Is the Velocity of Light Undefined?

  • Thread starter Thread starter avito009
  • Start date Start date
  • Tags Tags
    Light Velocity
AI Thread Summary
The discussion centers on the velocity of light and its relationship to kinetic energy and mass. While light travels at a constant speed of 299,792,458 m/s, the concept of velocity becomes complex when considering massless particles like photons. The traditional kinetic energy formula, Ek = 1/2 mv², is not applicable for photons since their mass is zero, leading to confusion about defining their velocity. Instead, the energy of photons is described by E = hf, where frequency governs their energy, not speed. Ultimately, light does have a defined velocity, which is constant in a vacuum, and the discussion highlights the importance of using appropriate equations for massless particles.
avito009
Messages
184
Reaction score
4
Firstly I know that light has a speed of 299 792 458 m / s. But does it have a velocity. Can this be plotted on graph that is (The velocity)?

Next if you derive the formula V2= 2Ek/ m. Where V2 is the velocity squared and Ek is kinetic energy and m is the mass. This is derived from Ek= 1/2 mv2.

So if we try to find velocity of a photon then the m would be zero and anything divided by zero is undefined. So why is the velocity of light undefined? V2= 2Ek/ 0= undefined.
 
Science news on Phys.org
Certainly, light has a velocity since it's definitely moving some direction. Light (in vacuum) has a velocity ##\vec{v}=c\hat{v}## which is really easy since its speed is constant, the only thing you have to specify is the direction of travel ##\hat{v}##.

The formula ##E_k=\frac{1}{2}mv^2## is only valid for ##v \ll c##, and is only valid for massive particles where ##m\neq 0##. For massive particles with speed close to ##c## the correct formula is ##E_k=(\gamma-1) mc^2 ##. For a massless particle like the photon, we have to look towards quantum mechanics to give us the energy ##E_k = hf##. The energy won't be speed dependent since all massless particles travel at the same speed, but it is rather the frequency of the massless particles that govern its energy.
 
Yes, light has a velocity. I take a laser, I point it towards the ground, the photons have velocity c towards the ground.

Now, you cannot use Ek = 1/2 mv2. That is only correct in when v<<c, which, by your very question, is not the correct use of the definition. You must use the full relativistic equation for kinetic energy, namely:

Ek = sqrt(p2c^2 + m2 c4 ) - mc2

Be also aware that the kinetic energy is defined with speed not velocity - energy is a scalar, and my kinetic energy is the same if I'm traveling at 10 m/s south, or 10 m/s north.

ETA: Sniped by Matterwave.
 
e.bar.goum said:
Yes, light has a velocity. I take a laser, I point it towards the ground, the photons have velocity c towards the ground.

Now, you cannot use Ek = 1/2 mv2. That is only correct in when v<<c, which, by your very question, is not the correct use of the definition. You must use the full relativistic equation for kinetic energy, namely:

Ek = sqrt(p2c^2 + m2 c4 ) - mc2

Be also aware that the kinetic energy is defined with speed not velocity - energy is a scalar, and my kinetic energy is the same if I'm traveling at 10 m/s south, or 10 m/s north.

ETA: Sniped by Matterwave.

The reason I did not use the equation ##E_k = \sqrt{p^2c^2+m^2 c^4}-mc^2## is that for ##m=0## this reduces obviously to ##E_k=pc##. Of course this equation is true, but the momentum of a mass-less particle will perhaps lead to just more confusion. And of course the next question is probably going to be "what's the momentum of a photon?" (which will get you right back to ##p=h/\lambda=hf/c=E_k/c##).
 
Matterwave said:
The reason I did not use the equation ##E_k = \sqrt{p^2c^2+m^2 c^4}-mc^2## is that for ##m=0## this reduces obviously to ##E_k=pc##. Of course this equation is true, but the momentum of a mass-less particle will perhaps lead to just more confusion. And of course the next question is probably going to be "what's the momentum of a photon?" (which will get you right back to ##p=h/\lambda=hf/c=E_k/c##).

Amusingly, I used ##E_k = \sqrt{p^2c^2+m^2 c^4}-mc^2## for precisely the reason you didn't.
 
Matterwave said:
but it is rather the frequency of the massless particles that govern its energy.

You must note that the concept of frequency applies to waves only and not particles.
 
avito009 said:
You must note that the concept of frequency applies to waves only and not particles.

This is not true, as was shown by de-Broglie in the early 1900's.

Light certainly has a frequency associated with it!
 
avito009 said:
You must note that the concept of frequency applies to waves only and not particles.

Matterwave said:
This is not true, as was shown by de-Broglie in the early 1900's.

Light certainly has a frequency associated with it!

I believe there's also that small caveat about light being an EM wave too.
 
Matterwave said:
This is not true, as was shown by de-Broglie in the early 1900's.

Light certainly has a frequency associated with it!
De Broglie talks about the Wavelength of a particle and not a Frequency.
 
  • #10
sophiecentaur said:
De Broglie talks about the Wavelength of a particle and not a Frequency.

I believe where you have a wavelength, you have a frequency ##f=c/\lambda##. But even if that part of my argument is somehow wrong, light certainly has a frequency associated with it, and the formula ##E=hf## is certainly valid!
 
  • #11
E = hf applies to Photons but not to particles in general. For particles, P = h/λ is the relevant expression and that's what de Broglie was talking about - not involving frequency. The velocity of de Broglie waves cannot be c because there is mass involved.
 
  • #12
sophiecentaur said:
E = hf applies to Photons but not to particles in general. For particles, P = h/λ is the relevant expression and that's what de Broglie was talking about - not involving frequency. The velocity of de Broglie waves cannot be c because there is mass involved.

Ah, I see where my post was unclear. Sorry about that. I meant only for ##E=hf## to apply to light as a particle.
 
  • #13
Matterwave said:
I believe where you have a wavelength, you have a frequency ##f=c/\lambda##. But even if that part of my argument is somehow wrong, light certainly has a frequency associated with it, and the formula ##E=hf## is certainly valid!

Ironic. Matterwave is not merely justifying an argument, but his very username!
 
  • #14
Does the equation V2= 2Ek/ m for v<<c prove that, for the velocity to be higher either the mass has to be less or the kinetic energy has to be higher for the object to attain high velocity?

So does that prove that the more massive the object, the lesser will be its speed (Velocity) and if the speed has to be more then the object has to have more kinetic energy?
 
  • #15
avito009 said:
Does the equation V2= 2Ek/ m for v<<c prove that, for the velocity to be higher either the mass has to be less or the kinetic energy has to be higher for the object to attain high velocity?

The equation is just showing a relationship between velocity, mass, and kinetic energy. If two objects are moving at the same speed then the one with more mass will have more kinetic energy. Two objects can have equal kinetic energy, yet have wildly different velocities if their masses are also very different.
 
  • #16
avito009 said:
Does the equation V2= 2Ek/ m for v<<c prove that, for the velocity to be higher either the mass has to be less or the kinetic energy has to be higher for the object to attain high velocity?

So does that prove that the more massive the object, the lesser will be its speed (Velocity) and if the speed has to be more then the object has to have more kinetic energy?

Mass is not something that will change in that equation, for a given particle. Only the kinetic energy and velocity may change. So higher velocity = higher kinetic energy. That's basically all that equation is saying.
 
Back
Top