Dr Ben Yelverton
Dr Ben Yelverton
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Why does pair production occur near nuclei?
By considering the conservation of energy and momentum, we explain why pair production can only take place when another particle - often a nucleus - is nearby.
To support the channel: ko-fi.com/benyelverton
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #quantum #pairproduction #photon #antimatter #antiparticle #electron #speedoflight #nucleus #nuclearphysics #charge #conservationlaws #energy #momentum #zeromomentumframe #relativity #specialrelativity #dispersion #physicsproblems #maths #math #science #education
Переглядів: 475

Відео

Fusion temperature: importance of quantum tunnelling
Переглядів 1 тис.14 днів тому
Estimating the temperature required for nuclear fusion to occur, using two simple models, one classical and one quantum. Comparing the models shows that we need to invoke quantum tunnelling to obtain a realistic value for the temperature! To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my...
Estimating the temperature of a star
Переглядів 1,3 тис.21 день тому
A quick energy-based method to estimate the temperature of a star given only its mass and radius. Although based on a simple model, the equation we derive shows surprisingly good agreement with the temperature of the Sun! To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spen...
Pressure inside a planet
Переглядів 1,2 тис.6 місяців тому
Finding the pressure inside a spherical planet as a function of radial distance from the centre, by integrating the equation of hydrostatic equilibrium. We assume a uniform density here, but the method can be generalised easily to the non-uniform case. Deriving the hydrostatic equilibrium condition: ua-cam.com/video/rqE6psWXQLA/v-deo.html To support the channel: ko-fi.com/benyelverton About me:...
Hydrostatic equilibrium: force-based derivation
Переглядів 6666 місяців тому
Deriving the relationship between pressure gradient, density and gravitational field for a fluid in hydrostatic equilibrium, by considering the balance of forces acting on a small Cartesian fluid element. To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teac...
Fibonacci numbers and resistor networks
Переглядів 8166 місяців тому
Exploring the relationship between the Fibonacci numbers and the effective resistances of ladder-like resistor networks, using recurrence relations. Full derivation of nth effective resistance: ua-cam.com/video/8-Ow1KY0_0A/v-deo.html And the nth term of the Fibonacci sequence (note that this starts with 1, 1 instead of 1, 2, but is easily adaptable): ua-cam.com/video/14pBNzI-yuw/v-deo.html Abou...
Finite resistor ladder: n repeating units
Переглядів 8526 місяців тому
Finding the effective resistance of a ladder-like resistor network with a finite number of repeating units, n. We need to solve a non-linear recurrence relation, and along the way we make use of matrix diagonalization. At the end, we take the limit as n becomes infinite and see that the golden ratio makes an appearance. Simple derivation in the case of infinite n: ua-cam.com/video/7i_PGNp7i1o/v...
Infinite ladder of resistors: general case
Переглядів 2,6 тис.6 місяців тому
Deriving an expression for the effective resistance of an infinite ladder-like resistor network, in the general case where the repeating unit contains three arbitrary resistors. We finish with some special cases, and see how the golden ratio turns up in the solution. About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent...
When is angular momentum not parallel to angular velocity?
Переглядів 1,9 тис.6 місяців тому
Developing some intuition about the direction of a rigid body's angular momentum, with the aid of some examples. We focus particularly on the case when angular momentum and angular velocity are in different directions, and discuss how to understand this scenario in terms of applied torque. The discussion is purely in terms of vector cross products, without referring to the inertia tensor. The t...
Understanding surface tension in liquids
Переглядів 1,6 тис.7 місяців тому
Here we discuss two different (but equivalent) ways of understanding surface tension in liquids, in terms of both forces and energy. We finish by explaining how surface tension causes liquid droplets to take on a spherical shape in the absence of external forces. To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed on to get a Ph...
