Another Physics Misconception

[@Fif0l]

Physics misconception: did you know the equation you learned in school is only accurate for stuff you'll deal with outside of hadron collider?

[@JohnNeves]

This is the best comment I have seen today.

[@theunknownblock5942]

what the heck am I gonna do outside of a hadron collider!?!?

[@Fif0l]

***** Whoa, calm down. My comment was supposed to be a joking summary of the video. Besides, can you say "backwater cave" about a place with hadron colliders?

[@YTEdy]

Or flip a burger.

[@junebugsolaris]

#firstworldproblems

[@bogdan1207]

that's not a misconception, it's just a classical approximation.

[@hayttman]

bogdan1207 but when you are unaware that its an approximation it's a misconception

[@pibroch]

Momentum is NOT mass times velocity so it IS a misconception. Saying the momentum of light is mass times velocity is not a classical approximation!

[@Ekvitarius]

Same as Newton’s law of gravitation, a misconception because people don’t know it’s an approximation.

[@accountii6382]

bogdan1207 No it’s definitely not

[@blauwbeer556]

@@hayttman fair enough

[@Mica_T]

Some day this will be in my physics test, and I will probably get it wrong because I followed the supposedly correct equation...

[@dullminecraft]

+INDIGO BLUEoO probably not, he'll give you a bonus point then

[@Mica_T]

Anthony Kovari Lol. it will be PERFECT xD

[@epicmeeltime]

+INDIGO BLUEoO No, the root part of the equation only matters when dealing with speeds approaching the speed of light. Think about it like this, two cars are coming at each other at a constant rate of 50km/h. What happens? They hit each other with a net force of 100km/h relative to someone outside. So now, what happens when two cars are coming at each other, one is traveling at 50km/h and the other is traveling at the speed of light. Is the resulting net force relative to someone outside going to be the speed of light + 50km/h? No of course not, general relativity tells us we can't do that. So our original premise that the two cars will hit each other at 100km/h relatively, is wrong. They hit each other at a marginally smaller number (it really only changes by some decimals). The only time it will have real effect is when you're dealing with speeds approaching the speed of light.

[@Mica_T]

Kkfm Harry And you look like me :D

[@ohnoitsjasmin]

+INDIGO BLUEoO No.... it won't. Because you aren't going to be dealing with things moving near the speed of light.

[@Ropsuguy]

I'm surprised that you didn't talk about in either episode about how we are taught only 3 forms of matter. Although understandable because most people learn that when they grow up.

[@MultiPoiu]

plasma is a bit contreversial

[@Ropsuguy]

sam loi that has nothing to do with my comment.

[@theunknownblock5942]

+sam loi there's still colloids, bose-einstein condensates....

[@Ant22v]

GangpanSpock my teacher told us we learn about those because they are the main states of matter, there's other but can't beat them

[@TheGuywithnolife]

GangpanSpock but plasma is just mixture of gaseous positive ions and electrons though, we learnt that in Chem

[@MatesakCZ]

I feel so damn smart everytime he talks about something I already know.

[@backyard282]

mv is what people originally defined momentum. People knew what's mass, what's velocity, and then they simply created momentum by multiplying them in order to obtain linear proportionality. Similarly, I can multiply mass and time and call it apple and give it variable y. It's just that that quantity isn't of some especially particularly big use and significance.

[@534General]

*What is the large hadron collider? And has it destroyed the world yet?*

[@pibroch]

It has destroyed the world in a parrrallel universe. We're all dead over there.

[@534General]

But somehow you have miraculously survived?

[@pibroch]

Mᴀsᴋᴡᴏʀᴛʜ I've survived in this universe, not in the countless other universes where everything was exactly the same as this one up until the point where they were destroyed by the LHC. Your body is rotting in that universe too - the condition of your once lovely fur is unknown.

[@534General]

o.O

[@hansvertriest9586]

Mᴀsᴋᴡᴏʀᴛʜ He's talking about a theory that says that there are countless universes, every single one just slightly different than the others. So say you made a decision this morning to make eggs instead of pancakes, there will be a universe completely the same as this one, but instead of eggs you've made pancakes.

[@slaptaszaidimas1444]

if i use this formule to exams to they cound this ?

[@fairylightsarepretty6391]

minutephysics i will forever support you. none of us will probably need it yet but is nice to know how it actually works. thank you for giving us this knowledge

[@kcwidman]

if you plug in that for light, you get mv/sq. root 0 or mv/0 or undefined. How does light have momentum if this is the correct equation?

[@alexanderchippel]

Reality is subjective.

[@mooncowtube]

Light has no mass, so mv/0 is 0/0 and this formula simply doesn't help us find its momentum. If you substitute m=0 into the first equation, or v=c into the second, you will obtain p=E/c. The momentum of a photon is based entirely on its energy, ie the frequency of the light. See also hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c2.

[@Veexliat]

Was wondering the same thing a while back, here you go /watch?v=M-VZdJu0bLU

[@pascalos99]

Thx for pointing this out! :) , so only if the velocity isn't c you can use this, or because objects with mass can't have a velocity of c; if an object has mass.

[@pascalos99]

+Dave Clark

[@firecatflameking]

I am 14 and i just LOVE this kind of stuff! I have watched nearly all your videos :) Thanks so much!

[@hristaki99]

80% of his subscribers are like you, including me.

[@firecatflameking]

I just love how the internet can bring together people like us!

[@coolsvilleowner]

***** Im 19. Got a physics exam in a few days lol! As you can see, at the moment I am procrastinating (which is bad)...

[@user-74652]

coolsvilleowner Better to procrastinate with this than with Memebase, for example. Since the comment timestamp claims to be a week old, I'll assume that those few days have already passed and you have already had your exam. How did you do?

[@firecatflameking]

***** :)

[@cloroxbleach1200]

Oh shit I love memorizing physics formulae

[@EtihwMaleGGT]

"The equation they taught you in school isn't true" It's okay, I never learned shit in school to begin with

[@riuphane]

Thank you Henry for continuing to re-educate us against what we were taught in school.

[@blaze9525]

Yeah, I am actually a bit impressed myself at how well this channel gets complex ideas relatively thoroughly explained in such a short amount of time.

[@AhnafAbdullah]

Why is Momentum p? I know why light is c, but p doesn't make sense to me. Yet

[@MissAbbyH]

***** Thank you for that, Dylan :) So thoughtfully enlightening.

[@AhnafAbdullah]

***** Yup, sorry I didn't see you there, Thanks a lot. I didn't find a good reason even after searching it on Google. :)

[@obhutara]

Ahnaf Abdullah there werent any letters left.

[@buckyball2003]

‘m’ is mass so they had to pick a different letter for momentum

[@PinkCammy]

Would be nicer if you said "Common Physics Approximations" ... We HAVE to teach them the basic/ Newtonian approximation when they are young, just for them to understand initially (Like when they are 13 years old and can't do maths all that well to treat square roots properly) ... So It would just be nicer if you didn't make it out to be that the educators are ignorantly teaching p=mv even though we know that it's just an approximation. Please? and Thanks :)

[@bobbo8131]

i dont understand what your saying but i nod because it makes me feel smart!

[@Adrastos42]

Well, you can actually just pick when it's close enough to zero. If you use p=mv, you're accepting that you're making an approximation and that the result will be not exactly right (though very close to it for everyday speeds). In many cases, this is perfectly fine, for example if the difference between p=mv and the full equation is less than your experimental error, or for anytime you're dealing with speeds where you know the difference will be fairly small and just want a ballpark figure.

[@alaaali7534]

man.. i loooooooooove this .

[@sirawesomehat8814]

Whats the momentum of light? please answer!

[@youtubeepicuser4209]

a billionth of a pound of pressure is exerted per photon.

[@DarkSt3ve]

sirawesome hat p = E/c

[@sirawesomehat8814]

DarkSt3ve This is an old comment, I now know that. Thanks for replying and not calling me a dumb fuck or something like that.

[@kanavkohli794]

sirawesome hat Dumb Ducking Fuck

[@sirawesomehat8814]

Kanav Kohli :)

[@0s0sXD]

this video makes me SO FUCKING ANGRY AND I HAVE NO IDEA WHY😬 DID LIKE

[@Cjaypanda111]

+Mohamed Osama too much math

[@0s0sXD]

+Cj it makes me feel like everything i learned before is wrong by 0.00000000000001 -_-

[@SciShorts]

+Mohamed Osama LOL hahahahah. Our knowledge is not perfect, we still have a lot to learn and we don't exactly know if it's perfectly correct, that's why everything is still a theory, because we might not know.

[@themagicbush1208]

Take that schooling system!!

[@Linkario86]

I think I understood like... A bit less than half of it, but given it's so short I still learned more than in a lesson at school back in the days

[@ThatGuyMason]

I love the awkward bass line on all of these so much, you have no idea

[@crimsonfoxonusae4472]

so i passed an exam with a false formula......

[@anotherKyle]

+The_cute _Wolfie it is not false it is simplified because the added part only effects the outcome in specific situations.

[@andrewrobertson1473]

+anotherKyle Not true. The relativistic form of momentum applies at low speeds as well, but the difference between the classical and relativistic forms is so minute at low speeds that within the margin of error of our measurements it means nothing.

[@WhopperBu]

lol i am a mechanical engineer so dont give a shit.

[@hristaki99]

Then don't watch science videos. It's that simple.

[@KartikayKaul]

Uhm... Engineers are able to solve problems just because physicists make formulaes.

[@mechwarreir2]

hristaki99 Yo its a joke referring to how mechanical engineers don't use special relativity equations. We use equations for materials, fourier transforms, lagrangians, but no modern physics needed to build a car xD.

[@jeremysmith6828]

Hey, here's hoping that someday in the distant future, mechanical engineers WILL need it - that would be a pretty amazing world to live in!

[@mechwarreir2]

Material engineers branch from mechanical and they do use this.... but on a nanoscopic scale.

[@ashtingillurd4751]

The God and Rock Paradox: If a god is considered to be able to do anything, then he could create a rock that he couldn't pick up, right? Well since made it so he could not pick it up, then he couldn't do it, even though he SHOULD be able to do anything.

[@TheSteamGamer99]

Solution: God dosen't exist.

[@deepstudios3832]

Solution.. why the hell wud he need to create a rock other than throwing it at a dumbasss like you

[@jjjuanig]

Physics misconception: There are less actual 1-minute physics videos in the channel than the total videos

[@alexschannel8883]

Well p still equals mv, but m is actually m0/(1-v^2/c^2)^(1/2) where m0 is the rest mass of the object, right?

[@jamieanderson7757]

***** In physics (both Newtonian and relativistic) the most common convention is that m represents "rest mass". This convention is used in this video. Semantics aside your equation is of course correct - you are just using a different meaning for m than the one in the video.

[@TheEgg185]

lol. wut?

[@IamGrimalkin]

I wouldn't call his a "misconception". It's an approximation.

[@martijnvanweele6204]

I think the misconception is that we forget that it is an approximation.

[@thecaptain0003]

3,640,555 people subscribed even though they don't understand a thing your saying XD

[@jackdaniels4975]

I mean I understood pretty well what he said.

[@ravinduwijayarathna6238]

me too lol

[@ayushgupta3234]

they understood pretty much

[@GratedArseCheeks]

I think it depends on the reference frame:) in your car you will see your headlights working as usual (Although you wont actually be able to see at all because of how light waves will never be able to catch you and everything will be shifted so far into the gamma ray spectrum/radio spectrum) But to an outsider there will be a buildup of photons in the headlights, kinda like a sonic boom, which would consist of an infinite frequency wave of light.

[@youknowwhenitsreal7]

You have a point, and the formula given in the video only works if m doesn't equal 0. There is another equation for the momentum of massless particles: p=E/c where p is momentum, E is energy and c is the speed of light.

[@SCarboni]

Just because its close to 1, doesn't mean it is one. He said if v< c ie. v = 2 c = 5 V^2/C^2 = 4/25 = .84 square root of 1-.84 = .4 That's not at all close to 0

[@a.j.coldwell6909]

Hey Henry I have seen a lot of your videos and now have two questions, #1 What would it take for us to reach the speed of light or something near it to travel to out of this solar system, #2 Would we be able to use some sort of cryogenics or something like that to freeze or sustain ourselves for extended space flight if we were able to reach light speed or a speed near it?

[@Cutiee457]

Question: even if you take all of the above about the misconception of momentum into account, if you're talking about a massless (like a photon of light) wouldn't p still equal 0? Mass is in the numerator and 0 divided by anything equals zero. If this is the case, then light is massless and has no momentum which would mean it has Energy = 0 and I know that isn't the case. What am I missing? (I really want to understand)

[@pibroch]

Look at the video again and you will see that the equation you are referring to applies to "objects with mass" (00:20). The first equations Henry puts up at 00:12 are can be used for massless particles.

