Why do we assume that the end point term vanishes at the end of the video and the minus term remains, could you please elaborate on that? Because for del(s)=0 both the endpoint term and the integral should vanish.
Venugopalan, the variation of S here is formally the same as varying the action in classical mechanics. The usual way this is done is to vary among paths (or histories) where the coordinates have fixed endpoint values. That is, if the coordinates are x and y, then we assume that delta(x) and delta(y) vanish at the initial and final values of lambda. If you do not impose these endpoint conditions, then delta(S) = 0 says (in addition to the geodesic equations or equations of motion) that the coefficients of delta(x) and delta(y) at the endpoints must vanish. In classical mechanics these coefficients are the momenta conjugate to the coordinates. So if you do not impose the condition that the coordinates are fixed at the endpoints, then the variational principle implies that the conjugate momenta must vanish at the endpoints.
@@PhysicsUnsimplified UNDERSTANDING A TWO DIMENSIONAL SURFACE OR SPACE AS invisible AND VISIBLE ON BALANCE, AS E=MC2 IS F=ma: E=MC2 AND F=ma IN BALANCE do also FUNDAMENTALLY represent (on average) what is a two dimensional surface OR SPACE (as what is a BALANCED MIDDLE DISTANCE in/of SPACE), AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma. The stars AND PLANETS are POINTS in the night sky. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. It is CLEARLY proven ON BALANCE. E=MC2 IS CLEARLY F=ma ON BALANCE. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. TIME dilation ULTIMATELY proves (ON BALANCE) that ELECTROMAGNETISM/energy is gravity, AS E=MC2 IS F=ma ON BALANCE !!! Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=ma IN BALANCE; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. "Mass"/energy is gravity. ELECTROMAGNETISM/energy is gravity. (BALANCED inertia/INERTIAL RESISTANCE is fundamental.) Objects AND MEN fall at the SAME RATE (neglecting air resistance, of course), AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity. By Frank DiMeglio
@@konev13thebeast Yes. They're not the same unless the parametrization is affine. You can check David Tong's lecture notes for GR. I think he explains it in the first chapter. Or you can check any GR text by looking up the index for affine parameters.
@abrarfaiyaz6503 Ill have to check it out. Ive been on a hunt to learn as much about calculus of variations as I can without shoving everything into the euler equation like every book and youtube video seems to do lol. Thanks!
Start with x = r*cos(theta). Then ask for the change in x due to small changes in r and theta. They are related by x + dx = (r + dr)*cos(theta+dtheta). Now carry out a series expansion on the right--hand side, keeping first order in small quantities, and then subtract x = r*cos(theta). You're left with dx = dr*cos(theta) - r*sin(theta)*dtheta. As you can see, the result of applying "d" is just like taking a derivative using the usual product rule, chain rule, etc.
@@PhysicsUnsimplified While you find dx in the equation x = r.cos (theta), do you take partial derivatives with respect to r and theta because you have both dr and d (theta)? So according to dr dr.cos (theta) d (theta) -r.sin (dtheta) and the result is dx = drcos (theta) -r.sin (dtheta). taking partial derivatives is correct. because the point there depends not only on r but also on theta angle.
@@fratkaymak1271 To obtain dx = dr*cos(theta) - r*sin(theta)*dtheta from x = r*cos(theta): It's as if x, r and theta all depend on some parameter, say, lambda. Then you just differentiate the expression x = r*cos(theta) with respect to lambda. Then replace dx/dlambda with dx, replace dr/dlambda with dr, and replace dtheta/dlambda with dtheta. (You "cancel out" the factors of dlambda from the denominators.)
![SAFEPLANET - ดาวเคราะห์ [ THE PLANET ]](http://i.ytimg.com/vi/4X1XL3I_k3c/mqdefault.jpg)
Could the solutions to the practice problems be released? It would be very helpful
how can you use Pythagoras theorem on sphere . it is a non Euclidean surface right ?
Why do we assume that the end point term vanishes at the end of the video and the minus term remains, could you please elaborate on that? Because for del(s)=0 both the endpoint term and the integral should vanish.
