How is a light photon affected by Gravity?

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  • Опубліковано 28 вер 2024
  • How is a light photon affected by Gravity? Without invoking General Theory of Relativity, we can use an alternate method to see how the energy of a light photon is affected by gravitational field. Since light has Energy, by virtue of Mass-Energy Equivalence Principle, therefore light has 'mass' due to its Energy. This mass can experience gravity.
    When such a photon is moving upwards, it will loose Energy and thus its frequency will reduce.
    When light escapes the gravitational pull of a massive star, its frequency is considerably reduced (for visible radiation, it's shifted towards the red end of the spectrum). This phenomenon is known as Gravitational redshift.
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КОМЕНТАРІ • 298

  • @FortheLoveofPhysics
    @FortheLoveofPhysics  4 роки тому +22

    If you liked this video, here are a few more suggestions :)
    1) Why pair production cannot occur in vacuum? ► ua-cam.com/video/fngJOOiAj5k/v-deo.html
    2) Can you Prove E=MC² ? ►ua-cam.com/video/VZEhmWFlrnM/v-deo.html
    3) What is a Black Hole? ►ua-cam.com/video/WuYpI3EoCnM/v-deo.html

    • @nikhilbhathi3778
      @nikhilbhathi3778 4 роки тому

      Wonderful job sir👌🏻

    • @aseemanandmishra4598
      @aseemanandmishra4598 4 роки тому +1

      Sir please answer as u said everything that has energy that has mass than how photon have energy but no mass how it's possible please tell sir

    • @MM-ci5mb
      @MM-ci5mb 3 роки тому

      answer mi ,, why there is no neutrons in hydrogen atom...and present in its isotopes.

    • @soumapriyamondal3808
      @soumapriyamondal3808 3 роки тому

      Sir, Can you also discuss advance topics topics like Maxwell-Boltzmann, Fermi-Dirac and Bose-Einstein statistics..

    • @souvikmandal6491
      @souvikmandal6491 Рік тому +1

      @@aseemanandmishra4598 according to Einstein E^2 = (M°C^2)^2 + (PC)^2. here p is the momentum and m° is the rest mass...
      Now for photon the rest mass is zero but still the photon will have energy as putting m° =0 in the equation we get..
      E = PC
      This things may not have mass but can have momentum like photon.. put it simply like a chemistry teacher it's an exception brooo..😂

  • @TheCosmicDestroyer....
    @TheCosmicDestroyer.... 2 роки тому +7

    I haven't been able to understand the gravitational redshift mathematically before...but thanks to you ,now i know it in a very simplest manner🙏

  • @girirajsharma3399
    @girirajsharma3399 4 роки тому +64

    your lectures are extremely convincing, crystal clear and unmatched. You are indeed doing a wonderful service to the Physics Community and I really thank you for these wonderful lectures on behalf of Physics lovers. I feel that you are a born teacher. Keep it up and uploading more and more lectures. Bye the way I could not see your name anywhere. The physics lovers must be quite curious to know about you. So please let us know or introduce yourself through a video and I firmly believe that it will inspire a lot of budding physicists.

    • @Ihab.A
      @Ihab.A 3 роки тому +2

      I can't but echo your comment and thank the professor for this marvelous explanation!

    • @shibhanlalpandita6975
      @shibhanlalpandita6975 3 роки тому

      This is all wrong. If earth pulls an upgoing mass then it won't go up. In religion we have one God. Here many many supposition r taken as true.

    • @Jamesdavey358
      @Jamesdavey358 3 роки тому +1

      Youve inspired me mate

    • @Jamesdavey358
      @Jamesdavey358 3 роки тому

      @@shibhanlalpandita6975 what are you on about

  • @akshatsrivastava9396
    @akshatsrivastava9396 5 років тому +36

    I humbly request you to please discuss about the general theory of relativity

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  5 років тому +11

      All in good time :)

    • @Just_mind_______
      @Just_mind_______ 3 роки тому +2

      @@FortheLoveofPhysics Sir I think you were wrong about the formula E=MC² . It's actually wrong . This formula is actually E= MC²+ PC². Any with rest mass we can measure it energy by E= MC² . But anything with out rest mass we can't measure it's mass but we can only measure it's energy so the formula for finding energy of a photon is E= PC . I hope you understand it.

