Need Help understanding Griffiths [Vector Analysis]

In summary, the conversation discusses difficulties with understanding the derivation of equation 1.18 in the context of electromagnetism and vector calculus. The participants suggest using the cyclic nature of the scalar triple product to solve the problem, but also acknowledge the need for a strong foundation in mathematics to fully understand the material. Some recommended resources for improving mathematical skills in this area include Griffiths' textbook and seeking guidance from a mentor.
  • #1
Slimy0233
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TL;DR Summary
Need help with Vector Triple Product mentioned in Griffiths Introduction to Electrodynamics
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I don't understand anything about how 1.18 came to be (Red Rectangle), any help would be greatly appreciated!
 
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  • #2
The second equation in (1.18) follows from (1.17) by first applying it to [itex]\mathbf{A} \times (\mathbf{B} \times \mathbf{E})[/itex] and then setting [itex]\mathbf{E} = \mathbf{C} \times \mathbf{D}[/itex].

For the first, you can use the cyclic nature of the scalar triple product [tex]A \cdot (B \times C) = B \cdot (C \times A) = C \cdot(A \times B)[/tex] to get [tex]
(A \times B)\cdot (C \times D) = (B \times (C \times D))\cdot A[/tex] and then apply (1.17).
 
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  • #3
pasmith said:
The second equation in (1.18) follows from (1.17) by first applying it to [itex]\mathbf{A} \times (\mathbf{B} \times \mathbf{E})[/itex] and then setting [itex]\mathbf{E} = \mathbf{C} \times \mathbf{D}[/itex].

For the first, you can use the cyclic nature of the scalar triple product [tex]A \cdot (B \times C) = B \cdot (C \times A) = C \cdot(A \times B)[/tex] to get [tex]
(A \times B)\cdot (C \times D) = (B \times (C \times D))\cdot A[/tex] and then apply (1.17).
applied 1.17, arrived at the line before 1.18, can't get to 1.18, can you please solve the last step?

edit: You are not supposed to arrive at 1.18 from the line before it. They are separate examples. This problem seriously made me re-evaluate my life choices. Did the author think the reader would be able to tell, yeah, I see no discontinuity here. I mean, i feel like he should have included more steps.

And sorry, but what was all this for??

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got it [Mathforums Link]! Thank you @pasmith
 
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  • #4
Slimy0233 said:
Did the author think the reader would be able to tell, yeah, I see no discontinuity here. I mean, i feel like he should have included more steps.
The semi-colon (;) at the end of the first equality indicates that the two equalities are not related.

Griffiths' textbook is on the topic of electromagnetism. While he does start with vector analysis, the topic i presented quite succinctly.
 
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  • #5
The math chapters at the beginning of Griffiths are intended as a review of material that the student has previously studied in math courses, partly so the student can become familiar with the notation that he uses. They are not intended as a substitute for separate math courses.

At most colleges and universities in the US, students taking an E&M course at the level of Griffiths are expected to have previously studied introductory physics at the level of Halliday/Resnick/Walker Fundamentals of Physics, and single- and multivariable calculus (3 semesters). Probably also linear algebra and ordinary differential equations. At the college where I taught, it was normally a third-year course.
 
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  • #6
I think Griffiths should have given each equation its own number. That might have usefullly emphasized the lack of connection. Otherwise Yep.
 
  • #7
Slimy0233 said:
TL;DR Summary: Need help with Vector Triple Product mentioned in Griffiths Introduction to Electrodynamics

View attachment 327595View attachment 327596

I don't understand anything about how 1.18 came to be (Red Rectangle), any help would be greatly appreciated!
If you are struggling on this level of mathematics you are probably not ready for Griffiths EM. The vector calculus coming up next is way more advanced than this introductory material.
 
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  • #8
I remember you wrote in another thread of yours how easy Griffiths is to understand and wanted similar books and here we are struggeling with intro chapter...

Just prove the relations using brute force if you can't think of anything else. It is straight forward but tedious
 
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  • #9
PeroK said:
If you are struggling on this level of mathematics you are probably not ready for Griffiths EM. The vector calculus coming up next is way more advanced than this introductory material.
If there a book you would recommend to hone my skills in mathematical physics generally? I am currently using Mary L Boas.
 
