# Mass of Light (photons) at C

1. Jan 31, 2007

### antariksh

Just a thought, according to Einsteins relativity mass changes with speed and tends to become infnite as it approaches 'c'. Since photons too have mass, why doesnt their mass become infinite since they travel at c?

2. Jan 31, 2007

### mathman

Photons have zero "rest mass", so that the Lorentz transformation can't be used - ie. m= (0/0)*c2.

3. Jan 31, 2007

### Gyroscope

How is it possible for photons to have momentum if they are massless?

4. Jan 31, 2007

### ranger

5. Jan 31, 2007

### ranger

6. Jan 31, 2007

### robphy

Rather than the phrase "rest mass", it might be more appropriate to use the term "invariant mass" or (up to factors of c) "invariant norm of the momentum 4-vector". With this term, then it is easier to see that one can apply the Lorentz Transformation to the photon's [necessarily non-timelike] 4-momentum.

One sees that its square-norm being zero is true in all inertial reference frames. In addition, its temporal component is essentially the relativistic energy (up to constants, the frequency) of the photon. Similar to the "relativistic mass" (or better, up to constants, "relativistic energy"), the relativistic energy of the photon can be boosted toward infinity. (Of course, the "factor" is different... for the photon, it is "k" (the doppler factor), which is $$\gamma(1+v)$$.)

Of course, what you can't do is to boost from the frame of a timelike particle (where that particle is at rest) to one for a null (or lightlike) particle.

Last edited: Jan 31, 2007
7. Jan 31, 2007

### Staff: Mentor

That's not the Lorentz transformation. The Lorentz transformation equations for position and time are

$$x^\prime = \gamma (x - vt)[/itex] [tex]t^\prime = \gamma \left( t - \frac {vx}{c^2} \right)$$

for position and time, and

$$p^\prime = \gamma \left( p - \frac {vE}{c^2}\right)$$

$$E^\prime = \gamma (E - vp)$$

for momentum and energy. As far as I know, they are valid for light (photons) as well as for particles with nonzero "rest mass".

8. Feb 2, 2007

### bernhard.rothenstein

photon momentum and energy is a frequent topic on the forum. consider a tardyon (u<c) the momentum of which transforms as
p=gp'(1+V/u') (1)
E=gE'(1+Vu'/c^2) (2)
state that special relativity theory ensures a smooth transition from the properties of the tardyon to the properties of a photon and make in (1) and (2) u=u'=c in order to obtain in its case
p(c)=gp'(c)(1+V/c) (3)
E(c)=gE'(c)(1+V/c) (4)
Is there more to say?
sine ira et studio

9. Feb 2, 2007

### nakurusil

Doesn't look correct. Here are the correct ones.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

For the photon you can further simplify the above by using the fact that energy and momentum are related by $$E=c*\sqrt<p,p>$$.

Last edited: Feb 2, 2007
10. Feb 8, 2007

### duordi

We are not sure that light has zero rest mass or that it travels 100% of the speed of light.

But we do know it is very close to it. Make that very very close.

11. Feb 8, 2007

### yenchin

Light does not travel at the speed of light? Then why do you call it "the speed of light"?

12. Feb 8, 2007

### duordi

Is 99.9999999999 equal to 100?

No it isn't, but can you tell? That is the point.

If a person is accelerated to 99% of the speed of light and that person measures the speed of the photon passing him he will determine the photon is traveling past him at the speed of light.

So if two photons are traveling in parallel paths what speed do they measure of each with respect to the other?

13. Feb 8, 2007

### rbj

or, looking at it another way, instead of mapping rest mass to "relativistic mass" (or "inertial mass" or whatever it is you get when you divide momentum by velocity),

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

map it back the other way:

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$

so, if the photon has a finite inertial mass

$$E = m c^2 = h \nu$$

or

$$m = \frac{E}{c^2} = \frac{h \nu}{c^2}$$

and the momentum is

$$p = m v = \frac{h \nu}{c^2} v$$

but if the velocity of the photon is $c$, then

$$p = m c = \frac{h \nu}{c}$$

no matter what that finite value is, the rest mass (or "invariant mass") is still zero when $v = c$.

$$m_0 = m \sqrt{1 - \frac{c^2}{c^2}} = m \sqrt{1 - 1} = 0$$

that's my oversimplistic spin on it.

14. Feb 9, 2007

### yenchin

The speed of light is, well, the speed of light. If you think that photon travel at 0.999999999999c where c is the current value of the speed of light, we can just define the "correct" speed of light as c'=0.999999999999c. :rofl:

15. Feb 9, 2007

### pervect

Staff Emeritus
Photons don't experience time, and thus they can't measure speed. The closest thing to "experiencing time" is that photon geodesics can be parametrized in terms of an affine parameter, which however is neither like time (timelike) nor like space (spacelike), but null.

There are a number of FAQ's and threads on this

and there's a lot more I've skipped over, including one by robphy that was particularly good that I can't find.

16. Feb 9, 2007

### duordi

You are correct pervect

For this reson if a photon has zero rest mass it may also be an unstable energy unit but without the time to decay for certainly it would have decayed in billions of years of travel if it contained a time component.

As to the light speed.

If a photon has a rest mass particle then it can not attain a speed of C but will only attain a velocity of C ‘ which is slightly less then the theoretical speed of light C.

In this case the measured speed of light is C ’ and the true speed of light C must be calculated or by other means to be known.

The photon must also be a stable energy unit to prevent decay.

17. Feb 9, 2007

### rbj

the quantity we call $c$ is the wavespeed of electromagnetic propagation in a vacuum that you get from solving Maxwell's Equations. you know:

$$c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$$

it would be more precise to say that "We are not sure that photons have zero rest mass or that they travel at 100% of the wavespeed of light." BTW, this was something totally new to me a year ago. i still have trouble believing it. (there is the aesthetic part of me that wants the dogma that the speed of photons are $c$ which means they must have zero rest mass.)

18. Feb 9, 2007

### duordi

Interestingly C was defined by definition in 1983.
Permittivity (E)of free space is defined by C and the Maxwell equations.
Permeability (u) is measured.

Which means C can not be determine by ( or is not determined ?) by the Maxwell equations.
Unless (E) can be measured.

http://en.wikipedia.org/wiki/Speed_of_light

19. Feb 9, 2007

### nakurusil

Correct, as per wiki:

"In metric units, c is exactly 299,792,458 metres per second (1,079,252,848.8 km/h). Note that this speed is a definition, not a measurement. Since the fundamental SI unit of length, the metre, has been defined since October 21, 1983 in terms of the speed of light; one metre is the distance light travels in a vacuum in 1/299,792,458 of a second."

You may wonder why is c defined and not measured to be 299,792,458 metres per second .
The reason is that any physical measurement has certain error attached to it, so that a number had to be chosen. To recap:

1. c is chosen
2. $$\epsilon_0$$ is defined based on c
3.$$\mu_0$$ is derived based on the values chosen at 1 and 2

So, all of the above guarantees that c is exactly 299,792,458

Last edited: Feb 9, 2007
20. Feb 9, 2007

### MeJennifer

Actually the reason c is a constant has to do with how the meter is defined.