[@PyrguiinY]

photons have mass, just such a small one it is basically 0.

[@IamGrimalkin]

Naoki Yahashi Nope. As far as we know photons have no mass. The momentum for massless particles is p=E/c where E is the energy and c is the speed of light. For light E=hν where ν is frequency and H is the plancks constant.

[@IamGrimalkin]

***** No they don't. Or at least, not so far as we know.

[@Tetrahelix]

Obviously, when you use units where c=1, relativistic mass is the same as total energy. However, I have a number of reasons that I prefer to have M in my toolbox.: 1) I was trained with it. 2) Much of my rest mass is the kinetic energy of the gluons in my nucleons. 3) When using non-natural units (c=/=1), having M makes certain metrics more elegant. The distance metric is l^2 = (c^2)*(t^2) - r^2, where l is the same in all frames; similarly, (m*c)^2 = (c^2)*(M^2) - p^2, where m is invariant.

[@adrian9545S]

Thanks for that. now it makes sense why P = mv in textbooks, I had read about special relativity and the concept that states mass as non-constant, but didn't really get why the equations still represented it as constant

[@SerAnthos]

p=(mv) / sqrt(1-(v²/c²)) is used for particle of mass moving near the speed of light. For massless particles such as light, p= hf/c is used instead.

[@nesslig2025]

No matter how hard I try, I just could not get those two equations at 0:19 to change into the one that shows with the square root of 1 minus velocity squared divided by speed of light squared. Can anyone help me?

[@DenyDenDenzel]

The best thing is that, in our college, we have 4 courses of Physics.. My professor actually said in advance that Einstein's theories, which is taught as the last course, is going to disprove almost EVERYTHING established on the first three courses, mostly about Newton's theories - and why there's a reason that why it was best to use momentum P in most equations than resorting to using MV, or anything dervied from that.

[@simoncrabb]

I've got another physics misconception: minutephysics videos are rarely 1 minute long.

[@GottaRantSomewhere]

you sir bring out the child of me not the bad side, i mean the "i wanna learn MOAR" side. i seriously crave for physics videos of you and 1veritasium since i started watching...

[@pureevilskaters]

I don't understand anything they talk about in these video, yet i keep watching them

[@Pursuit99]

Defined by that that momentum equation, an object with mass can never reach the speed of light. If you were driving at .99c, everything will be blueshifted past the visible frequencies and you won't see anything. If you could travel at c, you would essentially push a cloud of photons that you would never see because it wouldn't be a wave in your frame of reference. There's more going on, such as aberration and time dilation but try reading up on the sound barrier to understand what would happen.

[@Tangylives]

That took my first year lecturer at Warwick an entire semester to go over when you covered it in 50 secs.

[@pibroch]

You're right. And Iif you replay the video you will hear him say that the formula applies to "objects with mass". (The energy-momentum-mass relation appears briefly: E^2 = m^2c^4 + p^2c^2. (p is momentum, m is mass) For light m=0 so this equation reduces to E=pc so p=E/c. In other words the momentum of a photon of light is its total energy (divided by the constant c).

[@BlueCosmology]

No one at CERN thought so either, there was a paper released once by people running an experiment in conjunction with CERN that was released as they needed help figuring out what they had done wrong with their experiment as it was indicating faster than light particles. Then after I think it was less than a week the experiment was fixed.

[@pibroch]

If you look at the video again, at 0.20 he is referring to "objects with mass" - the equation which follows does not apply to light, which has no mass. Momentum of a photon is Planck's Constant divided by its wavelength - there are no zeros involved.

[@12tman12]

I think you mean massLESS particle is p=E/c, just to clarify for readers :). Eeppixx1, look at his other video 'E=mc² is incomplete'. Or quickly. E²=(mc²)² + (pc)² (so for zero mass (m=0)) E²= (pc)² E=pc p=E/c If anyone want to mess with the above for a non zero mass, go ahead. I'm WAY out of practice messing around with the units :P

[@AdevbRS]

this is why i love physics so much

[@ThisNameIsBanned]

Why ? because they are mostly incomplete ? Any math rule we have is probably incomplete, but it does work good enough for our problems.

