# C and frames of reference.

1. Aug 18, 2010

### Aeodyn

Assume you have two objects, with nothing else, moving directly away from one another, each at 51% of the speed of light. But, from the frame of reference of one object, the other is going 102% of c, an impossibility. How does this work?

And: Those same two objects, which has more energy, and, thus greater time dilation? If you use ether of them as a reference point, the other seems to have more energy, thus greater time dilation. How does this work?

2. Aug 18, 2010

### starthaus

No: $$w=\frac{v_1+v_2}{1+v_1v_2/c^2}=\frac{2v}{1+v^2/c^2}<c$$

3. Aug 18, 2010

### Passionflower

Ok.

Each?

Remember all movement is relative, you express speed in terms of absolute speed. Ask yourself: 51% with respect to what?

4. Aug 18, 2010

### Aeodyn

:tongue:
But what about the second scenario?

Last edited: Aug 18, 2010
5. Aug 18, 2010

### Aeodyn

Say an object between each, in the middle.

6. Aug 18, 2010

### Passionflower

In that case the relative speed between the moving objects is calculated by the formula that Starthaus provided.

7. Aug 18, 2010

### Fredrik

Staff Emeritus
Then the speed of one one of them relative to the other is what Starthaus said.

In the rest frame of (either of) the objects, the other one has a large speed and therefore a large kinetic energy. Therefore, they would both describe each other's clocks as ticking slower than normal, by a factor of

$$\gamma_w=\frac{1}{\sqrt{1-\frac{w^2}{c^2}}}$$

where w is the result of Starthaus's calculation.

Last edited: Aug 18, 2010
8. Aug 18, 2010

### Aeodyn

Ah...
But if you have another object, in between the other two, then is one of the objects energy determined by using both of the other 2 as a reference frame, or just one, and which?

9. Aug 18, 2010

### Staff: Mentor

The energy is frame variant, just like the speed is. It is different in each frame, there is no one number.

10. Aug 18, 2010

### Aeodyn

What if the observed object had a clock on it? It would measure time depending on it's observed energy level, observed by which observing object?

11. Aug 18, 2010

### Fredrik

Staff Emeritus
What it measures is the proper time of the curve in spacetime that represents its motion. That's a coordinate independent property of the curve, so it doesn't depend on who's observing it. However, if an observer wants to calculate that clock's ticking rate in the coordinate system associated with his own motion, he would have to use the clock's speed in that coordinate system, i.e. the clock's speed relative to him.

Edit: Oops. Thank yossell.

Last edited: Aug 18, 2010
12. Aug 18, 2010

### yossell

err... coordinate independent?

13. Aug 18, 2010

### Passionflower

I realize this is a little of a nitpick but just to avoid confusion a time interval observed in the moving frame must be multiplied by the Lorentz factor, that implies that to get the ticking rate one uses $1/\gamma$

In the above mentioned scenario $\gamma = 1.703069715$ so the moving clock rate is 0.587175024 times the local clock rate.

Last edited: Aug 18, 2010
14. Aug 21, 2010

### Aeodyn

It actually DOES matter what is observing it, as of energy = matter = curve-of-spacetime.
the reason an object cannot move past the speed of light, why it takes more and more energy to speed it up the faster it goes, why A = F/M is wrong, is that the more energy something has, the heavier it is. Matter has high concentration of energy, and so warps the fabric of space time more than a laser. Depending on what is observing it, it has more, or less, energy, thus different magnitude curves, thus different speeds of time.
I think...

Last edited: Aug 21, 2010