    • @Just_mind_______
      @Just_mind_______ 3 роки тому +1

      @@FortheLoveofPhysics And sir it is scientifically proven that light will lose its when it is travelling in space energy because space itself is expanding .

    • @sankalpshandilya6382
      @sankalpshandilya6382 3 роки тому +1

      @@Just_mind_______ yes u are right , for a proper treatment we should not comsider a photon as massive partice where its relativistic mass comes due to motion. Rather the real reason for red shift is simply the different rate of flow of time as photon moves away from G-field ,since rate of time flow changes so does the frequency. However for a simpler explanation sir has Incorporated the effect into the relativistic energy E = /gamma m c^2 where / gamma takes care of everything and then he has defined mass as /gamma into rest mass. Since rest mass is zero and /gamma is infinity the moving mass may still be finitr and thus the trick works

    • @debtanupatra3385
      @debtanupatra3385 2 роки тому +3

      @@sankalpshandilya6382 so you are dealing with 0/0. Photons energy comes from relativistic momentum. Nothing to do with rest mass...and its 0.

  • @thechemistrysection552
    @thechemistrysection552 4 роки тому +2

    Perfect crystal clear explanation in a simple and sophisticated way.

    • @saturnus8187
      @saturnus8187 4 роки тому +1

      The Chemistry Section you know that simple and sophisticated are opposites yeah?

  • @HighEnergyScientist
    @HighEnergyScientist 3 роки тому +11

    Bro E=mc^2 is applicable for stationary objects , how you can applied for photons . For photons E = √(m^2C^4+p^2c^2) . Please reply

    • @shrikant8446
      @shrikant8446 3 роки тому +4

      Exactly and instead of taking E=mc^2 we should take E=pc as mass of proton is zero

    • @asce4999
      @asce4999 2 роки тому +4

      *photon
      And photon’s rest mass is zero i.e m^2c^4 term in E^2 is zero leaving E= pc

    • @enricobelvisi3880
      @enricobelvisi3880 2 роки тому

      @@asce4999 momentum of a photon should be p=mc, so in the end you still have E =mc^2, where m is not the rest mass as he said in the video, but the one which cames from his energy content

  • @niteeshmishra388
    @niteeshmishra388 5 років тому +4

    very lucid explanation. excellent

  • @joserobing661
    @joserobing661 6 місяців тому

    Beautiful presentation. Thanks.

  • @manjitpalahe4129
    @manjitpalahe4129 2 роки тому

    Superb made much easy to understand a complex topic

  • @ImUpsetThatYouStoleMyUsername
    @ImUpsetThatYouStoleMyUsername 4 роки тому +6

    Can you actually solve for average redshift ? I'm curious to see how many nm shift occurs as a photon leaves earth and/or the sun

  • @prabhakerpandey5905
    @prabhakerpandey5905 3 роки тому +1

    big fan of your work sir..............
    but I think i have one more explanation that photons have the influence of gravity on it ....
    for this first consider the equation relativistic energy from special relativity i.e E^2=(M'C^2)^2 + p^2c^2 ...
    where m' is the rest mass of the particle (0 in the case of a photon). Hence this reduces to E=pc. Einstein also introduced the concept of relativistic mass (and the related mass-energy equivalence) in the same paper; we can write it as:
    mc^2=pc
    m=relativistic mass, hence:
    m=p/c
    in simpler word a photon does have relativistic mass proportional to is momemtum.
    (II) = De Broglie's relation, an early result of quantum theory (specifically wave-particle duality), states that :
    λ=h/p ; h = planks constant = 6.62607*10^-34 m^2 kg/s
    we can write is as:
    p=h/λ
    hence combining this with equation of relativistic energy; we get:
    E/c^2=m=p/c=h/λc; m= relativistic mass.
    so form this we can conclude that , photons have 'mass' inversely proportional to their wavelength! Then simply by Newton's theory of gravity, they have gravitational influence.
    I hope that you would consider it sir
    thank you......

  • @nnfefe9451
    @nnfefe9451 Рік тому

    You are a great teacher.

  • @sadrashyasingh1265
    @sadrashyasingh1265 3 роки тому +1

    The eq. E=mc^2 is valid for rest mass. For photon ( no rest ) this is zero , so how can you use it here? Kindly explain.

