- #1

- 1

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter antariksh
- Start date

- #1

- 1

- 0

- #2

mathman

Science Advisor

- 7,924

- 467

- #3

Gyroscope

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

- #4

ranger

Gold Member

- 1,676

- 1

- #5

ranger

Gold Member

- 1,676

- 1

The momentum is related to wavelength:How is it possible for photons to have momentum if they are massless?

[tex]p = \frac{h} {\lambda}[/tex]

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c2

- #6

- 5,824

- 1,129

^{2}.

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 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 [tex]\gamma(1+v)[/tex].)

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:

- #7

jtbell

Mentor

- 15,751

- 3,948

^{2}.

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

[tex]x^\prime = \gamma (x - vt)[/itex]

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

for position and time, and

[tex]p^\prime = \gamma \left( p - \frac {vE}{c^2}\right)[/tex]

[tex]E^\prime = \gamma (E - vp)[/tex]

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

- 991

- 1

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

- 329

- 0

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)

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 [tex]E=c*\sqrt<p,p>[/tex].

Last edited:

- #10

- 78

- 0

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

- #11

- 541

- 3

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

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

- #12

- 78

- 0

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

- 2,226

- 9

^{2}.

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),

[tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

map it back the other way:

[tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]

so, if the photon has a finite inertial mass

[tex] E = m c^2 = h \nu [/tex]

or

[tex] m = \frac{E}{c^2} = \frac{h \nu}{c^2} [/tex]

and the momentum is

[tex] p = m v = \frac{h \nu}{c^2} v [/tex]

but if the velocity of the photon is [itex]c[/itex], then

[tex] p = m c = \frac{h \nu}{c} [/tex]

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

[tex] m_0 = m \sqrt{1 - \frac{c^2}{c^2}} = m \sqrt{1 - 1} = 0 [/tex]

that's my oversimplistic spin on it.

- #14

- 541

- 3

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?

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

- 9,978

- 1,155

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

of light.

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

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

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html

https://www.physicsforums.com/showthread.php?t=132528

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

- #16

- 78

- 0

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

- 2,226

- 9

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

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

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

[tex] c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} [/tex]

it would be more precise to say that

- #18

- 78

- 0

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

- 329

- 0

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

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

You may wonder why is c

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. [tex]\epsilon_0[/tex] is defined based on c

3.[tex]\mu_0[/tex] is derived based on the values chosen at 1 and 2

So, all of the above guarantees that

Last edited:

- #20

- 1,997

- 5

Actually the reason c is a constant has to do with how the meter is defined.You may wonder why is cdefined and not measuredto 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. [tex]\epsilon_0[/tex] is defined based on c

3.[tex]\mu_0[/tex] is derived based on the values chosen at 1 and 2

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

- #21

jtbell

Mentor

- 15,751

- 3,948

Before 1983, the second was defined in terms of the period of a certain atomic transition, and the meter was defined in terms of the wavelength of another atomic transition. Those definitions were made because they could be reproduced in laboratories with the highest degree of precision possible at the time.

At some point, measurements of the speed of light became intrinsically more precise than the precision of the definition of the meter. Since there was (and still is) no experimental indication that the speed of light is not constant, defining the meter in terms of a constant, defined value of the speed of light maximizes the precision of measurements overall.

If at some point the speed of light is shown not to be constant, then the definition of the meter will surely be changed to reflect this.

- #22

- 35,847

- 4,676

If at some point the speed of light is shown not to be constant, then the definition of the meter will surely be changed to reflect this.

Actually, that need not necessarily be the case.

A second is defined using the frequency of Cs atoms right now. However, we know that the period of time is frame dependent (i.e. Cs atom in another frame would not have the same frequency). Yet, we still use this as our standard definition of a second.

So based on this, I think we can still use c to define a meter, even if we find (a very big if) situations where it isn't a constant. We just have to clearly define under what conditions this definition is to be used, just like most of our other constants.

Zz.

- #23

- 78

- 0

Because the terms are defined from one another it is impossible to determine if indeed light photons have slightly slower velocity then a theoretical maximum mass speed of light C.

It would be interesting to use the plank constant values to determine the minimum size of a mass particle and accelerate the mass until it had a photon equivalent relativistic mass.

The percentage of the speed of light the photon would achieve could then be calculated.

Different wavelengths would mean different relativistic mass quantities and different velocities however small these velocities differences may be.

It would be interesting to calculate the difference in velocity to determine if the velocity difference is measurable.

The effect would cause different wavelengths from a distant star to reach us with time delays.

Does anyone know if a pulsar has been checked to see if the lower and higher wave lengths are received without a time delay when viewed at the greatest distance possible?

If one photon had one smallest mass particle it would cause a quantum condition because photons could only have a whole number of smallest mass particles.

Last edited:

- #24

- 329

- 0

c is constant by definition in relativity. In addition, there is ample experimental confirmation, so I don't think that c being a constant has much if anything to do with the meter is defined.Actually the reasonc is a constanthas to do with how the meter is defined.

Actually, it is exactly the other way around, the definition of the meter is dependent on c:

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

and on the definition of the second.

Last edited:

- #25

- 1,997

- 5

That is fine, and you seem to be hard to convince , so I won't bother.I do not think so.

For others, the meter is defined in terms of the speed of light. One meter is the distance traveled by light in a vacuum in 1/299,792,458 of a second. So obviously c must be 299,792,458.

Share: