light beams
- Thread starter jameswade
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Darby
Epochal Historian
Maybe I don't understand the question.
If the observer is in the rest frame and looking at two photons approaching him from opposite directions he'll observe each, individually, to be moving at the speed of light. He can add their speeds c + c and get a closing rate of 2c from his prespective. There's no problem and no contradiction with Special Relativity. Nothing is observed to be moving at a speed greater than c from his frame of reference.
If he hitched a ride on Photon A and observed Photon B the situation is different. From his frame, as he observes the other photon, it reduces to:
w = relative velocity
u = photon A velocity = c
v = photon B velocity = c
c = speed of light = 1
Addition of velocities for parallel moving bodies
w = (v + u)/ 1+ (vu/c^2)
w = (c+c)/ 1+c^2/c^2 = 2/(1+(1/1) = 2/2=1=c
The observer at A observes the closing rate to be c, the speed of light. If the photons were moving in opposite directions, say after they passed each other, just reverse the signs on everything. You end up with -2/-2=1=c. The photons are then receding from each other at the speed of light from the perspective of the observer on photon A. The stationary observer would see them receding at 2c.
In scenarios involving velocities at or near the speed of light you have to make sure that you are clear on the frame of reference for the observer.
Did this address your question or did I indeed not understand it? /ttiforum/images/graemlins/smile.gif
reactor1967
Quantum Scribe
The distance between two beams of light does not by itself move mass. Both beams are still traveling at the speed of light. Currently the scientific thinking for moving mass pass the speed of light is to move space-time around the mass thus not breaking the laws of physics for the speed of light as a constant.
The next time you take a bath put one of your very small toys in the water that floats and drop a little bit of soap behind it. Now watch the toy move. The toy would be mass, the water would be space-time, the soap would be some kind of energy causing space-time to move the toy.
The toy itself thinks it is sitting still but you as the observer are watching it move. This is how travel past the speed of light would have to be accomplished. Don,t take the toy thing personally I was just explaining it to you.
RainmanTime
Super Moderator
James,
Your question is not complete, from a scientific standpoint. All velocities (speeds) are relative to some reference frame. When you ask your question "we would in fact be approaching each other @ twice the speed of light right?" you are not stating which reference frame you are measuring the speed of light with respect to. Reference frames are very important, especially because of Relativity. That is what Darby's answer is trying to tell you.
It is OK to admit if you do not undertand Darby's answer... he or I can walk you through it with greater explanation if you just ask. Do you understand what he wrote?
RMT
Your question is not complete, from a scientific standpoint. All velocities (speeds) are relative to some reference frame. When you ask your question "we would in fact be approaching each other @ twice the speed of light right?" you are not stating which reference frame you are measuring the speed of light with respect to. Reference frames are very important, especially because of Relativity. That is what Darby's answer is trying to tell you.
It is OK to admit if you do not undertand Darby's answer... he or I can walk you through it with greater explanation if you just ask. Do you understand what he wrote?
RMT
RainmanTime
Super Moderator
It would appear to be twice the speed of light closing velocity as measured by a "non-moving" observer on the face of the earth, yes. And that is exactly what I mean by "reference frames"... because when it comes to velocity it matters where (what reference frame) you are measuring the velocity.
But measuring the velocity from other reference frames (like one attached to you, or one attached to me) is different, right???? Forget the light speed issue for a minute and convince yourself that the velocity you measure depends upon your frame of reference. Do you understand what we are saying now about reference frames?
Now let's move on (no pun intended) to a reference frame attached to either you or me. Would you agree that both of these reference frames attached to you and me are accelerated with respect to someone standing on the ground? In other words, we are taking the person on the ground as being "fixed" (inertial is the word we use for this reference frame). Then, since we are both moving with respect to the ground, BOTH of our reference frames are accelerated with respect to the observer on the earth.
So for accelerated (i.e. non-inertial) reference frames, especially ones that are traveling at/near the speed of light, you must apply the corrections that Darby has shown above. This is a fundamental principle of Relativity.
Keep asking questions....you are on your way to a major understanding of how the universe works beyond the simplistic models that are taught under Newtonian rules. It is a very good thing!
RMT
Angleochoas
Quantum Scribe
I agree, it's a more exact language then at first glance. What appears to be say, 'velocity' in the mind of a person whom has not checked it's actual function, will have problems communicating how it's used in translating anything observed (regardless if the thought is correct, it's about effective communication, I'm gathering).
The more I learn, the less I know - but it becomes that much more interesting /ttiforum/images/graemlins/smile.gif
ps
I'm not posting trying to sound like 'an expert' - I'm just learning and finding it fascinating.
The more I learn, the less I know - but it becomes that much more interesting /ttiforum/images/graemlins/smile.gif
ps
I'm not posting trying to sound like 'an expert' - I'm just learning and finding it fascinating.
