Time interval between intermediate passing of two rods....

[Apashanka]

Homework Statement

Length contraction

Relevant Equations

$L_{rest}=\gamma L_{moving}$

The first part I have calculated is as follows::
Length of A seen by S =30m
Length of B seen by S=40m
In S frame,
Time for front of A and B to come in same line 0.8ct=40+0.6ct...t=200/c
From the above position time for back end of A and front of B to come in same line 0.8ct=30+0.6ct...t=150...
therefore total time for overtake is 350/c...

For the second part ...
From B frame ,A is approaching towards it with velocity 5c/13
The length of A as seen by B is 12*50/13=46.153...
Therefore time taken to completely overtake A as seen by B is coming to be...
1..Time taken by front of A for traveling to the front of B is 50/(5c/13)=130/c...
From the above position time taken by back end of A for traveling to the front of B 46.153/(5c/13)=120/c..
Therefore total time taken for overtaking is 250/c...
Are these two calculations done correct??
If so then can we find the interval in B frame directly from the interval in S frame...in that case the time interval from time dilation is coming to $\sqrt(1-0.6^2)*350/c$ which is 280/c...but the time interval in B frame is coming 250/c??

 

[PeroK]

You ought to use the Lorentz transformation here.

 

[PeroK]

... your mistake was to use length contraction and time dilation and forget about the relativity of simultaneity.

PS There ought to be a part c) which asks for the time interval in the A-frame.

 

[Apashanka]

Post ##1
Therefore calculation of 350/c in S frame and 250/c in B frame is correct??

 

[PeroK]

Post ##1
Therefore calculation of 350/c in S frame and 250/c in B frame is correct??

Yes.

 

[spacelover]

Yes.

No it is wrong
0.8ct - 0.6ct=40 and 0.8ct -0. 6ct =30
Now imagine the are moving in opposite direction then the equation will become
0.8ct +0.6ct=40
t=40m/1.2c
But hiw can v=1.2c >c
You are solving it in wrong way
Correct way, t for A to completely cross B as see from. S or wrt s frame= distance which A has to travel for complete overtake divide by velocity of A seen from s with which it has to trave this distance
t=Length contraction if A + length contraction of B that is length seen from s divide by 0.8c velocity of A with which it has to travel this distance as seen from s
t=30+40/0.8c=70/0.8c= 29.1 x 10 power minus 8=291 nank second. You can check the answer key..

 

[PeroK]

No it is wrong
0.8ct - 0.6ct=40 and 0.8ct -0. 6ct =30
Now imagine the are moving in opposite direction then the equation will become
0.8ct +0.6ct=40
t=40m/1.2c
But hiw can v=1.2c >c

I'm not sure what this means. $1.2c$ would be perfectly valid as the speed at which A and B are converging or separating, as measured in frame S.

You are solving it in wrong way
Correct way, t for A to completely cross B as see from. S or wrt s frame= distance which A has to travel for complete overtake divide by velocity of A seen from s with which it has to trave this distance
t=Length contraction if A + length contraction of B that is length seen from s divide by 0.8c velocity of A with which it has to travel this distance as seen from s
t=30+40/0.8c=70/0.8c= 29.1 x 10 power minus 8=291 nank second. You can check the answer key..

I'm not sure what that means either.

The calculation for part a) is simply $\Delta t_S = \dfrac{30m + 40m}{0.2c}$, where $0.2c$ is the speed with which A is overtaking B, as measured in frame S. That gives $\Delta t_S = 1.17 \times 10^{-6}s$.

The calculation for b) involves either the Lorentz transformation or velocity addition. In either case, I think the answer of $\Delta t_B = \dfrac{250m}{c} = 8.33 \times 10^{-7}s$ is correct.

 

[TSny]

Are these two calculations done correct??

I concur with @PeroK that @Apashanka's calculation of the time intervals 350 m/c (for S) and 250 m/c (for B) is correct.

If so then can we find the interval in B frame directly from the interval in S frame...in that case the time interval from time dilation is coming to $\sqrt(1-0.6^2)*350/c$ which is 280/c...but the time interval in B frame is coming 250/c??

The time dilation formula relates the time interval between two events, as measured by two different inertial observers, for the case where the two events occur at the same location for one of the two observers. (This observer measures the "proper time" between the two events.) But you can see that the two events of interest in this problem do not occur at the same place for any of the observers (S, A, or B).

[For example, if one of the two events is chosen to be the event when the front of ship A passes the rear of ship B and the second event is when the front of ship A passes the front of ship B, then the observer in ship A would say these events occurred at the same place. So the time interval between the two events as determined by observers A and B would be related by the time dilation formula. Observer A would measure the proper time interval.]

 

[spacelover]

Homework Statement: Length contraction
Relevant Equations: $L_{rest}=\gamma L_{moving}$

View attachment 262753

The first part I have calculated is as follows::
Length of A seen by S =30m
Length of B seen by S=40m
In S frame,
Time for front of A and B to come in same line 0.8ct=40+0.6ct...t=200/c
From the above position time for back end of A and front of B to come in same line 0.8ct=30+0.6ct...t=150...
therefore total time for overtake is 350/c...

