# Light Speed and Space

1. Aug 13, 2008

### Gear300

Lets say that some object (object 1) is traveling at 60% of light speed. An identical object (object 2) is also traveling at 60% of light speed towards object 1. That would imply that the space between the two objects is closing up at a speed faster than the speed of light...but is this vision right (based on what I know thus far, I'm somehow sure it isn't...I just don't know how it isn't)?

Last edited: Aug 13, 2008
2. Aug 13, 2008

### Staff: Mentor

From the viewpoint of the inertial frame that those speeds are with respect to, those objects are closing the distance between them at a rate of 1.2 times the speed of light. (This is sometimes called the closing speed or mutual velocity.)

But realize that this is not the speed of one object with respect to the other--that speed would be less than the speed of light (about 88 % of light speed). Nothing is moving with a speed greater than light.

3. Aug 13, 2008

### Gear300

By that do you mean the how one object would view the other object? If so, then why is that different from closing speed?

4. Aug 13, 2008

### Staff: Mentor

One has 3 objects in it, the other has two. Speed is one object wrt a second.

5. Aug 13, 2008

### DieCommie

Because they are undergoing time dilation. Time changes, and that "corrects" the speed.

6. Aug 13, 2008

### Staff: Mentor

Yes. That's usually called the relative speed of the objects.
Because closing speed is measured in a different reference frame--a third frame, not one of the two objects. In that third frame, the objects both move at a speed of 0.6 c. But the distance between them changes at a rate greater than c--according to that third frame.

7. Aug 13, 2008

### Gear300

I see...that explains it well...now I understand what russ_watters was saying...thanks.

8. Aug 13, 2008

### Gear300

Is the time dilation a local effect on space-time or is it just something the object experiences?

9. Jul 24, 2011

### valjok

Couldn't you please show the energy balance? Because sum of two .6c objects is E_total = 2x m c^2/sqrt(1-.6^2) = 2.5 mc^2
From the point of view of one of the objects, E_total = mc^2 (1 + 1/sqrt(1-.882^2)) = 3.12 mc^2. Something is wrong.

I understand that mentioning that we can exceed c is not what we should speak about. Yet, if closing speed can be > c then we can accelerate a space base toward a distant planet to 0.6c. Cannot we? The base inhabitants launch a shuttle at additional 0.6c see the Mother planet stays behind at 0.6c while the shuttle moves forward at 0.6c. The shuttle moves away from the Earth (and approaches distant planet) at 1.2c! We can do that!

Isn't it that the time dilation just delays the traveling processes but no dilation happens in reality if we look objectively, from a neutral point of view? The observers on the earth will not see the superluminal speed but they will know how quickly the shuttle goes. Likewise we know that the Universe expands at its visible borders faster than light - we just cannot see it because light cannot deliver it faster. We cannot observe it, but we know that it goes faster and reaches the distant planet faster!

Last edited: Jul 24, 2011
10. Jul 24, 2011

### Staff: Mentor

Nothing is wrong. Your calculations are correct. Energy is conserved, but it is frame-variant.

11. Jul 24, 2011

### Staff: Mentor

Energy is frame-dependent; it will be different in different frames.

OK.
No. The separation rate of Earth and shuttle is 1.2c as seen from the base station. But that's not the speed of anything. The speed of the shuttle with respect to Earth will be 0.88c. As I stated earlier, there is no frame in which anything (Earth, base station, or shuttle) is moving faster than c.

12. Jul 24, 2011

### Drakkith

Staff Emeritus
Valjok, an object cannot exceed c with relation between that objects frame of reference and another objects frame. The space station is a THIRD frame. You are not comparing the speed of the shuttle to the station, but to the Earth, so you MUST use either the shuttle's frame or the Earth's frame.

13. Jul 25, 2011

### valjok

> But that's not the speed of anything

The speeds of particles in accelerator must be measurable from the accelerator frame. You take two accelerators firing to each other. The speed of distance closure between approaching particles is measurable. Why is it "nothing"?

Why the "closing speed" between two particles is downgraded to "nothing" when viewpoint is changed to a 3rd party?

> Your calculations are correct. Energy is conserved, but it is frame-variant.

OK. It is because even one object alters its mc^2 when we observe it from another frame. The sum of its parts changes accordingly.

14. Jul 25, 2011

### Staff: Mentor

The closing 'speed' is definitely meaningful, but it's not the speed of anything. It's a calculated rate. Example: I shoot two light beams towards each other. As measured by me, the beams close the distance at a rate of 2c. Yet nothing is moving at twice the speed of light.

If the accelerated particles approach each other with a speed of v with respect to the lab, their closing speed is 2v (in the lab frame). But that's not the relative velocity of the particles.