Venugopalan, the variation of S here is formally the same as varying the action in classical mechanics. The usual way this is done is to vary among paths (or histories) where the coordinates have fixed endpoint values. That is, if the coordinates are x and y, then we assume that delta(x) and delta(y) vanish at the initial and final values of lambda.
If you do not impose these endpoint conditions, then delta(S) = 0 says (in addition to the geodesic equations or equations of motion) that the coefficients of delta(x) and delta(y) at the endpoints must vanish. In classical mechanics these coefficients are the momenta conjugate to the coordinates. So if you do not impose the condition that the coordinates are fixed at the endpoints, then the variational principle implies that the conjugate momenta must vanish at the endpoints.
@@PhysicsUnsimplified Thanks a lot! Now it is crystal clear! :)
@@PhysicsUnsimplified UNDERSTANDING A TWO DIMENSIONAL SURFACE OR SPACE AS invisible AND VISIBLE ON BALANCE, AS E=MC2 IS F=ma:
E=MC2 AND F=ma IN BALANCE do also FUNDAMENTALLY represent (on average) what is a two dimensional surface OR SPACE (as what is a BALANCED MIDDLE DISTANCE in/of SPACE), AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, AS E=MC2 IS F=ma. The stars AND PLANETS are POINTS in the night sky. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. It is CLEARLY proven ON BALANCE. E=MC2 IS CLEARLY F=ma ON BALANCE. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. TIME dilation ULTIMATELY proves (ON BALANCE) that ELECTROMAGNETISM/energy is gravity, AS E=MC2 IS F=ma ON BALANCE !!! Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=ma IN BALANCE; AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. "Mass"/energy is gravity. ELECTROMAGNETISM/energy is gravity. (BALANCED inertia/INERTIAL RESISTANCE is fundamental.) Objects AND MEN fall at the SAME RATE (neglecting air resistance, of course), AS E=MC2 IS F=ma ON BALANCE; AS ELECTROMAGNETISM/energy is gravity.
By Frank DiMeglio
Isn't extremizing with ds^2 the same as extremizing with ds? So isn't it easier to do that with ds^2?
Did you find an answer to this? I'm curious as well
@@konev13thebeast Yes. They're not the same unless the parametrization is affine. You can check David Tong's lecture notes for GR. I think he explains it in the first chapter. Or you can check any GR text by looking up the index for affine parameters.
@abrarfaiyaz6503 Ill have to check it out. Ive been on a hunt to learn as much about calculus of variations as I can without shoving everything into the euler equation like every book and youtube video seems to do lol.
Thanks!
17:32 dx=dr.cos (teta)-r.sin d (teta) how did you find this equation ???? Can you explain, please.
Start with x = r*cos(theta). Then ask for the change in x due to small changes in r and theta. They are related by x + dx = (r + dr)*cos(theta+dtheta). Now carry out a series expansion on the right--hand side, keeping first order in small quantities, and then subtract x = r*cos(theta). You're left with dx = dr*cos(theta) - r*sin(theta)*dtheta. As you can see, the result of applying "d" is just like taking a derivative using the usual product rule, chain rule, etc.
I emailed you on the subject. excuse me. (david_brown@ncsu.edu)
@@PhysicsUnsimplified While you find dx in the equation x = r.cos (theta), do you take partial derivatives with respect to r and theta because you have both dr and d (theta)? So according to dr dr.cos (theta) d (theta) -r.sin (dtheta) and the result is dx = drcos (theta) -r.sin (dtheta). taking partial derivatives is correct. because the point there depends not only on r but also on theta angle.
@@fratkaymak1271 To obtain dx = dr*cos(theta) - r*sin(theta)*dtheta from x = r*cos(theta): It's as if x, r and theta all depend on some parameter, say, lambda. Then you just differentiate the expression x = r*cos(theta) with respect to lambda. Then replace dx/dlambda with dx, replace dr/dlambda with dr, and replace dtheta/dlambda with dtheta. (You "cancel out" the factors of dlambda from the denominators.)
Excellent work. Thank you for making these videos.
Is the strong force another form of gravity?
no, not even close
Unsimplified physicist man explain me the physiccs of gravity emmiters described by Bob lazar?
It's pseudoscience. No explanation would do.
I did not know there are types of coordinates
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