Probably Simple Physics Question

[flere-imsaho]

I find relativity & time dilation fascinating, but don't really understand it that well, which is the reason I post this word problem:

Point A and Point B are 100 light-years apart.
Assume for this problem that there are no acceleration & deceleration times (the spaceships in question can get to full speed, and come to a full stop instantly).

Spaceship A travels from Point A to Point B at .9 c.

Spaceship B travels from Point A to Point B at .8 c.

1. What are the lengths of the trips from the viewpoint of the passengers on the spaceships?

2. How long will the people from spaceship A wait for the people from spaceship B to show up?

3. What are the lengths of the trips from the viewpoint of someone not on either ship (i.e. at rest)?

Note: This isn't for a "homework" problem - I assure you I'm not in school, just interested.


Addendum - I don't understand relativity. I don't understand why time seems to pass slower for someone traveling at relativistic speeds. People have tried to explain that this is because at relativistic speeds it takes "longer" for a clock to count a second, but this concept doesn't make sense. Surely time is a constant?

So, if someone's bored and has free time, I'd love an explanation.

[Samdari]

Originally Posted by flere-imsaho

I don't understand relativity.

Originally Posted by flere-imsaho

Surely time is a constant?

Your refusal to let go of the second explains the first.

[rkmsuf]

Shoot the hostage.

[Mr. Wednesday]

A super off-the-cuff explanation (and therefore, one that is likely to be wrong)...

...time is not the constant, speed of light is the constant. If you consider time that way (by inverting velocity), I think you can answer the question, although I'm not quite sure how you figure out the velocity of A relative to B or vice versa.

[flere-imsaho]

Originally Posted by Samdari

Your refusal to let go of the second explains the first.

Ah. So, time is elastic. So the faster you travel, the more time you appear to take, according to outside observers? Why?

[Samdari]

I don't know about elastic - perhaps RELATIVE would be a better term.

[Tara]

So, if someone's bored and has free time, I'd love an explanation.

i've just come back from my jogging. Let me have a shower and I'll explain everything (where I can...), unless someone is faster than me...

[General Mike]

Originally Posted by flere-imsaho

Ah. So, time is elastic. So the faster you travel, the more time you appear to take, according to outside observers? Why?

No you have it backwards. I think.

[Mr. Wednesday]

Something for you to keep in mind... suppose that instead of both ships traveling from point A to point B, they travel in opposite directions. If you add their velocities (presuming point A as the reference), then you would think that the velocity of ship A relative to ship B would be 1.7c. However, this is not actually the case, and this AFAIK is the main source of the time differences. Again, could be wrong... my understanding of this topic is very spotty.

[flere-imsaho]

I'm doing some reading. Perhaps the key is that the speed of light remains constant no matter what the frame of reference?

[QuikSand]

Originally Posted by flere-imsaho

I'm doing some reading. Perhaps the key is that the speed of light remains constant no matter what the frame of reference?

A little easier said than done, but getting your head around that idea is central to grasping relativity, I think.

[flere-imsaho]

Originally Posted by QuikSand

A little easier said than done, but getting your head around that idea is central to grasping relativity, I think.

I think that's the issue. It's hard for me to accept this concept.

How about this: It isn't that time is passing more slowly for the Person at Relativistic Speed, it is that their time appears to pass slower for the Person B (not at Relativistic Speed) observing them.

Thus the time dilation effect.

[Wolfpack]

I generally understood relativity through the following example:

If a train is riding along the tracks and the tracks are struck in front of and behind the train at some distance by simultaneous lightning strikes, then technically, the strike in front of the train is visible before the one behind it because the viewer is moving towards the strike, compressing the time it takes to reach him (however slightly) while the strike behind comes in later (however slightly). Thus, while the strikes occured simultaneously, the viewer's relative position dictates which becomes visible to him first. At least I think that's how I remember it going.

[Surtt]

Originally Posted by flere-imsaho

I think that's the issue. It's hard for me to accept this concept.

