How Do Relativity and Time Dilation Affect the Time Between Events?

[RodolfoM]

Homework Statement

Two rockets, both with lenght L, travel at opposite directions with speed 0.8c relative to the distant stars. Calculate how much time it will pass, relative to an observer that is moving along with one of the rockets, since the moment the rockets cross each other until they have completely surpassed one another.

Relevant Equations

ϒ=sqrt(1-v²/c²)
Length contraction, time dilation

Firstly, I calculated the relative speed between the two rockets, finding v=1.6c/1.64. Then, I applied the length contraction: the length of the moving rocket will be 0.6L due to this phenomena, so the total distance traveled by one rocket, with respect to the the other, will be 1.6L. Therefore, the time they will take to pass one another will be
t=1.6L/(1.6c/1.64)=1.64L/c.
Is this correct? How would I solve the problem finding the time with respect to the distant stars and then applying the time dilation?
Thanks!

 

[Orodruin]

Homework Statement: Two rockets, both with length L, travel at opposite directions with speed 0.8c relative to the distant stars. Calculate how much time it will pass, relative to an observer that is moving along with one of the rockets, since the moment the rockets cross each other until they have completely surpassed one another.
Homework Equations: ϒ=sqrt(1-v²/c²)
Length contraction, time dilation

Is this correct?

Your reasoning looks fine. I did not check the numerical computations.

Homework Statement: Two rockets, both with length L, travel at opposite directions with speed 0.8c relative to the distant stars. Calculate how much time it will pass, relative to an observer that is moving along with one of the rockets, since the moment the rockets cross each other until they have completely surpassed one another.
Homework Equations: ϒ=sqrt(1-v²/c²)
Length contraction, time dilation

How would I solve the problem finding the time with respect to the distant stars and then applying the time dilation?

This is quite a bit more tricky because the events are not co-located in the rocket frames. The easiest way would be to compute the time difference in the star frame and then Lorentz transform the coordinates of the events to one of the ship frames. The time in the ship frames would be the difference between the resulting time coordinates.

 

[Doc Al]

Firstly, I calculated the relative speed between the two rockets, finding v=1.6c/1.64.

Looks good.

Then, I applied the length contraction: the length of the moving rocket will be 0.6L

How did you get this result?

 

[Janus]

Homework Statement: Two rockets, both with length L, travel at opposite directions with speed 0.8c relative to the distant stars. Calculate how much time it will pass, relative to an observer that is moving along with one of the rockets, since the moment the rockets cross each other until they have completely surpassed one another.
Homework Equations: ϒ=sqrt(1-v²/c²)
Length contraction, time dilation

View attachment 251664
Firstly, I calculated the relative speed between the two rockets, finding v=1.6c/1.64.

Okay, so v = ~0.9756c

Then, I applied the length contraction: the length of the moving rocket will be 0.6L due to this phenomena,

That's the length contraction as measured in the frame that the rockets are traveling at 0.8c relative to.
You've already determined that, as measured from either rocket, the relative speed is 0.9756c. This is what you need to use to determine the length contraction of the other rocket as measured by either of the rockets.

so the total distance traveled by one rocket, with respect to the the other, will be 1.6L. Therefore, the time they will take to pass one another will be
t=1.6L/(1.6c/1.64)=1.64L/c.
Is this correct? How would I solve the problem finding the time with respect to the distant stars and then applying the time dilation?
Thanks!

 

[TSny]

How would I solve the problem finding the time with respect to the distant stars and then applying the time dilation?

The two important events are the event where the noses of the rockets pass each other and the event where the tails of the rockets pass each other.

In what frame of reference do these two events occur at the same place?
What is the time interval between the two events in this frame?

 

[PeroK]

Homework Statement: Two rockets, both with length L, travel at opposite directions with speed 0.8c relative to the distant stars. Calculate how much time it will pass, relative to an observer that is moving along with one of the rockets, since the moment the rockets cross each other until they have completely surpassed one another.
Homework Equations: ϒ=sqrt(1-v²/c²)
Length contraction, time dilation

View attachment 251664
Firstly, I calculated the relative speed between the two rockets, finding v=1.6c/1.64. Then, I applied the length contraction: the length of the moving rocket will be 0.6L due to this phenomena, so the total distance traveled by one rocket, with respect to the the other, will be 1.6L. Therefore, the time they will take to pass one another will be
t=1.6L/(1.6c/1.64)=1.64L/c.
Is this correct? How would I solve the problem finding the time with respect to the distant stars and then applying the time dilation?
Thanks!

