Calculating Time Elapsed in Rocket Collision

[Abhishek11235]

Consider an observer on Earth (Neglect any effect of gravity). Call him A. Let 2 rockets be moving in opposite direction along x-axis (x-axis coincides with the x-axis of A) with uniform velocities. Call them B and C. At t=0, in A's frame, the rockets are separated by length $l$ . Let $V_a$ and $V_b$ be there respective velocities. Calculate the time elapsed from t=0 to the time when B and C collide in the refrance frame of A. What is time elapsed according to B?

---

Attempt:
It is easy to calculate time elapsed according to A. It is given by:
$$t=\frac{l}{V_a+V_b}$$

It is also easy to calculate time elapsed according to B. It is just dilated time calculated above:
$$t_b=t/\gamma$$

However, I was thinking on how to explain the above result (time elapsed on the clock of B) according to B himself. The relative velocity of C w.r.t B is:
$$V_{rel}=(V_a+V_b)/(1+V_aV_b/c^2)$$

Now, the length between B and C as observed by B is given by length contraction:
$$l_b=l/\gamma$$

The time elapsed on the clock of B is therefore:
$$t_b=l/(\gamma V_{rel})=l/(\gamma( V_a+V_b)/(1+V_aV_b/c^2)$$

This does not matches with the time elapsed calculated above using Time dilation formula. What am I missing?

 

[Nugatory]

What am I missing?

Relativity of simultaneity

 

[Abhishek11235]

Relativity of simultaneity

Can you please explain it in more detail?

 

[ergospherical]

Now, the length between B and C as observed by B is given by length contraction:$$l_b=l/\gamma$$

Check this result; use the Lorentz transformations.

 

[Ibix]

Can you please explain it in more detail?

Length contraction is a special case of the application of the Lorentz transforms that applies only to distances that are not changing with time, when you can disregard the effects of the relativity of simultaneity on the measurement of length. It is not appropriate for the distance between the rockets because the distance is changing. Ergospherical has pointed you in the correct direction.

 

[Abhishek11235]

Check this result; use the Lorentz transformations.

Ok. Here is my attempt:
Let Event 1 be when the 2 rockets are separated by length l,t=0 and Even 2 be when they collided.

Now, the length observed by B is by defination happens when $\Delta t'=0$ (primes denotes measurement in B, unprimed in A). Then, we have:
$$\Delta t'=0 \Rightarrow \Delta t=V_a\Delta x/c^2$$

Using this, we have:
$$\Delta x'=\gamma(\Delta x-V_a\Delta t)=\gamma \Delta x(1-V_a^2/c^2)=\Delta x/\gamma$$

 

[Abhishek11235]

Length contraction is a special case of the application of the Lorentz transforms that applies only to distances that are not changing with time, when you can disregard the effects of the relativity of simultaneity on the measurement of length. It is not appropriate for the distance between the rockets because the distance is changing. Ergospherical has pointed you in the correct direction.

So, how we will calculate the time elapsed in B's frame using B's referance frame?

 

[Ibix]

So, how we will calculate the time elapsed in B's frame using B's referance frame?

Given the question as stated I don't think you can (edit: at least, there's some guesswork required). Have you quoted all of the question exactly? Where did you get it from?

 

[Abhishek11235]

Given the question as stated I don't think you can (edit: at least, there's some guesswork required). Have you quoted all of the question exactly? Where did you get it from?

The question was taken from the book A Guide to Physics Problem, Vol-1, Mechanics, Relativity, Electrodynamics by Cahn and Nadgorny. This is just 2nd question under Relativity section. The question to find the time elapsed in the reference frame of B was not asked but I thought about that question. I am attaching the Screenshot of the Problem

 

[Ibix]

Ok - that question is answerable. Your paraphrase is not. That's why it's important to state the question exactly.

The recipe for doing any special relativity problem is to write down the coordinates of every interesting event in a frame where you can, then transform into the frame you want and derive any quantities you want to find.

What are the interesting events for B's measure of time between b and collision?

 

[Nugatory]

Can you please explain it in more detail?

You have stated that at time $t=0$ B and C are separated by a distance $l$ using the frame in which A is at rest. That is tantamount to saying that B passes through the point $x=X_0$ at the same time that C passes through the point $X_0+l$, and B and C both start measuring the time to collision as they pass through those points in space at the same time.

But when we use the frame in which B is at rest or the frame in which C is at rest the two “start measuring” events do not happen at the same time (and neither happens at the same time that A starts measuring the time to collision). That’s relativity of simultaneity and you were mistakenly not allowing for it.

Note also that the A and B in the textbook problem are not the A and B in your original post. I’ve gone with the wording in your original post, but be alert to possible confusion in other answers.

