# Conceptual questions about electromagnetism inside

1. Jul 13, 2013

### tolove

The question I'm wanting to ask is, Why does light travel through a vacuum? This is too big of a question to start with, though. Before making this thread, I read around and tried to make a starter question.

If I'm not confused by the information I've found, a fundamental reason of why light can travel through a vacuum has to do with the energy-momentum relation:

E$^{2} = (pc)^{2} + (mc^{2})^{2}$

I don't understand this equation much at all. I think the first question I need to ask has to do with momentum.

How is relativistic momentum different than classical momentum? What is relativistic momentum?

Thank you very much for reading!

2. Jul 13, 2013

### Drakkith

Staff Emeritus
There is no known reason WHY light can travel through a vacuum. The conditions of our universe simply allow for energy and momentum to be removed from an object and transferred through space to another object in the form of EM waves that are quantized into photons.

There are many rules that tell us how this works, but none that tell us WHY.

I'll leave it to someone else to explain the relativistic momentum stuff.

3. Jul 13, 2013

### tolove

I can feel my brain being twisted into a knot. I'm going to ask a series of clueless questions for anyone who can answer them:

When does a photon have mass? Actual mass, that is... a classical sense of mass. Can I ever compare a photon to a baseball?
I've read the wiki article on invariant mass, but am still confused. What is invariant mass?
At what moment does the photon lose mass?
Are EM waves photons? Is there an equation for this?

My mental picture of a photon is so very confused. This is more or less how I am imagining a photon: A photon is a strange thing that behaves as either a particle or wave. When it is traveling through vacuum, the photon is a self-propagating wave that requires no medium and experiences no internal resistance (eg, continues forever).

Also, if anyone knows of a book that's heavy on conceptual pictures for this subject, I'd love to read it.

Last edited: Jul 13, 2013
4. Jul 13, 2013

### Drakkith

Staff Emeritus
If you are referring to invariant mass, aka rest mass, then never. Photons are ALWAYS massless.

One way to describe it is that invariant mass is the energy content of an object that remains the same to all observers. This is different from something like kinetic energy, which will be more or less depending on the observers motion relative to the object.

Invariant mass can be converted to an amount of energy using the famous equation E=MC2

Photons can never lose mass, as they never had mass to begin with.

The energy content of an EM wave can only be transferred to another object in small packets. Aka the energy of the wave is quantized. The size of this packet, the energy content per packet, is dependent on the wavelength of the EM wave, with smaller wavelengths having more energy content per packet than larger wavelengths. This "packet" is called a photon.

There are many equations that cover various aspects of light and other electromagnetic waves. I don't know which one you'd like.

Unfortunately, it is VERY VERY confusing to most people when they begin to mix Quantum and Classical physics. Things just don't work the same and may not make any sense. The photon is simply the quantized interaction (transfer of energy) of the EM wave with anything else. Since the EM wave interacts only in small packets, it is vaguely similar to firing a bunch of bullets at an object. (Very very vaguely. Sophiecentaur may have my head if I don't say this) This is why the photon is called the "particle" of light.

5. Jul 13, 2013

### tolove

Thanks for your lengthy reply!

I'm going to type some things. Please correct me if any of this is wrong. I am still very doubtful about the entire idea:

Photons never have invariant mass. Invariant mass is a classical idea? A 1kg ball has 1kg of invariant mass within its inertial frame of reference.

If we add an observer, from a non-inertial frame of reference, then the mass of that ball, to the observer, will increase or decrease according to the ball's observed momentum:

$m = \sqrt{ \frac{E^{2} - (ρc)^{2}}{c^{2}} }$

Momentum, ρ, here is in a classical sense? (invariant) Mass * Velocity. This equation, when used for photons, is simply E = mc2, because a photon does not have invariant mass.

A photon is a pure energy chariot thing, then? Once it hits a mass-particle, the energy is transferred, and then can be converted to invariant mass?

And for the quantization part... Let's take red light. One photon of red light. What is the meaning of the question: "What is half the wavelength of one photon of red light?"

6. Jul 13, 2013

### kevinferreira

Invariant mass is as classical as quantum idea. The quantum of electromagnetic energy, aka photon, has the invariant mass of zero, always, for everyone. Mass is a very tricky property, I prefer to look at it as an 'energy content' particles have (or don't, as the photon).

This is the idea, roughly, except that we are always talking here about INERTIAL frames of reference, where these ideas and equations are valid, we don't want to go to non-inertial, as strange things happen to energy.

Nope, it's the relativistic momentum $\vec{p}=\frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$.
For photons, the equation becomes E=pc.

One photon of red light in certain experiences finds an appropriate description in terms of an object with wavelike properties, with a certain wavelength. Take the half.
This is the tricky thing about the wave-particle duality: it is a duality, i.e. you can describe exactly the same (conceptual) object in two very different ways depending on which situation you are (which 'regime'). Sometimes you will do it perfectly thinking of light as a particle, in other regimes this will not be good and we will use an electromagnetic description. This is all physics can give you nowadays.

7. Jul 13, 2013

### Drakkith

Staff Emeritus
No, as mass means "invariant mass" when we say it nowadays. The mass would remain the same, but the momentum would change. (I think. I'm not familiar with non-inertial frames.)

As Kevin said, a photon would have zero mass and thus the equation reduces to E=pc. Which itself can be restated was E=hv, where h is planck's constant and v is the frequency of its associated EM wave.

It can, but it doesn't have to be converted to mass. A lone particle hit by a photon would simply accelerate and would not gain any mass.

It is asking what is one half wavelength of the photon's associated EM wave.

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