Ummmm... momentum changes the shape of matter at a very small scale. It's like going from a disc to a cone. But more complex and pretty. /ttiforum/images/graemlins/smile.gif As another poster noted, if the fluid of which a vortex is composed is also moving, then the shape of the vortex need not be deformed. However, the particles making up the fluid will be deformed. Energy is always stored at some level.
Twighlight
Quantum Scribe
It's easier to understand the responses you got, if you imagine that you and I are the only things in the universe. Now.....if we get closer together......is it because I alone am moving, or because you alone are moving, or because we both are moving ?
Without an external reference point it is impossible to say.....it is impossible to determine who is in motion.
All motion requires at least 3 reference points in order to determine not only who is moving, but the manner of any relative motion. For example, it is often stated that 'the Moon goes round the Earth'. What is not often realized is that that is ONLY true relative to an observer on Earth. However much it might defy 'common sense'...in fact relative to an observer looking down on the solar system from above....the Moon does not 'go round' the Earth in a circle at all...but weaves a wave-like path back and forth to one side and the other of the Earth's orbit. This is because both the Earth and Moon are themselves moving....around the sun. That's a good example of how all motion is relative.
If you want to measure the true relative motion of two objects....the very least you need is a third object as a reference point. Ideally, you need a complete 'reference frame'...a set of x,y,z co-ordinates....and the motion is relative to that. Specifically, for motion, one needs what is called an inertial reference frame. This is where special relativity comes in.
Really, your initial question is best answered if you look up the Michelson-Morley experiment...as that is the starting point for the speed of light being constant no matter how you move relative to it.
Darby
Epochal Historian
James,
You still have to define a frame for the observer. In the example that I gave above just insert yourself on Photon B and observe Photon A (where the other observer is riding). Each of you will calculate that the closing rate to be c and a third observer in the rest frame between the two of you would calculate the closing rate as 2c. None of the three observers would calculate that anything in this system was independently moving at a speed greater than c - exactly as predicted by Special Relativity.
Darby
Epochal Historian
James,
In Special Relativity the observer is assumed to be an inertial observer - traveling at a constant speed, in a straight line and not rotating (which can all be summed up as traveling at a constant velocity if you consider that velocity is a vector).
Multiple inertial observers can have different velocities with respect to each other but they will all observe a photon to be moving at 300,000 km/sec. It's not necessary for there to be some universal inertial frame of reference that they would point to to make their calculations.
In Special Relativity the observer is assumed to be an inertial observer - traveling at a constant speed, in a straight line and not rotating (which can all be summed up as traveling at a constant velocity if you consider that velocity is a vector).
Multiple inertial observers can have different velocities with respect to each other but they will all observe a photon to be moving at 300,000 km/sec. It's not necessary for there to be some universal inertial frame of reference that they would point to to make their calculations.
KerrTexas
Super Moderator
I believe I can see the source of this discussion. I may be wrong, but the first thing that came to mind when I was reading this thread, was two cars on the highway, each traveling from opposite directions towards each other.
If car A is traveling at 60 miles per hour, and car B is also traveling at 60 miles per hour, if they crashed into each other, the combined "energy" of the impact would be 120 miles per hour.
I don't know if this is leading to the same concept as presented in this thread regarding particles of light.
If the particles of light collide when they meet, would the impact be similar in dynamics to the impact of two vehicles ?
The way I see this, is that we can assign values to each vehicle, but the space between the vehicles will always have value of zero.
No matter what happens with the velocity of the vehicles, the value between the two, remains zero.
No matter where I'm standing, no matter what the velocity of the vehicles may be, the space between each of the vehicles remains as "zero".
Could you say that there is a value of 120 miles an hour "anywhere" between the two vehicles; as the two vehicles rush towards each other ? or no ?
When the light particles crash into each other, would the value of the impact ( energy ) equate to twice the speed of each of the particles of light, just as with the vehicles ?
It seems that even if they did, that the energy released would still not exceed the speed of light, if this was where this discussion was headed.
If car A is traveling at 60 miles per hour, and car B is also traveling at 60 miles per hour, if they crashed into each other, the combined "energy" of the impact would be 120 miles per hour.
I don't know if this is leading to the same concept as presented in this thread regarding particles of light.
If the particles of light collide when they meet, would the impact be similar in dynamics to the impact of two vehicles ?
The way I see this, is that we can assign values to each vehicle, but the space between the vehicles will always have value of zero.
No matter what happens with the velocity of the vehicles, the value between the two, remains zero.
No matter where I'm standing, no matter what the velocity of the vehicles may be, the space between each of the vehicles remains as "zero".
Could you say that there is a value of 120 miles an hour "anywhere" between the two vehicles; as the two vehicles rush towards each other ? or no ?