 

[ergospherical]

Let Event 1 be when the 2 rockets are separated by length l,t=0 and Even 2 be when they collided. Now, the length observed by B is by defination happens when $\Delta t'=0$ (primes denotes measurement in B, unprimed in A). Then, we have:$$\Delta t'=0 \Rightarrow \Delta t=V_a\Delta x/c^2$$Using this, we have:$$\Delta x'=\gamma(\Delta x-V_a\Delta t)=\gamma \Delta x(1-V_a^2/c^2)=\Delta x/\gamma$$

This isn't right because the $\Delta t'$, $\Delta t$ and $\Delta x$ in your first equation don't actually correspond to the same pair of events as the $\Delta x'$, $\Delta x$, and $\Delta t$ in your second equation.

 

[Abhishek11235]

Ok - that question is answerable. Your paraphrase is not. That's why it's important to state the question exactly.

The recipe for doing any special relativity problem is to write down the coordinates of every interesting event in a frame where you can, then transform into the frame you want and derive any quantities you want to find.

What are the interesting events for B's measure of time between b and collision?

First, from the frame of Earth observer (whom I called A):
Event 1: t=0, Coordinates of Rocket B=0 (Only X-Coordinates as the problem is 1D), Coordinates of Rocket C= $l$

Event 2: t=$l/(V_a+V_b)$, coordinates of B= $V_al/(V_a+V_b)$, coordinates of C=$l(1-V_a/(V_a+V_b))$

From the frame of Rocket B:
Event 1: Coordinates of Rocket C= $l/\gamma$ (See Post 6) (Edit: This is wrong!), t=0 (We assume that Clock of Earth observer and Rocket observer are initially Synchronised at the origin)
Event 2 (Collision): Coordinates of C= 0, t=?

Is this correct?

 

[ergospherical]

It's a bit hard to follow your work; write $\mathbf{x}_1 = (0,0)$ and $\mathbf{x}_2 = \dfrac{l}{v_A + v_B} \left( 1 , v_A \right)$ then putting the frames $K$ and $K'$ in standard configuration, use the Lorentz transformations to work out the $t'$ coordinate of $\mathbf{x}'_2$.

 

[Abhishek11235]

This isn't right because the $\Delta t'$, $\Delta t$ and $\Delta x$ in your first equation don't actually correspond to the same pair of events as the $\Delta x'$, $\Delta x$, and $\Delta t$ in your second equation.

Ok. I think I have confused myself in taking events.

It's a bit hard to follow your work; write $\mathbf{x}_1 = (0,0)$ and $\mathbf{x}_2 = \dfrac{l}{v_A + v_B} \left( 1 , v_A \right)$ then putting the frames $K$ and $K'$ in standard configuration, use the Lorentz transformations to work out the $t'$ coordinate of $\mathbf{x}'_2$.

Using this, we get (Lorentz Transformation to go from A to B):
$\Delta x'= \gamma(lV_a/(V_a+V_b)-V_al/(V_a+V_b)=0$
$t'=\gamma(lV_a/(V_a+V_b)-V_a(lV_a/(V_a+V_b))/c^2 = lV_a/(\gamma (V_a+V_b)$

 

[pervect]

The way I would solve this is to use frame independent invariants.

We have three frames, the Earth frame, A's, frame, and B's frame.. So we need some notation, perhaps unprimed for Earth, primed for A's frame, and double primed for B's frame.

We need to find the time and space coordinates of all the events in the Earth's frame to proceede. The events will be left, collision, and right.

I'll leave you to work those out in terms of the problem parameters, I'll just generalize the point I'm making.

Earth frame, left event. t=0, $x=x_{left}$
Earth frame, middle (collision) event. $t=t_{collide}$, $x=x_{collide}$
Earth frame, right event. t=0, $x=x_{right}$

It's your choice what spatial coordinates to use in the unprimed Earth frame. One can make a case for setting the spatial coordinate of the collision event equal to zero, in which case $x_{left} < 0$ and $x_{right} > 0$. But you might also want to make the coordinate of the left event 0. It's up to you. It won't matter to the end answer.

The main thing you haven't worked out yet is the collision point's spatial coordinate.

The invariant space-time interval between the left event and the collision event as expressed in the Earth frame is
$$c^2 \Delta t^2 - \Delta x^2$$
where
$$\Delta t = t_{collision}-0 \quad \Delta x = x_{collision} - x_{left}$$

This must equal the invariant interval between the left event and the collision event in A's frame, which we are representing in the prime frame:

$$c^2 \Delta t'^2$$

because in A's (primed) frame, the x-coordinate of A is always zero.

