Special Relativity (AQA Turning Points) - A-level Physics

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  • Опубліковано 31 січ 2025

КОМЕНТАРІ • 77

  • @jazzysingh6770
    @jazzysingh6770 7 років тому +130

    I liked the joke at the end

  • @XxJack893xX
    @XxJack893xX 6 років тому +104

    Ur getting a like for that joke at the end ;)

    • @AJ5711
      @AJ5711 Рік тому +3

      Alright lad chill out yeah 😲

  • @jacobseitler6836
    @jacobseitler6836 5 років тому +16

    At 3.32, the time delay parallel will be longer than the time delay perpendicular

  • @Ibat_countryI
    @Ibat_countryI 5 років тому +9

    Was trying to figure out how to get to the time dilation equation, completely forgot factorization was a thing. Thank you

  • @zahin.
    @zahin. 2 роки тому +13

    3:25 Wouldn’t the ray travelling parallel to the Earth’s velocity be *slower* than the other. Not “faster”.
    Light moving *parallel* to the motion of the Earth should take *longer* to travel to the mirror and back than light travelling at *right angles* to the Earth’s motion.

    • @dagalabable
      @dagalabable 7 місяців тому +1

      no it could be either, depending on whether it is parallel or antiparallel (they are often used interchangeably, rather confusingly)

  • @challengerchris5204
    @challengerchris5204 5 років тому +5

    Thanks for the help dude, without ur vids id have no chance of passing

  • @adanmalik1554
    @adanmalik1554 Рік тому +2

    haha love the pun at the end

  • @riajenson5485
    @riajenson5485 2 роки тому +1

    science shorts you are my hero

  • @adityamisra8869
    @adityamisra8869 5 років тому +7

    The moment after 14:28 scared me alot.

  • @elle123cat
    @elle123cat 5 років тому +14

    For the michelson and Morley's experiment shouldnt the light travelling parallel to the earth's rotation take longer to come back than the perpendicular one?

    • @eddyb4105
      @eddyb4105 3 роки тому +4

      Literally just did a paper with a 6 mark question on the michelson Morley experiment and I got part of it wrong because of this video😐 let’s just say I’m not very thankful for the explanation in this video

    • @joshplayz474
      @joshplayz474 3 роки тому +1

      @@eddyb4105 what did you get wrong?

    • @eddyb4105
      @eddyb4105 3 роки тому +10

      @@joshplayz474 according to AQA, the perpendicular ray is faster because the parallel ray is opposed by the aether. According to this video, the opposite is true and so I incorrectly stated which ray was faster

  • @davidplanet3919
    @davidplanet3919 4 місяці тому

    Time dilation can be derived from the “light clock” but it’s a very contrived example. Time dilation occurs even when there are no photons bouncing around between mirrors.I prefer using k calculus to derive it but I don’t know if this is taught at A level. I did learn this in university first year but didn’t really get it until much later,

  • @joshplayz474
    @joshplayz474 3 роки тому +3

    2 questions:
    1. In MM experiment, doesn't an interference pattern only occur if we have point sources of light like slits, here it looks there is a single beam of light?
    2. In muon decay, aren't we in the same inertial reference frame as the muon since it is not accelerating relative to us, so why do you say we are in a different reference frame to it?
    Thanks

    • @cucumbercow7379
      @cucumbercow7379 2 роки тому

      For the second question, the muon is in a different inertial frame as it’s moving relative to us. Inertial frame is based on velocity not acceleration.

  • @lauravegad9819
    @lauravegad9819 11 місяців тому

    Great video,, thank you, It made things sooooo much simpler!

  • @tomf2405
    @tomf2405 6 років тому +15

    anything on Bertozzi's experiment?

  • @Stewartthorp
    @Stewartthorp 6 років тому

    Awesomely presented.

  • @yoyoyoc3po
    @yoyoyoc3po 6 років тому +10

    Sorry, great vid, but I've got no clue what you did at 12:25.
    Also, is it the half life that has stretched or is it simply the time taken to travel between two points has decreased (meaning the number of decays is lower than expected)?

    • @sakondese-savage-376
      @sakondese-savage-376 9 місяців тому

      i dont understand that part aswell . did u find anything about it ?

  • @the_hasnat
    @the_hasnat 4 роки тому

    Change of bass: 14:27

  • @the_hasnat
    @the_hasnat 4 роки тому +9

    you didnt explain "Bertozzi’s experiment as direct evidence for the variation of kinetic energy with speed."

  • @Hatsoffsip
    @Hatsoffsip 6 років тому

    You're a legend.