Pressure required to inflate a balloon
Переглядів 2,9 тис.7 місяців тому
Why does it become easier to inflate a balloon once it grows beyond a certain size? Here we develop a simple model to find the excess pressure required to inflate the balloon to an arbitrary radius r, and use the result to gain some understanding of this effect. Deriving the equation Δp = 2γ/r: ua-cam.com/video/CtUetjUX-yY/v-deo.html To support the channel: ko-fi.com/benyelverton About me: I st...
Laplace pressure in a bubble: derivation using forces
Переглядів 1,6 тис.7 місяців тому
By considering the balance of forces on a surface area element, we derive expressions for the excess pressure inside a spherical bubble in terms of the surface tension. We consider two different cases - a gas bubble surrounded by liquid, and a bubble surrounded by a thin liquid film floating in a gas. To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University o...
Single-slit diffraction using phasors
Переглядів 7067 місяців тому
Using phasors to investigate the interference pattern I(θ) produced when a wave diffracts through a single slit of finite width a. To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a priva...
Double-slit interference with phasors
Переглядів 1,1 тис.8 місяців тому
Using phasors to investigate the interference pattern I(θ) produced when a wave diffracts through two narrow slits separated by a distance d. To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working...
Temperature of a planet with an atmosphere: the greenhouse effect
Переглядів 6198 місяців тому
Finding the equilibrium temperature of a planet heated by a star, including the effects of reflection and re-emission of radiation by its atmosphere (i.e. the greenhouse effect). The result depends on the albedo and emissivity of the atmosphere in appropriate wavelength ranges. To support the channel: ko-fi.com/benyelverton About me: I studied Physics at the University of Cambridge, then stayed...
Dust grain heated by a star: equilibrium temperature
Переглядів 5458 місяців тому
Dust grain heated by a star: equilibrium temperature
Collision between spheres: general result
Переглядів 1 тис.8 місяців тому
Collision between spheres: general result
Why do particles of equal mass rebound at 90 degrees?
Переглядів 3,2 тис.8 місяців тому
Why do particles of equal mass rebound at 90 degrees?
Drag forces and the Reynolds number: intuitive understanding
Переглядів 1,2 тис.9 місяців тому
Drag forces and the Reynolds number: intuitive understanding
Why can orbital and spin angular momenta be added?
Переглядів 1,9 тис.9 місяців тому
Why can orbital and spin angular momenta be added?
Why does Kapitza's pendulum oscillate upside down?
Переглядів 3 тис.10 місяців тому
Why does Kapitza's pendulum oscillate upside down?
Bar electret: when D is not εE
Переглядів 1,6 тис.10 місяців тому
Bar electret: when D is not εE
What is the electric displacement field?
Переглядів 9 тис.10 місяців тому
What is the electric displacement field?
Bound charge density: why does ρ = -∇⋅P?
Переглядів 2,6 тис.10 місяців тому
Bound charge density: why does ρ = -∇⋅P?
Polarisation and surface charge: why does σ = P⋅n?
Переглядів 2,9 тис.10 місяців тому
Polarisation and surface charge: why does σ = P⋅n?
Angle at which particle leaves sphere: force-free method
Переглядів 1,7 тис.10 місяців тому
Angle at which particle leaves sphere: force-free method
Ball and chain projected up a rough inclined plane
Переглядів 1,5 тис.11 місяців тому
Ball and chain projected up a rough inclined plane
Coiled chain falling off the edge of a table
Переглядів 2,1 тис.11 місяців тому
Coiled chain falling off the edge of a table
Chain sliding off an inclined plane
Переглядів 1,3 тис.11 місяців тому
Chain sliding off an inclined plane
Chain dropped onto scales: finding the effective mass
Переглядів 1 тис.11 місяців тому
Chain dropped onto scales: finding the effective mass

КОМЕНТАРІ

  • @juniorcyans2988
    @juniorcyans2988 12 годин тому

    I really enjoy such kind of problems, because I can learn a lot by just solving one single problem. Thank you very much!

  • @Math1729_
    @Math1729_ 18 годин тому

    What if the particle was on a hemisphere instead such that the hemisphere is kept on a frictionless surface so the hemisphere would also move, so how can we approach that problem?

