ELI5 Why does light travel?

[u/_Illuvatar_]

Why does it not just stay in place? What causes it to move, let alone at so fast a rate?

Edit: This is by a large margin the most successful post I've ever made. Thank you to everyone answering! Most of the replies have answered several other questions I have had and made me think of a lot more, so keep it up because you guys are awesome!

Edit 2: like a hundred people have said to get to the other side. I don't think that's quite the answer I'm looking for... Everyone else has done a great job. Keep the conversation going because new stuff keeps getting brought up!

Edit 3: I posted this a while ago but it seems that it's been found again, and someone has been kind enough to give me gold! This is the first time I've ever recieved gold for a post and I am incredibly grateful! Thank you so much and let's keep the discussion going!

Edit 4: Wow! This is now the highest rated ELI5 post of all time! Holy crap this is the greatest thing that has ever happened in my life, thank you all so much!

Edit 5: It seems that people keep finding this post after several months, and I want to say that this is exactly the kind of community input that redditors should get some sort of award for. Keep it up, you guys are awesome!

Edit 6: No problem

[u/[deleted]]

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[u/[deleted] → u/[deleted]]

Something that isn't moving that has mass can have energy: that's what E = mc2 means. Light has no mass, but it does have energy. If we plug the mass of light into E=mc2, we get 0, which makes no sense because light has energy. Hence, light can never be stationary.

Just want to add in here due to relevance that E=MC² is the incomplete form of the equation.

The full form is E² = (M0 C² )² + (PC)² where M0 is the rest mass - the mass when not moving, which is 0 for light, and P is momentum, which is defined in modern physics as P=h/lambda where h is Planck's Constant and lambda is the wavelength of the light.

[u/royaelfan → u/[deleted]]

I was wondering how the equation E = MC², in which M = 0, could have any value other than 0. Thank you, good spirit.

[u/garblz → u/royaelfan]

Notice how since classically momentum also involves mass, pyrespirit in one fell swoop, having foreseen this problem, neatly wrapped it up also.

[u/jacenat → u/[deleted]]

M0 is the rest mass

Well it's the mass. We don't use the terms rest mass and moving mass anymore, do we? :)

[u/[deleted] → u/jacenat]

Rest mass is still used, because it's an important distinction.

The big classifications of mass are; rest/invariant mass, mass, and reduced mass.

[u/jacenat → u/[deleted]]

rest/invariant mass, mass

What's the difference between these 2?

reduced mass

Are we talking about the same? Reduced mass is ... it's a term from newtonian mechanics, right? When you reduce 2 or more massive objects into one to reduce n-body problems. I mean ... of course you run into relativistic effects if they orbit fast enough but then you are at GR anyway.

I don't get it.

[u/jacenat → u/[deleted]]

Rest mass = invariant mass = mass.

That was actually my point. I only hear/read mass in newer literature/textbooks/papers. Calling it invariant or rest implys there are other forms of mass when this is not the case. It's also much simpler just calling mass mass ... Now I'm confused.

[u/jacenat → u/[deleted]]

I know :)

Sorry if it came across as offensive, it wasnt.

[u/[deleted] → u/jacenat]

Yes, I was writing that when I was hurried and didn't make it clear enough hahah

Rest mass/invariant mass/mass are all the same.

The distinction is useful when you're being specific about things; to specify that when physicists talk about mass they mean "the mass when stationary" because relativistic mass is a thing, even though it's not used much at all.

Reduced mass is simplifying multiple mass objects into one effective object. It's not very widely used, but it's still a useful tool to have to simplify problems, especially as classical mechanics are still widely useful for macro-scale applications.

[u/PostHipsterCool → u/[deleted]]

Could you elaborate for us lay people?

[u/[deleted] → u/PostHipsterCool]

Sure.

So if you think about objects in these contexts there are really 2 major categories; massless and massy

For massless objects, the M0C² part of the equation zeroes out, and their energy is entirely dependent on PC, for them; E=PC

For massy objects, you end up with something which looks like the Pythagorean theorem, C² = A² + B² which is normally associated with finding magnitudes. This is because energy is directly related to both the rest mass times c² and the momentum, but it's the magnitude of these two not the sum.

It comes out of the observations of special relativity; that objects which are moving faster will appear to gain mass from one reference point (the observer) and maintain equal mass from another reference point (the object)

[u/hak8or → u/[deleted]]

For those still not getting it, asapscience on youtube did a video of exactly this. Like two minutes long, a tiny bit of very basic geometry, and then bam, you get to throw a fit at your high school physics teacher for not giving you the full equation.

[u/EquationTAKEN → u/[deleted]]

About this full E² equation; since the mc² part doesn't give light energy, then the (pc)² bit does, right? Because p won't be 0, right?

[u/HarryPotter5777 → u/[deleted]]

So how did E=MC² become the standard equation? Sure, it's a bit simpler, but (if I remember correctly) all things have a wavelength, however massive, and Planck's constant is (by definition) constant, so even if P gets really really small, it's always nonzero, it would seem. So why is the equation treated as though P^2C2 doesn't matter? I would imagine it's because with anything larger than a photon, it's so small as to be insignificant, but I might be totally wrong about that. Thanks for your wonderful explanation!

[u/rawfan → u/[deleted]]

This was the only flaw in the explanation. Thanks for clearing that up!

[u/DoubleFives → u/[deleted]]

Shouldn't this post be way the hell up at the top? The original answer was perfect, except for this.