    • @GoGa_yt_
      @GoGa_yt_ 3 роки тому

      E=mc^2 is not valid for rest mass
      E=(m0^2c^4+p^2c^2)^1/2 is valid for rest mass .

  • @promitbiswas4771
    @promitbiswas4771 4 роки тому

    thank you so much sir.you are just making stuffs too much easy and realizable

  • @mrfinesse
    @mrfinesse 4 роки тому +2

    Thanks very much for another great video. What happens to the energy that the photon loses? Where does it go? The star is not a BH. Who gains the energy the photon loses?

  • @aravindrpillai
    @aravindrpillai Рік тому

    What a nice explanation.. 👏

  • @TheCosmicDestroyer....
    @TheCosmicDestroyer.... 2 роки тому +1

    Sir,if there is a shift in frequency then what happens to the energy lost by a photon during its escape...is energy conservation law doesn't hold up here....?

  • @ittayd
    @ittayd 2 роки тому

    E=mc^2 is for rest energy. The full equation is E^2=(mc^2)^2+(pc)^2 where p is the momentum. So a photon can have a zero mass and still have energy due to its momentum

  • @vanubatimajhi1010
    @vanubatimajhi1010 3 роки тому

    Every body says photons are mass less but photons have energy and mass that is very rare which is very difficult to measure in our nature though E=MC2,there is a balance of the nature what causes homogeneous mass rest at same place that is why photons run from one place to another place for a balanced placement.

  • @siddharthaghosh2150
    @siddharthaghosh2150 3 роки тому

    We need more in general theory of relativity by simple example

  • @mahimachaudhary1732
    @mahimachaudhary1732 5 років тому +5

    can you make vedios on superconductivity? plz sir 😇

    • @divyanshshrimali2651
      @divyanshshrimali2651 5 років тому

      I think the focus for now is on nuclear physics and high energy topics only

    • @mahimachaudhary1732
      @mahimachaudhary1732 5 років тому

      Divyansh Shrimali . you think whatever you want ...I was requesting him just to make vedios on that topic.😊

    • @divyanshshrimali2651
      @divyanshshrimali2651 5 років тому

      Mahima Chaudhary yeah, that’s why I wrote the word ‘I think’. Don’t get me wrong, I would also benefit if he makes a video on superconductivity!

    • @mahimachaudhary1732
      @mahimachaudhary1732 5 років тому

      Divyansh Shrimali no I m not taking you wrong 💗 😇

  • @surendrakverma555
    @surendrakverma555 4 роки тому

    Excellent lecture 👌👌👏👌👌👏👌👍😀😊😊👍👍👍👍👍👍👍👍

  • @JungleJargon
    @JungleJargon 2 роки тому

    I can get light to arrive instantaneously from the stars. What slows down light *from our perspective* is the gravity in the vicinity that slows down the *rate* of time. (It’s not that light slows down, it’s the rate of time that slows down for events to take place.)
    So what happens where there is no gravity is that the rate of time speeds up relative to our rate of time so that starlight travels the same (or greater) distance at a faster rate of time. It’s still traveling the *same relativistic frame speed* but an entire second passes by from our perspective at a much faster rate, in a fraction of a second *where there is no gravity* to slow down time. So for most of the time, starlight is traveling through space at a faster rate of time relative to our rate of time perspective.
    That means starlight arrives much more instantaneously because the rate of time is free from the effects of gravity for most of the way. With no mass of its own, starlight arrives instantaneously as it experiences no time of its own traveling at the speed of light. We see things in slow motion *where time is slowed down* by gravity.
    This relativistic effect allows for a young earth since it doesn’t take millions of lightyears for starlight to arrive through the *void* of space where there is no matter to slow down *time.*

  • @tapashnandy3594
    @tapashnandy3594 Рік тому

    Excellent 👍

  • @tonzzzblessan7757
    @tonzzzblessan7757 5 років тому +2

    can u make some basic videos on classical electrodynamics and relativistic electrodynamics and quantum mechanics

  • @hizbullahshah1776
    @hizbullahshah1776 2 роки тому

    Sir discuss general theory of relativity

  • @nripendrajha1995
    @nripendrajha1995 3 роки тому +1

    shouldn't the mass change due to change in frequency?

