Answer: Rockets Colliding: Time Dilation from Earth to A's Frame

[C_Ovidiu]

A person on Earth observes two rockets moving directly toward each other and colliding. At time t=0 in the Earth frame observer determines that rocket A, traveling to the right at [tex]v_a=0.8c[/tex] is at point a and rocket B traveling to the left at [tex] v_b=0.6c[/tex] is at point b. They are separated by a distance [tex]l=4.2*10^8[/tex].
How much time will elapse in A's frame of reference until the two rockets collide?

1. If we consider the time that will elapse in Earth frame until collision and then transform this interval in A's frame because of time dilation then we obtain :

[tex]t_A=\frac{l\cdot\sqrt{1-\beta_a^2}}{v_a+v_b}=0.6s[/tex]

2. Howerer, if we change our approach in the following way.
we transform the distance l between the two ships at [tex]t=t_A=0 [/tex] :
[tex]l_a=\frac{l}{\sqrt{1-\beta_a^2}}[/tex]
we consider then the speed of B in A's frame of reference [tex]v_b^a=-\frac{v_b+v_a}{1+\frac{v_bv_a}{c^2}}[/tex]

Now we get [tex]t_A=\frac{l_a}{v_b^a}[/tex] . As you can see this rationing yields a totally different result. What is wrong in this second approach?

 

[HallsofIvy]

A person on Earth observes two rockets moving directly toward each other and colliding. At time t=0 in the Earth frame observer determines that rocket A, traveling to the right at [tex]v_a=0.8c[/tex] is at point a and rocket B traveling to the left at [tex] v_b=0.6c[/tex] is at point b. They are separated by a distance [tex]l=4.2*10^8[/tex].
How much time will elapse in A's frame of reference until the two rockets collide?

1. If we consider the time that will elapse in Earth frame until collision and then transform this interval in A's frame because of time dilation then we obtain :

[tex]t_A=\frac{l\cdot\sqrt{1-\beta_a^2}}{v_a+v_b}=0.6s[/tex]

This is incorrect. The speed with which the two ships are approaching one another, in earth's frame of reference is not [itex]v_a+ b_b[/itex]. It is
[tex]\frac{v_a+ v_b}{1+ \frac{v_av_b}{c^2}}[/tex]

2. Howerer, if we change our approach in the following way.
we transform the distance l between the two ships at [tex]t=t_A=0 [/tex] :
[tex]l_a=\frac{l}{\sqrt{1-\beta_a^2}}[/tex]
we consider then the speed of B in A's frame of reference [tex]v_b^a=-\frac{v_b+v_a}{1+\frac{v_bv_a}{c^2}}[/tex]

Now we get [tex]t_A=\frac{l_a}{v_b^a}[/tex] . As you can see this rationing yields a totally different result. What is wrong in this second approach?

Of course the time is different in two different frames of reference.

 

[C_Ovidiu]

This is incorrect. The speed with which the two ships are approaching one another, in earth's frame of reference is not [itex]v_a+ b_b[/itex]. It is
[tex]\frac{v_a+ v_b}{1+ \frac{v_av_b}{c^2}}[/tex]

Why would that be so? The Earth is still and in Earth's frame the two velocities simply add.

Of course the time is different in two different frames of reference.

I compared the time [tex]t_a[/tex] yielded by the two types of rationing , that it the time until collision in A's frame of reference.

The first rationing mode is from "A guide to physics problems" -Cahn S., Nadgor , so I incline to think that the second solution has some holes in it.

 

[Ich]

What is wrong in this second approach?

This is wrong:

the distance l between the two ships at [tex]t=t_A=0[/tex]

The distance at t=0 is quit different from the distance at tA=0, in every frame.
Draw a spacetime diagram with the respective lines of simultaneity to see that both are totally different things.

...transform this interval... ...transform the distance...

At no point did you transform coordinates. Bear in mind that length contraction and time dilation are error-prone mathematical shortcuts, not transformations. Try to do the actual transformations, that's much easier and less confusing.

