Travelling faster than the speed of light - SR

[hmvince]

Picture yourself accelerating to ³/₄ the speed of light, taking your girlfriend with you, and leaving your best mate behind, back on Earth (with no speed!). At this speed you stop accelerating and continue to travel through space at 0.75c.
From this position, you see light traveling at the speed of light and would not know any better than to ASSUME you are stationary, with speed = 0 (even though you are moving away from your stationary friend back on earth).
You then send your girlfriend on her way, accelerating her in the same direction you began accelerating from your mate. Relative to you, she accelerates to 0.75c and keeps travelling.
But then your friend stops and thinks for a bit. He is not the brightest boy but does a quick calculation and finds that:
0.75c + 0.75c = 1.25c
"Wow" he says, that dudes girlfriend is traveling faster than the speed of light. "Is she?"
But faster than light travel requires infinite energy! It Isn't possible!

Really I can see nothing wrong with this, and if this is a problem, then it raises another question:

Couldn't we calculate how fast WE, on earth, and the solar system are traveling through space? As time dilation is not linear (from lorentz factor), we could put a clock on a plane (going really, really fast), and one on Earth and calculate the difference in time and from that work out how fast we are actually going! (obviously direction is going to be a big factor here, the Earth's spin on its axis, Earth's rotation around the sun, and suns rotation around milky way galaxy). But wouldn't this be possible?

 

[clamtrox]

0.75c + 0.75c = 1.25c
"Wow" he says, that dudes girlfriend is traveling faster than the speed of light. "Is she?"
But faster than light travel requires infinite energy! It Isn't possible!

First of all, that's 1.5c :)

Secondly, that's not how you move from one coordinate system to another. Space and time work differently in special theory of relativity, that's the whole point of it. In SR, you would get

$$v = 1.5c/(1+0.75^2) = 0.96c$$

(see http://en.wikipedia.org/wiki/Velocity-addition_formula)

 

[hmvince]

thats embarassing :(.

well that's very interesting. I'll have to look into it a bit more. Thankyou!

 

[phinds]

thats embarassing :(.

well that's very interesting. I'll have to look into it a bit more. Thankyou!

When you do, spent some time learning about frames of reference. You seem to believe that there is some absolute velocity that we could measure. This is nonsense, as you will find. Velocity is ALWAYS relative to something; it has no other meaning. What would you use as the basis for the speed of the earth? You can choose ANY reference point you like (and get different speeds accordingly).

 

[hmvince]

Just a quick, un-related topic. (I've been having fun with SR lately).
I have been trying to find the optimum speed to get from point a to point b in the fastest time possible, from an observers point of view.
I used the common time delation equation:
t^- = ^t/_{√(1 - (v²/c²)}
and replaced t with the distance formula, ^distance/_velocity.
now, when I pick a distance (won't matter what distance, let's say 1000m) and do this sum multiple times with different velocities (a spreadsheet and graph in excel helped here), I get the optimum speed is around 0.707 times the speed of light.

Obviously this is because any slower and you would just take longer to get there, simple! And any longer and time dilation would have a profound effect on the observers point of views time.

Is this correct? Or have I made some errors?

 

[Ich]

The observer's time is distance/velocity, obviously. Time dilation means that the measured rocket time is shorter by that factor. So there is no optimum velocity, you go as fast as you can and get as the lowest limit a time of distance/c (observer) or 0 s (rocket).

 

[Naty1]

no matter what you do, no matter how fast you accelerate, no matter how fast your velocity, light still leaves you at a velocity 'c'... sad, perhaps, but true.

 

[hmvince]

I don't think I am making myself clear!

I'm saying there is a dude standing at point B, with a timer and he can see point A. When the rocket leaves point A (dude starts the timer) it will need to travel at around 0.7c to optimize time FROM THE OBSERVERS REFERENCE FRAME (dude at point B). Any faster than that and the rocket would be traveling forward in time, and making the trip longer to complete from the observers reference frame but not from the reference frame of the rocket.

 

[DrGreg]

Your time dilation formula is the wrong way round. It should be $$t_{rocket} = t_B \cdot \sqrt{1-\frac{v^2}{c^2}} = \frac{d_B}{v} \sqrt{1-\frac{v^2}{c^2}}$$Both $t_{rocket}$ and $t_B$ get smaller as v approaches c. Ich's answer (#6) is correct.

