The discussion revolves around a hypothetical scenario involving a bus traveling at the speed of light and the implications of firing a gun from the back of the bus towards the driver. Participants explore concepts from special relativity, the nature of speed, and the frame of reference in physics.
Participants do not reach a consensus on the implications of the scenario, with multiple competing views on the nature of speed, reference frames, and the behavior of light. The discussion remains unresolved regarding the fundamental questions posed.
Participants highlight limitations in the assumptions made about speed and reference frames, as well as the complexities introduced by relativistic effects. The discussion reflects varying interpretations of the principles of relativity and the nature of light.
I think if you learn about theory of special relativity you will have the answer !brags said:
brags said:
HallsofIvy said:
No. The person on the platform sees the light exit the gun and travel at the speed of light (barely faster than the gun itself is moving).brags said:I thought the speed of light was a constant from any point of referrence?
That's exactly where I was going...relative to someone watching the train pass by, that laser would be traveling at twice the speed of light
brags said:I thought the speed of light was a constant from any point of referrence?
Only slower-than-light objects can have their own rest frame in relativity, light itself has no rest frame, so there's no frame where light is at rest but planets and galaxies are traveling at the speed of light.Schrödinger's Dog said:
"point of reference" is the same thing as a "rest frame", it's not a location in space, it's just an arbitrary choice of how you choose to define what is "at rest", with all other velocities defined relative to that. For example, if we are traveling at 50 mph relative to one another, we could pick a frame where I am at rest and your velocity is 50 mph to the right, or we could pick a frame where you are at rest and my velocity is 50 mph to the left.Schrödinger's Dog said:
If I start out at rest and you slam into me at 50 mph from the left, the effects will be exactly the same as if you were at rest and I slam into you at 50 mph from the right, in terms of the damage to each of us, and the forces and accelerations we each experience. For example, if you slamming into me from the left causes me to accelerate to the right, then it should also be true that me slamming into you from the right causes me to accelerate to the right by the same amount. All the known laws of physics work the same way in every inertial reference frame. This is true in both Newtonian mechanics and relativity.brags said:
According to modern physics, it is meaningless to ask what your "true" velocity is. If you analyze the same physical situation (like a collision between two objects in relative motion) from the perspective of different inertial reference frames, you will get precisely the same physical predictions about all measurable aspects of the situation (like the accelerations experienced by each object during the collision). So there is no empirical test that will pick out one frame's description as being more correct than the other.brags said:
Light does fall under the same laws as objects with mass - it's just that that ain't it! We use it at low speed because it isn't wrong by much, but it is still wrong.brags said:
Correct.
Read the fine print on those articles - they are not sending a signal at greater than C, just playing games with pulses, waves and interference. It ends up being akin to sweeping a laser beam past the moon - the point of the beam may be moving faster than C, but none of the photons in the beam are.
brags said:
Velocities don't add in relativity like they do in Newtonian mechanics--if I see you going to my right at velocity 30 mph, and Sue sees me going to her right at 50 mph, then in Newtonian mechanics that means Sue would see you going to her right at 30 + 50 mph, but it isn't a simple sum like this in relativity. If the two velocities are v and u, then instead of the third velocity being v+u, in relativity it's [tex](v + u)/(1 + uv/c^2)[/tex]. So if one of the two velocities is c--say, v=c--then the third velocity will be [tex](c+u)/(1 + u/c)[/tex], and if you multiply both the numerator and the denominator by c you get [tex]c(c+u)/(c+u)[/tex] so the (c+u) term cancels out, meaning the velocity as seen by the third observer will also be c, regardless of what u was. So if I see a light beam traveling to my right at v=c, and you see me traveling to your right at u=0.8c, then you will see that light beam traveling to your right at (c + 0.8c)/(1 + 0.8) = c. In other words, anything that is traveling at c in one frame will be traveling at c in all frames in relativity.brags said:
There isn't any easy conceptual answer to that question, it's just part of the fundamental laws of physics. Basically it has to do with the fact that all the fundamental equations of physics have a mathematical property called "Lorentz-symmetry" which insures that they will appear to work the same way in all reference frames.brags said:
This is another tricky part, there is no "objective" answer to which clock is ticking slower. In the rest frame of the galaxy, the moving clock is ticking slower than clocks at rest relative to the galaxy, but in the moving clock's own frame, the clocks at rest relative to the galaxy are the ones ticking slower. It might seem like this would lead to a contradiction--say I'm traveling past a row of clocks which are all at rest with respect to each other, and in the clocks' rest frame they are all synchronized to read 12:00 at the moment I pass the left end, and it takes me 25 minutes to pass them all, so they all read 12:25 at the moment I pass the right end. Now, say that my onboard clock also reads 12:00 at the moment I pass the left end, and that I'm moving at 0.6c relative to the clocks, so in their rest frame, my clock is only running at 0.8 the normal rate. An observer at rest relative to the clocks should predict that if the clocks measure the time for me to move from the left end to the right end as 25 minutes, then my own onboard clock should only tick forward by 0.8*25 = 20 minutes, so my clock will read 12:20 at the moment I pass the right clock. But if in my frame it's the row of clocks that are running at 0.8 the rate of my own clock, will I disagree with this prediction? The answer is no, because if the clocks are synchronized with each other in their own rest frame, they will not appear synchronized in my own rest frame, due to what is known as the "relativity of simultaneity". This means that different frames disagree on whether two distant events happened "simultaneously" or not, so if the clock on the left end of the row and the clock at the right end of the row both read 12:00 "at the same time" in their own rest frame, this is not going to be true in my own frame. In my frame, at the moment I passed the clock on the left end which was reading 12:00, the clock on the right end already read 12:09, and in the 20 minutes it took me to reach the right end, that clock only advanced forward by 0.8*20 = 16 minutes, so when I reached it, it read 12:09 + 16 minutes = 12:25. So I agree that when I reached the right clock, it read 12:25 and my clock read 12:20, but in my frame this isn't because the clock on the right end was ticking faster than mine, it's just because the clock on the right end had a head start, due to it being out-of-sync with the clock on the left end.brags said:
JesseM said:
