- #1

- 340

- 6

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter mertcan
- Start date

- #1

- 340

- 6

- #2

- 13

- 1

(Say what? ;)

If the student hasn't learned. The teacher hasn't taught. Seek out an alternative presentation.

Increase your knowledge base sampling rate by relying upon multiple sources for what you're striving to learn.

John

- #3

- 340

- 6

you want to say: Is suskind saying something wrong in this topic ????

- #4

- 340

- 6

- #5

- 340

- 6

- #6

jtbell

Mentor

- 15,765

- 4,009

According to my arithmetic, it was only one day and two hours, when you posted this.It has been 2 days

I'll move this over to the Classical Physics forum to see if it draws more useful responses. General Physics is for introductory-physics-type questions.

- #7

- 31,304

- 8,088

I read your question and watched the video and I am not sure what you are asking. Could you clarify?I can not get a satisfying answer to my question

In particular, are you asking how a quantity like ##A_u dx^u## is invariant, or are you asking how it can be integrated, or are you asking something else?

- #8

ChrisVer

Gold Member

- 3,381

- 462

How is that integral invariant:

Make a Lorentz transformation and you get the answer, I mean common, it's a SR course... in fact it's a scalar quantity by construction I'd say, since it's the Minkowski inner product. It will get you to an equation like [itex]\eta_{\mu\nu} \Lambda^\mu_\sigma \Lambda^\nu_\rho[/itex] which is equal to [itex]\eta_{\rho \sigma}[/itex].

https://en.wikipedia.org/wiki/Special_relativity#Metric

If you want another example not from the SR but from classical mechanics, why is the [itex]|\vec{v}|^2[/itex], with [itex]\vec{v}[/itex] the velocity, invariant under rotations? Or any type of inner product?

PS- Oh My God, it took me 7 days to answer I am so incapable.

Make a Lorentz transformation and you get the answer, I mean common, it's a SR course... in fact it's a scalar quantity by construction I'd say, since it's the Minkowski inner product. It will get you to an equation like [itex]\eta_{\mu\nu} \Lambda^\mu_\sigma \Lambda^\nu_\rho[/itex] which is equal to [itex]\eta_{\rho \sigma}[/itex].

https://en.wikipedia.org/wiki/Special_relativity#Metric

If you want another example not from the SR but from classical mechanics, why is the [itex]|\vec{v}|^2[/itex], with [itex]\vec{v}[/itex] the velocity, invariant under rotations? Or any type of inner product?

PS- Oh My God, it took me 7 days to answer I am so incapable.

Last edited:

- #9

- 17,813

- 8,778

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

- #10

- 340

- 6

ok guys thanks for your valuable responses:D

- #11

- 31,304

- 8,088

I am glad we could help, but what was your question?

- #12

- 340

- 6

I tried to ask how a quantity like $$A_udx^u$$ is invariant. Again Very thanks for your nice responses...I am glad we could help, but what was your question?

- #13

- 17,813

- 8,778

$$A[x]=\int_{t_1}^{t_2} \mathrm{d} t [-m c^2 \sqrt{1-\dot{\vec{x}}^2/c^2}-\frac{q}{c} A_{\mu} \dot{x}^{\mu}].$$

This is the action using the coordinate time ##t## of an inertial frame. Nevertheless this is a scalar action and thus the equations of motion can be forumulated in a manifestly covariant form by introducing an arbitrary scalar parameter which is monotonously increasing with ##t##. The action then reads

$$A[z]=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \left [-m c^2 \sqrt{\eta_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \lambda}} - \frac{q}{c} A_{\mu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \right].$$

The equations of motion leads to those for a charged particle in an electromagnetic field, represented by the four potential ##A^{\mu}##. Note that the variation of the action is also invariant under gauge transformations, i.e., changing the four-potential to ##A_{\mu}'=A_{\mu} + \partial_{\mu} \chi## with an arbitrary scalar field ##\chi## doesn't change the equations of motion, which depend only on the gauge-invariant field-strength tensor ##F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}##. All this makes the above Lagrangian a good guess for the correct force law for a charged particle moving in the electromagnetic field, and indeed experiment shows that this is a very good model. It's, however, incomplete since it does not take into account the energy loss by the radiation of electromagnetic waves when a charge is accelerated, but that's another (quite complicated) story.

- #14

anorlunda

Staff Emeritus

- 9,427

- 6,433

(Say what? ;)

If the student hasn't learned. The teacher hasn't taught. Seek out an alternative presentation.

Increase your knowledge base sampling rate by relying upon multiple sources for what you're striving to learn.

John

That's good advice. I did Susskind's courses, 15 minutes per sitting. When he said something I didn't understand, I could replay, or pause while consulting other sources. Wikipedia was frequently helpful. I subscribe to Sal Khan's theory of learning that it is easiest with 100% comprehension. With less than 100%, when you go to the following steps you are burdened by gaps in the underlying concepts.

- #15

ChrisVer

Gold Member

- 3,381

- 462

Are you sure that this course level targets you?

- #16

- 340

- 6

yes, of course. I have eagerness, passion, and devotion towards that kind of topics or physics in a general meaning.Are you sure that this course level targets you?

- #17

ChrisVer

Gold Member

- 3,381

- 462

I didn't imply you lack any of those (necessary) values...I implied that youI have eagerness, passion, and devotion

have you ever taken a course in Classical Mechanics? Or on linear algebra?

I mean he is obviously using a Lagrangian there, and he 'thinks' that it is known to the students (as mentioned in his Lec1, the people in the class are used to Lagrangians and the least action principle, basic quantum mechanics -although he doesn't want to get into quantizing anything-, etc)...

Share: