# Homework Help: Special Relativity (time/length)

1. Jan 30, 2017

### Nick Jarvis

1. The problem statement, all variables and given/known data
Spacecraft moving at 0.6c, length 180m with someone sitting at the back. He throws a cage with a mouse towards the front also at a speed of 0.6c. A stationary observer watches.

2. Relevant equations
What is length of spacecraft from observers POV? Length according to guy sitting at back? Length according to little mouse? How long does it take for cage to reach fron of ship according to mouse and observer? How fast (according to observer) does mouse travel

3. The attempt at a solution
According to Galilean velocity, I would say speed according to observer was 1.2c. However, as this is not possible then my answer would be c.
The length according to observer would be 144m ie 180 m x sq route of (1 - v^2/c^2)
Struggling to work out time/length as seen by mouse and how long it takes for mouse to reach the front.
Cheers

2. Jan 30, 2017

### Staff: Mentor

This is impossible as well.

It is a question about special relativity. You'll have to use special relativity to answer it.

144 meters is correct.
What did you try?
The different parts can be calculated in every reference frame, but often some frames are easier than others. If you get stuck, try a different reference frame.

3. Jan 30, 2017

### Nick Jarvis

Many thanks. Will take another look this evening and post what I have. Cheers :)

4. Jan 30, 2017

### Nick Jarvis

Thanks mfb. I have been through my notes properly and found the equation: u + v/1 - uv/c^2

With u and v = 0.6c, my answer is now 0.88c. So the observer views the speed travelled as 0.88c

Also, I take it from my notes that the astronaut on board will view the length as 180m. I now need to work out how long the craft is as seen by the flying mouse.

Cheers

5. Jan 30, 2017

### Staff: Mentor

Correct so far.

6. Feb 3, 2017

### Nick Jarvis

Morning

Still so confused as to what I need to do.
So far

Space craft (190m long at rest) moving at 0.6c so observer at rest will see the length as proper length / Lorentz constant. In this case the constant is 1.67 so:

Lenth = 190m / 1.67 = 113.78m

Therefore observer sees the length as 113.78m and sees the speed as 0.88c

The astronaut at the back of the craft will see the length as 190m

The astronaut fires an object also at 0.6c relative to the rocket. To calculate the time taken for object to reach the front,

Astronaut:

t = 190 / 0.6c = 1.06 x 10^-7 seconds

Observer:

He/she will calculate the time taken as length/velocity PLUS the the time taken according to astronaut, so t = 113.78m / 0.88c = 4.31 x 10^-7 PLUS 1.06 x 10^-7 seconds. Is this correct?

How long will the spacecraft be according to the object being thrown? Will this be the same as the observer's calculated length, as the object also moving at 0.6c??

I am finding this so difficult to grasp.

Thanks

7. Feb 3, 2017

### PeroK

Apart from the rest length of the spaceship changing from 180m to 190m, you were right on the first part.

Regarding the object moving within the spaceship: yes, it is an important and valid point that to them the spaceship is moving at 0.6c backwards, so their observations will agree with the "stationary" observer in terms of length contraction.

The final point is the time for this object to travel the length of the ship according to the stationary observer. The trick here is to note the following fact:

In SR (as in classical physics), if you stay within a single IRF, you may apply the normal addition/subtraction of velocities (for velocities in your frame).

If $v$ is the velocity of an object in your frame and $u$ is the velocity of another object in your frame, then the "separation" velocity of these two objects in your frame is simply:

$v_s = u - v$

In other words, SR requires no changes to this natural, classical calculation, as long as everything is measured in your reference frame. It's the same with distances. SR only requires changes when you consider velocities (or distances or times) measured in two different reference frames.

Note that the velocity addition formula relates velocities as measured in different frames, as follows.

If $V$ is the velocity of frame S' relative to your frame and $u'$ is the velocity of an object (as measured in frame S'), then the velocity of the object as measured in your fame is:

$u = \frac{V + u'}{1 + Vu'/c^2}$

8. Feb 3, 2017

### PeroK

I forgot to say. That's not valid in any form of physics. In classical physics, the time would be the same in both frames. In SR you can get the answer two ways:

1) Forget the astronaut's frame, which is irrelevant, and simply do the calculation in the observers frame (assuming you have all velocities and distances in the observer frame).

2) Have you studied the Lorentz Transformation yet?

9. Feb 3, 2017

### Nick Jarvis

Many thanks Perok. oops regarding the length of the ship :)

The Lorentz Transformation is confusing me. Observer's POV:

u=V+u′1+Vu′/c2 where V = 0.6c and u = 0.6c This is how I get my 0.88c

time taken from astronauts POV is distance / velocity = 180m / 0.6c = 1.0 x 10^-6 seconds

How do I start to work out the length of the craft from the object's perspective? And how can I work out how long it takes for the object to reach the front of the spacecraft from the astronauts perspective? I have been through my notes and I cannot find anything.

I know I work out the Lorentz Constant, which at 0.6c is 1.67

What should be quite easy is proving to be very troublesome. Thanks for any extra help you can provide. I have tried looking for examples online, but so far nothing

Many thanks

10. Feb 3, 2017

### PeroK

Okay, that's fine.

Let's do one question at a time.

1) From the object's perspective, describe the motion of the ship.

11. Feb 3, 2017

### Nick Jarvis

1) - At rest

12. Feb 3, 2017

### PeroK

Okay, I don't agree with that!

So, from the ship's perspective, what is the motion of the object?

13. Feb 3, 2017

### Nick Jarvis

0.6c?

14. Feb 3, 2017

### PeroK

Yes. So, you have a fundamental misunderstanding about relative velocity. If the velocity of the object in the ship's frame is 0.6c, then the velocity of the ship in the objects frame must be -0.6c.