  • @orla8195
    @orla8195 6 років тому

    Let’s say a you’re in a car travelling at half the speed of light (c/2) and you shine a torch directly in front of you. We’d assume that the velocity or the light, relative to the car is c/2 (which is c-c/2) BUT isn’t the case, because when we approach the speed of light (or when travelling at c/2) time appears to pass more slowly inside the car. This means that the light would appear to be travelling more distance per second because the second is stretched out for a longer period. And so the speed of light relative to the observer is C. This is relativity which states that the speed of light is absolute.
    My question is, if the same rules apply, what happens if we shine the light in the opposite direction to which the car is travelling? Surely time would pass more slowly inside the car because the car is travelling at c/2. And hence the speed of light relative to the car would appear to cover more distance in a given amount of time (eg. in one second), which would conclude that it’s actually travelling at more than c. Is this correct??

    • @guyguy1811
      @guyguy1811 4 роки тому

      @Justa Fool lol

    • @guyguy1811
      @guyguy1811 4 роки тому

      @Justa Fool Your a very angry individual. I didn't even critique you and only put lol because I wanted to see if you were really as off putting as you first seemed.

    • @guyguy1811
      @guyguy1811 4 роки тому

      @Justa Fool I could post something interesting. I'm in to maths and physics and spend a lot of my free time studying probability functions and interesting parts of Calculus, like that of the Gaussian integral and other such things but honestly you really come across as hostile and generally unlikable. Trust, I go to a school for 'gifted students', many of which are likely odd but most of them aren't actually unlikable as you pertray yourself on the internet . You wouldn't act like this in person so why do it over social media.

    • @guyguy1811
      @guyguy1811 4 роки тому

      @Justa Fool I don't even disagree or agree. I'm more of a maths person, hence me applying for pure maths at university. I don't have a educated opinion on relativity rather the basic level required to get a good grade for AQA physics. I just don't understand why you were so hostile when delivering your point the first time. If I ever do special relativity at university level (which i probably will given special and general relativity are options for my degree) I'll share my more educated opinion. Meanwhile, peace x

  • @emmacole2296
    @emmacole2296 4 роки тому +1

    8:53 How do you get from the two purple equations to the orange equation?

    • @the_hasnat
      @the_hasnat 4 роки тому

      did you find out? I had the same question.

    • @nevanmasterson46
      @nevanmasterson46 3 роки тому +3

      I know this is quite late after the original post, but I thought I'd share in case anyone else had the same question.
      starting with t = (2/c) sqrt((vt/2) ²+L²), you can begin by squaring the expression and expanding out:
      t² = (4/c²)(v²c²/4 + L²)
      = v²t²/4 + 4L²/c²
      Then you can group all expressions with t² in them onto one side and factor:
      t² = v²t²/4 = 4L²/c²
      t²(1-v²/c²) = 4L²/c²
      And after you take square roots, things will begin looking familiar:
      t sqrt(1-v²/c²) = 2L/c
      You can then equate this to proper time, t₀, by using the other equation of t₀ = 2L/c. All that’s left to do is to divide by gamma and you have our beloved formula:
      t₀ = t sqrt(1-v²/c²)
      t = t₀/sqrt(1- v²/c²)
      I feel that - given it was a lengthy derivation - it was worth leaving out of the video. But now it’s here just in case anyone was curious. Hope this helps someone!
      and just a little unrelated note i wanted to add: i wrote this once already pretty much to the end but i exited the youtube page by accident and it didn’t save. safe to say i was angry. a lot of pain has gone into this comment :)

  • @goldybindra8785
    @goldybindra8785 3 роки тому +1

    @science shorts what does "in a laboratory frame of reference" mean in question, is it the proper lenth or the external observers length?

  • @the_hasnat
    @the_hasnat 4 роки тому

    11:30 why is the proper time 1.5 x10^-6. I thought proper time is what is felt by the moving thing. So 1.5 x10^-6 is the observed time = t not t0?

    • @the_hasnat
      @the_hasnat 4 роки тому +2

      Unless you meant that 1.5 x10^-6 is the time felt by the muon

  • @nabin6781
    @nabin6781 6 років тому +1

    for the last equation for the energy why did you minus it by 1 at the end because on the equation they do not do that.

    • @azzbaz2495
      @azzbaz2495 6 років тому

      That equation at the end is for the kinetic energy. The formula given in the book is for total energy of the particle. E(tot) = E(k) + E(rest), so E(k) = E(tot) - E(rest) = The formula in the book - m(0)c^2, which factorizes to be what he has shown.

    • @nabin6781
      @nabin6781 6 років тому

      AzzBaz so it sbould be Ek = mc^2 - m(0)c^2

    • @azzbaz2495
      @azzbaz2495 6 років тому +1

      Yes

    • @nabin6781
      @nabin6781 6 років тому

      AzzBaz thank you that makes more sense now

  • @dunsel5887
    @dunsel5887 5 років тому

    so they zeroed out the experiment in the condition it was to be tested in, the rotating, orbiting Earth, isn't that like standing on a scale and setting it to zero to get your weight?

  • @User-ei2kw
    @User-ei2kw 4 роки тому

    3:43 Didnt you say one of the mirrors was moved back so the distance isnt the same?

  • @adzo1379
    @adzo1379 4 роки тому

    Legend !

