Length contraction and direction of tavel

[PWiz]

And if a rocket travels at 0.6c for example and it shine a laser at the front,
so the speed of the laser plus the speed of the rocket is...
$\frac{r+l}{1+(r*l)/c^2}$
r is the speed of the rocket
l is the speed of the laser, which is equal to the speed of light
c is the speed of light.
So we can rewrite the formula into...
$\frac{r+c}{1+(r*c)/c^2} = \frac{r+c}{1+r/c} = \frac{r+c}{c/c+r/c} = \frac{r+c}{(r+c)/c} = c*\frac{r+c}{r+c} = c$
Is that so?

Yes, but again it should be no surprise that this works - the velocity addition formula is derived from the Lorentz transformation, which in turn is based on the premise that $c$ is constant for all inertial observers. It's essentially a "reverse" argument.

I mean from the ground observer, the speed of sound is this?
$\frac{s+p}{1+(s*p)/c^2}$
S: the speed of sound
P: the speed of plane
C: the speed of light

Yes, that's correct.

 

[PWiz]

@Stephanus I can see that you have some confusion regarding multiple concepts in SR. I've tried to compile a checklist that you can use to understand stuff better. Try to understand the concepts in the order in which I'm describing them, and you will probably get some clarity on the subject.

Since you know about the velocity composition law, I recommend that you look up the relativistic momentum derivation. If you have the necessary background in calculus and a little bit of conics (check if your calculus knowledge is sufficient by using this link: http://tutorial.math.lamar.edu/Classes/CalcII/CalcII.aspx ), then you should be able to understand the derivation for $E=mc^2$ here: http://www.emc2-explained.info/Emc2/Deriving.htm#.VXq9OMu3TqB .

After this, try undestanding what a "manifold" really is and why a concept of invariant "distance" between two points on a manifold is useful. Since spacetime is a manifold on which each point is described as an "event", you should be able to prove for yourself that $ds^2 - dt^2$ works as an expression which computes the same spacetime interval (spacetime version of invariant "distance") between 2 points, regardless of the inertial frame from which it is viewed (the expression is invariant under the Lorentz transformation; additionally, $ds^2$ is the spatial distance between two events in the expression above, and it can take different forms depending on the type of coordinates you're using and has the familiar form of $dx^2 + dy^2 + dz^2$ when Euclidean coordinates are used).

Are you familiar with spacetime diagrams? If no, then learn them now. If yes, then draw the hyperbola represented by the spacetime interval equation for the case of one dimensional motion (1D because its easy). The twin paradox is most easily understood graphically, and the same applies to time dilation and length contraction. Search up older threads here at PF or do some Googling on it, and you'll marvel at the usefulness of spacetime diagrams. Twin paradox confusion down!

Now you're all set for the 4 vector formalization. Understand that the Lorentz transformation can be used for calculating the components of any 4 vector in another reference frame. There are plenty of notes on four-velocity, four-acceleration, four-momentum and four-force out there on Google. Drink them up - you'll bag in relativistic dynamics right there.

A simple application of four-momentum vectors for light will give you the relativistic version of the Doppler effect. Two down!

...whoops, I forgot what else was confusing you At any rate, I recommend that you do this first - I'm sure 99% of your problems will dissolve along the way! Of course, other users are welcome to suggest changes to this little chronology of concepts. I by no means proclaim that this list is comprehensive or exhaustive - the invaluable inputs of other users can help tailor this list, and you can finally say "I nailed SR!"
Remember that search engines are your friends.

Note: if explanations of key concepts online prove to be too difficult for you, buy an undergraduate level book on Relativity. I recommend Schutz's.

 

[Stephanus]

Thank you very much PWiz.
This the answer that I want!
You've given me the direction and what I have to learn.

 

[Stephanus]

The Lorentz transforms imply that space and time are one four-dimensional whole called space-time - [..]the equivalent of Pythagoras' theorem is $c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}$. [..]
It's then easy to see that if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) then to (10,0,0,0) (i.e., take five years to get three light years away in the x direction, then turn round and come back) that the $\Delta\tau$s are different - $\sqrt{10^2-0^2}=10$ years for me, $\sqrt{5^2-3^2}+\sqrt{5^2-(-3)^2}=8$ years for you.

Thanks Ibix for your explanations!.
It's clear (at least I can understand some ) for me.
Okay...
This equation: $c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}$
So...
$\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}$ is phytagorean hypotenuse in 3D.
$c\Delta t$ looks like spatial dimension. Speed (of light) multiplied by time.

Let me see if I understand this. Come on, I'm not Albert who could devise a very complicated equation and KNEW that it was (is) TRUE!

if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) [..]]

It implies that distance units here are in time light speed takes. Not in miles, yard, km, feet.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
How much speed should he take to reach 3 ly wrt me and come and in 8 years me?
What is his clock after he comes back?

Okay...
$8 = \sqrt{t^2-3^2} + \sqrt{t^2-3^2}$
$8 = 2 * \sqrt{t^2-9}$
$t = 5$

A. Is that HOW I should solve the question/problem?
B. What does it means? 8 years for me is 10 years (5+5) for him?
C. His speed is ... to travel 3 ly (3 ly in my frame) he takes 5 years (5 years in my frame), so it is 0.6c (wrt me)? or
D. His speed is ... to travel 3 ly (3 ly in his frame) he takes 5 years (5 years in his frame), so it is 0.6c (wrt me)?

or
-------------------------------------------------
For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
-------------------------------------------------
is the wrong question.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (4,3,0,0,0) comeback...
T should be locked to 4
And the question is...
What is his speed?
What is his time?
then...
His speed is 0.75c
His time...
$2 * \sqrt{4^2 - 3^2} = 5.291503$
Is that true?