That, in fact, is also true in classical physics. For example, if someone is walking down a train, then in their reference frame the train is moving backwards at walking speed. In the reference frame of someone sitting in a train, the train is at rest. And, to someone on the ground outside the train, the train is moving at 100km/h (or whatever).

For each pair of observers, their relative motion with respect to each other must always be equal and opposite. That's essentially a postulate of classical physics and SR.

15. Feb 3, 2017

### Nick Jarvis

Thanks PeroK. I am really confused though. If the train is moving forwards at 10mph and I am also walking forwards in the train at 10mph, to the observer at the platform I would be moving at 20mph. But...If the ship is moving at +0.6c in the +x direction AND the object is thrown in the +x direction at 0.6c, then they are both moving in the +x direction. Why is the velocity of the ship in the objects frame -0.6c?

Thanks

16. Feb 3, 2017

### PeroK

That's velocity addition involving velocities in two reference frames: one velocity in the ground frame (velocity of the train) and one velocity in the train frame (velocity of you relative to the train), which can be "added" to give the your velocity in the ground frame.

This is where classical physics and SR depart from each other.

In classical physics you have simple addition $u + v$.

In SR, you have the velocity "addition" using: $\frac{u + v}{1 + uv/c^2}$

But, whatever this adds up to, let's call it $v_{ug}$ for your velocity relative to the ground, then we have:

$v_{ug} = -v_{gu}$

I.e. the ground's velocity relative to you is equal and opposite to your velocity relative to the ground. The train is irrelevant.

17. Feb 3, 2017

### PeroK

PS

If $u, v$ are small compared to $c$, as in the case of a train, then we have:

$\frac{u + v}{1 + uv/c^2} \approx u + v$

18. Feb 3, 2017

### PeroK

PPS This is probably the most fundamental postulate of SR. As a first step, you must dispel any doubts about this.

19. Feb 3, 2017

### Nick Jarvis

Ah ok, I am seeing now that (+) velocity as seen by the ship, would be (-) velocity as seen by the ground?

I think I am still failing to grasp the negative velocity on the ship. If I was in the ship sitting down and holding the object, then the object would be moving at 0ms^-1 relative to me and the ship. If I then throw this towards the front at 0.6c, then the object will be moving at + 0.6c away from me, and would use the equation above and I work that out to be 0.88c. But this is what the observer on the ground would see as he is at rest. So if ship is moving at 0.6c and the object within the ship moves at 0.6c, then I still get 0.88c. So does this mean that object moves at 0.88c relaltive to the ship?

Cheers and sorry for dragging this out. I cannot grasp the difference between what the observer sees the ship moving at and what the object experiences relative to the ship

20. Feb 3, 2017

### PeroK

No, this is not right. Let me give you two arguments.

First, the stationary observer is irrelevant to the ship and object. They know nothing about any "stationary" ground observer. The astronaut fires an object at 0.6c relative to him/her and the object measures the astronaut's velocity as -0.6c. The rest of the universe is irrelevant.

Second, suppose we had another observer to whom the ship was moving at 0.5c. Then he would measure the velocity of the object as approx 0.8c (at a rough calculation). Now, would the speed of the object to the astronaut be 0.88c (as a result of the unseen influence of the first observer); or at 0.8c (as a result of the unseen influence of the second observer)?

You could even have a third observer, to whom the ship was moving at -0.6c and to whom, therefore, the object was at rest. So, would the astronaut now measure the speed of the object as 0 as a result of this observers measurements?

21. Feb 3, 2017

### Nick Jarvis

Right ok, so the object will see the length of the ship then based on a velocity relative to the object of -0.6c. So the object would view the length as:

L=Lo√(1-v^2⁄c^2 ) (sorry, cannot seem to work out how to format equations properly)

So even if v is + or -, once squared will be the same?? Or...is it now 1 PLUS (v^2⁄c^2)

Why am I finding this so hard!

22. Feb 3, 2017

### Staff: Mentor

The length of the ship will be seen as contracted for the observer, and this contraction is independent of the direction of motion. The direction has consequences for how the observer sees the mouse, however.

23. Feb 3, 2017

### PeroK

$v^2$ is always positive. In any case, length contraction is always contraction. The length of an object is always less than or equal to its rest length.

You have to be very careful, though, when you multiply two different velocities, as in $uv/c^2$. This term will be negative if $u$ and $v$ have opposiite signs.

You're finding this hard because a) it is hard and b) your knowldege of the basic concepts from classical physics is letting you down somewhat.

Anyway, I think we are on

2) And how can I work out how long it takes for the object to reach the front of the spacecraft from the observers perspective? I have been through my notes and I cannot find anything.

24. Feb 3, 2017

### Nick Jarvis

This is what I am struggling with even more. I have the constant at 1.67

I have calculated the time taken relative to the spacecraft as 1.0 x 10^-7 seconds so from the observers prospective is it:

Lorenz Constant x (1.0 x 10^-7 seconds PLUS (velocity of object x 180m))

Am I on the right track?

25. Feb 3, 2017

### PeroK

This is the wrong track. What the observer sees is:

An object moving at 0.88c (in his frame); aimed at a target (the front end of the ship) which is an initial distance of the contracted length of the ship away, and which is itself moving at 0.6c.

This is where we came in. You need to recognise this as a classical kinematics problem. You have the position and speed of all the objects in your reference frame. SR does not enter into the calculations now. It's done its job by giving you the length contraction of the ship and the velocity of the object in your frame.

Note that within a single reference frame, if you have all the data in that reference frame, then you can solve the problem normally.

Does that make sense?