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# Length contraction Watch

1. Could somebody please take a look at this worked example on page 10...

I understand part a).
I also understand how part b) works if I calculate the dilated time observed by the people on the planet/moon, and then use this dilated time to calculate the new distance

It seems so logical to me that the distance is in fact longer when using the time dilation formula.

But the use of the length contraction formula alone confuses me. On the page before it says
a rod moving in the same direction as its length appears shorter than when it is stationary.
It then quotes the length contraction formula, and says
Lo is the proper length measured by an observer at rest relative to the rod

Using this information, for part b) I can't understand why they haven't used Lo as 6.8x108m rather than using this as L. I also don't get why the distance is longer according to the observers, because I thought it would appear contracted, as the statement says a moving rod appears contracted.

I'm not sure if the source of my error may be because the original text was in the context of the length of an object, but the example is in the context of the distance between two points - nevertheless, I still can't logically see how this would alter the situation. Also the paragraph above the example discusses mesons which I can't understand using the formula, but can with time dilation.

As I said though, it makes sense using time dilation for me, but not length contraction. So frustrating. What am I missing, thanks...
2. ...
3. So basically this is a case of having to work out what the 'rest frame' is (ie where the observer is at rest). When we measure the distance from the point of view of the rocket, we can imagine that the rocket isn't moving, instead the planets are moving relative to the rocket (think about being on a train, when it starts moving it can seem like you're at rest as the platform moves away. So for a), just using the velocity, distance and time relation we can work out what the distance seems like to people on the rocket. But because the rocket is moving near the speed of light, the planets appear to be moving in the opposite direction at near the speed of light. So the distance appears contracted.

For part b) we are considering someone at rest in the same frame as the planets. in this case we can see the rocket is moving, with the planets as stationary. We use L0 as being the distance in the rest frame - and in this case the rest frame is where the rocket is moving. So L has to be the contracted length, which is the distance as seen by the rocket.

So the length is longer, as it's the distance between the planets as viewed by the person on the rocket that is contracted as that is when the planets seem to be moving.

Hope this helps, any questions just ask
4. (Original post by pianofluteftw)
So basically this is a case of having to work out what the 'rest frame' is (ie where the observer is at rest). When we measure the distance from the point of view of the rocket, we can imagine that the rocket isn't moving, instead the planets are moving relative to the rocket (think about being on a train, when it starts moving it can seem like you're at rest as the platform moves away. So for a), just using the velocity, distance and time relation we can work out what the distance seems like to people on the rocket. But because the rocket is moving near the speed of light, the planets appear to be moving in the opposite direction at near the speed of light. So the distance appears contracted.

For part b) we are considering someone at rest in the same frame as the planets. in this case we can see the rocket is moving, with the planets as stationary. We use L0 as being the distance in the rest frame - and in this case the rest frame is where the rocket is moving. So L has to be the contracted length, which is the distance as seen by the rocket.

So the length is longer, as it's the distance between the planets as viewed by the person on the rocket that is contracted as that is when the planets seem to be moving.

Hope this helps, any questions just ask
Thanks. I did consider it to be this way around.

However when would I know to take the planets as moving from the rocket or the rocket moving from the planets - this affects the calculation but I can't see why I should take it such a way around.

Although again as I say if I consider time it seems logical, but not with length contraction alone :s
5. Basically it's just a case of carefully labelling the situation. Work out which frame you are in for each measurement (so the first one, you are in the rocket). Then use the definition of L0 as the length when you are in the rest frame of the rod (ie, when you the planets aren't moving but the rocket is).

You can sort of try and imagine as if you were standing on the rocket or the planet - what appears to move (basically assume you stay still)? Then remember that L0 is the distance in the rest frame of the object (as if you were standing on the planet), and this L0 is always the same distance no matter which place/ frame you are in. In this case L0 is ALWAYS the distance when you're standing on the planet, and even if you do a further calculation when standing on the rocket this value doesn't change.

