Special Relativity: Intuition & Quantities Explained

[rakso]

TL;DR Summary

Special Relativity

Hi, I'm taking an introduction course to Special Relativity and encountered a fairly simple problem:

Dirac travels to alfa centauri, which is 4.37 lightyears away. He stays there one Earth year and then travels back, and when he comes back he has aged 5 years. At what speed did he travel?

Fairly simple task, the answer is about $v \approx 0.91c$.

What are different quantities according to Special Relativity from Earth's perspective vs Dirac?

Time - The trip took longer according to us on Earth than Dirac.
The distance, 4.37 LY, according to us on earth. Does Dirac measure something else?
His speed, to us he traveled at 0.91c, what would Dirac say his speed was?

 

[Ibix]

Does Dirac measure something else?

Yes. Why do you think that might be?

His speed, to us he traveled at 0.91c, what would Dirac say his speed was?

What frame would Dirac be using for this?

 

[phinds]

His speed, to us he traveled at 0.91c, what would Dirac say his speed was?

He would say his speed was zero. The question I think you are trying to ask is how fast would he see the Earth approaching him. As Ibix already pointed out, speed is a relative thing and depends on the frame of reference. In your own frame of reference, you are always at rest, by definition.

 

[rakso]

Alright, what speed is the Earth traveling away from Diracs spaceship?

 

[Ibix]

Alright, what speed is the Earth traveling away from Diracs spaceship?

What do you think? Hint: the answer to my first question will help.

 

[Nugatory]

Alright, what speed is the Earth traveling away from Diracs spaceship?

It is symmetrical; if A's speed relative to B is $v$ (more precisely, when we choose to use a frame in which B is at rest then A's speed is $v$) then B's speed relative to A is $-v$.

 

[rakso]

What do you think? Hint: the answer to my first question will help.

When Dirac arrives at Alfa Centuri and looks back at earth, the distance should be 4.37 LY.

How can the speed be symmetrical if the time it took differs but the distance dont? I mean when Dirac steps out on a planet, let's say it's not moving relative to earth, the distance should be the same whether you're on Earth or in Alfa centuri?

 

[PeroK]

Summary:: Special Relativity

Hi, I'm taking an introduction course to Special Relativity and encountered a fairly simple problem:

Dirac travels to alfa centauri, which is 4.37 lightyears away. He stays there one Earth year and then travels back, and when he comes back he has aged 5 years. At what speed did he travel?

This question is poor. In SR it is vital to be clear about which frame of reference things are measured in. The question should state explicitly that the Earth and Alpha Centauri remain approximately at rest with respect to each other and that the distance of $4.37$ light years is measured in their common inertial rest frame.

Likewise, asking the speed he traveled at only makes sense if you specify the reference frame.

 

[jbriggs444]

How can the speed be symmetrical if the time it took differs but the distance dont? I mean when Dirac steps out on a planet, let's say it's not moving relative to earth, the distance should be the same whether you're on Earth or in Alfa centuri?

In the rest frame of Dirac on his way to Alpha Centauri, the distance between the Earth and Alpha Centauri is length-contracted. It is less than 4.37 light years.

In the rest frame of Dirac at rest on Alpha Centauri, the distance between the Earth and Alpha Centauri is 4.37 light years.

In the rest frame of Dirac on his way back to Earth, the distance between the Earth and Alpha Centauri is length-contracted. It is less than 4.37 light years.

 

[rakso]

In the rest frame of Dirac on his way to Alpha Centauri, the distance between the Earth and Alpha Centauri is length-contracted. It is less than 4.37 light years.

In the rest frame of Dirac at rest on Alpha Centauri, the distance between the Earth and Alpha Centauri is 4.37 light years.

In the rest frame of Dirac on his way back to Earth, the distance between the Earth and Alpha Centauri is length-contracted. It is less than 4.37 light years.

I don't get this though. Dirac has a watch on him and takes a long nap and waking up when he arrives at Alpha. He looks at earth, in rest frame of Alpha, and measures Earth to be 4.37 LY away. He then looks at his watch and sees it took 5 years. He must then draw the conclusion that he traveled at $v = \frac{4.37}{5}c$ ?

Meanwhile, since he's Paul Dirac he has figured out a way to communicate simultaneously with his pal on earth. The pal says no no, the trip took you 10 years, and your travelspeed is half of what you just said.

What is wrong with this approach?

 

[PeroK]

I don't get this though. Dirac has a watch on him and takes a long nap and waking up when he arrives at Alpha. He looks at earth, in rest frame of Alpha, and measures Earth to be 4.37 LY away. He then looks at his watch and sees it took 5 years. He must then draw the conclusion that he traveled at $v = \frac{4.37}{5}c$ ?

The round trip is 5 years, including a year on Alpha Centauri. It only takes him 2 years to get there by his watch.

 

[jbriggs444]

I don't get this though. Dirac has a watch on him and takes a long nap and waking up when he arrives at Alpha. He looks at earth, in rest frame of Alpha, and measures Earth to be 4.37 LY away. He then looks at his watch and sees it took 5 years. He must then draw the conclusion that he traveled at $v = \frac{4.37}{5}c$ ?

Dividing a distance computed in one frame by an elapsed time recorded in another. tsk, tsk.

 

[rakso]

Dividing a distance computed in one frame by an elapsed time recorded in another. tsk, tsk.

I see, this is the wrong part.

So I confused Diracs frame while traveling and Diracs frame on Alpha?

 

[jbriggs444]

I see, this is the wrong part.

So I confused Diracs frame while traveling and Diracs frame on Alpha?

Yes. Also note that (as has been pointed out) in Dirac's rest frame, it is Alpha Centauri that is moving, not Dirac.