If you stick with the algebra instead of plugging in numbers, you might get a neat answer.

PS or use relativity of simultaneity and time dilation and length contraction (aka Lorentz transformation) in the original frame you get the same extraordinary answer.

 

[RodolfoM]

Okay, so v = ~0.9756c That's the length contraction as measured in the frame that the rockets are traveling at 0.8c relative to.
You've already determined that, as measured from either rocket, the relative speed is 0.9756c. This is what you need to use to determine the length contraction of the other rocket as measured by either of the rockets.

Thank you! But when I use the relative speed, some weird answer shows up. Let's say the rockets are moving not with $0.8c$, but with $\beta c$. The relative velocity will be
$$v_{rel}=\frac{\beta c+\beta c}{1+ \frac{\beta c\cdot\beta c}{c^2}}=\frac{2\beta c}{1+\beta^2}$$
In this reference frame, the moving rocket will have a length of
$$L'=L\sqrt{1-\frac{v_{rel}^2}{c^2}}=L\frac{(1-\beta^2)}{(1+\beta^2)}$$
Thus, the total displacement will be
$$\Delta S=L+L'=\frac{2L}{1+\beta^2}$$
Finally, the elapsed time will be
$$\Delta t=\frac{\Delta S}{\beta c}=\frac{2L}{1+\beta^2}\cdot\frac{1+\beta^2}{2\beta c}=\frac{L}{\beta c}$$
However, this result does not make sense, as it is simply the result for non-relativistic speeds, or does it?

 

[RodolfoM]

The two important events are the event where the noses of the rockets pass each other and the event where the tails of the rockets pass each other.

In what frame of reference do these two events occur at the same place?
What is the time interval between the two events in this frame?

In a fixed inertial frame of reference (distant stars), the events would occur at the same place, let's say, at the origin of the reference frame. In this frame, both rockets will have their lenghts contracted:
$$L'=L\sqrt{1-\beta^2}$$
As one of the rockets moves to the right with speed $\beta c$, the other moves to the left with the same speed, so the total displacement will be equal $L'$. Therefore, the time between events will be
$$\Delta t_0=\frac{L'}{\beta c}=L\frac{\sqrt{1-\beta^2}}{\beta c}$$
As in this reference frame the events are located in the same space coordinate, the time interval I just calculated should be the proper time, right? Hence, the time in the reference frame of one of the rockets would be
$$\Delta t=\frac{\Delta t_0}{\sqrt{1-\beta^2}}=\frac{L}{\beta c}$$
But this is the result expected by galilean relativity, is this right?

 

[TSny]

In a fixed inertial frame of reference (distant stars), the events would occur at the same place, let's say, at the origin of the reference frame. In this frame, both rockets will have their lenghts contracted:
$$L'=L\sqrt{1-\beta^2}$$
As one of the rockets moves to the right with speed $\beta c$, the other moves to the left with the same speed, so the total displacement will be equal $L'$. Therefore, the time between events will be
$$\Delta t_0=\frac{L'}{\beta c}=L\frac{\sqrt{1-\beta^2}}{\beta c}$$
As in this reference frame the events are located in the same space coordinate, the time interval I just calculated should be the proper time, right? Hence, the time in the reference frame of one of the rockets would be
$$\Delta t=\frac{\Delta t_0}{\sqrt{1-\beta^2}}=\frac{L}{\beta c}$$
But this is the result expected by galilean relativity, is this right?

That all looks good. Yes, a surprising result!

 

[RodolfoM]

That all looks good. Yes, a surprising result!

Shouldn't I obtain some expression such as when $\beta=1$ I'd get $\Delta t=0$?

 

[TSny]

Shouldn't I obtain some expression such as when $\beta=1$ I'd get $\Delta t=0$?

Letting $\beta = 1$ causes problems. The gamma factor blows up to infinity. So, formulas such as the time dilation formula become ambiguous.

 

[RodolfoM]

Letting $\beta = 1$ causes problems. The gamma factor blows up to infinity. So, formulas such as the time dilation formula become ambiguous.