For the second part ...
From B frame ,A is approaching towards it with velocity 5c/13
The length of A as seen by B is 12*50/13=46.153...
Therefore time taken to completely overtake A as seen by B is coming to be...
1..Time taken by front of A for traveling to the front of B is 50/(5c/13)=130/c...
From the above position time taken by back end of A for traveling to the front of B 46.153/(5c/13)=120/c..
Therefore total time taken for overtaking is 250/c...
Are these two calculations done correct??
If so then can we find the interval in B frame directly from the interval in S frame...in that case the time interval from time dilation is coming to $\sqrt(1-0.6^2)*350/c$ which is 280/c...but the time interval in B frame is coming 250/c??

I'm not sure what this means. $1.2c$ would be perfectly valid as the speed at which A and B are converging or separating, as measured in frame S.

I'm not sure what that means either.

The calculation for part a) is simply $\Delta t_S = \dfrac{30m + 40m}{0.2c}$, where $0.2c$ is the speed with which A is overtaking B, as measured in frame S. That gives $\Delta t_S = 1.17 \times 10^{-6}s$.

The calculation for b) involves either the Lorentz transformation or velocity addition. In either case, I think the answer of $\Delta t_B = \dfrac{250m}{c} = 8.33 \times 10^{-7}s$ is correct.

Same question same part same data
But what if spaceship A and Spaceship B are moving in opposite direction, then according to your logic we will get 0.8ct +0.6ct=3.6
t=3.6÷1.4c i. e d/v
1.4c=V>c...it violates the of special theory of relativity..

 

[PeroK]

View attachment 338272
Same question same part same data
But what if spaceship A and Spaceship B are moving in opposite direction, then according to your logic we will get 0.8ct +0.6ct=3.6
t=3.6÷1.4c i. e d/v
1.4c=V>c...it violates the of special theory of relativity..

If A and B are moving towards each other at $0.8c$ and $0.6c$ respectively, then their separation speed is $1.4c$. All as measured in some inertial reference frame. The maximum separation speed in SR is $2c$.

 

[kuruman]

If A and B are moving towards each other at $0.8c$ and $0.6c$ respectively, then their separation speed is $1.4c$. All as measured in some inertial reference frame. The maximum separation speed in SR is $2c$.

What happened to the relativistic addition of velocities?

If b moves relative to a with velocity $v_{ba}$ and c moves relative to b with velocity $v_{cb}$, then the relativistic addition of velocities says that $$v_{ca}=\frac{v_{cb}+v_{ba}}{1+\frac{v_{cb}~v_{ba}}{c^2}}.$$Here, we identify a = S, b = A and c = B. The velocity addition equation becomes $$v_{BS} =\frac{v_{BA} + v_{AS}}{1+\frac{v_{BA}~v_{AS}} {c^2}}.$$ We solve this equation to get the velocity of B relative to A,$$v_{BA} =\frac{v_{BS} - v_{AS}}{1-\frac{v_{BS}~v_{AS}} {c^2}}.$$ If we take $B$ to be moving in the negative direction, $v_{AS}=0.8 c~;~~v_{BS}=-0.6c$ and $$v_{BA} =\frac{-0.6c - 0.8c}{1-\frac{(-0.6c)(0.8c)} {c^2}}=\frac{-1.4 c}{1.48}=-0.946c.$$ Even if they move in opposite directions at the speed of light, their relative speed will be $c$.

 

[PeroK]

If A is moving away from the origin to the left with a speed of nearly $c$ and B is moving away to the right with a speed of nearly $c$, then they are separating at a speed/rate of nearly $2c$.

That's different from their relative speed, which involves a transformation to the rest frame of either A or B.

 

[kuruman]

If A is moving away from the origin to the left with a speed of nearly $c$ and B is moving away to the right with a speed of nearly $c$, then they are separating at a speed/rate of nearly $2c$.

That's different from their relative speed, which involves a transformation to the rest frame of either A or B.

What is this speed/rate and how is it defined mathematically and conceptually?

 

[PeroK]

What is this speed/rate and how is it defined mathematically and conceptually?

It's the speed at which the distance between two objects changes with time.

It's relevant to this problem, as that allows a calculation to be done in the original frame.

 

[jbriggs444]

What is this speed/rate and how is it defined mathematically and conceptually?

Begin by picking a coordinate system. "Separation speed" is the rate at which the coordinate distance between two point-like objects increases (or decreases) compared to coordinate time.

 

[kuruman]

So it's just $\frac{d}{dt}(x_A-x_B)$ where $x_A$ and $x_B$ are the coordinates, in this case, of A and B measured in the unprimed rest frame of S. Yes, it's relevant to this problem but I didn't know it had a name and this statement confused me

If A and B are moving towards each other at $0.8c$ and $0.6c$ respectively, then their separation speed is $1.4c$.