# Two Parallel Beams of Light Pass Each Other At What Speed?

## Main Question or Discussion Point

Two parallel beams of light going opposite ways pass each other. What speed do they pass at? I'm told they pass at the speed of light? How is that possible, I know in relational math you would add the two speeds but for light it is different right? What math does it use?

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Dale
Mentor
Their closing speed is 2c in any inertial reference frame.

You use Galilean Relativity if you are observing the interaction. If you are observing from the perspective of one of the beams of light, I learned not too long ago, you use Lorentzian Relativity which uses "relativistic velocity addition"

$$v_{13} = \frac{v_{12} + v_{23}}{1 + v_{12}v_{23}/c^2}$$

From a stationary observer the particles are both moving at C, closing the gap at twice the speed of light. But from the beam of light's perspective it is not approaching the other beam faster than C. The above formula is the math that is used.

It depends where you are viewing it from... are you the photon, or an outside observer?

$$c = \frac{c + c}{1 + (c*c)/c^2}$$

$$c = \frac{2c}{1 + c^2/c^2}$$

$$c = \frac{2c}{1 + 1}$$

$$c = \frac{2c}{2}$$

$$c = c$$

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Dale
Mentor
the perspective of one of the beams of light
If by "perspective" you mean "rest frame" (as is usually implied) then there is no such thing. If not, then what do you mean?

If by "perspective" you mean "rest frame" (as is usually implied) then there is no such thing. If not, then what do you mean?
Ok, from the Rest Frame of a stationary observer Photon A and Photon B are closing the gap between them at 2c. From the Rest Frame of Photon A OR Photon B they are approaching each other at c.

I may be a little off on the exact wording but this is how I thought it worked.

I just got done having this explained to me here: https://www.physicsforums.com/showthread.php?t=397851"

If I'm understanding it incorrectly please tell me how.

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If by "perspective" you mean "rest frame" (as is usually implied) then there is no such thing. If not, then what do you mean?
Aha, I did a bit of looking and I see that there is quite a bit of discussion about "rest frame" from something moving at C and there is a problem with it. I didn't understand you're statement at first but I see what you are trying to say. Sooo....

Change the values a bit... a stationary observer will see two object moving towards each other at .9c for a closing velocity of 1.8c, while the objects will view a closing velocity of only .994c.

I'll read up on the "rest frame" at c issue when I'm not so tired :-)

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Dale
Mentor
Aha, I did a bit of looking and I see that there is quite a bit of discussion about "rest frame" from something moving at C and there is a problem with it. I didn't understand you're statement at first but I see what you are trying to say. Sooo....

Change the values a bit... a stationary observer will see two object moving towards each other at .9c for a closing velocity of 1.8c, while the objects will view a closing velocity of only .994c.

I'll read up on the "rest frame" at c issue when I'm not so tired :-)
Exactly correct!

The basic problem with the rest frame at c issue is the http://en.wikipedia.org/wiki/Postulates_of_special_relativity" [Broken], which basically states that something that travels at c in one inertial frame travels at c in all inertial frames. This immediately means that it cannot travel at 0, which in turn means that it does not have an inertial rest frame. I am sure if you read up on the matter you can find lots of detail, including mathematical derivations where you get division by 0, but physically that is the crux of the matter. c is invariant so it cannot be 0 in any inertial frame.

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