Photon Momentum: Massless but Impactful

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Discussion Overview

The discussion revolves around the concept of photon momentum, particularly addressing the apparent contradiction of photons being massless yet possessing momentum. Participants explore the implications of relativistic physics on this topic, contrasting it with Newtonian mechanics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion over the idea that massless photons can have momentum, referencing Newtonian physics where momentum is defined as p=mv.
  • Others clarify that in relativistic physics, the momentum of a photon can be expressed as p=E/c, where E is the energy of the photon, indicating that its momentum arises from its energy rather than mass.
  • A participant emphasizes that particles traveling at the speed of light (c) must have zero mass, arguing that this is a requirement of relativistic physics and not merely a possibility.
  • There is a challenge to the interpretation of momentum in Newtonian mechanics, with a participant asserting that momentum is a vector quantity and should not be conflated with scalar definitions.
  • Another participant provides calculations related to the velocities of massive particles, arguing that any particle with mass cannot reach the speed of light, reinforcing the claim that only massless particles can travel at c.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the interpretation of mass and momentum in the context of relativity versus Newtonian mechanics. While some assert that massless particles must have zero mass, others question the implications of this on definitions of momentum.

Contextual Notes

There are unresolved assumptions regarding the definitions of momentum in different frameworks, and the discussion reflects varying interpretations of relativistic principles. Some mathematical steps and definitions remain unclear or contested among participants.

Who May Find This Useful

This discussion may be of interest to those studying physics, particularly in understanding the nuances of momentum in relativistic contexts and the implications of massless particles in theoretical frameworks.

Parbat
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Photon has momentum but is massless.Doesn't that seem strange?
 
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If you write down the fully relativistic equation for momentum (in pc units) for any particle (including photons) of kinetic energy T, the quation is

(pc)2 = (T + mc2)2 - (mc2)2

= T2 + 2(mc2)T

So, for a bullet use the second term; for a photon use the first (for mass = 0). So a massless photon has momentum.

Bob S
 
Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c can only have zero mass.
 
No it's not strange because photon has energy. So its momentum is [tex]p=E/c=h/\lambda[/tex]

(ok it is little bit strange...)
 
jcsd said:
Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c can only have zero mass.

No.

"Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c [STRIKE]can only have[/STRIKE] may have zero mass".

Also, in Newtonian mechanics, p = sqrt(2mE) where E = ½mv2

Bob S
 
Bob S said:
No.

"Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c [STRIKE]can only have[/STRIKE] may have zero mass".

Also, in Newtonian mechanics, p = sqrt(2mE) where E = ½mv2

Bob S

No, not 'may', 'must'. The four-momentum of a particle is tangent to it's wordline for a particle traveling at c this means it's four-momentum is always null. The mass of particle (in the absence of a rest frame to define it) can be taken as the norm of it's four-momentum which is zero for a particle with null four-momentum. Hence all particles traveling at c MUST have zero mass (unless we're going to allow particles to have undefined four-momentum and hence undefined momentum and energy).

In Newtonian mechanics a particles momentum is defined as p(t) = mv(t).

Your definition is clearly incorrect as momentum is a vector quantity whereas your definition defines a scalar quantity.
 
jcsd said:
No, not 'may', 'must'.

Exactly.

[tex]v = c \sqrt{1-\frac{m^2 c^4}{E^2}}[/tex]
 
From jcsd:...but relativity shows that particles that travel at c can only have may have zero mass".No, not 'may', 'must'. The four-momentum of a particle is tangent to it's wordline for a particle traveling at c this means it's four-momentum is always null. The mass of particle (in the absence of a rest frame to define it) can be taken as the norm of it's four-momentum which is zero for a particle with null four-momentum. Hence all particles traveling at c MUST have zero mass (unless we're going to allow particles to have undefined four-momentum and hence undefined momentum and energy).
You are correct.

From Bob S: "but relativity shows that particles that travel at c [STRIKE]can only have[/STRIKE] may have zero mass".

My pocket calculator shows that a 3.5 TeV proton in LEP has a v/c =

β = 1-1/2γ2 = 1-3.6·10-8 = 0.999 999 964

For the 1020 eV Oh My God cosmic ray proton

β = 1-1/2γ2 = 1-5·10-23 = 0.999 999 999 999 999 999 999 9 (more or less)

So any particle with mass >0 cannot have a velocity of exactly c. I was not careful enough in rounding off.

Bob S
 

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