the other way around. If velocities added vectorially (i.e., how we add velocity vectors in Newtonian mechanics) it would be easy to make something go faster than the speed of light. Use a two-stage rocket, each of which can accelerate it's payload to 3/4 c. The first stage takes the entire rocket (1st stage rocket + 2nd stage rocket + 2nd stage payload) from at rest with respect to some observer to 3/4 c with respect to that observer. The second stage starts firing when the first stage burns out. It takes the second stage (2nd stage rocket + 2nd stage payload) from at rest with respect to the burnt-out first stage to 3/4 c with respect to that burnt-out first stage. If velocities added in the Newtonian sense our payload would now be going at 3/2 c. That's not how things work in the relativistic world. Strictly speaking, it's not even how things work in our slow Newtonian world. Suppose you are driving along at 60 miles per hour and a motorcycle races past you at 40 mph relative to you. How fast is the motorcycle going with respect to the ground? Newtonian physics says 100 mph. Relativity theory says 99.999999999999466 mph, not 100. The difference between those sums per Newtonian mechanics versus relativity theory is immeasurably small at our slow paced (relative to c) world. The difference becomes significant once speeds reach a significant fraction of the speed of light. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now