[@Exevium]

But that is what makes Math so beautiful. By the time you've learned "everything", everything has become something.

[@Zenniverse]

Does this only apply to that which is near the speed of light?

[@AlchemistOfNirnroot]

It applies to everything but it's basically 1 for everything.

[@commandolam]

It applies to everything. The deviation from the approximation P = MV becomes exponentially larger the closer you are to the speed of light.

[@foxxravin]

it apply if something goes faster then 10% of C then you need to calculate relativistic

[@123eldest]

The energy for a photon is E = hbar*w , where hbar = (planck's constant)/ 2*pi and w = 2*pi*f where f is the frequency of the photon f = c/wavelength. E = hw = pc (m = 0) so p = hw/c = hf/c

[@hypotherima1]

slight correction. relative mass = m0/sqrt(1-v^2/c^2) where m0 is the rest mass. so p=m0v/sqrt*(1-v^2/c^2)

[@pibroch]

Wrong - there is nothing incorrect in this video. The video never mentioned "relative mass" as that construct was unnecessary. In physics the convention is that if the word "mass" is used without a qualifier it means what you have referred to as "rest mass" which is the same concept of mass used by Newton. Despite what you may have thought special relativity does not require the term "relative mass" as relative mass is not a type of mass as you seem to be implying: it is an energy term! In relativity, mass is invariant - only the energy of an object with mass changes with velocity.

[@Tetrahelix]

It can have momentum according to this equation because, for any massless particle, v=c, so the equation becomes p=0/0. Because p=0/0 is just another way of writing p*0=0, which is true for any value of p you select, this is the equations way of telling you that, while the photon may have some momentum, you need to use a different method to find what that momentum is.

[@localnebula]

Simple way to derive the p=h/λ. With no rest mass, the relevant equation, E²=(mc²)²+(pc)², reduces to E=pc or p=E/c. For a photon, E=hν=hc/λ, so p=E/c=h/λ.

[@HD-fy2wu]

Expand mv/sqrt(1-(v/c)^2) using Maclaurin series you get mv + mv^3/(2c^2) + (3mv^5)/(8c^4) + (5mv^7)/(16c^6) + ... So it is just an approximation, for values of v where mv^3/(2c^2) is not significant as compared to mv.

[@shpilzak]

That means to get momentum as function of other variables. In this case, square 2nd equation, put expression for E^2 into it and find what p equals to.

[@totallyUnimodular]

Henry, You should make a sequel of this video on momentum of massless particles and the momentum equation you mentioned here.

[@pibroch]

Short answer: don't use this equation for objects with no mass as it doesn't work, as you have discovered! A wave-packet such as a photon is a different beastie than an object with mass such as an electron. From E^2 = m^2*c^4 + p^2*c^2 (which can be derived from quantum physics) substitute 0 for m: E=pc, p=E/c

[@HellsMascot]

While it is true that p=ɣmu, it would be better to say using the de Broglie relations, the four-momentum, and the four-wavector that P=ħK

[@kevinmuhanji1614]

can you make a video about how light has momentum although light is massless

[@tsunami7394]

Its in his common physics misconceptions video.

[@falschgedenkt9086]

v=c×(pc)/E ==> p=E/c so it depends on its Energy

[@MrBeiragua]

You can show that with electromagnetism as well. Light is a wave on the electromagnetic field. This wave when bumping on a charged particle will interact with it, giving it a small momentum. Since momentum needs to be conserved, light was carrying this momentum before, and from this and from the equation of the electromagnetic force (Lorentz force), we can deduce that light carries momentum.

[@abysspanda]

Cont. Since the water is changing direction so many times, this is also the cause for things like diffraction, or the bending of light. Now, when referring to how light has energy, well light in itself is energy. As for momentum, refer to E=mc^2 or the relation of mass to energy, basically you can quantify a light particle's momentum by creating a surrogate term for mass and replacing it using the mass energy equivalence. Well, at least in simple terms, the real quantification require calculus.

[@vinoyletritas]

I love these. Please if you can do more, I'd be glad.

[@stamriel]

You are applying what I like to call sophistry reasoning. By that logic, we are virtually dimensionless. What you seem not to see is the importance of your frame of reference. From a cosmic being who omnisciently embodies the infinite knowledge, yes we all are infinitely lacking. BUT from a fellow mortal human being or a non-infinite sentient, we use OURSELVES as a reference point. In Physics your frame of reference is quite crucial.

