# Bus that is travelling at the speed of light

1. Jan 6, 2006

### brags

If I'm seated in the back of a bus that is travelling at the speed of light, and I fire a gun towards the driver...will the bullet ever reach him? Ignoring of course that nothing of mass can reach the speed of light..allegedly.

Last edited: Jan 6, 2006
2. Jan 6, 2006

### linhtm

I think if you learn about theory of special relativity you will have the answer !

3. Jan 6, 2006

### Staff: Mentor

Since no mass can reach the speed of light, we need to change the problem to say you are traveling at .99999...[add as many as you want]...9999% C. The answer is yes. You will notice nothing unusual about the bullet's trip.

4. Jan 6, 2006

### Janus

Staff Emeritus
If you ignore that nothing with mass can reach the speed of light, you are ignoring Relativity completely and then there is no point to your question.

5. Jan 6, 2006

### HallsofIvy

You cannot ask a physics question while telling people to ignore the laws of physics in their response!

Now, suppose you are sitting in a bus moving at 0.99999999c (relative to what frame of reference?) and fire a bullet that goes at 0.00000000001c (relative to the gun which is in the same frame of reference as the bus). Does the bullet reach the driver? Of course it does. Since you, the gun, the driver, and the bus are all in the same frame of reference, that is all that matters. How fast the bus is going, relative to some other frame of reference is irrelevant.

What you should be worrying about is the fact that you fantasize about shooting people!

6. Jan 6, 2006

### Staff: Mentor

To put it another way, if the windows of the bus are blacked out so that you can't see outside, there is no way that you can distinguish, by doing experiments inside the bus, whether the bus is "at rest" (relative to whatever) or moving at constant speed (whatever speed, < c). This is the fundamental principle of relativity.

7. Jan 6, 2006

### Schrodinger's Dog

Re:

Yeah to put it another way of course your assumption that your travelling at the speed of light relatively speaking is flawed in the first place. How do you tell how fast anyone is going in relation to anyone or anything else for that matter. I may be travelling at the speed of light and so may someone else, relative to me, how do we know this for sure? I may in fact be going twice the speed of light or half of it. Without a frame of reference I really wouldn't know? who's to say when I started I wasn't moving allready, and the other stars we're moving at varying speeds in relation to me. Where's the zero point, where is my point in space which is motionless that I may determine my speed?

Back to your question, so if I fire a laser gun at someone on a train assuming the train inside is in a vacuum,it will be travelling at the speed of light, is it travelling twice the speed of light you ask?

Last edited: Jan 6, 2006
8. Jan 6, 2006

### brags

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 travelling at twice the speed of light (if it reaches the driver it must be travelling faster than c because the observer could watch it travel from the back of the train to the front, and if the driver was fortunate enough to be missed by the laser, it would continue out through the window and beyond at twice c)
BUT I understand if you stick to the law that nothing of mass can reach c, then everything but the laser would be remain at .9999999999...c so c would remain a constant. Einsteins whole theory of relativity is relying on the fact that this .9 can never be eclipsed by objects of mass? Is this why?

Last edited: Jan 6, 2006
9. Jan 6, 2006

### DaveC426913

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).

The key is that the time frame inside the train is not the same as on the platform. Time is slowed down inside the train relative to the platform.

The guy on the platform might see the light take, I dunno, 10 years to reach the conductor (because he's travelling at .999999999c). Inside the train, the shooter sees the light take .001 seconds to reach the conductor. But what he doesn't know - until he looks out the window - is that, in that .001 seconds, the universe has aged around him by ten years.

Do you see how they can both see the same event differently when their speed messes with the passage of time?

10. Jan 6, 2006

### Schrodinger's Dog

OK put it this way, what if every visible star in the universe is travelling at the speed of light and so am I, how do I determine what speed I'm travelling at? The real speed of light is?

If every frame of reference in the universe is travelling at the speed of light, what then is the true speed of light? How can I know how fast I am really travelling without pinning down wether I am in fact motionless or not?

Read up on the twins paradox. It kinda makes the point. I would post a link but times got the best of me I'm sure someone can help.

Define a point of reference? what is your point and is it itself in motion? Is there any way to find this magical point of reference which is stationary or arbitrary? A centre of the universe if you will?

Last edited: Jan 6, 2006
11. Jan 12, 2006

### brags

I was mainly wondering though, does light not fall under the same laws as does an object with mass? ie a ball being tossed at 10 km/hr on a train moving 100 km/hour would be travelling at 110 km/hr to an outside observer.

Is light the maximum velocity? Superluminal velocities are impossible?
I googled and it turns out that they have produced through pulse lasers velocities that are 300 times c. Which brings me back to my question, light then is not a fixed speed, and should be affected by "point of view" velocity as is the ball on the train ?

12. Jan 12, 2006

### JesseM

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 travelling at the speed of light.
"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 travelling 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.

13. Jan 12, 2006

### brags

but the velocity is what it is..I am moving or not moving, you can't have it both ways. It may seem from my point of view that I am stationary and you are moving past me, but if I were to slam into you, the result would clearly show that I was the one in motion because you would be forced backwards after the collision, not the other way around. This is why I don't see the physical importance to referrence frames?

