Special Relativity and the Twin Paradox

[NanakiXIII]

I read an article about SR, and after alot of re-reading, I still don't comprehend.

 

My first question, it states that while moving at high speeds, time slows, but it doesn't say how or why, could someone please explain?

 

Secondly, it states that when you view something in motion from a reference fram in rest, it becomes contracted in the direction of the motion, but again, it doesn't really say how or why.

 

Finally, about the Twin Paradox. The article states that if two people, one standing on earth, the other flying away at 60% lightspeed, where the one flying away sends a message every hour, the one standing on earth receives a message every 2 hours. (I don't understand why he gets one every 2 hours, so if someone could explain that too, thanks.)

After 12 hours (and 12 messages) the one flying away turns around and flies back to earth, again sending a message every hour. Now the article states that at this point, the person on earth will have received 12 messages, so 24 hours would have elapsed. Is that right? It seems obviously wrong to me. If the person on earth received a message ever 2 hours, shouldn't he have received 6 messages by the time the second person turns around, and receive the other 6 while the other person was on the way back?

 

Please bear with me, I haven't studied physics or anything.

[blike]

for a really good explanation in terms for a general reader I recommend brian greene's book "The Elegant Universe". He gives a good, readable explanation of special and general relativity, along with a moderatly understandable explanation of quantum theory, etc.

 

I say moderately because its harder to visualize, it takes longer to understand the concept.

 

I think one of the best ways used to explain it is the bouncing mirrors analogy.

 

try here:

http://physics.miningco.com/cs/generalrelativit1/a/110703_4.htm

 

Basically it all hinges on the speed of light being a constant in every frame. No matter how fast you're moving, light is always moving at c relative to you.

 

This is counterintuitive, because you would think that as you move away from something at near the speed of light, it would appear to be moving slow, because the photons take longer to "catch up" to you.

 

According to a person at rest [relative to you], they do take longer. According to you, they don't.

[NanakiXIII]

I don't think I'd be able to find that book, and that article is even more confusing than the one I read. Do you think you (or someone else) could just explain it to me? I understand how the light clocks thing work, but that just affects the clock, I don't see how that affects real time. Also, I don't see how it could work with anything other than lightclocks or something similar.

 

Also, I have another question. If light moves at the same speed in all frames, wouldn't it be moving at twice the speed of light when you observe it while moving at the speed of light?

[blike]

NanakiXIII said in post #3 :

I don't think I'd be able to find that book, and that article is even more confusing than the one I read. Do you think you (or someone else) could just explain it to me? I understand how the light clocks thing work, but that just affects the clock, I don't see how that affects real time. Also, I don't see how it could work with anything other than lightclocks or something similar.

 

The book is readily available at pretty much any bookstore.

 

I'll paraphrase his explanation later this evening.

 

NanakiXIII said in post #3 :

Also, I have another question. If light moves at the same speed in all frames, wouldn't it be moving at twice the speed of light when you observe it while moving at the speed of light?

 

No, because any measurements you take will be measuring light's speed relative to you: 3.0xE8 m/s always. Its a funny phenomenon, and quite counter-intuitive.

[NanakiXIII]

I doubt I live wherever you live. I don't live in an english country anyway. But I might find it.

 

Yes, relative to me it will travel at the speed of light, but If I, moving at the speed of light, somehow measured light's speed, and someone standing still looked at whatever I was measuring it with. They would see something moving at lightspeed relative to them, measuring light's motion at the speed of light, so to them it would seem as if it were moving at twice the speed of light. Right?

[JaKiri]

Sum of velocities (A at a, B at b, c = speed of light in a vacuum) =

 

(a+b)/(1+ab/c^2)

[NanakiXIII]

I'm sorry, that doesn't make any sense to me.

[JaKiri]

add the velocities together, then divide them by ((the velocities multiplied together divided by the speed of light squared)+1), and you'll have how fast they move away from eachother.

[NanakiXIII]

Oh I see. Thanks. But can I ask why it works like that?

[aman]

I like to picture it with the analogy of a movie screen. When you first approach the speed of light you flatten relative to an outside observer. If you shine a light ahead near C it is like the character on the screen is shining a light ahead, still flat to the outside observer but moving at the speed of light to the movie character.