This application of the invariant space-time interval allows you to solve for the time in A's (primed) frame, $\Delta t'$. If your textbook doesn't discuss invariants, you might want to provet, using the Lorentz transform, first that
$c^2 t^2 - x^2 = c^2 t'^2 - x'^2$. This shows that the invariant interval between the points (0,0) and (t,x) is constant. It's a small step to generalize from the invariance of this quantity between the origin and any point to the invariance of this quantity between any two points by noting that nothing in the formula is dependent on the choice of the coordinate origins.

It's not strictly necessary, but you might want to apply the Lorentz transform to work out explicitly t and x in the primed frame (A's frame) and the double primed frame (B's prime) to gain some insight.

If you do so, the relativity of simultaneity is shown by the fact that in general, $t_{left}=0$ does NOT mean that $t'_{left}=0$ and $t''_{left}=0$. Details on what $t'_{left}$ and $t''_{left}$ are will depend on the details of your coordinate choices, though.

For more on the relativity of simultaneity, see any of the many threads about the phrase, and/or about "Einstein's train". Einstein's discussion of this can be found at https://www.bartleby.com/173/9.html. It is very likely that whatever text you are using will also have some discussion of the relativity of simultaneity. This is well worth looking up, as about 99 percent of the problems people have with special relativity involves not understanding the relativity of simultaneity. There is also a paper by Scherr and other authors about this topic, which describes how difficult it is to teach. See http://www.physics.umd.edu/perg/papers/scherr/ScherrAJP2.pdf, "The challenge of changing deeply held student beliefs about the relativity of simultaneity".

 

[Abhishek11235]

Ok. I think I have confused myself in taking events.Using this, we get (Lorentz Transformation to go from A to B):
$\Delta x'= \gamma(lV_a/(V_a+V_b)-V_al/(V_a+V_b)=0$
$t'=\gamma(lV_a/(V_a+V_b)-V_a(lV_a/(V_a+V_b))/c^2 = lV_a/(\gamma (V_a+V_b)$

I have made a mistake in typing. It should be:

$t'=l/(V_a+V_b)\gamma$
This the time of the collision as observed in the referance frame of Rocket A which does equal the way I calculated time elapsed in post #1. $x'$ is 0 as expected since the event takes place at the origin in the frame of A.

The way I would solve this is to use frame independent invariants.

We have three frames, the Earth frame, A's, frame, and B's frame.. So we need some notation, perhaps unprimed for Earth, primed for A's frame, and double primed for B's frame.

We need to find the time and space coordinates of all the events in the Earth's frame to proceede. The events will be left, collision, and right.

I'll leave you to work those out in terms of the problem parameters, I'll just generalize the point I'm making.

Earth frame, left event. t=0, $x=x_{left}$
Earth frame, middle (collision) event. $t=t_{collide}$, $x=x_{collide}$
Earth frame, right event. t=0, $x=x_{right}$

It's your choice what spatial coordinates to use in the unprimed Earth frame. One can make a case for setting the spatial coordinate of the collision event equal to zero, in which case $x_{left} < 0$ and $x_{right} > 0$. But you might also want to make the coordinate of the left event 0. It's up to you. It won't matter to the end answer.

The main thing you haven't worked out yet is the collision point's spatial coordinate.

The invariant space-time interval between the left event and the collision event as expressed in the Earth frame is
$$c^2 \Delta t^2 - \Delta x^2$$
where
$$\Delta t = t_{collision}-0 \quad \Delta x = x_{collision} - x_{left}$$

This must equal the invariant interval between the left event and the collision event in A's frame, which we are representing in the prime frame:

$$c^2 \Delta t'^2$$

because in A's (primed) frame, the x-coordinate of A is always zero.

This application of the invariant space-time interval allows you to solve for the time in A's (primed) frame, $\Delta t'$. If your textbook doesn't discuss invariants, you might want to provet, using the Lorentz transform, first that
$c^2 t^2 - x^2 = c^2 t'^2 - x'^2$. This shows that the invariant interval between the points (0,0) and (t,x) is constant. It's a small step to generalize from the invariance of this quantity between the origin and any point to the invariance of this quantity between any two points by noting that nothing in the formula is dependent on the choice of the coordinate origins.

It's not strictly necessary, but you might want to apply the Lorentz transform to work out explicitly t and x in the primed frame (A's frame) and the double primed frame (B's prime) to gain some insight.

If you do so, the relativity of simultaneity is shown by the fact that in general, $t_{left}=0$ does NOT mean that $t'_{left}=0$ and $t''_{left}=0$. Details on what $t'_{left}$ and $t''_{left}$ are will depend on the details of your coordinate choices, though.

