Register to reply 
I know that nothing can exceed the speed of light, but ... 
Share this thread: 
#1
Feb1813, 03:11 PM

P: 261

... can someone explain to me why the following thought experiment does not give the object a speed greater than 3*10^8 m/s.
Imagine that the experiment takes place in a very large vacuum. A very long toy train is moving at 100 m/s on a track from rocket propusion. On the top of this train, there is another smaller train that starts to run on a track on the top of the larger train. The smaller train now has a speed of 100 m/s relative to the larger train and 200 m/s relative to a still observer. Then, a smaller train does the same thing on the second train. There are 3*10^6 + 1 trains that are all on the backs of another doing 100m/s. Exactly what happens when the 300000th train fires up its rocket relative to a still observer? If possible, what happens when the 300001th train's rocket fires? 


#2
Feb1813, 03:15 PM

P: 784

Because velocities don't add like we think they do. Well, at low speeds them seem to increase through simple addition, but once you get up to relativistic speeds you have to start using the relativistic velocity addition formula ##V_{total} = \frac{v_1+v_2}{1+ \frac{v_1 v_2}{c^2}}##
http://en.wikipedia.org/wiki/Velocityaddition_formula Either reading the page or doing some quick checks yourself will show you that you can never get something moving faster than the speed of light with this formula. 


#3
Feb1813, 03:50 PM

P: 261




#4
Feb1813, 03:52 PM

P: 642

I know that nothing can exceed the speed of light, but ...



#5
Feb1813, 03:54 PM

P: 784

Rationalism isn't dead, far from it. All relativity (and quantum physics for that matter) does is limit the usefulness of common sense.
The relativistic velocity formula comes very directly and very rationally from a postulate that violates common sense. If you accept that common sense isn't always correct, then its very intuitive. If you can get a deep understanding of the initial postulates of relativity, then you can start seeing why relativistic physics makes sense, the trouble is the initial hurdle of "huh?" that is at the beggining of all relativity classes. 


#6
Feb1813, 04:27 PM

Sci Advisor
Thanks
P: 3,480

There's a long thread in the relativity forum recently: http://www.physicsforums.com/showthread.php?t=670253 


#7
Feb1813, 04:41 PM

Sci Advisor
Thanks
P: 3,480

Relativity just requires you to consider that the intuition you've developed in a lifetime of dealing with speeds much less than that of light might not be applicable at higher speeds. It's somewhat like the adjustment we go through as we learn that, all appearances and common sense notwithstanding, the world is not flat and "up" is a completely different direction in Australia and America. But quantum mechanics... That's bigleague counterintuitive. 


#8
Feb1813, 05:27 PM

P: 261




#9
Feb1813, 05:38 PM

P: 784

It's the taking the two postulates for granted that's the hard part. 


#10
Feb1813, 05:42 PM

Mentor
P: 16,989




#11
Feb1813, 06:08 PM

P: 351

The fact is they don't. I could only conclude that there is a fundamental measure distance/time together, that is invariant in all (nonaccelerating) reference frames. It's not a terrible sacrifice of intuition, and great deal of fruit falls out quite naturally from that tree, once I accept it exists. (Edit: I should hastily add that the d/t thing is not mine. It's an old variant of one of Einstein's postulates. I only decided it was true) 


#12
Feb1813, 06:59 PM

Sci Advisor
Thanks
P: 3,480

If we apply the first postulate to all the laws of physics, then we cannot leave out Maxwell's laws of electrodynamics. These laws predict electromagnetic radiation propagating at speed c in a vacuum; so I don't find t such a huge leap of common sense and intuition to accept that either different observers will have different electrodynamical laws in violation of the first postulate, or that all observers must get the same value for the speed of light. (Of course if I had studied Maxwell's electrodynamics in the 1870s instead of the 1970s common sense and intuition would have led me to an ether theory instead. Historical context matters). 


#13
Feb1913, 01:56 PM

P: 784

Your point is good, but I'd just say that it implies just that maxwell's laws also imply things that are contrary to common sense. Do you really think saying that an object moving past you that doesn't seem to slow down if you move in its direction of motion follows common sense? 


#14
Feb1913, 02:27 PM

Sci Advisor
Thanks
P: 3,480

When you say that "the relativistic velocity formula comes very directly and very rationally from a postulate that violates common sense" I agree with you that it follows directly and rationally from a postulate; I just don't think the postulate violates common sense. 


#15
Feb1913, 03:15 PM

P: 784

The first postulate I'd completely agree with you, the second seems to me to be completely against common sense. However, I think you're right that this seems a matter of individuality.



#16
Feb2213, 09:39 AM

P: 3,185

In contrast, you can intuitively understand that SR equation by accounting for two relativistic effects in combination with the synchronization convention; it then becomes common sense. However, this implies  if you are a normal human being  that you will need to do some calculation exercises to develop that intuition. It may be helpful to realize that this "velocity sum" equation technically isn't really an addition but a transformation; the fact that 2+2=4 isn't challenged! Note also that the two postulates of SR (in their original form) were considered both common sense and rather well established, although they seemed to be incompatible with each other. 


