Special Relativity with Three reference frames

In summary: So we can find the speed of rocket C relative to the laboratory frame by finding the distance between event 1 and event 2 in the laboratory frame and dividing by the time between them in the laboratory frame. The distance is 6 light-years and the time is 10 years. So the speed of rocket C relative to the laboratory frame is 6/10 = 0.6 c.In summary, the conversation discusses events occurring at different positions and times in various frames of reference. These events are related through the equations s^2 = c^2Δt^2-Δx^2 and t = (1/√1-v^2/c^2)t'. To find the position of an event
  • #1
scienceLilly75
4
0

Homework Statement


In the laboratory frame, event 1 occurs at x = 0 light-years, t = 0 years. Event 2 occurs at x = 6 light-years, t = 10 years. In all rocket frames, event 1 also occurs at the position 0 light-years and the time 0 years. The y- and z- coordinates of both events are zero in all frames.
a) In rocket frame A, event 2 occurs at time t’ = 14 years. At what position x’ will event 2
occur in this frame?
b) In rocket frame B, event 2 occurs at position x’’ = 5 light-years. At what time t’’ will
event 2 occur in this frame?
c) How fast must rocket frame C move if events 1 and 2 occur at the same place in this
rocket frame?

Homework Equations


s^2 = c^2Δt^2-Δx^2
t = (1/√1-v^2/c^2)t'

The Attempt at a Solution


For part A, I did c^2Δt^2-Δx^2 = c^2Δt'^2-Δx'^2. Because the everything in the rocket frames start at 0, then the entire left side becomes 0 and I get. √c^2 * (14)^2 = Δx' which is 4.2*10^9 or 14c.
I did the same thing for part B. The c^2Δt^2-Δx^2 = c^2Δt''^2-Δx''^2 and ended up with √5/c = 1.29*10^-4 = Δt''.
For part c, I know that I need to use t = (1/√1-v^2/c^2)t' . However, if I set t=0, then I get v rocket = c ---which shouldn't be right. It should be some fraction of c. My question is if if I am looking at a third reference frame, then do I use the values for time and position from the one before it? So to find the v in rocket frame C, I need to only use the data from rocket frame B?
 
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  • #2
Welcome to PF!

Hi scienceLilly75! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
scienceLilly75 said:
In the laboratory frame, event 1 occurs at x = 0 light-years, t = 0 years. Event 2 occurs at x = 6 light-years, t = 10 years. In all rocket frames, event 1 also occurs at the position 0 light-years and the time 0 years. The y- and z- coordinates of both events are zero in all frames.
a) In rocket frame A, event 2 occurs at time t’ = 14 years. At what position x’ will event 2
occur in this frame?

For part A, I did c^2Δt^2-Δx^2 = c^2Δt'^2-Δx'^2. Because the everything in the rocket frames start at 0, then the entire left side becomes 0 and I get. √c^2 * (14)^2 = Δx' which is 4.2*10^9 or 14c.

No …

i] if x is in light-years, and t is in years, then c = … ? :wink:

ii] the separation (c2t2 - x2) has to be between events 1 and 2, so it isn't 0 ! :smile:
 
  • #3
But at the beginning the time and position of the event is 0, so the left side has to be zero right?

Or do I need to find spatial separation between A and B, then spatial separation between B and C?
 
  • #4
Hi scienceLilly75! :smile:

(just got up :zzz:)
scienceLilly75 said:
But at the beginning the time and position of the event is 0, so the left side has to be zero right?

Or do I need to find spatial separation between A and B, then spatial separation between B and C?

sorry, but that doesn't even make sense …

A B and C are frames, and there's no separation between frames (spatial or otherwise)

separation is between two events

so which two events are you going to use? :smile:

(oh, and c = … ? :wink:)
 
  • #5
I am confused. I understand that there is no separation between frames but between events in frames. In that case, I am stuck. The only thing I know about the frame C is that event one occurs at t=0, x=0. If the events occur on the same place on the rocket, then x' also equals 0. But then that will leave me with t' = 0 using the spatial relativity equation. if t'=0, then it is not moving. Where am I going wrong?
 
  • #6
Hi scienceLilly75! :smile:
scienceLilly75 said:
I understand that there is no separation between frames but between events in frames.

Why do you say "in frames"? Separation is between two events, period. :wink:

Use event 1 and event 2. :smile:
 
  • #7
So to find the speed, I have to use info from event 1 and 2 how they appear in frame c? then I can use v = x/t?
 
  • #8
Make up a table. Label the rows Laboratory, Rocket A, Rocket B, Rocket C. Label the columns Event 1 Time, Event 1 Distance, Event 2 Time, Event 2 distance. Fill in the table with any information you have available. If there is information missing, fill in any algebraic variable name that you choose. Add a 5th column to the table in which you calculate the following quantity

((Event 2 Time) - (Event 1 Time))2 - ((Event 2 Distance) - (Event 1 Distance))2. This quantity must be the same for all rows of the table.
 
  • #9
scienceLilly75 said:
So to find the speed, I have to use info from event 1 and 2 how they appear in frame c? then I can use v = x/t?

why do you need for speed for part (a) ? :confused:

what is the separation between events 1 and 2 ?​
 
  • #10
Question 3 is a little ambiguous. I guess it is asking how fast rocket C is traveling relative to the laboratory frame of reference. The key to answering this question is recognizing that rocket C is physically present at both events.
 

What is the theory of special relativity?

The theory of special relativity, proposed by Albert Einstein in 1905, is a fundamental concept in physics that describes the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion and that the speed of light is constant in all inertial frames of reference.

What are the three reference frames in special relativity?

The three reference frames in special relativity are the rest frame, the moving frame, and the observer frame. The rest frame is the frame of reference in which an object is at rest, the moving frame is the frame of reference in which an object is moving at a constant velocity, and the observer frame is the frame of reference of an observer who is measuring the motion of the object.

How does time dilation work in special relativity?

According to special relativity, time dilation occurs when an object moves at high speeds relative to another object. This means that time on the moving object will appear to pass slower than time on the stationary object. This effect becomes more significant as the speed of the moving object approaches the speed of light.

What is the twin paradox in special relativity?

The twin paradox is a thought experiment in special relativity that involves two twins, one of whom stays on Earth while the other travels through space at high speeds. Due to time dilation, the traveling twin will age slower than the stationary twin, resulting in a paradox when they are reunited and have a significant age difference.

How does special relativity impact our understanding of space and time?

Special relativity has significantly impacted our understanding of space and time by changing the way we think about the relationship between the two. It has shown that space and time are relative to the observer and that they are not absolute concepts. This has led to a better understanding of the nature of the universe and has allowed for the development of technologies such as GPS that rely on the principles of special relativity.

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