  • @Write_with_me_gknotes
    @Write_with_me_gknotes 4 роки тому +1

    What actually bends...mass or energy??

  • @sweet_girl269
    @sweet_girl269 3 роки тому

    sir, Your explanation of this topic is awesome 👏,

  • @AamerPawarymyye
    @AamerPawarymyye 10 місяців тому

    Can you explain the Dopler red shift

  • @FoamyDave
    @FoamyDave 2 роки тому +1

    Thank you for the very practical explanation. This has been a question I've had for a while and you answered it in a way that I can grasp.

  • @pg7703
    @pg7703 8 місяців тому

    Thank-you so much...

  • @MM-ci5mb
    @MM-ci5mb 3 роки тому

    answer mi ..why there is no neutrons in hydrogen atom , and present in its isotope ????

  • @nadeemsaleem
    @nadeemsaleem Рік тому

    Hlo Dear Sir
    How are you
    I hope you will be fine dear
    Your every topic is well explained
    Please Sir make videos on
    LASER subject

  • @donol4828
    @donol4828 4 роки тому

    Excellent! Thank you!

  • @borntolive2097
    @borntolive2097 3 роки тому

    Can the last derived equation be used for black holes?

  • @priyabratadash381
    @priyabratadash381 4 роки тому +1

    Nice video but I am having few confusions here: in a gravitational field, which is a conservative force field any object except light photons, at certain hight can have zero kinetic energy and maximum potential energy until it escapes the field in to a force free region . However photon can never have zero kinetic energy in any gravitational field except within event horizon of a blackhole. Also if photon attains zero kinetic energy in any ordinary gravitational field, then its rest mass is zero so it will simply not be affected by any gravitational field & too photon can never be in rest. This is the main difference between ordinary objects and light photon in a gravitational field.
    Again you applied law of conservation of energy or sum of kinetic energy + potential energy is constant. This is only valid within conservative force field but how can you think that it is also valid at infinite distance from the force field, where the field is absent??

  • @martin.monfils
    @martin.monfils 2 роки тому

    Thank you for your interesting & clear lecture! I just wondered, so all incoming light observed at the surface of the earth is slightly blue shifted?

  • @richienkharbuli8475
    @richienkharbuli8475 5 років тому +1

    Now it make senses why light cannot escape once entered into a black hole. ☺

  • @angelkamboj2318
    @angelkamboj2318 4 роки тому

    But E=mc² is use for object at rest, in motion E²=m²c⁴+p²c², where p is momentum

  • @TheNishantGi
    @TheNishantGi 4 роки тому

    Very nice explanation !!

  • @tantaiphan9153
    @tantaiphan9153 3 роки тому

    It helped me a lot. Thank you.

  • @AnwarKhan-ni9dz
    @AnwarKhan-ni9dz 5 років тому +1

    Sir discuse relationship between matter and energy plz..

  • @life42theuniverse
    @life42theuniverse 4 роки тому

    thank you

  • @firozKhan-wo8dg
    @firozKhan-wo8dg 3 роки тому

    Can u advice me best book for GR...
    Like zettli and Griffith for QM....
    And
    Leonard suskind is better for special theory or some one best than this

  • @amirulislam9595
    @amirulislam9595 4 роки тому

    Thanks sir

  • @spookyaction
    @spookyaction Рік тому

    7:50 you are wrong mass of the photon is changing through the way so you should calculate the potential energy using an integral

  • @minimalisttraveler9337
    @minimalisttraveler9337 4 роки тому

    How do you know if you're observing gravitational redshift or redshift caused by motion?

  • @martinricharte7114
    @martinricharte7114 2 роки тому

    Photons are massless particles of spin 1 according wigner classification!

