First class presentation! I find the topic difficult but the way that you explain each & every step is immensely helpful. It avoids that problem when following similar discussions in text books where it only needs one step not to make sense - and one is stuck! Voice & graphics perfectly paced too, which is a great help 😀. Many thanks sir 👍
I think that's because even if that computation leads to the proper result, it's formally a bit lousy. Eg, in which frame are that F and a measured? Actually, a is measured in the accelerating frame, while all other parameters (force, time, speed and position) are relative to an inertial frame. It's not obvious one can write those equations without breaking SR logic. It actually works, but see how weak it is. A more formal approach is to use Rindler coordinates, and that less easy, but more rigorous.
I have a doubt in force applied in perpendicular direction. Won't the velocity change in the perpendicular direction - which chabges gamma. So dr/dt cant be zero? Or did you mean that change of momentum on the direction perpendicular to force is gamm.m.a and in the direction parellel to it is gamma^e.m.a ?
Overall, your presentation is OK: the Force applied in the direction of the velocity is good; the force perpendicular to the velocity would be more precise if you didn't just state the dv/dt is the acceleration perpendicular to the velocity. A vector presentation would improve the presentation; also to state that the force is either ALWAYS applied perpendicular to the velocity or ALWAYS applied in a constant direction, e.g. Y axis.
I'm no physicist, but your question inspired me to derive an expression for (relativistic) kinetic energy, where force is given by the proper time derivative of (relativistic) momentum (i.e. F=dp/dτ=γ⁴ma). By computing ∫Fdx, I deduced that KE=½mc²(γ²-1). It looked pretty neat. I then proceeded to perform a series expansion of this new KE formula and got KE=½mv²+½mv⁴/c²+...+o(v⁶). I noticed that the Newtonian-KE term appeared in the expansion, so that was reassuring. There is just one interesting caveat. After rearranging the equation, it states that ½γ²mc²=KE+½mc², implying that ½γ²mc² is the total relativistic energy expression and ½mc² is the rest energy expression. In other words, E₀=½mc². This is in direct contradiction with Einstein's famous mass-energy equivalence principle E₀=mc². Honestly, I think E₀=½mc², although not as elegant looking, is more consistent in the sense that it has the same structure as the Newtonian-KE. However, if one were to take E₀=mc² as sacred (which it kind of is by default), then to say F=dp/dτ is akin to heresy as it directly leads to E₀=½mc², further implying that E₀=mc² is wrong! So yeah... hopefully that clears some things up.
I just watched a video and the guy was saying it's going to be tricky to derive formulas for force components and indeed he made it difficult to understand I could tell he was telling me what he crammed and with you I can tell you truly understand what you are saying...This is proof that "If you can't explain it easily, then you don't understand it enough"😌.....and just to let you know I've developed a habit of saying "upon" instead of "divided by" after this video🤣🤣🤣
Sir something is needed to be added there,if some object is moving with velocity approachable to c then that object must have some change in its mass,so from that perspective,some thing is missing?
With increasing velocity, there's an increase in energy, and due to mass energy equivalence, we can say that there's an increase in inertia, but physicists these days avoid talking about mass when they can as there's no way of measuring mass of a moving object, only momentum and energy
Sir please use green board with your face. That is best . But here I have a question for you.. When we talk about length contraction Is it length contraction of the object which is traveling with very high speed or is it the contraction of the path which is it covering??? Please sir answer this ...? I'm very confused.
If I consider (see from) the frame of reference of the moving particle then from that the length of entire path is contracted But, If it take the frame of reference of a non moving observer wrt to path then from that the length of the moving object is contracted.
Since ¥ depend upon the magnitude of the velocity of an object and the force applied will acts as centripital force so the magnitude of the velocity doesn't change so it's differentiation w.r.t time will be zero.
@@akshaychoudhary1439 as force is perpendicular, so the speed of the object won't change but its velocity will change as it's a vector quantity. (So as per you, differentiation represent the change in magnitude and not directional changes can be accounted in differentiation) but at the same time he is assigning some value to dv/dt, ie "a". So if dv/dt exist, taking it further d¥/dt must also exist as it will ultimately go down to dv/dt form???
@@Mohi19 My dear, gamma(refer its formula) is only depend on the square of magnitude of velocity(not on its direction)which is constant in circular motion. Thus differentation of gamma(a constant) is zero. But suppose(as per your query),if in the formula of gamma vector v is present in place of v square or anywhere else then gamma will also depend on its direction and then its differentation will not be zero.
Relativistic Energy & Momentum ► th-cam.com/video/X4xLx8fxdy4/w-d-xo.html
First class presentation! I find the topic difficult but the way that you explain each & every step is immensely helpful. It avoids that problem when following similar discussions in text books where it only needs one step not to make sense - and one is stuck! Voice & graphics perfectly paced too, which is a great help 😀. Many thanks sir 👍
Great video! Too many people try making this way more complicated than it needs to be
Why my professor cannot explain things as clear as you, lol. Anyway, thanks for you detailed derivation and explanation! :)
I think that's because even if that computation leads to the proper result, it's formally a bit lousy. Eg, in which frame are that F and a measured? Actually, a is measured in the accelerating frame, while all other parameters (force, time, speed and position) are relative to an inertial frame. It's not obvious one can write those equations without breaking SR logic. It actually works, but see how weak it is.