  • @Grimreaper-zt3tt
    @Grimreaper-zt3tt 4 роки тому

    at 7.45, isn't the distance of the S/L triangle meant to be vt' not vt? as the time they see the train moving for is t' not t?

  • @roycebong3631
    @roycebong3631 4 роки тому

    14:29 , he was either trying to hold in a sneeze or a yawn

    • @the_hasnat
      @the_hasnat 4 роки тому

      nah it was edited in im sure

  • @MrMas9
    @MrMas9 7 років тому +4

    Amazing video! Any chance you can cover the AQA optional module Astrophysics? :)

  • @daniel117100
    @daniel117100 7 років тому +1

    i was wondering why it said video removed on the playlist

  • @ゆきだるま-h6n
    @ゆきだるま-h6n 6 років тому

    How do you get t=t0/(√(1-(v^2/c^2)) from 2/c[√(vt/2)^2+L^2] 9:01

  • @FP1Pod
    @FP1Pod 6 років тому

    If asked for the p.d. for those questions at the end is it okay to leave the answer in eV? Or does it want just V?

    • @partyDragon2000
      @partyDragon2000 6 років тому +3

      Um have u every done a past paper before dabby???

  • @matthewwaldron3712
    @matthewwaldron3712 5 років тому +2

    Is this in physics A level for AQA because I can't find it in the text book?

    • @ScienceShorts
      @ScienceShorts  5 років тому +2

      It's only in the Turning Points optional module.

  • @farwarizvi6188
    @farwarizvi6188 4 роки тому +3

    my mind,,,,,,,, woww

  • @mathebest07
    @mathebest07 7 років тому +2

    Where was the mistake? Was it anything major?

    • @ScienceShorts
      @ScienceShorts  7 років тому

      +mathebest07 Intensity using half life was wrong.

    • @mathebest07
      @mathebest07 7 років тому

      ah that is fine then, thank you

  • @kylestormeyes4976
    @kylestormeyes4976 6 років тому

    saviour.

  • @the_hasnat
    @the_hasnat 4 роки тому

    12:10 Can you explain this bit in more detail plz.

    • @the_hasnat
      @the_hasnat 4 роки тому

      what does the second equation started here mean: 12:18

  • @ST3VORSTUNNA
    @ST3VORSTUNNA 6 років тому

    but then doesn't the speed of light have absolute motion?

    • @flaccid4338
      @flaccid4338 5 років тому +4

      speed of light is constant if that's what you mean

  • @hoverton6541
    @hoverton6541 4 роки тому

    King

  • @new-knowledge8040
    @new-knowledge8040 7 років тому

    The fact that you can't measure absolute motion, does not mean that there is no such a thing as absolute motion ( 4:49 ).
    In fact, if you start with ......,
    1) An absolute 4 dimensional environment that is composed of 3 spatial dimensions, and 1 time dimension.
    2) All objects present within it travel at an equal magnitude of absolute motion.
    3) This magnitude of absolute motion is equivalent to the motion of light across space.
    4) Any change in an objects direction of travel within space-time also leads to a 4D rotation.
    ......., the outcome of this combination of absolutes results in the SR phenomena. So, in short, absolute motion ongoing within an absolute 4D space-time environment, produces the SR phenomena. Using a simple geometric representation of these two absolutes, one can derive the SR equations and the Lorentz Transformation equations in mere minutes.

    • @I3uzzzzzz
      @I3uzzzzzz 6 років тому +13

      shut up

    • @umarmohammad6841
      @umarmohammad6841 6 років тому

      lmao

    • @Sol-um1qn
      @Sol-um1qn 6 років тому +2

      Great reply but i think the video was only aimed at A level so he said it not to be too confusing, thanks though

    • @I3uzzzzzz
      @I3uzzzzzz 6 років тому

      thanks

  • @YY-lz5mn
    @YY-lz5mn 5 років тому

    please explain AQA 2017 question 5 i cant get the speed in c that they asked for

  • @StuMas
    @StuMas 6 років тому +1

    Am I missing something or, has no-one realised one crucial point about SR? The light-path illustrated at 7:20 is MISLEADING because, in reality, the distance that the beam of light has to travel is EXACTLY the same for BOTH clocks. The 'zig-zag' path that is always used to explain this theory is deceitfully confusing and WRONG! In truth, the beam of light that's in motion, IS STILL only going UP and DOWN - it is NOT zig-zagging! Therefore, there is NO difference in the distance it has to travel, or in the amount of elapsed time for BOTH observers. Where am I going wrong?

    • @maddocksjos
      @maddocksjos 6 років тому +3

      It is only moving up and down in its own frame of reference. To the people around it they see it has the velocity of the 'train' and the velocity up and down, hence the zig--zag.

  • @user-nk5qs8km3d
    @user-nk5qs8km3d Рік тому +2

    Love u more than my gf

  • @TwoTruthsOneLie
    @TwoTruthsOneLie 7 років тому

    Your channel icon is so distracting. I just posted a video about physics. Will you please check it out and let me know what you think?