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

In summary, based on the given information and equations such as length contraction and time dilation, we can calculate the relative speed between the two rockets to be v=1.6c/1.64 and the length of the moving rocket to be 0.6L. The total distance traveled by one rocket, with respect to the other, is 1.6L, resulting in a time of t=1.64L/c for the rockets to pass each other. To solve for the time with respect to the distant stars, we can use the Lorentz transformation to calculate the time difference in the star frame and then transform the coordinates to the ship frame. The resulting time in the ship frame would be the difference between the time coordinates
  • #1
RodolfoM
8
2
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
yjurty.png

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!
 
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  • #2
RodolfoM said:
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.

RodolfoM said:
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.
 
  • #3
RodolfoM said:
Firstly, I calculated the relative speed between the two rockets, finding v=1.6c/1.64.
Looks good.
RodolfoM said:
Then, I applied the length contraction: the length of the moving rocket will be 0.6L
How did you get this result?
 
  • #4
RodolfoM said:
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!
 
  • #5
RodolfoM said:
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?
 
  • #6
RodolfoM said:
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.
 
Last edited:
  • #7
Janus said:
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?
 
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  • #8
TSny said:
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?
 
  • #9
RodolfoM said:
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!
 
  • #10
TSny said:
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##?
 
  • #11
RodolfoM said:
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.
 
  • #12
TSny said:
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!
 
  • #13
RodolfoM said:
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!
 
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  • #14
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 ... :wink:
(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##.)
 
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  • #15
RodolfoM said:
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.
 
  • #16
RodolfoM said:
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##.
 
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FAQ: How Do Relativity and Time Dilation Affect the Time Between Events?

1. How does relativity affect the time between events?

The theory of relativity states that time and space are not absolute, but rather are relative to the observer's frame of reference. This means that the time between two events can be perceived differently by different observers depending on their relative speeds and positions.

2. Can time travel be possible with relativity?

According to the theory of relativity, it is possible for time to pass at different rates for objects in different frames of reference. This means that time travel could theoretically be possible by traveling at extremely high speeds or near massive objects like black holes. However, the feasibility of actual time travel is still a topic of debate among scientists.

3. Does time dilation occur in both general and special relativity?

Yes, time dilation occurs in both general and special relativity. Special relativity deals with the effects of relative motion on time, while general relativity includes the effects of gravity on time. Both theories predict that time will pass at different rates for objects in different frames of reference.

4. How does the speed of light factor into relativity and time between events?

The speed of light is a fundamental constant in the theory of relativity. According to the theory, the speed of light is the same for all observers, regardless of their relative motion. This means that the perception of time between events can be affected by an observer's speed, but the speed of light will always remain constant.

5. Can relativity explain the concept of time dilation in space travel?

Yes, relativity can explain the concept of time dilation in space travel. As an object approaches the speed of light, time appears to slow down for that object, meaning that less time passes for the moving object compared to a stationary observer. This phenomenon has been observed in astronauts who have traveled in space at high speeds for extended periods of time.

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