  • @WeissGuertena
    @WeissGuertena 2 роки тому

    7:47
    In my following question:
    " λ_bot " (lambda_bot) corresponds to " ν " (nu) in the video
    and
    " λ_top " (lambda_top) corresponds to " ν' " (nu_prime) in the video
    _________________________________________________________________________________
    Since:
    the equivalent mass of a single photon = m = [ (h) (λ) ] / (c^2)
    I then assume:
    mass of photon at the bottom = m_bot = [ (h) (λ_bot) ] / (c^2)
    and
    mass of photon on the top (that is, at height "H") = m_top = [ (h) (λ_top) ] / (c^2)
    Question:
    why is the total energy at the top ( at height " H " ):
    E_top = (h) (λ_top) + (m_bot) (g) (H) {Equation A}
    instead of:
    E_top = (h) (λ_top) + (m_top) (g) (H) {Equation B}
    or
    Gravitational Potential Energy in Integral form: integral of " (GMm)/(r^2) " with respect to " r ", where " m = m(r) " is the mass of photon as a function of radius " r " {Claim C}
    _________________________________________________________________________________
    Many other sources and textbooks do the same substitution so {Equation A} has to be CORRECT.
    I need to know:
    why {Equation A} is CORRECT.
    and
    why {Equation B} and {Claim C} are WRONG.

  • @shwetaphutela2481
    @shwetaphutela2481 2 роки тому +1

    The cosmos is written in form of science and maths

  • @manzoorhussain6523
    @manzoorhussain6523 4 роки тому

    Thx

  • @bigggggkev
    @bigggggkev 3 роки тому

    Surely this means if photons have mass in momentum energy, anything with mass can have particles attached to it, therefore enabling some form of speed of light travel for anything with mass ? In theory that has to be true right ?

  • @omarmohamad5499
    @omarmohamad5499 4 роки тому

    It's Really amazing

  • @priyamshukla3741
    @priyamshukla3741 3 роки тому

    hard..👌💝

  • @Diandredofus
    @Diandredofus 4 роки тому +3

    You've got to clarify that using the Energy-mass equivalence equation in the case of a photon yields its relativistic mass, not its mass at rest. The mass of a photon is and always will be zero. We use the energy-momentum equivalence instead to demonstrate that a photon's mass is always zero.

    • @jeanemare4116
      @jeanemare4116 4 роки тому +1

      Especially for laymen like myself who didn't enjoy higher education on these subjects. When it comes to mathematics and physics we were often miseducated with incomplete truths.

    • @shubhamkamble
      @shubhamkamble 4 роки тому

      Photons are considered massless and we cannot use energy - mass equivalence, also energy - mass equivalence was a special case where we are in a rest frame of reference and we know that we cannot be in a rest frame in this universe. Only way to find the energy of massless photons are by energy-momentum equation (putting m equals zero) which yield c times magnitude of momentum.

  • @Deepakfly
    @Deepakfly 3 роки тому

    Sir, You are using classical and quantum results at same time, why ? I think Quantum Gravity should work here !

  • @chritophergaafele8922
    @chritophergaafele8922 4 роки тому

    that is a wonderful question, now what if the light is inside the event horizon and is moving away from the center of the blackhole, does it come back ( because it cannot escape the gravitational pull of the blackhole).

  • @pratikbarsing6919
    @pratikbarsing6919 4 роки тому

    Please explain and derive general theory of relativity

  • @bishwajitbhattacharjee-xm6xp

    Physics is indeed in trouble due to Standard Model and Neutrino.
    Where special relativity is comfortable with quantum mechanics.

  • @sushreesatapathy6173
    @sushreesatapathy6173 4 роки тому

    Thnk u

  • @zzzoldik8749
    @zzzoldik8749 4 роки тому

    How about speed of light, we know light travel at the speed c. When it fall down into gravity like from sun to earth so light got acceleration, does it make light travel more than speed of c?

    • @alaididnalid7660
      @alaididnalid7660 2 роки тому

      No. Always c for all observers which is fundamental in relativity. Instead, the fréquency will be different according to différent observers instead of the speed of the photon as is mentioned in the video.

  • @AwaisKhan-uw9en
    @AwaisKhan-uw9en 4 роки тому

    zbardast...

  • @koshavlog2694
    @koshavlog2694 3 роки тому

    Sir does the speed of photon also decreases with frequency

    • @anhadrajkhowa5850
      @anhadrajkhowa5850 2 роки тому

      C is a universal constant. Hence the velocity of the photon will never change.

  • @nebertoketch322
    @nebertoketch322 Рік тому

    The potential energy of the photon is still 0 after distance H.