#17
Feb2313, 12:24 PM

PF Gold
P: 4,695

We can illustrate how this works with an example involving three inertial observers who we show as colocated at the origin of a spacetime diagram depicting a single ether rest frame. We refer to the observers by their color. The black observer is stationary in the ether frame. The green observer is traveling to his right at 0.6c and the blue observer is traveling to his left at 0.6c. Now we are going to see how each observer measures the speeds of each of the other observers using radar signals which will travel at c relative to the ether frame and not relative to each observer. However, because of the Principle of Relativity, we insist that when one observer measures the relative speed another observer, the other observer will get the same answer when measuring the relative speed of the first observer. And for that to happen we must come to the same conclusions that scientists came to prior to Einstein which is that clocks run slower when traveling through the ether. So on our spacetime diagram, we show dots for each observer that mark out oneyear intervals of time. Now a little review about radar signals. An observer aims a radar signal at another observer noting the time on his clock at which he sent it. Eventually that signal hits the other observer and reflects back to the first observer. When the reflected radar signal gets back to the first observer, he notes this second time. Now with these two time measurements, the observer does a little calculation. He adds the two times together and divides by two. This gives him the time at which he will say the "measurement" applies. He also subtracts the two times and divides that by two. This gives him a measurement of the distance (in compatible units where c=1). So he gets a distance applied at a particular time. To calculate the speed of the second observer, he uses that fact that they started out colocated so all he has to do is divide the distance by the applied time. So let's see how this works first for the black observer measuring the speed of the green observer: When the black observer's clock reads 1, he aims his radar gun at the green observer and he receives the echo back when his clock reads 4. The radar signals are shown in red and always propagate along 45degree angles with respect to the ether frame. Black calculates (4+1)/2=2.5 as the time at which the measurement was made and he calculates (41)/2=1.5 as the distance that the green observer was away from him at his time of 2.5 and we can see that since he is at rest with respect to the ether, his measurement matches what the ether frame portrays as shown by the yellow line. He can also calculate the speed of the green observer with respect to him as 1.5/2.5=0.6c. Although I don't show it, he can make the same measurement of the blue observer and will get the same results since blue is symmetrically a mirror image of green. Now let's see how green will measure the speed of black: We see that the exact same explanation used by black to measure green applies for green measuring black. In fact, since we know that they must get the same answer, that is the reason why scientists concluded that green's clock must be slowed down while he is traveling through the ether. Note that the distance of 1.5 for the yellow line does not correspond to any such distance in the ether frame (but the green observer can't tell that). Although I don't show it, we could also show that blue would get the same measurement of black's speed since green is symmetrically a mirror image of blue. Now we'll take a look at how green would measure the speed of blue using exactly the same radar method as before: Once again, green fires his radar signal at his clock time of 1 but this time, since he is waiting for the reflection from the blue observer, it takes until his clock time of 16 before he gets it. So his calculation of the time the measurement applies is (16+1)/2=8.5 and for the distance is (161)/2=7.5. So he concludes that at his time of 8.5, the distance to the blue observer was 7.5 and the speed of the blue observer is 7.5/8.5=0.882c. Note that the length of the yellow line does not correspond to anything in the ether frame. And again because blue and green are mirror images, blue will get the same measurement of green's speed. But now let's use the relativistic addition formula to see what speed we get when we add 0.6+0.6: (0.6+0.6)/(1+0.6*0.6) = 1.2/(1+0.36) = 1.2/1.36 = 0.882c This is the same answer we got using the ether frame and the radar measurement. So you see, it is not necessary to apply Einstein's second postulate to demonstrate the validity or the common sense of the relativistic velocity addition formula. 


#18
Feb2313, 02:01 PM

P: 73

I just happen to see the original question by student34 at Post 71. Here is one physicist's intuition.
1. I want you to recall Newton's 2nd law: Force vector = Rate of change of momentum vector. If you keep mass and the direction of velocity vector fixed, then force = (Inertial mass)(acceleration). 2. Only that which has an inertial mass, CAN be accelerated or decelarated. 3. Light (or em waves) do not have inertial mass. That means, light CANNOT be accelerated or decelerated. That means, poor light is stuck with whatever speed Nature assigned it. 4. The same rules apply when adding speeds. 5. A photon's spin is 1 but only two projections: +1 and 1; no zero. 6. There is no such thing as a travelling photon. Em waves travel; photons interact with whatever they can. I recommend you discuss these with your professor. 


Register to reply 
Related Discussions  
Exceed the speed of light  Special & General Relativity  24  
Can something exceed the speed of light?  General Physics  4  
Why can't a photon exceed the speed of light?  Special & General Relativity  18  
Nothing can exceed the speed of light  Special & General Relativity  4  
Catch the neutrino speed exceed light  General Physics  1 