When the light particles crash into each other, would the value of the impact ( energy ) equate to twice the speed of each of the particles of light, just as with the vehicles ?
It seems that even if they did, that the energy released would still not exceed the speed of light, if this was where this discussion was headed.
Darby
Epochal Historian
Their combined speed would be 120 mph.
Now we make an assumption for simplicity: both cars have the same mass.
Their combined velocity would be zero and their combined momentum would be zero - if it were anything else you would be getting a free energy lunch. Both velocity and momentum (mass times velocity) are vectors. They have magnitude and direction. One car has positive velocity and momentum because you arbitrarily assign the direction as +x. You then assign the direction of the other car, moving in the opposite direction as moving in the -x direction. In other words, if you graph the vectors they will be the same length and pointing in opposite directions toward each other. Do some vector addition (make it a unit vector so we only have to deal with the number "1"...which is perfectly valid because both cars are moving at the same speed and have the same mass:
Total velocity: (+1v)+(-1v) = 0
Total momentum (+1mv) + (-1mv) = 0
If the net velocity and momentum is zero why do they go "Crash!" and get all crumpled up when they collide at 120 mph? Because they trade velocity and momentum with each other over a very short period of time. That's a lot of kinetic energy shooting into and out of each car simultaneously. The poor electrons that are holding the "stuff" together get so excited that they simply take a hike and leave the nuclei behind.
In the end, however, both cars are at rest - smoking and deformed, but at rest. Net velocity? Zero. Net momentum? Zero. Exactly the same as it was before the crash. All is well - conservation of energy is apparently valid.
Here's a good question: assume the cars were made entirely of some clay-like plastic. They collided and fused and none of the plastic flew away. What is the total mass of the new ball of plastic?
Angleochoas
Quantum Scribe
I'm tossing that over in my mind trying to understand correctly.
So - if you know you have two bodies getting closer, then you must be able to have the points to measure regardless of the space perceived between.
If that's correct, then couldn't we individually use some sort of translation comparison?
KerrTexas
Super Moderator
If the mass was equal, and as you did, to an arbitrary value of 1 as thier mass;
(Vehicle A mass )1 + (Vehicle B mass ) 1 = (Vehicle C mass) 2
The combination of the two, without any loss of mass, results in the creation of a "new" 1.
That "new" 1 will have the mass equal to the sum of A and B. which we could say is now C.
....................
It is this description where the confusion is coming from...the "combined speed" of 120 mph.
It comes down to a matter of terminology, and trying to fit the terminology to describe exactly what is taking place as the vehicles approach each other ...
...and the equations really are the simplest way to define the dynamics of what is taking place, instead of the use of just the "words", especially for a layman not entirely familiar with the definitions of the terminology.
The combination of the terminology with the visual's fired up the light bulb, as it were.
(Vehicle A mass )1 + (Vehicle B mass ) 1 = (Vehicle C mass) 2
The combination of the two, without any loss of mass, results in the creation of a "new" 1.
That "new" 1 will have the mass equal to the sum of A and B. which we could say is now C.
....................
It is this description where the confusion is coming from...the "combined speed" of 120 mph.
It comes down to a matter of terminology, and trying to fit the terminology to describe exactly what is taking place as the vehicles approach each other ...
...and the equations really are the simplest way to define the dynamics of what is taking place, instead of the use of just the "words", especially for a layman not entirely familiar with the definitions of the terminology.
The combination of the terminology with the visual's fired up the light bulb, as it were.
Darby
Epochal Historian
Good job. That's why I used "combined speed". Speed is a scalar...it doesn't have any direction so the speed is just the simple addition of 60 + 60 = 120.
"Velocity" as a physics term combines speed with a specific direction - a vector. Without the vector your energy calculations won't be correct.
BTW: The combined mass of the ball of clay will be >2m, though at 120 mph combined speed it won't be measurable. Maybe 1 electron mass or so.
If after the collision you account for all the mass and energy "leaks" - sound, heat, electrons that escaped, kinetic energy that was transferred to ground, etc., measured the remaining mass with an impossibly accurate instrument it would be greater than expected. Some of the kinetic energy that you added to accelerate the vehicles didn't leak away in the collision and was retained in the nuclei of the atoms. You "rang their bell" when you collided the cars and some portion of that vibration was trapped by the protons and neutrons as internal kinetic energy. Dancing quarks. /ttiforum/images/graemlins/smile.gif
It's a different situation with a particle collider. When CERN gets LHC online and starts slamming protons together at near the speed of light and you look at the exotic particles that are sprayed out their rest masses will be much greater than the original rest mass of the protons. The gamma factor for the protons when they collide will be well over 2,000. When they collide they will have a relativistic mass greater than the combined mass >2,000 at rest protons. Their momentum will be truly huge considering that their velocity will be over 295,000 km/sec.