A more formal approach is to use Rindler coordinates, and that less easy, but more rigorous.
Thank you for such awesome videos, can you made a series like this but about general relativity
I have a doubt in force applied in perpendicular direction.
Won't the velocity change in the perpendicular direction - which chabges gamma. So dr/dt cant be zero?
Or did you mean that change of momentum on the direction perpendicular to force is gamm.m.a and in the direction parellel to it is gamma^e.m.a ?
Overall, your presentation is OK: the Force applied in the direction of the velocity is good; the force perpendicular to the velocity would be more precise if you didn't just state the dv/dt is the acceleration perpendicular to the velocity. A vector presentation would improve the presentation; also to state that the force is either ALWAYS applied perpendicular to the velocity or ALWAYS applied in a constant direction, e.g. Y axis.
Excellent explanation!
Hello sir...
What typing pad do you use
And software
Why wasn't the proper time derivative of momentum taken to find the force ?
To find the momentum, we had used the proper time derivative of position.
I'm no physicist, but your question inspired me to derive an expression for (relativistic) kinetic energy, where force is given by the proper time derivative of (relativistic) momentum (i.e. F=dp/dτ=γ⁴ma). By computing ∫Fdx, I deduced that KE=½mc²(γ²-1). It looked pretty neat. I then proceeded to perform a series expansion of this new KE formula and got KE=½mv²+½mv⁴/c²+...+o(v⁶). I noticed that the Newtonian-KE term appeared in the expansion, so that was reassuring. There is just one interesting caveat. After rearranging the equation, it states that ½γ²mc²=KE+½mc², implying that ½γ²mc² is the total relativistic energy expression and ½mc² is the rest energy expression. In other words, E₀=½mc². This is in direct contradiction with Einstein's famous mass-energy equivalence principle E₀=mc². Honestly, I think E₀=½mc², although not as elegant looking, is more consistent in the sense that it has the same structure as the Newtonian-KE. However, if one were to take E₀=mc² as sacred (which it kind of is by default), then to say F=dp/dτ is akin to heresy as it directly leads to E₀=½mc², further implying that E₀=mc² is wrong! So yeah... hopefully that clears some things up.
Sir which board (soft.) are you using??
Thank you so much Sir 😊🙏🙏
Thankyou sir, btw which tab are you using sir ?
the maths of special relativity is art
Great sir ji 🙏
I just watched a video and the guy was saying it's going to be tricky to derive formulas for force components and indeed he made it difficult to understand I could tell he was telling me what he crammed and with you I can tell you truly understand what you are saying...This is proof that "If you can't explain it easily, then you don't understand it enough"😌.....and just to let you know I've developed a habit of saying "upon" instead of "divided by" after this video🤣🤣🤣
Sir something is needed to be added there,if some object is moving with velocity approachable to c then that object must have some change in its mass,so from that perspective,some thing is missing?
With increasing velocity, there's an increase in energy, and due to mass energy equivalence, we can say that there's an increase in inertia, but physicists these days avoid talking about mass when they can as there's no way of measuring mass of a moving object, only momentum and energy
OK I am satisfied sir but Heisenberg uncertainty principle reopens the issue.
@@AmanDeepSingh-dd4uq this is related to quantum mechanics but it is an interesting idea
In this case why not we using proper time?
Sir please use green board with your face. That is best .
But here I have a question for you..
When we talk about length contraction
Is it length contraction of the object which is traveling with very high speed or is it the contraction of the path which is it covering???
Please sir answer this ...?
I'm very confused.
It depends on the frame of reference we have chosen
@@rekhabenrajpuriya9348 Sir this is not perfect answer.
Please clear that one thing is possible or both are possible?
The person who is observing only one thing is possible depending on his/her frame of Reference
@@rekhabenrajpuriya9348 can you guide please which Frame of reference will decide which phenomena?
If I consider (see from) the frame of reference of the moving particle then from that the length of entire path is contracted
But, If it take the frame of reference of a non moving observer wrt to path then from that the length of the moving object is contracted.
THANK YOU SO MUCH
Nice work 👍👍
Why d(gamma)/dt = zero in second case??
Since ¥ depend upon the magnitude of the velocity of an object and the force applied will acts as centripital force so the magnitude of the velocity doesn't change so it's differentiation w.r.t time will be zero.
@@akshaychoudhary1439 as force is perpendicular, so the speed of the object won't change but its velocity will change as it's a vector quantity. (So as per you, differentiation represent the change in magnitude and not directional changes can be accounted in differentiation) but at the same time he is assigning some value to dv/dt, ie "a". So if dv/dt exist, taking it further d¥/dt must also exist as it will ultimately go down to dv/dt form???
@@Mohi19 My dear, gamma(refer its formula) is only depend on the square of magnitude of velocity(not on its direction)which is constant in circular motion. Thus differentation of gamma(a constant) is zero. But suppose(as per your query),if in the formula of gamma vector v is present in place of v square or anywhere else then gamma will also depend on its direction and then its differentation will not be zero.
@@akshaychoudhary1439 thank you very much. Thanks for the patience.
Sir ap green board per pahle ke jaisa padaiy
Sir please use green board
Where is your face😐😐
Pls mention the source of reference book..