2 objects at 0.6c in opposite directions. Can outrun light?

[DirkMan]

So I'm reading this

"If you're moving at 0.6 the speed of light in one direction, and your friend is moving at 0.6 the speed of light in the other direction, how could you transmit information to one another? There's no mode that will make up the growing distance between the two of you. So, while many observers could record the information you beam out from your spaceship, your friend can't. It's not that information is or is not, but that information reception is relative."

Is this true? I thought you can't outrun light .

http://curious.astro.cornell.edu/ph...ravel-faster-than-the-speed-of-light-beginner

 

[phinds]

So I'm reading this

"If you're moving at 0.6 the speed of light in one direction, and your friend is moving at 0.6 the speed of light in the other direction, how could you transmit information to one another? There's no mode that will make up the growing distance between the two of you. So, while many observers could record the information you beam out from your spaceship, your friend can't. It's not that information is or is not, but that information reception is relative."

Is this true? I thought you can't outrun light .

http://curious.astro.cornell.edu/ph...ravel-faster-than-the-speed-of-light-beginner

When you friend emits light, it immediately starts moving towards you at c. His speed is utterly irrelevant. Since you are moving at less than c it will catch up with you.

 

[vanhees71]

Where is the problem, since the velocity of my friend from the point of view of my reference frame is still smaller than $c$ I can send a message with light signals. In this case the relative velocity is (in units of $c$)
$$\beta_{\text{rel}}=\frac{1}{1-\vec{\beta}_1 \cdot \vec{\beta}_2} \sqrt{(\vec{\beta}_1-\vec{\beta}_2)^2-|\vec{\beta}_1 \times \vec{\beta}_2|^2}=\frac{1}{1+0.6^2} \sqrt{1.2^2} \simeq 0.882.$$
For details, see Sect. 1.7 in

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Dale]

Is this true?

No, the quote is flat out wrong. Two observers, each traveling at .6 c in opposite directions relative to some inertial frame can certainly exchange information via light signals.

 

[robphy]

The trigonometric version of @vanhees71 's post is
http://www.wolframalpha.com/input/?i=tanh(2*arctanh(0.6)) = 0.882353... < 1.

 

[Jeronimus]

This can be answered easily by pointing at the second postulate of SR. Light always travels at c ~ 300000km/s in a vacuum absent of gravity. That's all you have to know.

Meaning that no matter how fast you see your friend traveling away of you, whenever he emits a light signal, you will always see that light signal travel at c towards you from the point he emitted it.

The speed at which you see your friend traveling away is irrelevant. The 2nd postulate alone suffices to answer your question. In reverse, your friend will always see light signals emitted by you, traveling at c towards(and away) of him.

 

[Stephanus]

Dear @DirkMan, perhaps I can help you.
Some other members have post very invaluable information. But, I think I'd like to give you a summary here.

..."If you're moving at 0.6 the speed of light in one direction, and your friend is moving at 0.6 the speed of light in the other direction,

You move 0.6c to the west wrt (with respect to Bob)
Alice moves 0.6c to the east wrt Bob.
So we have three parties here. Dirkman, Alice and Bob.
With Respect to Dirkman
Dirkman: 0
Bob: moves 0.6c to the east
Alice: moves $\frac{0.6+0.6}{1+0.6*0.6} = 0.882$ as @vanhees71 had pointed out.
See vanhees post

wrt Bob:
Dirkman: 0.6c to the west
Bob: 0
Alice: 0.6 to the east

wrt Alice:
Dirkman: 0.88c to the west
Bob: 0.6c to the west
Alice: 0

...how could you transmit information to one another? There's no mode that will make up the growing distance between the two of you. So, while many observers could record the information you beam out from your spaceship, your friend can't. It's not that information is or is not, but that information reception is relative."

Is this true? I thought you can't outrun light .

http://curious.astro.cornell.edu/ph...ravel-faster-than-the-speed-of-light-beginner

No!, you can't outrun light. Unless you are in a faraway galaxy which moves faster than c because of Hubble Law, but that belongs to cosmology.

No, the quote is flat out wrong. Two observers, each traveling at .6 c in opposite directions relative to some inertial frame can certainly exchange information via light signals.

Alice and Bob still can exchange signal, because their speed is below c (0.882)
Thanks Dale, I have known velocity addition inside out (in 1 dimension only ), but I've never thought the consequences that they can exchange signal.

 

[Mister T]

So I'm reading this

"If you're moving at 0.6 the speed of light in one direction, and your friend is moving at 0.6 the speed of light in the other direction, how could you transmit information to one another?

The person who wrote that is confused about the meaning of speed. It makes sense to say say you're moving at a speed of 0.6 c only if you state what you're moving relative to!

As others have indicated, you can remedy the confusion by stating that they're moving in opposite directions relative to, say, planet Earth. Because speed is the ratio of distance to time, and because measurements of both distance and time are relative, each person will not observe the other moving moving at speed 1.2 c.

 

[m4r35n357]

Is this true? I thought you can't outrun light .

OK, since you specifically used the word "outrun" rather than "exceed" I'll bite!

You can in theory outrun a pulse of light by continuously accelerating (you are never going faster than c though). I've tried to find a simple reference but none comes to mind just yet (try searching for "Rindler horizon"). There is an intense mathematical treatment here, but it's not for the faint-hearted.

The key quotation begins ($a$ is the acceleration): "In other words, the event of the emission of the flash of light is always in the present, and always a distance of 1/a behind the ship. As far as the passengers on the ship are concerned, the flash of light isn’t getting any closer."

 

[David Lewis]

… they're moving in opposite directions relative to, say, planet Earth. Because speed is the ratio of distance to time, and because measurements of both distance and time are relative, each person will not observe the other moving moving at speed 1.2 c.

How fast is one person moving with respect to the other in the Earth frame?

 

[Ibix]

How fast is one person moving with respect to the other in the Earth frame?

It's best to make a distinction between "relative speed" (the speed you measure me to be traveling at in your rest frame) and "separation rate" (the rate at which a third party measures the distance between us to be growing in their rest frame). I doubt that everyone makes that distinction or uses those terms, but if you keep the concepts separate in your head you are less likely to be confused.

What you are asking about here is a separation rate, and it is 1.2c. Separation rate cannot exceed $\pm 2c$, the limiting case being two light pulses traveling in opposite directions.

 

[PeterDonis]

Since the OP is long since gone and the thread was done in 2016 before the last two necroposts, it is now closed.