# Something I don't get about speed of light

Discussion in 'Physics & Math' started by a_ht, May 2, 2004.

1. ### a_htRegistered Senior Member

Messages:
158
Hi, what would happen if there is a stationnary object (from a particular frame of reference) and an object moves away from this object at the speed of light. Now, imagine a third object moving away from the stationnary object also at the speed of light but in the opposite direction. Would the two moving object move away from each other at twice the speed of light?

3. ### AgamemnoNRegistered Senior Member

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58
Well, i think it wouldn't be possible for the object to travel at the speed of light because of the need of infinite energy to acelerate it to the speed of light. If, the object(a particle) that are moving away is a foton, then, it does'nt matter from which frame of reference you are measuring the speed, it will always be the speed of light, not twice, not half, exactly the speed of ligh.

That's my bet...

This link, explays in a simple way this schema of frames of reference, and the constancy of speed of light.
http://science.howstuffworks.com/relativity2.htm

5. ### a_htRegistered Senior Member

Messages:
158
Can you explain this with maths please. That what I was really after.

Let me rephrase the problem more clearly;

you have three object, A, B and C. B is stationnary. A is moving away from B in a south-north direction. C is moving away from B from north-south direction. Both A and B are moving at the speed of light. The graphic below illustrates this:
North South
A <---- B ----> C

What is the difference in speed of A and C? Or, if you prefer, what is the speed of C measured from A?

Last edited: May 3, 2004

7. ### James RJust this guy, you know?Staff Member

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37,109
As seen by B: A and C are both moving at the speed of light, but in opposite directions.
As seen by A: B and C are both moving at the speed of light, travelling south.
As seen by C: A and B are both moving at the speed of light, travelling north.

The relativistic formula for the addition of velocities is:

v' = (v + u) / (1 + uv/c<sup>2</sup>)

where, for example:

v' is the speed of C as seen by A
v is the speed of C as seen by B
u is the speed of B as seen by A.

Plugging in v = u = c for your example, we find:

v' = (c + c) / (1 + c.c/c<sup>2</sup>) = 2c / 2 = c

8. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
Hi a_ht,

I'm in a car cruising along at exactly 100m/s (360kph!) relative to the road.
You're going in the opposite direction at exactly 200m/s relative to the road.

You have a very precise radar gun on your vehicle. You use it to measure my speed relative to you. What do you find?

300m/s, right?

Wrong! Velocities do not simply add up like that!

Using the real formula, my speed in your frame of reference is:

v' = (100 + 200) / (1 + 20000/300000000<sup>2</sup>)
v' = 299.99999999993 m/s

Of course, you need a pretty accurate speed measuring device to tell the difference at these speeds!