 

[Ibix]

It implies that distance units here are in time light speed takes. Not in miles, yard, km, feet.

Yes - if I measure distance in light years and time in years, then the speed of light is 1ly/y (also light seconds and seconds, light months and months etc etc). It's a "smart" choice of unit that makes calculations easier because multiplying by c is trivial.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...

Well, you're gaining dimensions as you go, but presuming you mean (8,0,0,0) and (t,3,0,0) then ok...

How much speed should he take to reach 3 ly wrt me and come and in 8 years me?
What is his clock after he comes back?

Okay...
$8 = \sqrt{t^2-3^2} + \sqrt{t^2-3^2}$
$8 = 2 * \sqrt{t^2-9}$
$t = 5$

A. Is that HOW I should solve the question/problem?

No. $\Delta\tau$ is the time on the traveller's clock; the quantity you have called t ($\Delta t$ would have been better) is the coordinate time in your rest frame when the traveller arrives at 3ly. So what you have calculated is that, if the traveller is to experience 8 years total, then you would have to experience 5 years on each leg (a 10 year trip, from your perspective).

B. What does it means? 8 years for me is 10 years (5+5) for him?

No - 8 years for him is 10 for you.

C. His speed is ... to travel 3 ly (3 ly in my frame) he takes 5 years (5 years in my frame), so it is 0.6c (wrt me)? or

This is correct - but not what you said above. He takes five years on each leg, according to you, so his speed is 0.6c.

D. His speed is ... to travel 3 ly (3 ly in his frame) he takes 5 years (5 years in his frame), so it is 0.6c (wrt me)?

No. In his rest frame, his speed is zero. His destination comes to him at 0.6c, and takes 4 years (half of the 8 year total trip time), which will let you calculate the length-contracted distance it travels.

or
-------------------------------------------------
For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
-------------------------------------------------
is the wrong question.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (4,3,0,0,0) comeback...
T should be locked to 4
And the question is...
What is his speed?
What is his time?
then...
His speed is 0.75c
His time...
$2 * \sqrt{4^2 - 3^2} = 5.291503$
Is that true?

This seems to be correct.

 

[Stephanus]

Thank you, thank you Ibix for your answer.
And thanks for your previous formula $c\Delta\tau = \sqrt{c\Delta t ^2 - \Delta x^2 - \Delta y^2 - \Delta z ^2}$

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)

Well, you're gaining dimensions as you go, but presuming you mean (8,0,0,0) and (t,3,0,0) then ok...

Wow, did I type four zeros?

No. $\Delta\tau$ is the time on the traveller's clock;

Ahhh, thank you, thank you... If only I had pondered two more minutes before I started calculating!

No. In his rest frame, his speed is zero. His destination comes to him at 0.6c[...]

But, A is always stationary in its own reference frame. A cannot travel at 0.8c in its own reference frame. You need to rethink your questions based on that.

Need confirmation here. Must all the calculation in SR be based on the obeserver REST FRAME? Must the observer assume or see that EVERYTHING is moving, and he STAYS?
Because I haven't got confirmation, yet.

There's no way to measure absolute motion. No observer can say "I'm travel absolutely at 100m/s or 0.8c or whatever". All an observer can say is that they are moving with respect to something else

Should we change to this?
"All an observer can say is that they everything else are moving with respect to something else the observer"

(4,3,0,0)
This seems to be correct.

Of course! I do understand now!

Thanks for your help so far. I know these questions are not important. I'm just a computer programmer, not a physicist, teacher much less scientist. But these twins paradox and universe absolute frame of reference are always haunting me at night. Before I go to sleep, it's always, why, why, why
Perhaps, if an amateur like me can understand this theory, those mentors, professors would know how to teach their students.
What path should they take to make their students understand.
Math are important of course. NGT( Neil Degrasse Tyson) once (only once?) said that "Math is the language of the universe". And so far with all these questionings and answers, I haven't met differential integral, yet. Perhaps the integral nightmare will pop up somewhere later

 

[harrylin]

[..]
Need confirmation here. Must all the calculation in SR be based on the obeserver REST FRAME? Must the observer assume or see that EVERYTHING is moving, and he STAYS?
Because I haven't got confirmation, yet.Should we change to this?
"All an observer can say is that they everything else are moving with respect to something else the observer"

No, the observer can choose to be in a "moving" frame. You can choose any inertial frame you like and pretend that to be "rest frame". For example GPS calculations are the easiest by choosing the "Earth Centered Inertial" frame. Therefore, when you look at your GPS coordinates when you are sitting in your garden chair, your GPS receiver assumes that you are in a "moving frame". It gives your position and velocity (zero) relative to your moving frame.

 

[Stephanus]

@harrylin
Yep, Perok has confirmed me.
Thanks!

And, in fact, when doing classical problems, choosing the best reference frame often makes the problems easier to solve. For example, if you have two objects moving towards each other, you could study this from the point of view (rest frame) of either object; or, the reference frame where their centre of mass is at rest - which is often very useful. Or, of course, from your reference frame as a "stationary" observer.