It makes sense! I'm still at the bargaining stage but after plugging some numbers I'm heading towards acceptance. The result is really impressive, and it only happens because of the symmetry of the problem.
Consider that the left rocket is moving towards right with a speed of $\beta c$, and the other in the opposite direction with a velocity of $\alpha c$. Therefore, the relative speed will be:
$$u=\frac{c(\alpha+\beta)}{1+\alpha\beta}$$
And the Lorentz's factor:
$$\frac{1}{\gamma}=\sqrt{1-\frac{(\alpha+\beta)^2}{(1+\alpha\beta)^2}}=\frac{\sqrt{(1-\alpha^2)(1-\beta^2)}}{1+\alpha\beta}$$
The total displacement:
$$\Delta S=L+L'=L\Bigg[1+\frac{\sqrt{(1-\alpha^2)(1-\beta^2)}}{1+\alpha\beta}\Bigg]$$
Finally, the time:
$$\Delta t=\frac{L}{c}\frac{1+\alpha\beta+\sqrt{(1-\alpha^2)(1-\beta^2)}}{\alpha+\beta}$$
In the special case where $\alpha=\beta$, we have
$$\Delta t=\frac{L}{c}\frac{1+\beta^2+\sqrt{(1-\beta^2)^2}}{2\beta}=\frac{L}{\beta c}$$
If $\alpha=0.6$ and $\beta=0.8$, we will have
$$\Delta t=\frac{L}{c}\frac{1+0.6\cdot 0.8+\sqrt{0.64\cdot 0.36}}{0.6+0.8}=\frac{1.96L}{1.4c}=1.4\frac{L}{c}$$
If $\alpha=1$ (a beam of light), we'd obtain $\Delta t=L/c$, as expected.
Thank you very much!

 

[TSny]

It makes sense! I'm still at the bargaining stage but after plugging some numbers I'm heading towards acceptance. The result is really impressive, and it only happens because of the symmetry of the problem.
Consider that the left rocket is moving towards right with a speed of $\beta c$, and the other in the opposite direction with a velocity of $\alpha c$. Therefore, the relative speed will be:
$$u=\frac{c(\alpha+\beta)}{1+\alpha\beta}$$
And the Lorentz's factor:
$$\frac{1}{\gamma}=\sqrt{1-\frac{(\alpha+\beta)^2}{(1+\alpha\beta)^2}}=\frac{\sqrt{(1-\alpha^2)(1-\beta^2)}}{1+\alpha\beta}$$
The total displacement:
$$\Delta S=L+L'=L\Bigg[1+\frac{\sqrt{(1-\alpha^2)(1-\beta^2)}}{1+\alpha\beta}\Bigg]$$
Finally, the time:
$$\Delta t=\frac{L}{c}\frac{1+\alpha\beta+\sqrt{(1-\alpha^2)(1-\beta^2)}}{\alpha+\beta}$$
In the special case where $\alpha=\beta$, we have
$$\Delta t=\frac{L}{c}\frac{1+\beta^2+\sqrt{(1-\beta^2)^2}}{2\beta}=\frac{L}{\beta c}$$
If $\alpha=0.6$ and $\beta=0.8$, we will have
$$\Delta t=\frac{L}{c}\frac{1+0.6\cdot 0.8+\sqrt{0.64\cdot 0.36}}{0.6+0.8}=\frac{1.96L}{1.4c}=1.4\frac{L}{c}$$
If $\alpha=1$ (a beam of light), we'd obtain $\Delta t=L/c$, as expected.
Thank you very much!

Very nice analysis!

 

[Orodruin]

The second ship is a red herring. As the relevant events are colocated in the star frame, let us put the spatial origin of the star frame in that point. The question becomes: "How long does it take the origin of the star frame to pass the rocket as seen in the rocket frame?"
I assume nobody has conceptual problems with this one ...
(And if so, the solution is the length of the rocket divided by the speed of the origin of the star frame, which is $L/v$.)

 

[Orodruin]

The result is really impressive, and it only happens because of the symmetry of the problem.

Not only. Regardless of the lengths and relative speed of the rocket, there is always going to be one frame where the relevant events are colocated (see previous post). The symmetry of the problem (having equal length rockets) just ensures that this frame is the frame where both rockets have the same speed, but I could have the exact same result with different speeds if I changed the rest lengths of the rockets accordingly.

 

[Orodruin]

Shouldn't I obtain some expression such as when $\beta=1$ I'd get $\Delta t=0$?

What was said in #11 is true. You can however consider the physics of limiting case $\beta \to 1$. In that case the relative speed goes to $c$ and the second ship's length goes to 0 in the rest frame of the first. It takes an object of speed $c$ and 0 extension a time $L/c$ to pass a stationary object of length $L$ so the limit should be $L/c$.