[@ThePondus430]

according to your equation the momentum of light should be undefined as far as i am concerned. light is massless so that means p=0/0 and that is undefined. what? is there something I'm missing?

[@pibroch]

Back in Black You're missing what he said at 00:20 "objects with mass".

[@lgasc]

Isn't 0/0 equal to infinity ?

[@MrKorrazonCold]

Matter is opposed motion simulating rest and balance now through violent motion. The locational spherical inward absorption and outward emission of electromagnetic waves is forming all forces of nature, antimatter matter annihilation forming + and - electric charge and electromagnetic fields and the forward passage of time. Thus limited range Spherical Inward and outward waves and Doppler causes a redshift. Redshift with distance a consequence of less energy exchange, less EM-wave interactions.

[@ManintheArmor]

The closer you travel to the speed of light, the more dilated time becomes. Chemical, nuclear, and subatomic processes take longer to occur. Therefore, your perception of time would slow down relative to someone else's frame of reference, because it takes longer for signals in your brain to travel, even if the distance itself never changed. So, from your perspective and frame of reference, the speed of light never changed.

[@dhvsheabdh]

You end up with a lorentz factor (which is what you times mv by to get momentum) of infinity. Hence, you have 0*infinity, which is indefined, yet has a value.

[@petegarofallou6266]

The "wavy" equals sign is read as "approximately equal to." Not equal. For velocity v=0.1c (one-tenth the speed of light, or 18,600 miles/sec, or 30,000 meters/sec) v/c = 0.1 , (v/c)^2 = 0.01 , 1 - (v/c)^2 = 0.99 , square root of (1 - (v/c)^2) = 0.995. So, treating a velocity of v = 1/10 c introduces an error of one-half of one percent. For velocities we are likely to encounter in non-relativistic settings, the error is a tiny fraction of one percent, and may safely be ignored.

[@seancohan5219]

Even if it takes a couple repeated views, I love being able to understand this concept.

[@MoreKevinLiang]

also E = mc^2 is actually: E = (mc^2) / the same square root value from the momentum equation the square root value ( 1 / sqr(1 - v^2/c^2)) is, in most textbooks, denoted by the greek lowercase γ (gamma)

[@Rigardoful]

I mean root of 1 - c^2 / c^2 equals 1 - 1 equals 0, right? And then we have massless particles, so m = 0 which means, m times v = 0. So with this we have : (m times v) / squareroott[1 - (c^2 / c^2)] = 0 / 0.

[@g0rsk13g4n9st4]

The equation [p=mv] is valid for everyday objects. The "Lorentz Factor" value (that whole equation inside the square root) is virtually 1 up until speed of objects reaches 1/2 of the speed of light.

[@Tetrahelix]

Actually, that depends on how you define "mass." MinutePhysics uses the word "mass" to mean "rest mass," which is an object's mass in its own frame of reference. Photons have a rest mass of 0, so they are called "massless." However, "massless" particles do have another type of mass - "relativistic mass." That is the mass that an observer measures the thing to have. Also, you can divide zero by zero - it just lacks a specific answer. E.g., contrast 5=0/0 (i.e., 5*0=0) with x=2/0 (i.e., 0*x=2).

[@pibroch]

That's right, you are not missing anything: Let c =1 and m=1 (each is invariant). In reference frame where p=1, E2 = (mc2)2 + (pc)2 = c2( m2c2 +p) = 1(1+1), so E = sqrt2. In reference frame where p=0, E2=mc2 = 1, so E=1.

[@faneb]

Well, actually light has a definite momentum that depends on its wavelength/frequency via p = h/f, where h is Planck's constant and f is frequency. So for light, you can just use that relation to find the momentum. For other particles (as far as I know) there isn't a convenient formula like this.

[@moisesbessalle]

as an object with mass approaches the speed of light you will see that the energy you need to do so becomes huge and when you do the limit with v approaching c you get that the amount of energy you need is infinite. or if you don't do the limit you get a division by zero, basically telling you that you can't move an object with mass at the speed of light.