For more on the relativity of simultaneity, see any of the many threads about the phrase, and/or about "Einstein's train". Einstein's discussion of this can be found at https://www.bartleby.com/173/9.html. It is very likely that whatever text you are using will also have some discussion of the relativity of simultaneity. This is well worth looking up, as about 99 percent of the problems people have with special relativity involves not understanding the relativity of simultaneity. There is also a paper by Scherr and other authors about this topic, which describes how difficult it is to teach. See http://www.physics.umd.edu/perg/papers/scherr/ScherrAJP2.pdf, "The challenge of changing deeply held student beliefs about the relativity of simultaneity".

The problem can be solved using this or else the method I gave in Post #1 (using time dilation). However, I don't want to use any other frame (Like applying Time Dilation formula or using Lorentz Invariants). I want to answer it in the Rocket's frame on how he will explain the time elapsed on his clock. But the PDF you have was really useful. Thanks!

 

[robphy]

In general, a spacetime diagram is useful tool analyze the problem.
It organizes and helps give meaning to the various quantities in the problem.
Often, it suggests what boils down to
a trigonometry problem (or two) involving hyperbolic functions
analyzed as you might a free-body diagram.

Upon recognizing that $4.2\times 10^8{\rm\ m}=1.4\mbox{ light-seconds}$ and you were given $v_A=0.8c$ and $v_B=-0.6c$,
it turns out that the diagram is easy to draw, with relatively-simple arithmetic.
(These velocities are "nice" velocities because they are associated with Pythagorean triples and rational Doppler factors.)
The relative velocity formula might be the most complicated,
but the arithmetic isn't bad if one works with fractions.

 

[pervect]

To find the answer in A's frame, one approach, which I'll call the usual approach, is to first adopt a specific coordinate system to describe the Earth. Then, one find the coordinates of all the events in the Earth's frame using these coordinates. You've done most (though not all) of the work on this already, I think.

Letting unprimed coordinates represent Earth frame coordinates, and primed coordinates representing rocket A's coordinates, you can find the coordinates of all events in rocket A's frame S' via the Lorentz transform. Given that you know the coordinates of all events in the unprimed Earth frame, you take each event of interest as described by a pair of coordinates (t,x) of an event in frame S, then use the Lorentz transform to computes the corridantes of this event in frame S'.

Your text probably has this, but the Lorentz transform can be written as:
$$t' = \gamma (t - \beta x/c) \quad x' = \gamma(x - \beta c t)$$
here $\beta = v/c$, $\gamma = 1/\sqrt{1-\beta^2}$, v being the velocity of the origin of frame S' in frame S.

Some versions replace the normalized velocity $\beta$ with the unnormalized velocity v, but I prefer to use the normalized velocity.

In your diagram, you have two starting events, one occurring at position a on the diagram at a time t=0 in the Earth frame, and another occurring at position b on the same diagram at a time t=0 in the Earth frame. We will call these events a and b, respectively. And of course you have the collision event as well. Not that events a and b occur at the same time in frame S, but do not occur at the same time in frame S' (A's frame), or frame S'' (B's frame)!

Getting the answer to the problem is just going through the math to apply the formula, then interpreting the results.

Plotting the results on a space-time diagram can also be helpful.

The relativity of simultaneity enters the picture in the fact that events a and b occur at the same time (t=0) in frame S, but do not occur at the same time in the frame of rocket A, frame S', as has been mentioned.

In words:
The Lorentz transform can be thought of as composed of three parts, length contraction, time dilation, and the relativity of simultaneity. The last we've discussed, and is usually the roadblock.

Length contraction means the spatial distance between a and b is shorter in frame S' than it is in S.

Time dilation means that in frame S, A's clock appears to run slower, while in frame S', the Earth clock seems to run slower.

You may also want to work things out from the perspective of rocket B - the concepts will be the same, but the numbers slightly different since B is moving in the opposite direction (sign differences in the velocity) and has a different speed.

There are also graphical techniques you can use using light clocks if you don't like doing all the math. But that does requires you to be comfortable with drawing and interpreting the space-time diagram of a light clock, and also comfortable with light clocks, for that matter. The Lorentz transform method requires you to be comfortable with the approach of assigning coordinates, transforming them, then interpreting the physical picture given the coordinates.

If you think you are more comfortable with drawing and interpreting diagrams than assigning coordinates, doing the math to transform them, then interpreting these coordinate results, there's a paper by one of our PF posters, robphy that I rather like, "Relativity on Rotating Graph paper". The preprint version is available at https://arxiv.org/abs/1111.7254, as far as I know the published version, which is slightly different, is still paywalled. He also has an insight article on PF, I believe.

You don't necessarily have to limit yourself to one approach, but you need to see at least one approach through to the end to gain understanding. Which approach appeals to you more is up to you, along with possibly changing your mind and/or spending the time to learn both.