There is a fairly simple proof to why it works, I'll try and write it down tomorrow. But the important thing is labelling each quantity carefully
6. hey i was just reading this today the aqa notes and had the same dilemma , when can we consider the rocket to be stationary and earth to be moving relative to it ? ( like the flip-side situation)because i can understand it clearly with time dilation. I approached it with time dilation as i thought that the observer relative to planet and moon will observe that rocket is time dilating where as in the frame of reference of the moving frame - rocket it will have value t(o) which would be 2.3 s so the distance relative to spacecraft is L= 0.99c*2.3 ( would this be L(o) or L ) ? like can i both consider the fact that an observer will also measure a contracted length and also time dilation in whatever situation it asks for length or time ?or is length considered an event in itself rather than the rocket now so this would mean it is moving ?
7. (Original post by PhilosopherKing)
hey i was just reading this today the aqa notes and had the same dilemma , when can we consider the rocket to be stationary and earth to be moving relative to it ? ( like the flip-side situation)because i can understand it clearly with time dilation. I approached it with time dilation as i thought that the observer relative to planet and moon will observe that rocket is time dilating where as in the frame of reference of the moving frame - rocket it will have value t(o) which would be 2.3 s so the distance relative to spacecraft is L= 0.99c*2.3 ( would this be L(o) or L ) ? like can i both consider the fact that an observer will also measure a contracted length and also time dilation in whatever situation it asks for length or time ?or is length considered an event in itself rather than the rocket now so this would mean it is moving ?
I'm still uneasy with length contraction, so I just use time dilation and then convert it using the speed to get the new length, or at least use time dilation to check I am doing things correctly...
8. (Original post by fayled)
I'm still uneasy with length contraction, so I just use time dilation and then convert it using the speed to get the new length, or at least use time dilation to check I am doing things correctly...
This is an explanation I gave someone else but it's essentially all about working out what frame you are in - I always have to derive the time dilation formula to get it...

Part 1:
it's to do with what frame the observer is in and what frame proper time/length are defined in. For example, if I need to solve in the frame of the detectors and the time is defined as a half life for the particles in their rest frame then t0 and l0 would be for anything in the particles frame.

I always got confused with which to use until I derived the time dilation from the invariance of the speed of light and some fairly basic trigonometry (pythagoras) with examples of a stationary and moving light clock - have a go yourself

Part 2:
I drew you a picture to help illustrate it.

I'll try to think up an example.

Say an event takes place over 5 seconds in it's rest frame. An observer travelling at 0.6c witnesses the event, how long does the event take in his frame.

So from the question we have t0 defined as in the frame of the event so t0=5s

so our lorentz factor is 1/sqrt(stuff)=1.25

so if we are sitting in his frame it is like he is staying still but the other frame is the moving one so we can apply our stuff and say that his time is 't'=1.25t0=6.25s

does that help at all?

(the diagram is for time dilation btw but the same rules for frames still applies.)
Attached Images

9. (Original post by PhilosopherKing)
hey i was just reading this today the aqa notes and had the same dilemma , when can we consider the rocket to be stationary and earth to be moving relative to it ? ( like the flip-side situation)because i can understand it clearly with time dilation. I approached it with time dilation as i thought that the observer relative to planet and moon will observe that rocket is time dilating where as in the frame of reference of the moving frame - rocket it will have value t(o) which would be 2.3 s so the distance relative to spacecraft is L= 0.99c*2.3 ( would this be L(o) or L ) ? like can i both consider the fact that an observer will also measure a contracted length and also time dilation in whatever situation it asks for length or time ?or is length considered an event in itself rather than the rocket now so this would mean it is moving ?
I had a question like this in my first yea uni exams. There was a spaceship coming to dock at a space station and the ship appears shorter than it is so the space station picks a docking bay that is too short. The the situation is reversed and we were asked to find how long the docking bay appeared to the spaceship which was even shorter!

It's like the skier going very fast over a ravine - do his skis get longer or does the ravine get shorter - the answer is kind of neither and both... but what happens is tha even though his skis now look longer than the ravine he still clips the other edge.

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