If I was traveling at 51% the speed of light

[glassjester]

and you flew in a spaceship going the opposite direction, also at 51% the speed of light...

Would each of us be breaking the light barrier, relative to the other?

If not, why not?

 

[Pengwuino]

No, velocities do not add the way you'd expect when you start talking about speeds near the speed of light. Roughly speaking, the speed you two would see is:

$$v = {{v_A + v_B} \over {1+{{v_a*v_b} \over {c^2}}}}$$ where $$v_A$$ is your velocity and the 'B' is the spaceship.

This speed is never beyond the speed of light.

 

[glassjester]

No, velocities do not add the way you'd expect when you start talking about speeds near the speed of light. Roughly speaking, the speed you two would see is:

$$v = {{v_A + v_B} \over {\sqrt{1-{{v_a*v_b} \over {c^2}}}}}$$ where $$v_A$$ is your velocity and the 'B' is the spaceship.

This speed is never beyond the speed of light.

So what would the speed of the spaceship APPEAR to be to an observer in the other spaceship?

 

[Pengwuino]

Just plug in the values you want. The fastest way is to say $$v_a = v_b = 0.51c$$ where the speeds are 51% of the speed of light. And this is a very rough way of doing it. What people typically do is say you have a spaceship going in one direction relative to someone on the ground, then the spaceship has something shoot out at another extremely high speed. Then you can immediately use that equation... but your situation is a bit different but gives the same result.

 

[Dusty_Matter]

The two speeds would not add up anyway as they are separate velocities in different directions. You can travel at 51% the speed of light, and the other ship can do the same as well, just like the particle beams in a particle accelerator. They generally have two separate beams of particles traveling in opposite directions, both going at close to the speed of c. Then they collide them.

With both of you traveling at such speeds your view of what's in front of you would be blue shifted. Viewing the other ship however might not be possible as not only would your view of the other be blue shifted because of just your motion, but also due to the motion of the other ship. The blue shift might put your view and detection of each other out of range, unless you were to collide with each other.

 

[glassjester]

The two speeds would not add up anyway as they are separate velocities in different directions. You can travel at 51% the speed of light, and the other ship can do the same as well, just like the particle beams in a particle accelerator. They generally have two separate beams of particles traveling in opposite directions, both going at close to the speed of c. Then they collide them.

With both of you traveling at such speeds your view of what's in front of you would be blue shifted. Viewing the other ship however might not be possible as not only would your view of the other be blue shifted because of just your motion, but also due to the motion of the other ship. The blue shift might put your view and detection of each other out of range, unless you were to collide with each other.

At what velocity would the two ships collide?

 

[Pengwuino]

EDIT: Sorry, i screwed up the first equation I posted... I fixed it though!

 

[russ_watters]

At what velocity would the two ships collide?

Please plug the numbers into the equation yourself! It isn't hard to use and you'll learn more by doing it yourself than you will by people spoon-feeding you the answer!

 

[Dusty_Matter]

The particles in a particle accelerator collide at close to the speed of light. You would collide with the other ship at 51% the speed of light, and the other ship would do the same with you. The energies involved though would go up if your speed were to be increased, but as stated previously your separate velocities are not in themselves added together to give you a speed that is greater than light.

 

[DarioC]

There is a similar situation that is perhaps more interesting: if there is a separate observer, "off to the side," timing the closing rate according to his clock and his time frame, of the two ships, what would be his result in time? Would it equal out to a FTL closing velocity?

Re: the original question, no one mentioned the effect of relativity on the on-board clocks necessary for determining the velocity of the other ship.

 

[Pengwuino]

There is a similar situation that is perhaps more interesting: if there is a separate observer, "off to the side," timing the closing rate according to his clock and his time frame, of the two ships, what would be his result in time? Would it equal out to a FTL closing velocity?

Re: the original question, no one mentioned the effect of relativity on the on-board clocks necessary for determining the velocity of the other ship.

Yah my explanation was a little weak. The prototypical example for velocity addition has a person "to the side" at rest. He sees a spaceship going by near the speed of light. The spaceship then fires something ahead of it at a speed, relative to the spaceship, near the speed of light. Then you calculate what the speed of the thing fired is relative to the guy on the ground and see that it isn't faster than light speed.

For the situation you're asking, if you had a ship coming from the left at say, 0.9c and one coming from the right at 0.9c, you would see them approaching at 1.8c. Things can approach and recede at faster than the speed of light because their approach and recession is not something that can carry "information"

 

[yuiop]

and you flew in a spaceship going the opposite direction, also at 51% the speed of light...

Would each of us be breaking the light barrier, relative to the other?

If not, why not?

You have not made it clear who measures the spaceships to be traveling at 51% of the speed of light (0.51c). Let us assume that ship A is moving at 0.51c according to observer C and ship B is moving at -0.51c also according to observer C which we can state as:

vAC = velocity of ship A according to observer C.
vBC = velocity of ship B according to observer C.

Now what you want to know is what is the speed of ship A according to ship B (vAB) or vice versa. The regular equation of relativistic velocity addition can now be written int terms of these variables as:

$$vAC = \frac{vAB+vAC}{1+vAB*vAC/c^2}$$

This equation can be rearranged to isolate the relative velocity we are interested in:

$$vAB = \frac{vAC-vBC}{1-vAC*vBC/c^2}$$

Plugging in the numbers using units of c=1 now gives:

$$vAB = \frac{0.51-(-0.51)}{1-0.51*(-0.51)} = 0.809c$$

Of course if you look at the above equations it can be seen if two particles are moving in opposite directions according to a single common observer (C), then the relative velocity of one particle according to an observer at rest with the other particle can be expressed in terms of velocity magnitudes (speed) as:

$$|vAB| = \frac{|vAC|+|vBC|}{1+|vAC|*|vBC|/c^2}$$

Ether way, if the speeds of any two particles are individually less than the speed of light according to any given observer, then the speeds of the particles relative to each other (or relative to any other inertial observer) is always less than the speed of light.

Also, just to confirm what Pengwuino said, objects can have "closing velocities" that are apparently greater than the speed of light, but this does not violate relativity because information is never transmitted from one observer to another faster than the speed of light and the velocity of one observer relative to another observer is never greater than the speed of light.