  • @deepaksainseth9265
    @deepaksainseth9265 4 роки тому

    How it is possible that light don't has mass but has energy which is not possible according to mass energy equivalence.

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  4 роки тому

      Photon doesn't have rest mass. But there is a relativistic mass due to its energy

  • @GenerateIDEA
    @GenerateIDEA 5 років тому

    If light has no mass. And not accelerating from up to down. Then why lightning always strike down and not up?

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  5 років тому +1

      Lightning is due to the excess accumulation of e- charges in clouds. The answer is self evident. It has nothing to do with photon mass

    • @GenerateIDEA
      @GenerateIDEA 5 років тому

      @@FortheLoveofPhysics ok but i just want to know if lightning is affected by gravity? If not. Then why is always going downward

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  5 років тому +1

      Its because the excess electrons accumulate in clouds and gets transferred to ground. Not reverse

  • @suvajitpaul9385
    @suvajitpaul9385 2 роки тому

    You cannot define mass of photon since its' rest mass is taken zero so it won't be affected by gravity as a classical particle do.

  • @user-ff1ii1po2w
    @user-ff1ii1po2w 2 роки тому

    ❤️

  • @physicslessons869
    @physicslessons869 5 років тому

    If this is true so when light beam collide to a body why body isn't feel any force or push????

    • @physicslessons869
      @physicslessons869 5 років тому

      My other question is what is limitations of einstein's equation hv=mc^2

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  5 років тому

      Because it is extremely small. Do u feel push from 1-2 atoms colliding on ur body?

    • @physicslessons869
      @physicslessons869 5 років тому

      @@FortheLoveofPhysics sir I agree with this that mass is extremely small but velocity of partical is 3×10^8 and it is not in finger countable it is in 10^30 per second

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  5 років тому

      Lets calculate it then. Momentum transported by light of energy E=hv is p=E/c which is equal to hv/c (for visible light of wavelength 500nm lets suppose, if u plug in the values) it comes out to be equal to 1.3 X 10^(-27) Newton. Compare this with the weight of a human who is (200-900) N

  • @ericsu4667
    @ericsu4667 4 роки тому

    The actual gravitational deflection predicted by Schwarzschild metric is about 2.1 arc seconds from a star to the earth.
    It is not 1.74 claimed by Albert Einstein.
    Click on "70. Gravitational Deflection from Schwarzschild's Metric" on this website.
    sites.google.com/view/physics-news/gravitation

  • @JustinBenjaminOnline
    @JustinBenjaminOnline 3 роки тому

    19 people don't understand the concept of mass equivalence.

  • @chrisstanford3652
    @chrisstanford3652 2 роки тому

    🤗🤗 💡 🕯🔦

  • @SeshachalamMalisetti
    @SeshachalamMalisetti 3 роки тому

    as far as we can tell there is no infinity

  • @ASHOKLIGHT
    @ASHOKLIGHT 4 роки тому

    What shift we have to call ,if the light is going towards the star

  • @sabrango
    @sabrango 4 роки тому

    XD!

  • @arijeetdutta7499
    @arijeetdutta7499 4 роки тому +2

    This video is nonsense, when we discuss about photon we should not use newtonian mechanics because it fails there. That is what quantum physics is based upon. We should not solve this type of question with mass. It is wrong explanation.

  • @duckyies
    @duckyies 4 роки тому +13

    Wtf I came here outta of nowhere and now I know how* gravity affects light 😂

  • @JenFoxBot
    @JenFoxBot 4 роки тому +6

    Very clear math and excellent commentary along the way. Thank you!!

  • @unitedspacepirates9075
    @unitedspacepirates9075 3 роки тому +1

    Expansion of the universe is not actually acceleration, it's an illusion caused by a gravitational dopplar shift. For the same reason that gravitational lensing happens, gravity distorts the frequency of light emitted from distant galaxies.
    Gravity causes a time-space distortion that redshifts its photons throughout their journey to you.
    The further away a galaxy is, the more time its photons are distorted at an exponentially reduced value as their distance from the galaxy's gravitational field increases.
    Galaxies are much further away than measurements indicate, although they typically aren't accelerating away from you.
    Dark energy is an illusion caused by a spacial distortion caused by the gravitational fields of the distant galaxies emitting the light, there is no mysterious energy accelerating galaxies away from you, no dark energy causing acceleration, no unaccounted for dark matter, and certainly, no big bang. This universe is eternal, without beginning or ending. Cosmic background microwaves are but distant galaxies who's light has been gravitationally distorted beyond your visible spectrum.

