Do photons have inertia?

**wellwisher** · 04-10-12, 10:35 AM

I think they did an experiment that showed this. If I remember, the experiment used a thin foil of a shiny noble metal, attached to a fine thread, suspended in a vacuum. They then shined a light at the foil and the foil started to rotate. The light bounces off or reflects from the foil creating and equal and opposite force.

Another way to demonstrate this is in the link below:
https://wiki.brown.edu/confluence/display/physlecdemo/4D20.10+Crooke's+Radiometer

**Robittybob1** · 04-11-12, 11:13 PM

Originally Posted by waitedavid137

For anything massive or massless the three momentum is given in terms of a three dimensional description of wavelength, the wave number three vector,
$\vec{p} = \hbar \vec{k}$
It yields a magnitude for the momentum of
$p = h / \lambda$
This was discovered by those doing particle physics in association with the advent of quantum mechanics. I don't know that there is any kind of intuative way to explain it, thats just the underlying universal case whether your talking about something massive like an electron or massless like a photon.

How do you use the second equation for an electron? What is the wavelength of an electron? That would be an electron associated with an atom (bound).

**Tach** · 04-12-12, 01:15 AM

Originally Posted by Robittybob1

How do you use the second equation for an electron?

same way, $\lambda=\frac{h}{p}$ where , $p=mv$ , $m=\gamma m_0$

**waitedavid137** · 04-12-12, 01:25 AM

Originally Posted by Robittybob1

How do you use the second equation for an electron? What is the wavelength of an electron? That would be an electron associated with an atom (bound).

So for example you could do a Bohr atom electron energy easy enough. This is undergrad intro level but I doubt you'd get anything from the full modern quantum analysis including spherical harmonics.
For something in circular orbit the centripital force is the electric force
-kZe²/r² = -mv²/r
-kZe²/r² = -(mv)²/mr
mkZe²/r = (mv)²
p² = mkZe²/r
p = h/λ
h²/λ² = mkZe²/r
The modes exist as
nλ = 2πr
(nh)²/(2πr)² = mkZe²/r
r = n²(h/2π)²/(mkZe²)
The energy of the orbit is
E = -kZe²/r + (1/2)mv²
E = -kZe²/r + (1/2m)(mv)²
From above
E = -kZe²/r + (1/2m)mkZe²/r
E = -(1/2)kZe²/r
From above
E = -(mkZe²)(1/2)kZe²/[n²(h/2π)²]
E = -(kZe²)²(1/2)mc²/[n²(hc/2π)²]

E = -Z²α²(1/2)mc²/n²

α ≈ 1/137

These energy levels are what is observed in nature aside from small energy level splitting corrections coming from things like relativity and magnetic spin interactions.

**Robittybob1** · 04-12-12, 03:17 AM

Thank you Tach and David.

**prometheus** · 04-12-12, 03:37 AM

Originally Posted by Tach

...where , $p=mv$ , $m=\gamma m_0$

This makes my toes curl. Strictly it is correct, but I would much rather see $p = \gamma m v$

**waitedavid137** · 04-12-12, 04:05 AM

Originally Posted by prometheus

This makes my toes curl. Strictly it is correct, but I would much rather see $p = \gamma m v$

Mine too. I can't figure for the life of me why people still do that.

**waitedavid137** · 04-12-12, 04:09 AM

Originally Posted by Tach

same way, $\lambda=\frac{h}{p}$ where , $p=mv$ , $m=\gamma m_0$

Wiki is not a valid reference.
$p=mU = \gamma mu$
$m^{2} c^{2} = |g_{\mu \nu} p^{\mu} p^{\nu}|$

**Tach** · 04-12-12, 09:14 AM

Originally Posted by prometheus

This makes my toes curl. Strictly it is correct, but I would much rather see $p = \gamma m v$

Because wiki is not quite correct on the subject, so I needed to make it clear that the $m$ in their $p=mv$ is really $\gamma m_0$ .

**Tach** · 04-12-12, 09:15 AM

Originally Posted by waitedavid137

Mine too. I can't figure for the life of me why people still do that.