How about this: It isn't that time is passing more slowly for the Person at Relativistic Speed, it is that their time appears to pass slower for the Person B (not at Relativistic Speed) observing them.

Thus the time dilation effect.

No, time does not change for the observer, but it does slow for the travler.
If you measure something going .9 the speed of light it is going .9 the speed of light.


It isn't speed its mass.
Time runs slower the closer you are to a heavy mass.

As you go faster your mass increases, until it reaches infinity at the speed of light.
It is on a logarithmic curve, so it doesn't show until you start getting close.
Obviously it would take infinite energy to accelerate infinite mass, that's why objects can never reach the speed of light.

Anyway, at .9% of the speed of light your mass has increased to the point that time has slowed.

[flere-imsaho]

Wolfpack: Nice example.

Surtt: Why does your mass increase?

[Surtt]

Originally Posted by flere-imsaho

Why does your mass increase?

E = mc2
The speed of light is constant.

The faster something goes the more energy it has.
The more energy it has the more mass it has.

[Tara]

I'm Back...
I start with a little introduction.

For relativity theory is classically considered the theory of relative motion, namely the coordinate transformation between a system that is considered in relative motion respect to another.

Classic relativity, introduced by Newton and Galileo, deals with the coordinate transformation between two systems in linear uniform motion. Classical mechanics is based on this trasformation.
Consider a still system ( the definition of still is obviously relative... )and another one that moves respect this with costant velocity V. Consider a particle that moves in the moving system with velocity v.
For a person in the moving system particle travels with velocity 'v'. For the still sistem the particle travels with velocity ----> v+V. You simply add the two velocity. This is quite easy.
What stand under everything is that here time is an absolute concepts. Everyone, doesn't matter if in motion or still will always measure the same time intervals.
And for the most of mechanical problems this works.


The need of introduce a new Relativity theory, came out in 19th century, when from Maxwell equation arised that speed of light is an absolute constant. If you masure a ray of light travelling on a system in motion you will alwayas measure 'c' (in vacuum).
Light doesn't care about classical additional law for velocity.

Simply speaking,( The thing is much more complex), if you want to mantain Speed of light costant in ccordinates transformation, you must define again the concept of time---> For writting tranformation equation in wich 'c' is costant in every coordinates system, time must different from system to system and in the same way space---->people moving will measure different time intervals respect to who is still.

I think this is usefull.

In my next post I'll answer your questions. Sorry but I'm a little slow when I write in english.

[flere-imsaho]

Well, I'm certainly finding the answers to be interesting reading.

[Whar]

Short answer

Each day at .9C = 2.29 days for a person at rest or standing still.
Each day at .8C = 1.67 days

Long answer

Lorentz determined a contraction of length based on the velocity of the observer versus the observed. The reciprocal of this equation can be used to determine the time difference between to points of reference. The equation is ...

t(prime) = t/SqrRoot(1 - (v^2/c^2))

So ...

X days = 1 day/SqrRoot(1 - (.9^2/1^2))
X days = 1 day/SqrRoot(1 - (.81/1))
X days = 1 day/SqrRoot(.19)
X Days = 1 Day/.436
X Days = 2.294 Days

Just for giggles the fella travelling .9C would observe Shaq as being 3'1" if the traveller passed Shaq while he was laying down.

[Tara]

Second part

Everything is based on these equations:

(for one dimentional problems-a system is still the other one moves along the x axis on a three dimentional frame, with speed V. The coordinates with ' are related to the moving system)

x'=(x-Vt)/radq(1-(v^2/c^2)
y'=y
z'=z
t'=(t-(V/c^2)x)/radq(1-(v^2/c^2)

and the inverse ones:

x=(x'+Vt)/radq(1-(v^2/c^2)
y'=y
z'=z
t=(t'+(V/c^2)x)/radq(1-(v^2/c^2)

velocity transformation for a particle moving in the travelling system with speed equal to w, you have that from the fixed frame the speed is W, given by:

W=V+w/(1+(Vw/c^2)

try to put w=c, namely you are considering somethin that travel at the speed of light (it is only possible for null mass particle such as photon---> so light), well you have W=c.
Speed of light is a constant.