[@FrameSticker]

Infinity isn't a number you can just put into your everyday algebra. It's an idea, a concept of something simply greater than anything else. If you allow infinity in common algebra you get, for example: 1/0=infinity 2/0=infinity 1/0=2/0 1=2 As a concept, you could say that something "approaches," infinity, or the limit of something is "equal," to infinity, but you can't say a number is infinity. Not in common algebra, at least :)

[@DaniPaunov]

ok, but how do you find out the momentum of light? all the equasions here have mass, so they must be 0. and 0+0=0, so the energy of light is 0, yet somehow light can push on, accelerate and heat up things (like a solar sail), which would violate the first law of thermodynamics (E(before)=E(after)). something i'm missing here?

[@pibroch]

p=E/c, E=hf , where h is Planck constant,f is the frequency of the photon.

[@andymcl92]

May have been covered before but a video on F=d/dt(mv) rather than ma could be good

[@cleanunderyournails]

That's basically what I was trying to say just way less advanced. Thanks for the more in depth description.

[@McTheWarhammer]

Almost to 1,000,000. Good job Henry.

[@DustinNiehoff]

I know this is late, but yes you do since the speed of light is the same in all frames. Therefore, you would see your light beams traveling away from you at the speed of light (since you yourself are not moving at all relative to yourself, this is fine). An external observer, would simply see an pancake car (0 width in the direction you are travelling) and no beams ahead of the car. Though this isn't a very good scenario since nothing with mass can have speed exactly c.

[@matacusa]

Taking into account the formula you showed was correct for momentum, if we assume that an object is going at the speed of light, that would mean square root of 1-1=0, which would also mean we would need to get the result for P=(mv)/0. The only way to travel at the speed of light is to be massless, thus (mv)= (0v) or 0. This results into p=0/0. Taking into account that if one divides by zero the universe asplodes, how do we calculate the momentum of a particle that goes at the speed of light?

[@Cyborg2587410]

@minutephysics I tried to look it up but i dont get it so im coming to you for help, about quantum mechanics and . How do we know if a particle is in a superposition or has been in one, if every time we observe it, it has collapsed to one position or the other(whatever the position maybe).

[@Tetrahelix]

First, what is an "innate number"? Anyway, x=(0.999...) can be rewritten as x=(0.9+0.09+0.009+...). The proof then proceeds as normal: 10*x=10*(0.9+0.09+0.009+...)=(9+0.9+0.09+0.009+...), etc.

[@pro7gr]

There are two objects that are in contact. One object has a velocity of zero in the x,y and z direction. What is the velocity of the other object? The velocity of the bb must hit zero at some time when it turns around correct? Therefore, if one velocity is zero, what is the velocity of the train?

[@estebansingh9411]

now I wonder how many equations are incomplete or can have a "no longer accurate" label at some point. excluding of course the fact that every equation can be incorrect in mysterious environments such as the interior of a black hole

[@m3nm3nm3n]

Using m = 0 in the energy-momentum relation, you get the momentum for the photon: p = E/c= h/lambda.

[@TsiSiFa]

Alternatively, say you have a long tube (say 186 miles) of successive, powerful, ring-shaped magnets. If you program them (like the cameras in the Matrix, sorta) to activate in such a way that the time between the first and last magnet activation is equal to or less than 1/1000th of a second, and then placed a minute piece of metal in the tube, would it travel at or near to the speed of light?

[@selachimorpha2000]

so v2 is the same as v in the equation, right? Or did we introduce a new value of velocity?

[@mrwho995]

You are correct in that they can act as both waves and particles, but being a particle doesn't necessarily mean they have to have mass (believe it or not).

[@TsiSiFa]

I have a question. If moving the beam of a laser across the moon is provides the appearance of exceeding the speed of light (guess where I got that), what would happen if you replicated that experiment with a physical object? Hypothetically, if you took a structure long enough and firm enough, you should be able to make the very tip move faster than the speed of light, no?

[@lolageez1]

The class seemed to enjoy them, though some of the content was a bit higher than the level we're at now they were interested in it. I'd already told my teacher about MinutePhyiscs before, but this was the first time he'd used it in a lesson.

[@pibroch]

You're completely correct and if you listen again carefully Henry states that the formula applies "for objects with mass".

[@youtubepeace]

Please make another video discussing more about "Shrodinger's Cat", the last one was really interesting.

[@MrJimshot1990]

So the faster the particle moves the less momentum it has as it approaches the speed of light? Where a particle traveling at the speed of light, regardless of its mass would have no momentum? ... or an undefined momentum because it would be MV/the square root of 0?