 

[hmvince]

This is the bit that I don't understand:

The observer's time is distance/velocity

Shouldn't that be the other way around (rockets time is distance / velocity). I'm trying to find the optimum speed to get somewhere according to the observers reference frames time, not the rockets reference frame. (In the rockets reference frame, of course optimum time is going to occur at maximum velocity, but he will arrive at his destination at a later date, according to the observer).

Just so I know I have the formula correct:

t_{observers reference frame} = ^t_rocket/_{√(1-v²/c²)}

So the observers time will always be greater than the rockets time for speeds between 0-c.

 

[Ich]

So the observers time will always be greater than the rockets time for speeds between 0-c.

Right.
And the observer's time is definitely distance/velocity. Think about it. It could not be different, as that is the definition of velocity.
That means that the rocket time is smaller than distance/velocity.

 

[hmvince]

That means that the rocket time is smaller than distance/velocity.

Exactly! The rocket time is smaller by a factor of "the lorentz factor?" (if that's how you say it ). And that factor slowly increases as v increases.
ie. the faster the rocket travels, the more time passes (relative to the rocket) on earth. By that logic you wouldn't want to travel from point A to point B at near the speed of light, because although from the rockets reference frame less time passes, on Earth more time passes.
That is why you need the balance, and I graphed it and found optimum to be approximately 0.707c.

 

[Doc Al]

That is why you need the balance, and I graphed it and found optimum to be approximately 0.707c.

Optimum for what? What do you mean by 'balance'?

 

[hmvince]

Optimum for getting from point A to point B in the least time possible (from the rest reference frame)
By balance I mean getting the least time with
t = distance/velocity
while keeping the effect of time dilation (as the faster the rocket travels, the more time is going to pass in the rest frame) to a minimum.

 

[Doc Al]

Optimum for getting from point A to point B in the least time possible (from the rest reference frame)
By balance I mean getting the least time with
t = distance/velocity
while keeping the effect of time dilation (as the faster the rocket travels, the more time is going to pass in the rest frame) to a minimum.

Well, what criteria did you use? (Obviously not minimum time. For that you just go as fast as possible.)

 

[hmvince]

to get minimum time (according to the rest frame) I used this formula:
where γ is the lorentz factor:

t' = tγ
t' = (distance/velocity)γ

graphing that you get a minimum at v = 0.707c

 

[Ich]

t' = (distance/velocity)γ

t' is smaller than t by a factor of γ.
Try
t' = (distance/velocity)/γ
instead.

 

[Doc Al]

t' = tγ

I understand this one:
Earth time = spaceship time * gamma

t' = (distance/velocity)γ

But what's this mean? Whose distance/velocity did you use?

 

[hmvince]

I used the spaceship's distance/velocity
Is that incorrect?
Please explain why, I really don't understand.

 

[Doc Al]

I used the spaceship's distance/velocity
Is that incorrect?
Please explain why, I really don't understand.

If you used the spaceship's distance/velocity that would be correct. But then you'd end up with the same equation as the first one, since that distance/velocity would equal t:
t' = (distance/velocity)γ = tγ

And you're right back where you started.

 

[Ich]

I understood
t Earth time for the journey (=earth distance/velocity; spaceship distance is zero anyway)
t' Spaceship time for the journey
and, incorrectly,
t'=tγ,
leading to the alleged optimal travel time at v=sqrt(1/2)c.

 

[Doc Al]

spaceship distance is zero anyway

I was interpreting 'spaceship distance' as the distance from Earth to the destination as measured by the spaceship. So 'spaceship distance'/velocity would give the time as measured by the spaceship. (But I see that he had the Earth and spaceship times mixed up.)

 

[hmvince]

If you used the spaceship's distance/velocity that would be correct. But then you'd end up with the same equation as the first one, since that distance/velocity would equal t:
t' = (distance/velocity)γ = tγ

And you're right back where you started.

precisely. I changed tγ to (distance/velocity)γ as t = (distance/velocity). That is the time it takes the rocket to travel from A to B, which is calculated from the distance IT has to travel / its velocity.

I understood
t Earth time for the journey (=earth distance/velocity; spaceship distance is zero anyway)
t' Spaceship time for the journey
and, incorrectly,
t'=tγ,
leading to the alleged optimal travel time at v=sqrt(1/2)c.