  • @sushobhitjakhmolaAdmin
    @sushobhitjakhmolaAdmin 3 роки тому +1

    Okay, so if we apply this to a light photon, trying to escape black hole's gravity, what would be the case? ? As u said,speed won't change but freq would decrease. So if freq decreases, wavelength must increase.
    Concluding that momentum of individual photon would decrease? ( P=H/lambda) where p=mc
    So basically mass of each photon decreases so maybe the mass of photon tends to zero and wavelength tends to infinity. So maybe light escapes but it's mass is so small and it loses the characteristics of light

  • @himanshusinha2161
    @himanshusinha2161 3 роки тому +3

    Wow…. Speechless ! Amazing to see how you have been explaining complex topics in such lucid manner 🙏

  • @biswajitray8420
    @biswajitray8420 4 роки тому +3

    You perceive physics the way it is projected in your mind. Tx for ur fantastic explaination

  • @waqasahmad5887
    @waqasahmad5887 3 роки тому +1

    good.. change in frequency has effect on energy (hf`=E`) SO further change will effect mass and speed of light (E`=m`c^2) so mass has to change itself to keep speed of light constant..then tell me is mass infinitly large enough to cause the light not to escape from the blackhole..what will be the speed of light against blackholes surface .is it constant or zero? rediculs

  • @ROHANDATAR1
    @ROHANDATAR1 3 роки тому +1

    Light bend around sources with high mass due to gravity. This is not because the mass pulls on the photons directly, but instead because the mass warps the space.

  • @vibha4397
    @vibha4397 4 роки тому +4

    this video was sO helpful, im doing a project and this video is frickin beautiful, ty

    • @praptiupadhayay5305
      @praptiupadhayay5305 3 роки тому

      What is your topic name? I am also looking for quantum mechanics project.

  • @irwinvillalobos2837
    @irwinvillalobos2837 3 роки тому +2

    I read that gH/c^2 is an approximation for the relative change of the frequency of the photon in Earth's surface, where H is the difference in altitude and g is Earth's surface acceleration. Do you know how can I prove that this is a good approximation? I ask that because I get very different results if I proceed as you explain in this video.

  • @naginabahadurofficial6923
    @naginabahadurofficial6923 4 роки тому +2

    It's crazy I came here because I wanted to listen to John Mayer's Gravity and gave this video a go. Not at all regretted clicking this.

  • @GoGa_yt_
    @GoGa_yt_ 3 роки тому +1

    You are better physicist then einstein

  • @shatanikbhattacharya9056
    @shatanikbhattacharya9056 5 років тому +3

    At 6:30 mins of this video, you stated that the P.E. at height H is mgH... But isn't it supposed to be m'gH where m'=h*(nu')/c^2 instead of h*nu/c^2 because the relativistic mass at H is not the same as it was at reference h=0...
    Then the final equation becomes (nu')=nu/[ 1+gH/c^2 ] which on binomial approximation gives the formula you have deduced at 8:10 mins. of this video
    Please do clarify this doubt.
    Thanks in advance 😊

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  5 років тому +2

      Hi Shatanik, you are correct in saying that the mass at H is not m. I didnot want to make the calculations complicated, so took the easy way and wrote m instead, as I was anyways more interested in the second part of the video. To add to the complexity, even the acceleration due to gravity is changing over a height H !! haha, see what i am talking about. I knew this was a problem, therefore mentioned at 08:50 that I am only talking about small distances. You are however correct in your approach.

    • @randomstuff9960
      @randomstuff9960 4 роки тому +1

      But i was thinking of something different.Because you can see that mass of the photon is a function of height so therefore the total work done on the photon by the earth's gravitational field when it reaches a height h is actually smaller than m'gh right?Because see, it would have been m'gh had the mass been constant at m' throughout the journey.But actually we should be taking the mass as f(h)(a function of height), then actual work done should have been integration from 0 to H of f(h)gdh which gives a result less than m'gh because the mass will continually decrease. So what do you say???Amn't I right???🤔

    • @sverkere
      @sverkere 4 роки тому +1

      @@randomstuff9960 Very good thinking! One should integrate for an accurate/correct answer.

