# Relative speeds of photons

1. Jul 7, 2006

### nighthelios

i dont understand a problem
if i two photons A & B travel in a straight line but in opposite directions
at the ultimate velocity , what is the speed of B w.r.t. A . The answer maybe is the ultimate velocity but how?:surprised

2. Jul 7, 2006

### MeJennifer

The speed between them is c (the speed of light).

3. Jul 7, 2006

### nighthelios

I know that the speed is c .
I want to know why

4. Jul 7, 2006

### MeJennifer

Well such is the structure of space-time.
Why, well good question, I do not believe anybody knows. :uhh:

As Einstein postulated it, the speed of light is constant for each frame of reference.

Last edited: Jul 7, 2006
5. Jul 7, 2006

### pervect

Staff Emeritus
A photon does not have a reference frame as such.

See for instance the sci.physics.faq I am going at the speed of light and ....

Read the above link, I'm quoting parts of it to attempt to motivate the OP to go and read the entire link for this frequently asked question (FAQ) in full.

Last edited: Jul 7, 2006
6. Jul 7, 2006

### JesseM

As pervect says, because a photon does not have a valid rest frame in relativity, it doesn't make any sense to ask what the speed of one photon is relative to another. But suppose you have some object moving at some high velocity v--say, 0.99c--in your frame, and a photon moving at c in the other direction. You can then ask, what would the speed of that photon be in the object's own rest frame? It would still be c, and this is because velocities don't add in relativity in the same way they do in classical mechanics. See this page, which gives us the relativistic velocity addition formula w = (u+v)/(1 + u*v/c^2), where u and v are the two velocities you want to add. With u=0.99c and v=c, you can see that the sum of the velocities would be (0.99c + 1c)/(1 + 0.99*1) = (1.99c)/(1.99) = c. Basically, the reason that velocities don't add in the normal way in relativity is because each reference frame uses its own set of rulers to measure distance and its own set of clocks to measure time, defining speed in that frame as distance/time, and of course we know that different frames perceive each other's rulers to be shrunk and each other's clocks to be running slow (and clocks that are in sync in their own frame are perceived to be out-of-sync in other frames).

7. Jul 8, 2006

### actionintegral

You mean the speed between them is 2c

8. Jul 8, 2006

### HallsofIvy

Staff Emeritus
?? Did you not read the other responses? The speed of a photon relative to any other thing, even another photon, is c. That is one of the basic "postulates" or relativity, derived directly from experiment. JesseM gave a very nice calculation showing that in the example of two photons: the formula for addition of velocities is not "u+ v". That's a low speed approximation. It is
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
In the simple case that u= v= c, the numerator is 2c but the denominator is
$$1+ \frac{c^2}{c^2}= 2$$
so the velocity of one photon relative to another is 2c/2= c.

9. Jul 8, 2006

### actionintegral

You are right. I misread the original post. But the other answers are incorrect because the use "special relativity" to answer a question about a
frame of reference moving at the speed of light.

10. Jul 9, 2006

### kvantti

You can always use c as a limit, so the use of special relativity is justified.

11. Jul 9, 2006

### JesseM

But there is no unique way to take the limit in the case of two photons. Imagine two photons travelling in the same direction--one way to take the limit would be to look at how a sub-lightspeed object would see a photon moving in the same direction as its own velocity approached c, in which case it'd see the photon moving at c in this limit, but you could also imagine two sub-lightspeed objects moving in the same direction and look at the limit as both their speeds approached c, in which case each one would see the other at rest in this limit. In both cases the limiting case is two objects moving at c as seen in your inertial frame, but the answer to how they see each other in the limit is different.

12. Jul 9, 2006

### robphy

It seems to me that there is, in the context of special relativity, an invariant way to address questions of relative speed between any two inertial particles (i.e. traveling on future-directed non-spacelike geodesics) that met at a common event.

Use relative-rapidity (i.e., the Minkowski-angle, the spacelike-arclength of an arc cut by those geodesics on the unit hyperbola centered at that common event). [The Euclidean analogue is to use the relative-angle (i.e. the arclength of an arc on a unit circle).] The relative-speed is "c times the hyperbolic tangent of the relative-rapidity".

In this way of thinking, a photon makes an infinite Minkowski-angle with any other geodesics that it met in the past. In terms of relative-speed, that photon has a relative-speed of c with any other inertial particle (including non-spacetime-collinear photons) that it met in the past.

This interpretation is independent of any discussion of reference frames [including when no reference frame exists for any of the particles involved].

13. Jul 10, 2006

### actionintegral

Ow! You hurt my brain. Where can I learn about relative-rapidity ( I mean a book - not a website)

14. Jul 10, 2006

### robphy

Sorry...
no pain... no gain.

Check out the classic 1966 maroon "Spacetime Physics" by EF Taylor and JA Wheeler... the best edition has the worked solutions appended to the book. [Rapidity was removed from the newer 1992 light-blue edition... apparently, from what I was told, because it wasn't used by instructors.]

Here's a classic article which appeared in the American Journal of Physics, 1979

15. Jul 12, 2006

### actionintegral

Hi robphy,

The definition I found says tanh(r)=v/c. But I can't seem to find a
value of r that sets tanh(r)=1 ! So I don't know how to find the
rapidity of a photon.

Sorry - I flunk

16. Jul 12, 2006

### robphy

There is no such finite value of r.
As a said above,
That is, r is infinite. One implication is that a finite value of rapidity (say from a boost) added to infinite-value of rapidity is infinite... that is, the photon travels at c to all observers. Another implication of this is that no massive particle [with a finite value of r, whose particular value depends on the observer] can ever attain an infinite value of r... that is, travel at the speed of light.