# B Classical equations and light

1. Jan 1, 2018

### JulianM

But momentum = mass x velocity so that part of the equation is E2 = c.(mv)

If the mass is zero then that formulation also yields zero energy (which we know is not true)

Last edited by a moderator: Jan 1, 2018
2. Jan 1, 2018

### DrStupid

This definition is based on another concept of mass. Today "mass" means invariant mass and that results in another equation for momentum.

3. Jan 1, 2018

### PeroK

The momentum of a photon is given by $p = hf/c$. Not by $p = mv$.

In fact, $p = mv$ is a non-relativistic approximation of the momentum of a massive particle. The correct equation is:

$p = \gamma mv$, where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

4. Jan 1, 2018

### JulianM

So lightarrow was wrong?

5. Jan 1, 2018

### DrStupid

No, he was and is still right.

6. Jan 1, 2018

### JulianM

Regardless of whether you include gamma this equation still contains mass.
Anything multiplied by zero is still zero.

7. Jan 1, 2018

### PeroK

For a photon $E = cp$. So, he is right and you are not.

$p = mv$, as I have already explained, is a classical equation for the momentum of a particle that is not valid in relativity - for any particle, massless or otherwise.

Last edited by a moderator: Jan 1, 2018
8. Jan 1, 2018

### DrStupid

Again, this "m" is an outdated concept of mass. It dates back to classical mechanics and is not used anymore because it is frame dependent in relativity. With the modern concept of mass (invariant mass) the equations for momentum and energy are

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

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

Of course they cannot be used for photons (because v=c), but they result in equations like

$p = \frac{{E \cdot v}}{{c^2 }}$

and

$E^2 = m^2 \cdot c^4 + p^2 c^2$

which are valid for v=c and m=0.

Last edited by a moderator: Jan 1, 2018
9. Jan 1, 2018

### PeroK

That equation is for a massive particle. It doesn't apply for a photon. In fact, it is not $0$ for a photon, it is undefined. Since the photon has zero mass and speed $c$ you would get:

$p = \frac{mc}{0} = \frac00$

Which is undefined. So, you have to look elsewhere for the equation that gives the momentum of a photon.

10. Jan 1, 2018

### JulianM

So what is the equation in relativity? It can't be just multiplying momentum by gamma as PeroK did (above). That still gives zero energy.

11. Jan 1, 2018

### PeroK

The general equation for momentum is:

$p^2c^2 = E^2 - m^2c^4$

That works for both massive and massless particles.

For a massive particle the energy is $E = \gamma mc^2$

And, for a photon the energy is $E = hf$.

12. Jan 1, 2018

### DrStupid

But the limit for $v \to c$ (and therefore $m \to 0$) is defined.

13. Jan 1, 2018

### JulianM

How does

$p^2c^2 = E^2 - m^2c^4$

work for a massless particle if p= m.v

That would give (for v = c)
( mc^2).c^2 = E^2 - m^2.c^4
is equivalent to
E^2 = m^2.c^4 - m^2.c^4

There's something wrong, surely it cannot apply to a massless particle which posses energy

14. Jan 1, 2018

### PeroK

What will it take for you to accept that $p = mv$ is not a valid equation?

15. Jan 1, 2018

### Orodruin

Staff Emeritus
Again, as you have been told repeatedly, p = mv is not a valid equation relativistically. Repeating the same mistake does not make it true.

16. Jan 1, 2018

### JulianM

Well now you have me very confused. How do we know when to apply relativistic equations or not?

17. Jan 1, 2018

### phinds

For a massless particle, you have no choice. For massive particles, the relativistic equation is always true to but the slower the particle is, the less difference there is between the results of the classical equation and the relativistic equation. At human-scale speeds, the classical equation always gives results that are useful.

18. Jan 1, 2018

### PeroK

In principle, the relativistic equations are always valid and the classical equations are an approximation. If that approximation is good enough, then you can apply the classical equations.

If the velocities involved are a significant proportion of the speed of light, then you definitely need the relativistic equations.

In practice, it depends how accurate you need your answer. Time calculations for GPS satellites must be so accurate that relativistic equations are needed despite the modest speeds. But, normal engineering calculations are usually more than accurate enough with classical equations.

19. Jan 1, 2018

### ZapperZ

Staff Emeritus
No, in THIS case, momentum is defined via the wavenumber, i.e.

p=ħk

So there is no requirement for a "mass" here.

Zz.

20. Jan 1, 2018

### Staff: Mentor

You can always apply the relativistic equations, and you will always get the right answer. They're more exact than the non-relativistic ones.

However, the relativistic equations are also more complicated and using them is more work, so we don't use them when the relativistic effects are small enough to ignore. For example: what is the kinetic energy and the momentum of a one-kilogram mass moving at ten meters per second? Try calculating that using the classical formulas and using the relativistic ones. Which was more work? And what difference did it make?

The classical formulas can be used any time the speeds involved are small compared with the speed of light (equivalently, $\gamma$ is so close to 1 that you're OK with the approximation $\gamma=1$). Clearly this is not the case for particles travelling at the speed of light, so you know that the classical formulas can't be used here.