Last edited: Jan 12, 2006
14. Jan 12, 2006

### JesseM

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.

15. Jan 12, 2006

### brags

yes..for every action there is an equal and opposite reaction...or I see you could look at it this way, but what I'm saying is relative perception is just a perception of a real and single event. If I am travelling 50 km/s in one driection, then that is the event, not matter how it is percieved, from what point of referrence.

Last edited: Jan 12, 2006
16. Jan 12, 2006

### JesseM

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.

Consider the question of which of two objects has a larger x-coordinate. Obviously this depends on where you choose to put the origin of your spatial coordinate system, and how you orient the x-axis, and there's no physical reason to prefer one choice over any other. You don't think there's a single objective physical answer to this question, do you? If not, why do you think the question of which object is at rest, which also depends on your choice of coordinate system, should be any different?

17. Jan 12, 2006

### Staff: Mentor

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.
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.

18. Jan 12, 2006

### HallsofIvy

No, that's the whole point of relativity (even Gallilean relativity that precedes Einstein by several hundred years). Velocity is NOT "what it is". Velocity only exists relative to some frame of reference. It is a freshman physics experiment to show that the result of a collision does not depend upon which one was moving. If you really believe "It may seem from my point of view that I am stationary and you are moving past me, but if I were to slam into you, the result would clearly show that I was the one in motion because you would be forced backwards after the collision, not the other way around" then I suggest you take a good freshman physics course!

If object B, having mass M, is stopped at a point and object A, also having mass M, moves toward it from the left with velocity V (measured in the frame of reference in which B is stationary) then after the collision, A will be stopped and B will be moving with velocity V.
The same experiment, viewed from A's frame of reference has, before the collision, A stationary and B moving from the right with velocity -V. After the collision, A is now moving with velocity -v and B is stopped. Both viewpoints are correct and it doesn't matter which we think of as moving.

19. Jan 12, 2006

### brags

thanks all...I know when I am in over my head. I only have minimal math...was just curious though. I will research this in my spare time and hopefully get a better grasp of the fundamentals.

20. Jan 12, 2006

### brags

one last thing that I just don't understand. How can we determine what the speed of light is if it depends on our point of reference while calculating it? If we are travelling toward the light source, or away from it, then the speed relative to us would be different?

Last edited: Jan 12, 2006
21. Jan 12, 2006

### JesseM

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 $$(v + u)/(1 + uv/c^2)$$. So if one of the two velocities is c--say, v=c--then the third velocity will be $$(c+u)/(1 + u/c)$$, and if you multiply both the numerator and the denominator by c you get $$c(c+u)/(c+u)$$ 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 travelling to my right at v=c, and you see me travelling to your right at u=0.8c, then you will see that light beam travelling to your right at (c + 0.8c)/(1 + 0.8) = c. In other words, anything that is travelling at c in one frame will be travelling at c in all frames in relativity.

The reason velocities don't add the same way in relativity that they do in Newtonian mechanics basically boils down to the idea that each observer measures velocity in terms of distance/time as measured on rulers and clocks which are at rest relative to himself, but clocks which are moving relative to one another don't tick at the same rate in relativity, and rulers which are moving relative to one another are shrunk in each other's rest frame, and clocks which are synchronized in one observer's rest frame will appear out-of-sync in another observer's rest frame. The Newtonian formula assumes all observers agree about time and distance measurements, but this won't be true in relativity.

22. Jan 12, 2006

### bernhard.rothenstein

consider that the inside of the bus is a cabine under a ship's desk and that you are one of the people Galileo invite to take there a sit. Then Galileo convinces you that the bullet hits the target (I am a pacifist!) located next the driver.
i started learning special relativity reading the celebrated invitation of Galileo you can find in many introductory textbooks (see Leo Sartori)

23. Jan 13, 2006

### bernhard.rothenstein

consider that the inside of the bus is a cabine under a ship's desk and that you are one of the people Galileo invite to take there a sit. Then Galileo convinces you that the bullet hits the target (I am a pacifist!) located next the driver.
i started learning special relativity reading the celebrated invitation of Galileo you can find in many introductory textbooks (see Leo Sartori)

24. Jan 13, 2006

### brags

OK, I read your reply JesseM like 10 times and I think I see your point. Interesting....and totally beyond me for the most part. BUT, you mentioned the clocks. I have heard of this before, and I believe they measured atomic clocks on a space shuttle, or something to that affect, and they actually showed a slight time variation to those on earth. This blows my mind because clocks are mechanical/digital devices. How can they be affected by point of reference?

Besides, even if we ignore that, from the point of referrence of clock X which is nearing the speed of light, it would be standing still, and the universe around it would be travelling at near the speed of light towards it..so essentially to the clock its velocity would be nill, so why would it be ticking slower in referrence to an outside observer (clock Y) who is at a relative speed to the universe which clock X perceives is moving towards it? The observers clock Y should slow down relative to clock X, but clock X should also slow down relative to clock Y. If you compared the two clocks would they show the same time if sychronized before the experiment?

Last edited: Jan 13, 2006
25. Jan 13, 2006

### JesseM

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.
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 travelling 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.