  • @nushratmahjabeen4305
    @nushratmahjabeen4305 День тому

    Thank you for the splendid explanation. Well, where would I make changes if the ball rolls AND slides too?

  • @rummyrummyrum76
    @rummyrummyrum76 День тому

    Can you conserve angular momentum with the force of gravity acting on the ball's center of mass a distance R away from the step edge axis of rotation, would that not count as torque? and void the conservation of angular momentum about that axis?

  • @geon79
    @geon79 2 дні тому

    Very interesting video, but there is a catch: I don't agree with one approximation you are making. It is true that ω is small, but it also true that the radius of Earth is much larger than the height of the tower! It means that the centrifugal acceleration isn't negligeable at least at mid latitudes. Let's calculate the centrifugal force in function of the latitude, considering that the height of the tower is negligable compared to the radius of Earth (Re) and taking Earth as a perfect sphere. The distance from the axis of Earth |r|=Re·cos(λ), so the vector r is [0,−Re·sin(λ)·cos(λ),Re·cos²(λ)]. Plugging the vectors r and ω into the expression of the centrifugal acceleration we obtain a(centr)= −ω x (ω x r)= [0,−Re·ω² ·sin(λ)·cos(λ),Re·ω² ·cos²(λ)] Let's plug in some numbers: |ω|=7.292E-5 rad/s, Re≈6.37E6 m and let's consider the case of λ=45°, then a(centr)=[0,−0.0169, 0.0169] m/s². The x component is always 0 and the z component is much smaller than g, so it can be ignored. We are left with a negative y component, though, that accelerates the particle towards the equator. How does it compare to Coriolis acceleration? For a tower 100 m high, vz at the end of the fall is −44.3 m/s, so the x component Coriolis acceleration reaches aₓ(Cor)=−2·ω· vz · cos(λ) =0.00457 m/s², which is actually ~3.7 time smaller than the y component of the centrifugal force! If my calculations are correct, at mid latitudes, a falling particle is more deflected towards the equator by the centrifugal force than it is to the East by the Coriolis force. Notice, however, that the y component of the centrifugal force tends to 0 as we get closer to the equator, while the Corriolis force does not, so at low latitudes the latter prevails and becomes the only source of deflection at the equator.

  • @omarazami7377
    @omarazami7377 2 дні тому

    Terrific video!! Griffiths only allow us to visualize uniform polarization (and therefore only surface bound charges). It makes perfect sense that if the polarization is changing (I.e. has a derivative) then charges “don’t cancel out” internally. I’m going to show this to a student I’m tutoring. Thank you again!!

  • @ivarszickus4570
    @ivarszickus4570 3 дні тому

    I understand the case where the sphere is grounded. There you can set V=0 and you have clear boundary conditions for your Poisson equation. However I completely fail to understand the case where the sphere isn't grounded. What are the boundary conditions here? I thought the point was to take some image charges that yield the same boundary conditions for the Poisson equation but these conditions are nowhere mentioned. How would we know what the potential on the surface of the non-grounded conductor is in advance? Isn't this essential information that we need before we can even start using the method of images? I've breen trying to understand this for days and it's really driving me mad.

  • @paxshild4924
    @paxshild4924 4 дні тому

    Hello Dr. Ben, an interesting hypothetical question for you, if we were to teleport a chunk of neutron star mass the size of a golf ball here on earth , would it sink down to the core of the earth, creating earthquakes as its gravitational tidal forces rip through the crust , or will it explode? If it were to explode, it would be interesting for you to make a video on it. In a sense, how do you calculate the energy of that explosion? Its a question for fun. I think the audience might enjoy it.

  • @shrivatsa8604
    @shrivatsa8604 5 днів тому

    Hello, sir. I have a question related to the concepts in your recent video. If we take an electron-positron pair as point particles, without a definite radius, their collision doesn’t happen in a classical way, and calculating the annihilation time becomes challenging. Using Coulomb’s law and calculus, we could derive a time-to-distance relation as they accelerate towards each other. However, as they approach close distances, the velocity derived from potential approaches the speed of light, and near the Planck length, Coulomb's law breaks down due to high uncertainty, making exact calculations difficult. Could you explain how to determine the time taken for such a particle-antiparticle annihilation, addressing these relativistic and quantum aspects?