Yes. Yes. And is t'=tγ incorrect? I got this equation from: http://www.thebigview.com/spacetime/timedilation.html
And yes, v=sqrt(1/2)c, is the exact optimum time that I get!

I was interpreting 'spaceship distance' as the distance from Earth to the destination as measured by the spaceship. So 'spaceship distance'/velocity would give the time as measured by the spaceship. (But I see that he had the Earth and spaceship times mixed up.)

I see what you're saying, but I don't understand how I got them mixed up. Why should I set
t' = distance/velocity (time on earth)
and not
t = distance/velocity (time on spaceship)
What is the difference and how should I know which way to do it in the future.

 

[Doc Al]

And is t'=tγ incorrect?

If you are using t' to represent the spaceship time, then you have the formula backwards. Earth observers will measure the time (t) to be greater than the spaceship time (t').
Thus: t' = t/γ or t = t'γ

I see what you're saying, but I don't understand how I got them mixed up. Why should I set
t' = distance/velocity (time on earth)
and not
t = distance/velocity (time on spaceship)
What is the difference and how should I know which way to do it in the future.

The travel time according to the the Earth is:
t = distance measured by earth/velocity

The travel time according to the the spaceship is:
t' = distance measured by spaceship/velocity

The distance measured by the spaceship is smaller by a factor of γ.

 

[hmvince]

Ok, now we're getting somewhere!

I have just had a quick read through this thread:
https://www.physicsforums.com/showthread.php?t=599015
and shatteredmerc seems to be thinking pretty much exactly the same thing as me!
reading through the last few posts in this thread and the posts in that thread, I will definitely be able to understand the concepts behind it!

The two main things I think I think I didn't think through is firstly making the mistake with the formula and having it in reverse! And ignoring length contraction!

Thankyou so much. If you don't mind answering one more question?

If there is an observer on earth, and a traveller at the speed v = 0.99c (from the observers reference). Space around the traveller will seem to shrink (in his reference frame), and in the observers reference frame the travellers length will expand?
is this correct?

and while all this time changing and length contraction is happening, mass is also deciding to mix it up a bit!?

 

[Doc Al]

If there is an observer on earth, and a traveller at the speed v = 0.99c (from the observers reference). Space around the traveller will seem to shrink (in his reference frame), and in the observers reference frame the travellers length will expand?
is this correct?

Here's how I would put it: The distance between points that are at rest in one frame will always be shorter as seen by any other frame. So imagine Earth and planet X a distance D apart (as measured by earth). The distance from Earth to planet X will equal D/γ as measured by the spaceship.

It works both ways, of course. Say the length of the ship is L (as measured by the ship). The length of the ship will equal L/γ as measured by earth.

and while all this time changing and length contraction is happening, mass is also deciding to mix it up a bit!?

Thinking of mass as increasing with speed will only muddy the waters. Generally, we speak about invariant or rest mass, which doesn't change. It does get harder to accelerate something the faster it goes, which is what you care about.

 

[m4r35n357]

Here is a very important and basic principle to remember. A traveller will not see his ship change size, nor his time run slower/faster, nor his own mass changing. Relativistic effects such as time dilation and length contraction are only observed for things that are moving relative to the observer.

 

[Doc Al]

Here is a very important and basic principle to remember. A traveller will not see his ship change size, nor his time run slower/faster, or his own mass changing. Time dilation and length contraction are only observed for things that are moving wrt the observer.

Right!

 

[m4r35n357]

Right!

Pardon me for editing my text a bit whilst you were commenting; I wanted to make better use of repetition in my wording!

 

[hmvince]

sweet as guys. Thanks for all your help!

So back to the equation at the start and finding minimum time (observers reference frame) I would use the equation (to find minimum time over a distance of say 100km):
t'=tγ
since t = D/V and distance(moving reference frame) = 100000/γ
t' = ((100000/γ)/velocity)/γ

is this correct?

 

[hmvince]

which once graphed, shows that optimum time is to go as fast as possible.

 

[Ich]

is this correct?

No.

You can use Earth distance D, then
t=D/v
t'=t/γ.

Or you use spaceship distance D', then
D'=D/γ
t'=D'/v
t=t' γ.

Either way, t'=(D/v)/γ.