 

[michelcolman]

This is incorrect. The speed with which the two ships are approaching one another, in earth's frame of reference is not [itex]v_a+ b_b[/itex]. It is
[tex]\frac{v_a+ v_b}{1+ \frac{v_av_b}{c^2}}[/tex]

No, in Earth's reference frame the speeds simply add up. In one second, one rocket will travel 0.8 lightseconds while the other will travel 0.6 lightseconds in the other direction. This brings them 1.4 lightseconds closer together, you don't need rocket science for that ;-)

Your formula gives the speed of one rocket as seen from the reference frame of the other, which is an entirely different matter.

 

[michelcolman]

What is wrong in this second approach?

The problem is in simultaneity. Suppose, from Earth's point of view, rocket A is passing a space station named Alpha which is stationary relative to the Earth and at a distance dA to the left of the point of impact. At the same time (in Earth's reference frame), B is passing space station Beta which is at a distance dB to the right of the point of impact. dA + dB = 4.2*10^8.

Now, from the point of view of A, as it is passing space station Alpha, B has already passed Beta! They don't pass the stations simultaneously in A's reference frame. It's like the "simultaneous" lightning strikes on the train, where someone in the train will say the front was struck before the back.

The length-contracted distance you used corresponds to the distance between the two space stations, but this does not equal the distance between A and B at the time A is passing Alpha, because at that time B has already passed Beta.

(and obviously, in B's reference frame, A will pass Alpha before B passed Beta)

 

[C_Ovidiu]

As much as I tried to understand all the responses posted I could not. Firtly, I do not know how to draw space-time diagram. Secondly, I would like to know how to solve this problem using the second method and using only Lorentz transformations.
Can someone explain to me how this can be done please ?
Sorry for being such a bother.

 

[michelcolman]

OK, I'll give it a try:

Since relativity deals with "events" (something happening to some object) but the actual coordinates depend on the observer, we have to make sure we have some well-defined "events" for this problem. Spaceships being a certain distance apart doesn't quite cut it, since distances vary and objects may be at different locations at different times depending on who's watching. So we'll introduce position markers in the form of space stations Alpha and Beta that are stationary relative to the earth. A passing Alpha is a much more well-defined event than A being a certain distance away from B.

So imagine that, in Earth's reference frame, at time t=0, spaceship A is passing space station Alpha and spaceship B is passing space station Beta. The two space stations are 4.2*10^8 meters apart. Since A and B are going straight towards each other, they will collide at t=1 second.

Now let's look at everything in A's coordinate system. The distance between Alpha and Beta is now only 2.52*10^8. Supposing A passes Alpha at time t'=0, at what time did B pass beta? Just fill in gamma=1.667, t=0, v=0.8c, x=4.2*10^8 and you'll get t'=-1.87 seconds. So as A passes Alpha, B has already passed Beta nearly two seconds ago!

Multiply this time with B's speed relative to Beta as seen by A (relativistically add 0.8c to 0.6c and then classically subtract 0.8c, so (0.8 + 0.6)/(1 + 0.8*0.6) - 0.8 = 0.146c) to get about 8.2*10^7 meters. This means B is currently (at time t' = 0) 2.52*10^8 - 8.2*10^7 = 1.70*10^8 meters away from A. Now divide this by the speed of B as measured by A, and you get... 0.6 seconds.

Which makes sense because, as seen from earth, A's clock is ticking at a rate of 0.6 seconds per Earth second, and the collision happens 1 second after A passed Alpha!

Now you may be wondering why we considered A's clock to be running slower... why not Earth's clock?

Well... supposing the Earth is near the collision, A will say that people on Earth started their timer more than a second too soon, when A was not at Alpha yet, so obviously their timing is off. In fact, when A passed alpha, Earth's timer (t'=0, x=2.4*10^8) was already indicating t=0.64 seconds. So if they would not have cheated, they would have measured only 0.36 seconds, which proves their clock is running slowly by a factor of 0.6.

Getting dizzy yet? ;-)