    • @randomstuff9960
      @randomstuff9960 4 роки тому

      @@sverkere 😊😊😊 Thanks for the appreciation.

    • @sverkere
      @sverkere 4 роки тому

      @@randomstuff9960 Look here for a gravitational redshift of a point mass. 145.239.0.91/static/gravredshift.pdf

  • @Komal-1401
    @Komal-1401 Рік тому +1

    Sir your lectures are extremely good. such a nice explanation of the things.
    Sir keep uploading the videos🤩

  • @jameslawrence6068
    @jameslawrence6068 5 років тому +6

    Love your videos! Love from South Africa

  • @jamesholbert8127
    @jamesholbert8127 4 роки тому +2

    Very clear presentation, Thank You.

  • @advikawalia1914
    @advikawalia1914 5 років тому +3

    Nicely explained.

  • @interstellarconveyance4865
    @interstellarconveyance4865 3 роки тому +1

    That was AMAZING!
    Your second interpretation of red shift was wonderful. Can we interpret the final dispersion of red shift by removing Plank's constant and substitute Plank's with another formula? Perhaps Mass to energy, -E+Kv=0/P -G+em? What distance does the last photon particle travel and what is the final principle of dispersion?
    Are angular and rotational kinematics of photaic mass considered in the gravitational model? Can we measure the final dispersion pattern from an absorption plate and reverse calculate the primary Mass of the 0 point?
    Thank you! 🙏💖🌱🌍🌱

  • @discoveringthegardenofeden7882

    Can the frequency of light from a planet, and its redshift, be used as a direct measure of the amount of gravity and mass associated with a planet?

  • @mushfiqurrahmanshishir8055
    @mushfiqurrahmanshishir8055 3 роки тому +2

    This is a man, who know what he does
    Lots of love 💗

  • @mikel4879
    @mikel4879 8 місяців тому

    There's some big mess in your cognitive prowess regarding this subject.
    If the so-called "energy" of the so-called "photon" is equal to h•v ( where v is frequency; sorry, I'm writing from my cellphone ) then what is 'h' in REALITY? ( NOT IN THEORY! ).
    Is 'h' a REAL "mass" phenomenon or is it a REAL 'frequency' phenomenon, when you enter it in the notion of the so-called "energy"?
    Be careful in your thinking, because I don't play here a semantic game.
    The true REALITY is in only ONE way. It can't be one time "mass" and other time "frequency".
    TRUE REALITY is never dual, because you can't create through a theoretical hocus-pocus math model a highly causal non-existent phantasmagory and say that is absolutely TRUE-REAL.
    What I mean here is that any aggregated entity, in any way, shape and form, for example , a "solid" object ( apparently 'solid' ) like, let's say, a golf ball, something not true "solid" like an "atom", an "electron", a "photon", a ( so-called) "unit of quanta" of the value of exactly "h", etc, has exactly the same CAUSAL entropic emergent dynamic, the same real dynamic causal process is behind it.
    Therefore, to say that a so-called "photon" has only "frequency" and no "mass", and the golf ball has the so-called"mass", is something very unintelligent.
    Somebody here is thinking exactly like a perfect parrot, parroting other parrots' wrong ideas and their erroneous theories, without thinking at all with their own personal brain.
    What if 🙄🤔😯😏the so-called "photon" and the "golf ball" and etc ( = absolutely everything else in the Universe from the micro to macro, ad infinitum ) is all in fact the same REAL "thing", the same real, universal dynamic? What if any real aggregated dynamic takes place CAUSALLY exactly the same, at any scale of the Universe, ad infinitum?
    Etc.
    So, in the end, please tell me, if you can, what is the 'h' made of in REALITY, without postulating stupidities again.
    It is made of something, isn't? 🤔😯😏

  • @kushagrakinra4505
    @kushagrakinra4505 2 місяці тому

    Sir if frequency changes constantly as we go up due energy conservation, then mass of photon should also change, then how can we calculate it directly by energy conservation

  • @saftheartist6137
    @saftheartist6137 2 місяці тому

    Thank you for this educational lecture!