# Speed of 2 photons traveling in opposite directions

1. Jun 16, 2014

### iwant2beoz

If two photons ( A and B ) are traveling in opposite directions from a point in a vacuum will the relative speed of photon A to photon B be grater then C?

2. Jun 16, 2014

### Orodruin

Staff Emeritus
There is no inertial frame in which either photon is at rest so the question is not well defined.

3. Jun 16, 2014

### Matterwave

The separation distance as seen by a third observer for the two photons, would be increasing at a rate of 2c. However, there is no way to go into the frame of reference of a photon, so you can't say what "photon A" sees, or what "photon B" sees. The relative speed of photon A to photon B is undefined.

4. Jun 16, 2014

### mathman

Try a thought experiment involving two material objects going in opposite directions, each of which is traveling at close to the speed of light, according to an outside observer.

5. Jun 16, 2014

### iwant2beoz

I guess my question boils down to this. If nothing can travel faster then light from a given point, then what happens if 2 objects are traveling near the speed of light away from each other? wont they be traveling faster then the speed of light relative to one another? Am I just thinking about this all wrong?

6. Jun 16, 2014

### DrGreg

In relativity, relative velocities don't add, instead they combine like this:$$w = \frac{u+v}{1+uv/c^2}$$If u and v are slightly smaller than c, then so is w.

See Velocity-addition formula

7. Jun 16, 2014

### iwant2beoz

Ok now I understand, thank you. Does this apply to 2 masses traveling near the speed of light colliding as well? Or is it additive in that case?

8. Jun 16, 2014

### Staff: Mentor

No. When we say "velocity of A relative to B" we mean "velocity of A in the reference frame in which B is at rest". To calculate this, we use DrGreg's formula. It always gives a result which is less than or equal to c.

What you are thinking about is mathematically the difference between the velocities of the two objects in some reference frame in which they are both moving. This is sometimes called the "separation velocity" of the two objects: $\vec v_{sep} = \vec v_A - \vec v_B$. This can be larger than c, up to 2c in fact for two photons.

9. Jun 16, 2014

### pervect

Staff Emeritus
IMO it's good to ask what happens if two objects each travel at nearly the speed of light away from a third object because it leads to some important insights.

If we call the middle object B, the situation is that the velocity of A relativity to B is equal to v, the vleocity of C relative to B is -v.

Using the fact that the velocity of B relative to C is just the negative of the velocity of C relative to B, we can recast this as what is descried as a "velocity addition" question. Specifially, we have that:

The velocity of A relative to B is v, and the velocity of B relative to C is also v. We call the velocity of A relative to C the combined velocity, or the sum of the velocities, and it is given by the velocity addition formula

http://en.wikipedia.org/w/index.php?title=Velocity-addition_formula&oldid=605439341

(v + v) / ( 1 + v*v/c^2)

This turns out to be always less than c if v<c, one can prove this for instance by considering that v = c-$\epsilon$ and evaluate the above expression in the limit in which $\epsilon$ approaches zero (this proof requires the concept of limits, which is taught in calculus).

This approach also explains why you can't reach "c" by constantly accelerating. Acceleration is just repeated velocity addition, but no matter how many times you add sublight velocities to each other, you'll never get a velocity greater than or equal to the speed of light.

It does remain, however to understand where the velocity addition formula came from. Wikki doesn't do a terribly clear job of presenting this. That seems to me to be a topic best left for another thread, though.

10. Jun 16, 2014

### ghwellsjr

Do you really need calculus and limits to evaluate the formula when v=c? Can't you just evaluate it directly?

(c + c) / ( 1 + c*c/c^2) = 2c / (1 + c2/c2) = 2c / (1+1) = 2c/2 = c

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