Because wiki is not quite correct on the subject, so I needed to make it clear that the $m$ in their $p=mv$ is really $\gamma m_0$ . Now that you understand, I hope that you no longer have a problem with that.

**Tach** · 04-12-12, 09:18 AM

Originally Posted by waitedavid137

So for example you could do a Bohr atom electron energy easy enough. This is undergrad intro level but I doubt you'd get anything from the full modern quantum analysis including spherical harmonics.
For something in circular orbit the centripital force is the electric force
-kZe²/r² = -mv²/r

You realize that the "centripital" force (sic!) that you are using is not correct , the correct form is $\gamma \frac{mv^2}{r}$ .

**waitedavid137** · 04-12-12, 09:29 AM

Originally Posted by Tach

You realize that the "centripital" force (sic!) that you are using is not correct , the correct form is $\gamma \frac{mv^2}{r}$ .

If you'd actually read the whole post you might understand some of it. Just to make you do so, quote for us here what the last sentence of that post of mine was.

**Tach** · 04-12-12, 09:33 AM

Originally Posted by waitedavid137

If you'd actually read the whole post you might understand some of it. Just to make you do so, quote for us here what the last sentence of that post of mine was.

There was no point in reading past your mistaken "centripital" force, at the speeds involved, $\gamma$ is not 1.

**waitedavid137** · 04-12-12, 09:45 AM

Originally Posted by Tach

There was no point in reading past your mistaken "centripital" force, at the speeds involved, $\gamma$ is not 1.

I didn't make a mistake. You just don't know what you're talking about or you are lieing. So, as I said,
... quote for us here what the last sentence of that post of mine was.

**Tach** · 04-12-12, 09:51 AM

Originally Posted by waitedavid137

I didn't make a mistake. You just don't know what you're talking about or you are lieing.

Why would I be "lieing"? I simply pointed out a basic mistake at the very beginning of your derivation. The speeds involved disallow neglecting the relativistic effects.

**waitedavid137** · 04-12-12, 09:57 AM

Originally Posted by Tach

Why would I be "lieing"? I simply pointed out a basic mistake at the very beginning of your derivation. The speeds involved disallow neglecting the relativistic effects.

You are deliberately lieing as proven by your refusal to quote the last line of my post which puts it all in context. I hope you get banned for it.

**prometheus** · 04-12-12, 11:33 AM

Originally Posted by Tach

The speeds involved disallow neglecting the relativistic effects.

If that were true, the solution to the Schrodinger equation for the electron in the hydrogen atom would be pointless. I suggest you stop picking fights with people for the sake of it.

Also $m \neq \gamma m_0$ .

**Tach** · 04-12-12, 11:34 AM

Originally Posted by prometheus

Also $m \neq \gamma m_0$ .

Really? What made you change your mind? You had it right earlier.

If that were true, the solution to the Schrodinger equation for the electron in the hydrogen atom would be pointless.

The Schrodinger equation is not valid in the relativistic domain, it has long been replaced by the Dirac equation in the QED formulation. About 84 years ago, in 1928.

**Emil** · 04-12-12, 11:39 AM

Originally Posted by waitedavid137

You're kidding right?
Like essentially every peice of technology we use, right down to the computer your typing into. Even the very existence of the magnetic field. Everything at some level at large or small scale verifies the validity of Maxwell's equations which themselves are Lorentz invariant. Everything points toward Lorentzian spacetime geometry, not Euclidean. No doubt people have already pointed to aspects of technology that are obvious demonstrations of one aspect or other of the validity of Lorentz transformation, but the truth is relativity is needed to truely understand nearly all technology.

A simple question,
Did you understand my demonstration and logic, why SR is not valid in a medium different from vacuum?

**AlphaNumeric** · 04-12-12, 12:04 PM

Originally Posted by Emil

A simple question,
Did you understand my demonstration and logic, why SR is not valid in a medium different from vacuum?

SR is valid but it isn't just SR, it's things like QFT. QFT includes SR but also explains why light can move slower than c in a medium.

When are you going to learn some physics? That's a pretty simple question.