****

more to come....

[Tara]

Equation that express time dilatation and space interval contarction, that are simply obtained frome the ones above are:

space contraction:

L=L°*radq(1-(V^2/c^2))

time dilatation:

T=T'/(radq(1-(V^2/c^2)))

keep always in mind that
radq(1-(V^2/c^2)) is less than one--->it follows from the fact that V
well these are the 'hot' formulae for your problem. The frist states that if L° is the measure of an object at rest, you will find it contracted by an amount of [radq(1-(V^2/c^2))] when it moves with speed V.

The second states that if T' is a time interval measured in the system in motion, T is the same interval measured from the fixed frame. It results dilatataded by an amount 1/radq(1-(V^2/c^2)).

More to come...wait some times, It's fuckin hard write well in english!!!

[flere-imsaho]

Jeepers!

[MrBigglesworth]

Flere, if you are interested in this stuff I would recommend "The Universe in a Nutshell" by Stephen Hawking. It's written for people that don't have physics degrees.

[Tara]

1 So think about that:

consider first as the two spaceship both moving in the space.
In this situation space is the fixed frame while the two spaceship are two moving system each one with his speed (V in the formulae).

If the speed of the frame is V, the time that the frame takes for making the 100 Ly distance is the classical T'=s/V.

To find how long the travel takes for the observer, see at the formula relating T and T'----> you should find a longer time.
When you talk about time dilatation, what's dilating is the time for the one who's looking--->clock move slover when moving.
I can make you a real example.

There are particles named Muons that travel to the earth from the space. They travel for us at a speed of 0.96c (their mass is really small so they can).
Mouns takes on avarage 1,1E-5 seconds to arrive. The courious fact is that their life time in their sytem is 2.2E-6. So speed make us see them live much more than they actually does.

Sorry if I didn't calculation, i don't like...you can do it.
...

[sabotai]

Nothing turns me on more than a girl who understands physics.

And don't worry about not writing english well. You write better than most people I meet on the internet (who learned english as their 1st language).

[Tara]

wait tomorrow for the answer to #2 question. I'm going to sleep. Gotta wake up at 6:00 am.

[Tara]

If the two ship travel 0.9c and 0.8c, the key problem is what the relative speed of the two system.
You have to use the formula for composition of velocity.
Take W the speed of the faster frame (ship A), take V the speed of slower frame (ship B). The A ship see the B ship travel at w speed:

w = V-W/(1-(VW/c^2))

this mean that A see B moving at -0.19c (B is lower so seem to move back , sign "-")
at the same way B see A moving at a speed of 0.19c.

If you are in A you mesure a time T'=L/W (for complete the trip)

If you are in B you measure a time T''=L/V

if you are in A and want to measure how long takes B from your point of view use:

T=T''/(radq(1-((w^2)/(c^2))) , where w is the relative velocity

if you are in B and want to know how long takes A from your poin of view use:

T=T'/(radq(1-((w^2)/(c^2)))

if you are outside You'll see A travel in

T=T'/(radq(1-((W^2)/(c^2)))

and B in

T=T'/(radq(1-((V^2)/(c^2)))

****************************

Now, I've changed something on my previous post. I've done a mistake.
If you ask someone else and give you a different answer, maybe he is right. I'm graduating in Physic but I'm a 'sperimental physics' who needs reletivity just analize photon and electron collision.
'Easy' relativity problems are not easy at all, above all if you rarely study that. I've never studied special relativity in a deep way, always considered that a great instrument.

For every problem I'm here. I hope I've written a lot of bullshit.

[Mr. Wednesday]

A possibly stupid question... what is radq? Is that square root, fourth root, or something else entirely?

[MrBigglesworth]

Originally Posted by Mr. Wednesday

A possibly stupid question... what is radq? Is that square root, fourth root, or something else entirely?

Square root.