  • @sinecurve9999
    @sinecurve9999 5 днів тому

    So the particle absorbs a photon then emits a virtual photon that then decays into a particle anti-particle pair? What is meant by "in the presence"? These equations are independent of distance between the pair production and particle X. The particle could just as well be in the same atom or halfway across the observable universe, but as long as it somehow receives an impulse from the pair production, conservation of momentum is satisfied. This seems suspect because you would think that these events should be causally connected. The restrictions on what X can be is extremely loose in this framework. I could substitute X with "Milky Way Galaxy" and the math would still be the same. Could X be another photon? It seems all you have shown here is that pair production cannot occur in a universe only containing a single photon if conservation of energy and momentum are to be satisfied.

    • @DrBenYelverton
      @DrBenYelverton 5 днів тому

      The photon decays into a particle-antiparticle pair, then either the particle or antiparticle exchanges a virtual photon with what I called particle X. See e.g. Wikipedia's article on pair production for a Feynman diagram illustrating this more clearly. You're correct that there is lots of physics missing in this analysis! It's a quantum-mechanical interaction between particles, and for a full treatment we'd need to use QED. In your example of a galaxy, the photon would really be interacting with a specific particle somewhere in the galaxy, not the entire galaxy at once. About the distance between the photon and particle X - there is always a probability for the interaction to happen, but of course the probability decreases with separation and eventually becomes negligible. In the case of a nucleus, classically we'd say that the electric field it produces has infinite extent and influences particles at arbitrarily large distances, but in practice the field is so weak at large distances that it's negligible. So, the phrase "in the presence of" does indeed mean that there must be another particle somewhere in the universe, it's just that the process is far more likely to happen the closer that particle is to the photon. Finally, pair production could indeed happen with a second photon instead of a nucleus, provided that it has enough energy. The Feynman diagram is similar, except there is now no particle X and the second photon is not virtual.

  • @Nxck2440
    @Nxck2440 6 днів тому

    Woah, I always wondered about this. Never thought I'd see an explanation on exactly this, thanks! Also, do you happen to have a video about how magnetic fields are kind of like electric fields but in a different reference frame? I heard there is a link based on special relativity but it's hard to understand.

    • @r2k314
      @r2k314 5 днів тому

      search Y.T.: How Induction Helped Einstein Discover Relativity! Physics - problems and solutions

    • @DrBenYelverton
      @DrBenYelverton 5 днів тому

      Excellent, this is certainly a very specific topic and I'm glad it was helpful! I haven't covered the relationship between electric and magnetic fields in relativity yet, but it's something I've been meaning to do for a while.

  • @NerdZEY
    @NerdZEY 7 днів тому

    "Why are you learning this?" A robotics enthusiast: Nothing really important (CiWS)

  • @mausamkalita9375
    @mausamkalita9375 7 днів тому

    Oh man, what you did!

  • @Anselm-wb8we
    @Anselm-wb8we 7 днів тому

    thank you

  • @darwinvironomy3538
    @darwinvironomy3538 9 днів тому

    can you do about geothermal model of the earth probably? using heat equation on steady state with a core that generates heat and a mantle. is it reasonable?

    • @DrBenYelverton
      @DrBenYelverton 6 днів тому

      I suppose the complication is that you probably need to account for convection in the mantle!

    • @darwinvironomy3538
      @darwinvironomy3538 4 дні тому

      @@DrBenYelverton i see!

  • @mrcooper3139
    @mrcooper3139 11 днів тому

    How would the transmission coefficient, which shows the probability of penetration for nuclei, can be used to solve this problem?

    • @kocahimself
      @kocahimself 10 днів тому

      That’s a good question !

