[SR] Relativistic runner and two relativistinc trains....

[tobix10]

Homework Statement

A train of length L moves at speed 4c/5 eastward, and a train of length 3L moves at speed 3c/5 westward. How fast must someone run along the ground if he is to coincide with both the fronts-passing-each-other and backs-passingeach-other events?

Homework Equations

Velocity addition formula
u = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}
Lorentz contraction
L = \frac{L_0}{\gamma}
Time
t = \frac{s}{v}
Lorentz factor
\gamma (v) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

The Attempt at a Solution

I took the point of view of a running man (its frame). Initial point is when both fronts and a man are in line. Then I calculated speed of both trains in man's frame.
Train B is the one going westward and C eastward.
u_B = \frac{\frac{3}{5}c + v}{1+\frac{\frac{3}{5} v}{c}}
u_C = \frac{-\frac{4}{5}c + v}{1-\frac{\frac{4}{5} v}{c}}
Then calculate the length of trains in a frame of reference related to man
L_B = \frac{3L}{\gamma (u_B)}
L_C = \frac{L}{\gamma (u_C)}
I think that the passaging time of the trains should be equal
t_B = \frac{L_B}{u_B} = \frac{L_C}{u_C} = t_C
but this equation gives solution v = c or v = -c so something is wrong.
I would expect that the velocity v is less than 4/5c.

 

[kuruman]

There are two different gammas here. What did you use for them? When you wrote $\gamma(u_B)$, you meant "gamma as a function of u_B" so your last equation must be $$t_B = \frac{L_B}{\gamma(u_B)} = \frac{L_C}{\gamma (u_C)} = t_C$$.

 

[tobix10]

There are two different gammas here. What did you use for them? When you wrote $\gamma(u_B)$, you meant "gamma as a function of u_B" so your last equation must be $$t_B = \frac{L_B}{\gamma(u_B)} = \frac{L_C}{\gamma (u_C)} = t_C$$.

With this notation I meant gamma as a function o speed.

I think your equation is incorrect. Dimensions are wrong.

 

[kuruman]

I think your equation is incorrect. Dimensions are wrong.

Yes, of course. The expression should also have u_B and u_C in the appropriate places. What is the expression you got? I got a fourth order polynomial. I plotted it and it has only one solution between -c and c, two solutions at -c and +c (like you got) and one solution less than -c (speed greater than c).

 

[tobix10]

I just put equation to Mathematica and get those results. Could you post full equation that you think is valid?

 

[TSny]

I find it much easier to analyze this problem by staying in the Earth frame. No need to use the velocity addition formula, just length contraction and the fact that the gap between the rears of the trains is closing at a rate of v_A + v_B in the Earth frame. Here, v_A and v_B are the speeds of the trains relative to the earth.

This approach is only a suggestion.

 

[kuruman]

I just put equation to Mathematica and get those results. Could you post full equation that you think is valid?

I too used Mathematica and initially got the same answer as you. Then I realized that the correct way to put it in is to solve
$$ \frac{L_B}{u_B~ \gamma(u_B)} + \frac{L_C}{u_C~\gamma (u_C)} =0$$.
That's because one of the ratios is positive and the other is negative since one velocity is positive and the other negative. Mathematica gives four solutions only one of which is a speed less than c.

 

[kuruman]

I find it much easier to analyze this problem by staying in the Earth frame. No need to use the velocity addition formula, just length contraction and the fact that the gap between the rears of the trains is closing at a rate of vA + vB in the Earth frame. Here, vA and vB are the speeds of the trains relative to the earth.

But the simultaneous events of the fronts and rears coinciding occur in the reference frame of the runner not the rails.

 

[Hiero]

But the simultaneous events of the fronts and rears coinciding occur in the reference frame of the runner not the rails.

What do you mean? Those events are not simultaneous. Rather they are coincident with the runner, which is true in every frame...

I agree with TSny, the problem should not be very difficult in the Earth frame... You just need to know the time and distance between the events (front passing/backs passing). The ratio of this time and distance is of course the speed someone would need to move (relative to Earths frame) in order to coincide with both events.

 

[kuruman]

What do you mean? The events are not simultaneous. Rather they are coincident with the runner, which is true in every frame...

I interpreted the statement "if he is to coincide with both the fronts-passing-each-other and backs-passing each-other events" to mean that the observer is the runner not someone sitting on a bench at the train station. I mean that the runner sees the two fronts and the two backs to be where he is (as measured by his meter stick) and at the same time (as measured by his clock.) In other words, the fronts, the backs and the runner are at two different positions simultaneously in the runner's frame of reference. This simultaneity does not happen in any of the other frames including the Earth's. I may be wrong, and it wouldn't be the first time. What is your interpretation?

 

[Hiero]

I interpreted the statement "if he is to coincide with both the fronts-passing-each-other and backs-passing each-other events" to mean that the observer is the runner not someone sitting on a bench at the train station.

The runner is an observer. There is no 'the' observer, as if it changes the answer. In the runners view, the events happen in the same place but different times. But of course, in the runners view the runners speed is trivially zero.

What is your interpretation?

The problem states, "how fast must someone run along the ground" so I think that in itself is a big hint that the ground frame is the easiest frame to work in.

 

[kuruman]

The problem states, "how fast must someone run along the ground" so I think that in itself is a big hint that the ground frame is the easiest frame to work in.

Indeed we are looking for the runner's speed relative to the ground. However, the runner observes the coincidence of the fronts and backs; not the ground. Their simultaneous coincidence occurs in the frame of reference of the runner; he sees the fronts and backs at the same position at the same time as he, not the ground. If this simultaneous coincidence occurs in the frame of reference of the runner, it cannot be simultaneous in the frame of reference of the ground and vice-versa.

On edit: If the simultaneous coincidence were to happen in some frame other than the runner's, why is the runner needed?

 

[Hiero]

If this simultaneous coincidence occurs in the frame of reference of the runner, it cannot be simultaneous in the frame of reference of the ground and vice-versa.

That is just not true... Simultaneity is only frame dependent for spatially separated events... The simultaneity of events which also occur at the same location is a universally agreed upon, frame-independent statement... So what (relevant) events occur at the same time but different place in the runners view?

 

[kuruman]

So what (relevant) events occur at the same time but different place in the runners view?

None. He sees the fronts meet at t = 0 and the backs meet at some later time. I was wrong to imply in post #10 that he has moved relative to his frame.