# Will 2 spaceship flying at opposite direction with the speed of 0.51c be able to see each other?

I understand the relativistic addition formula which means the sum is around 0.8c, but is it safe to conclude that they will be able to see each other?

• why would you think not? – ZeroTheHero Mar 25 '18 at 17:24
• Because the real speed is 1.02 c, it's just that relative to their frame of reference the speed is 0.8c – Kenneth Kho Mar 25 '18 at 17:28
• The "real speed" in which frame? The speed of one seen by the other is $0.8c$. – ZeroTheHero Mar 25 '18 at 17:39
• Ah okay, i'll assume the correct frame is on the spaceship – Kenneth Kho Mar 25 '18 at 18:49
• There is no "correct" frame. All frames of reference have equal "standing" - there is no preferred frame ( except in the sense that you choose one because calculations might be easier in it because e.g. some values are zero in that particular frame :-) ). – StephenG Mar 25 '18 at 19:23

## 1 Answer

To put this another way, for two objects to not see each other they would need to have a speed relative to each other that is greater than the speed of light.

So can we do that ? No.

There's no way to make that happen no matter what velocity they move at relative to some other observer as long as that relative velocity does not exceed light speed.

We start from the relativistic velocity addition formula :

$$V_{CA}=\frac { V_{BA} + V_{CB} }{ 1 + V_{BA}V_{CB} }$$

Where all the values are fractions of the speed of light, and e.g. $V_{CA}$ means the velocity of C relative to A.

To not be visible to each other (and hence traveling at a relative velocity greater than light ) we would need $V_{CA}>1$ which means we need :

$$V_{BA} + V_{CB} > 1 + V_{BA}V_{CB}$$

Which is :

$$V_{BA}(1-V_{CB}) > 1 - V_{CB}$$

Which means we need :

$$V_{BA} > 1$$

which is saying that at least one observer B would need to see A moving faster than light.

As you start out with velocities seen by an observer less than the speed of light, you can't have that.

The simplest way to see that a massive object can never be seen traveling faster than light relative to any observer is to use energy. In special relativity kinetic plus rest mass energy is always given by :

$$E = \frac {mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

No matter how much energy the object has relative to me, it cannot ever travel faster than or even as fast as $c$ because :

$$v = c \sqrt{ 1-\left(\frac{mc^2}{E}\right)^2 }$$

I can make that as close to $c$ as I like by increasing $E$, but I can never reach $c$.

So no object with mass can ever reach the speed of light for any observer and (not surprisingly) applying the velocity addition formula does not let any other observer see a speed greater than $c$.