  • @mohsenrezaei5965
    @mohsenrezaei5965 11 днів тому

    Can you tell what force keeps the bead on the top point of equilibrium? theta=pi. thank you for this solution

  • @shrivatsa8604
    @shrivatsa8604 12 днів тому

    Hello sir, I have read your website and found the thesis of yours on the topics the influence of planetary and stellar companion and debris disks. Quite interesting. So, could you please tell me an overview on what is this about?

    • @DrBenYelverton
      @DrBenYelverton 11 днів тому

      So, in other planetary systems it is often much easier to detect debris discs than it is to detect the planets themselves. Essentially this is because the asteroids that make up the discs gradually grind themselves down into small dust particles through collisions, and the total cross-sectional area of the many dust particles is much greater than the cross-sectional area of a typical planet. By studying the structures of the discs (e.g. gaps and asymmetries) we can infer things about what sort of otherwise-undetectable planets might be present. I spent a lot of time during my PhD thinking in particular about how multiple-planet systems can open gaps in a disc through a mechanism called secular resonance, which occurs when the orbits of the asteroids and planets precess at the same frequency and the asteroids' orbits become very eccentric as a result.

  • @adityk8138
    @adityk8138 12 днів тому

    Why we use this relation in band brake and rope pulley problem both while in band brake, pure slipping is there but in case of rope-pulley pure rolling is there between rope and pulley. This result is better suits for band brake(because pure slipping is there i.e maximum value of friction u×R is generated). Please help through video. Please. louuu from india.

  • @phyarth8082
    @phyarth8082 13 днів тому

    Nice property of wave-particle duality de Broglie equation to reduce approx. d=1 fm result. I calculated static distance exactly for hydrogen atom using Casimir effect and this distance for hydrogen atom is d=0.583 fm.

  • @gesucristo0
    @gesucristo0 13 днів тому

    Why didn’t you take into account any relativistic effect?

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      A quick calculation using the equipartition theorem shows that typical velocities of nuclei in the Sun are less than 1% of the speed of light, so not fast enough that relativistic effects would change our result at this level of accuracy.

  • @Nxck2440
    @Nxck2440 13 днів тому

    The quantum analysis reminds me of the Gamow theory of radioactive alpha decay, by predicting the chance of an alpha particle tunneling out of the strong nuclear potential well. Very cool!

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      Indeed! I believe it was also Gamow who came up with this model for fusion, see e.g. Wikipedia's article on the Gamow factor.

  • @MissPiggyM976
    @MissPiggyM976 13 днів тому

    Welcome back, many thanks!

  • @notsodope834
    @notsodope834 13 днів тому

    7:32 3/2kT term is definded as KE per gas molecule , here hydrogen is diatomic so M/2mH h2 molecules ( where mH= mass of hydro. atom)

    • @notsodope834
      @notsodope834 13 днів тому

      This makes 3/4kT( M/mH) T= 4GMmH/5kR

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      Hydrogen exists in monatomic form in a star because the temperature is so high. If it were diatomic, we'd have to use (5/2)kT instead of (3/2)kT as diatomic molecules have rotational kinetic energy as well as translational.

    • @notsodope834
      @notsodope834 13 днів тому

      Yes, I did a mistake but why multiplied by 2 still can't get that

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      The hydrogen atoms are ionised due to the high temperature, so each hydrogen atom contributes two separate particles, a proton and an electron.

  • @joshualeft
    @joshualeft 14 днів тому

    Fantastic explanation and video!!!

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      Great to hear, thanks for your comment!

  • @PeacefulAnxiety
    @PeacefulAnxiety 14 днів тому

    Been missing your videos, cool as always

  • @MathwithMing
    @MathwithMing 14 днів тому

    More thermodynamics and statistical mechanics please!

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      So many things I would like to cover! Statistical mechanics was one of my favourite courses during my time as a student, definitely want to revisit this.