Does my analogy make sense?

Just aman

[NanakiXIII]

I'm sure it makes sense, but I don't get it.

[JaKiri]

NanakiXIII said in post #9 :

Oh I see. Thanks. But can I ask why it works like that?

 

It's a consequence of the speed of light being the same for all observers.

 

All the maths comes from that statement.

[NanakiXIII]

Yes, but why is it that you have to

 

add the velocities together, then divide them by ((the velocities multiplied together divided by the speed of light squared)+1)

[JaKiri]

Lets have two light particles moving in opposite directions.

 

Under Newtonian physics, they would recede at c + c = 2c, which breaks the rule that c is the greatest speed available relative to any rest frame.

 

If we put it in the equation, we get

 

a = c = b

 

(c+c)/(1+c*c/c*c) = 2c/2 = c

 

Maintaining the speed limit.

[NanakiXIII]

(c+c)/(1+c*c/c*c) = 2c/2 = c

 

I can see the 2c but c*c/c*c is c^2, isn't it?

c*c=c^2

/c=c

*c=c^2

[JaKiri]

you multiply first and then you divide.

 

(Multiplication, division, subtraction, addition is the order in which things are done)

[NanakiXIII]

I was taught multiplying and deviding in the order it states, then adding and substracting. Sorry.

[JaKiri]

There's implied brackets round the second c*c, as well. Remember, I quoted it as c^2 earlier.

[JaKiri]

Anyway, this is all beside the point.

[NanakiXIII]

True. And I still don't understand Special Relativity.

[Janus]

NanakiXIII said in post #3 :

I understand how the light clocks thing work, but that just affects the clock, I don't see how that affects real time. Also, I don't see how it could work with anything other than lightclocks or something similar.

 

Okay, the first thing you probably need to do is re-think the concept of "real time".

 

Time is merely a measurement used to separate events, it is no more "real" than that.

 

As far as the clocks go: Remember, as each observer is concerned, the light clock next to him and any other method of time measuring are in perfect sync. And, each observer also has to see that the other set of light clock and time measuring device are in sync with each other. (even though they won't be in sync with his.)

[Sayonara]

Nanaki == Red XIII?

[NanakiXIII]

I know that much about time. Maybe if I call it aging. How does one object "age" slower than another when moving at high velocities?

 

As far as the clocks go: Remember, as each observer is concerned, the light clock next to him and any other method of time measuring are in perfect sync. And, each observer also has to see that the other set of light clock and time measuring device are in sync with each other. (even though they won't be in sync with his.)

 

*confuzzled*

 

Yes, Nanaki == Red XIII

[Sayonara]

\o/

 

I knew I had heard that before

[Janus]

NanakiXIII said in post #23 :

I know that much about time. Maybe if I call it aging. How does one object "age" slower than another when moving at high velocities?

 

 

 

*confuzzled*

 

Yes, Nanaki == Red XIII

 

Okay, you have two people and two light clocks. each prson is standing next to his light clock.

 

For each person, the light clock next to him takes one second for the beam to travel to the mirror and back. This is true no matter what manner he uses to measure the time. (stop watch, his own pulse etc. )

 

Now these two people are moving relative to each other, and each watches the other. By their measurement, the other light clock takes more than one second to "tick".

 

By extension, they have to see the timepiece used by the other person run slow by the same amount. (If he didn't, then all kinds of paradoxes would arise.)

 

Each sees the other's clock as running slow. So this answers your first question; what happens is not that high veloctiy causes one to age more slowly, but that high relative velocity causes each to measure the other as aging more slowly.

 

Now in the twin paradox, at the end, one twin has aged less than the other. This will be the twin that experienced an acceleration. This is because experinced acceleration also effects how you measure things. At one point of the trip he will see the other twin age much more rapidly than he does. (but only for part of the trip, for the rest of the time he sees his brother age more slowly. )

 

The twin who doesn't accelerate, merely sees his brother age more slowy during the trip.

 

At the end of the trip both brother's will agree as to their relative ages, but they won't agree as to how that age diffence came about. And neither twin is more correct as to what happened.

 

As far as each is concerned, they are the ones who measured "real" time and their brother experienced "altered time". And they are both right.

 

This what is meant by "time is Relative".