Consistent with special relativity?

ZephyrWest

Here's the problem:

Two identical spacecraft are moving in opposite directions each with a speed of 0.80c as measured by an observer on the ground. The observer on the ground measures the separation of the spacecraft as increasing at a rate of 1.60c.

Explain how this observation is consistent with the theory of special relativity.
I know that nothing can exceed the speed of light c. So is it consistent with special relativity because it is not a physical object that is moving at 1.60c? I'm asking because I'm really not sure if I'm correct.

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ZephyrWest

I understand relativistic velocity addition but am not sure how to apply it to this situation. Can you please explain?

Thrice

Explain how this observation is consistent with the theory of special relativity.
Are you asking how the space can be increasing by that much & yet none of those two spaceships measures a superluminal speed ... ? Ordinary velocity addition doesn't apply.

russ_watters

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Well, the simple answer here is that "separation rate" is not a speed. Speed is measured between two objects, not three. So it is perfectly fine to say that according to the 3rd observer, the two spacecraft are separating at 1.6C. That isn't contradicting SR because it isn't saying anything about a speed.

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ZephyrWest

Hmmm... ok. From the frame of reference of one of the spacecrafts, is the seperation rate, or speed, this?

$$\frac{0.80c + 0.80c}{1 + \frac{(0.80c)(0.80c)}{c^2}} = \frac{1.60c}{1.64} = 0.976c$$

nakurusil

Here's the problem:

I know that nothing can exceed the speed of light c. So is it consistent with special relativity because it is not a physical object that is moving at 1.60c? I'm asking because I'm really not sure if I'm correct.

1. SR precludes one massive object from moving at a speed equal or larger than c.(photons having zero mass move at c)

2. In your example you are dealing with two separate objects closing on each other at a net speed v>c. SR does not preclude this, actually speeds as large as 2c are acceptable in SR.

nakurusil

Hmmm... ok. From the frame of reference of one of the spacecrafts, is the seperation rate, or speed, this?

$$\frac{0.80c + 0.80c}{1 + \frac{(0.80c)(0.80c)}{c^2}} = \frac{1.60c}{1.64} = 0.976c$$

This example above simply says that if a rocket flies in the +x direction at 0.8c and you are observing it from another rocket flying at 0.8c in the -x direction, you would measure the other rocket receding from you at 0.976c.
The observer in your original problem sees you and the other rocket receding away from each other at 1.6c.
The difference is quite clear now, right?

ZephyrWest

Yes, so if I was in one of the rockets, I would see the other moving traveling away away from me at 0.976c.

The external observer on the ground would see the two rockets traveling away from each other at 1.6c. This is possible because special relativity permits two separate objects to have a net speed greater than c, but not one object.

Did I get it right? :)

nakurusil

Yes, so if I was in one of the rockets, I would see the other moving traveling away away from me at 0.976c.

The external observer on the ground would see the two rockets traveling away from each other at 1.6c. This is possible because special relativity permits two separate objects to have a net speed greater than c, but not one object.

Did I get it right? :)
perfect :-)

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