    • @MathwithMing
      @MathwithMing 13 днів тому

      @@DrBenYelverton please! Actually it was something I didn’t take due to competing interests so I’m very much looking forward to it

    • @Cameouo
      @Cameouo 11 днів тому

      ​@@DrBenYelvertonGood Day Doc, I just wanted to ask If you could make a video on all kinds of math and science courses, I'm confused what to pick and teachers are of no help, thanks

  • @MathwithMing
    @MathwithMing 14 днів тому

    Great to have you back!

    • @DrBenYelverton
      @DrBenYelverton 13 днів тому

      Nice to hear from you! I've enjoyed getting back to making videos.

  • @drscott1
    @drscott1 14 днів тому

    👍🏼

  • @wassserfast6605
    @wassserfast6605 14 днів тому

    👍👍

  • @jaspertenberge1730
    @jaspertenberge1730 15 днів тому

    short and clear, thanks!

  • @FALLENINLOVE-c5i
    @FALLENINLOVE-c5i 15 днів тому

    Very basic question, Hope you would answer me shortly. In your video at the last part you wrote f which is F/m=gt but the force of gravity being applied on the rocket changes continiously as the rocket is gradually losing mass . so how can you write F/m=g should not it be f=(m-dm)g/m??

    • @DrBenYelverton
      @DrBenYelverton 14 днів тому

      All objects in a gravitational field of strength g experience a force per unit mass of g, independent of their mass. The rocket is indeed losing mass over time, this changes its weight but not the applied force per unit mass.

  • @ISRAELCHERINET
    @ISRAELCHERINET 16 днів тому

    Tnx

  • @paxshild4924
    @paxshild4924 18 днів тому

    Hey Dr. Ben, it's great to see you back! I really enjoyed your recent video on the temperature of a star, especially after such a lon break. Your content always brings out some of the most interesting problems in physics! I was thinking, as a follow-up to this thermodynamics topic, what if you did a video exploring how temperature changes over time for an object in a completely empty vacuum? Specifically, if we consider an object with a certain initial temperature in an ideal vacuum (with no surrounding objects), how does its temperature evolve over time as it radiates heat away? I've been trying to come up with a satisfying equation for this, but it seems like Newton's law of cooling might not fully apply in this scenario. I feel like this could be a really fascinating continuation of your recent video. Also, I wanted to say that I'm deeply sorry for your loss, and I admire your resilience in continuing to share such amazing content. Looking forward to more thought-provoking problems from you!

    • @umylten4142
      @umylten4142 18 днів тому

      Just for fun (as almost all assumptions are incorrect), assume: - constant density r - constant specific heat capacity c - uniform temperature T(t) - black body radiation Then, thermal energy is (up to a constant) r•c•V•T(t) (where V is the volume) and radiated power is s•A•T(t)⁴ (where s is Stephan-Bolzmann constant and A is the area of the object). Equating the loss of thermal energy with the radiated power: r•c•V•(dT/dt) = -s•A•T⁴ This is a simple differential equation which provides T(t) :^) If I didn't mess up, the result is T(t) = T(0)/[(1 + bt)^(1/3)] where b = 3•s•A•T(0)³/(r•c•V).

    • @DrBenYelverton
      @DrBenYelverton 14 днів тому

      Thanks for your kind words, I'm always happy to see your comments! I've just recorded another video today and hope to get it uploaded soon. Cooling of a black body is on my to-do list, this could of course become arbitrarily complicated but I was thinking of doing it using a simple model like the one proposed by the commenter above.

  • @Herogamergaming59
    @Herogamergaming59 19 днів тому

    Sir after doing some analysis using the Stefan-Boltzmann law I found out that this model suggests that the intensity released by a star is proportional to the mass of the star raised to the power of 4 ! I find that quite interesting

    • @DrBenYelverton
      @DrBenYelverton 19 днів тому

      Apparently, that is actually a pretty good approximation for Sun-like stars! See e.g. en.wikipedia.org/wiki/Mass%E2%80%93luminosity_relation

  • @yahavhazut
    @yahavhazut 20 днів тому

    Great video that shows the power of estimation theory! Thanks!

    • @DrBenYelverton
      @DrBenYelverton 19 днів тому

      Thank you - it's amazing how well simple models work sometimes!

  • @mohsenrezaei5965
    @mohsenrezaei5965 20 днів тому

    brilliant! your work in this channel is priceless, thank you

    • @DrBenYelverton
      @DrBenYelverton 19 днів тому

      Thanks for your kind words! Glad you are enjoying the videos.

  • @shrivatsa8604
    @shrivatsa8604 20 днів тому

    Great to have you back, sir. I had previously worked with the same problem, which after solving yielded the same result. And then tried to solve it in a non-classical, relativistic way .my initial goal was to find the potential energy of the entire system using relativistic way calculating energy (Swartzchild interior solution) , which was complicated. Based on your expertise in this field, I believe working on relativistic physics in your upcoming videos will be very brilliant and intresting like this one.

    • @DrBenYelverton
      @DrBenYelverton 20 днів тому

      Thanks for your support. Solving this one relativistically certainly sounds complicated!

  • @mxminecraft9410
    @mxminecraft9410 21 день тому

    Sir I have been doing a lot of physics problems some of them are very hard where can I share the problems with you?

    • @DrBenYelverton
      @DrBenYelverton 20 днів тому

      Feel free to email me at ben.yelverton@cantab.net!

    • @mxminecraft9410
      @mxminecraft9410 20 днів тому

      @@DrBenYelverton ok

    • @mxminecraft9410
      @mxminecraft9410 20 днів тому

      @@DrBenYelverton question sent

    • @DrBenYelverton
      @DrBenYelverton 20 днів тому

      @@mxminecraft9410 Got it, looks interesting! I will have a go at solving it when I get the chance.

    • @mxminecraft9410
      @mxminecraft9410 20 днів тому

      @@DrBenYelverton ok can you also upload a solution too?

  • @drscott1
    @drscott1 21 день тому

    👍🏼

  • @logician1234
    @logician1234 21 день тому

    Prof takes a break and then the first video he posts is calculating temperature of the star 💀

    • @DrBenYelverton
      @DrBenYelverton 20 днів тому

      I had to think of something impressive for my return!

  • @gustavobagu7156
    @gustavobagu7156 21 день тому

    ONE QUESTION: MINUTE 8:35: WHY IS IT THAT VECTOR "r" DOES NOT HAVE Z-COMPONENT??

    • @DrBenYelverton
      @DrBenYelverton 20 днів тому

      That's because the plane that we're intersecting the cone with is the x-y plane, in which all points have z = 0.

  • @BlockTechnology
    @BlockTechnology 22 дні тому

    I think I must finish high school first (and not try to understand it in middle school 💀)

    • @DrBenYelverton
      @DrBenYelverton 22 дні тому

      I certainly would not have understood this at your age!

  • @shalalalalalalalalalalala
    @shalalalalalalalalalalala 22 дні тому

    cheers m8

  • @aadih.n3438
    @aadih.n3438 24 дні тому

    hello why isn't t1>t2?

  • @timaryalllll_physics
    @timaryalllll_physics 26 днів тому

    Amazing❤thanksssss

  •  28 днів тому

    IPhO 2024 T1, youre the goat for anyone whos prepping for a physics olympiad sir

    • @DrBenYelverton
      @DrBenYelverton 27 днів тому

      Just had a look at the paper - interesting to see that this came up! I'm glad the videos are helpful.

  • @shlokishan443
    @shlokishan443 28 днів тому

    Sir what if we throw the mass from the top of the incline (here, point P), will the angle change ?😊

    • @DrBenYelverton
      @DrBenYelverton 14 днів тому

      I don't think we've assumed anything about the sign of α here, so the equations derived should still work for a downwards-sloping plane, you'd just need to make α negative!

  • @mr.guardian4491
    @mr.guardian4491 29 днів тому

    Very elegant and simple derivation. Thanks!

  • @ujjwal335
    @ujjwal335 29 днів тому

    Thank you sir love from india