  • #10
Do the hard work yourself, no book can help you with that. When you write "I dont understand anything" you have completely wrong mindset. Did you try anything before you asked? You cant read yourself to math and physics skills
 
  • #11
malawi_glenn said:
Did you try anything before you asked? You cant read yourself to math and physics skills
For hours... I did try that problem. But it didn't work out. It might not have worked out because It was kinda my first exposure to this kind of vectors, but still. I really don't post stuff which I haven't tried myself for at least an hour.
 
  • #12
Slimy0233 said:
For hours... I did try that problem. But it didn't work out.
This is an indicator that you need to take the time and effort to learn the fundamentals. Repeatedly trying the hardest problem in a book is not a productive way to learn the subject matter covered therein.
Honestly assess your level of understanding and proceed accordingly. It is useful to enlist the aid of a mentor to do this.
 
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  • #13
Slimy0233 said:
For hours... I did try that problem. But it didn't work out. It might not have worked out because It was kinda my first exposure to this kind of vectors, but still. I really don't post stuff which I haven't tried myself for at least an hour.
Then why did you not post your attempt?

This kind of vectors? What kind of vectors other than vectors in R^3 have you done earlier?
 
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  • #14
Let me reiterate. Griffiths is a very good text, and while I applaud it as a target for your efforts, you need to ruthlessly assess what you don't know if you desire to be well educated. It is important to approach this with efficiency and not just bluster. For instance, if you cannot manipulate vectors with some facility you need to develop that skill. Intellect and force of will are not sufficient.
Playing a Stradivarius badly does not make you a violinist
 
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  • #15
hutchphd said:
This is an indicator that you need to take the time and effort to learn the fundamentals. Repeatedly trying the hardest problem in a book is not a productive way to learn the subject matter covered therein.
Honestly assess your level of understanding and proceed accordingly. It is useful to enlist the aid of a mentor to do this.
Thank you. I always welcome advice from y'all!! God knows I need it.

What do you mean by
"enlist the aid of a mentor to do this."
edit: @hutchphd If you were in my shoes would you prefer Boas or Weber, I like Boas but I feel like Weber is much more beginner friendly.

Also, picking up Griffiths was my attempt at learning vectors. Do you think if my only intention is to learn vectors, I should pickup Boas or Weber?
 
  • #16
Slimy0233 said:
And sorry, but what was all this for?
You need to prove these to yourself by starting with (ax,ay, az)X(bx,by,bz) and then crossing with (cx, cy, cz) to find (AXB)XC, then do it the other way to find AX(BXC).

The reason is so that later you can work with the equations in vector form, without writing out all of the components.
 
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  • #17
Slimy0233 said:
Also, picking up Griffiths was my attempt at learning vectors. Do you think if my only intention is to learn vectors, I should pickup Boas or Weber?
Sorry but I don't know either book well enough to have an opinion. I have heard others speak highly of Boas. I learned from Sokolnikoff and Redheffer and it was pretty good.
 
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  • #18
You basically only need this for vectors in ##\mathbb{R}^3##.

##\vec u = (u_x, u_y, u_z)##

Definition of scalar product ## \vec u \cdot \vec v = u_xv_y + u_yv_y + u_zv_z = |\vec u | |\vec v | \cos \theta ##, where ##\theta## is the angle bewteen the two vectors and ##|\vec u | = \sqrt{u_x^2 + u_y^2 + u_z^2}## is the norm (lenght) of a vector in ##\mathbb{R}^3##.

Definition of cross product ##\vec u \times \vec v = \vec w ## where ##w_x=u_y v_z - u_zv_y##, ##w_y = u_z v_x - u_xv_z## and ##w_z = u_x v_y - u_yv_x##.

As a first set of excersices, show that
1) ## \vec u \cdot \vec u = |\vec u |^2##
2) ## \vec u \cdot \vec v = |\vec u | |\vec v | \cos \theta ##, the law of cosines is helpful here.
3) ##\vec u \times \vec v = - (\vec v \times \vec u)## (the cross product is anti-commutative)
4) ##\vec u \times \vec u = \vec 0## where ##\vec 0 = (0, 0, 0)## is the zero-vector in ##\mathbb{R}^3##.
5) ## \vec u \times (\vec v + \vec w ) = \vec u \times\vec v + \vec u \times \vec w ##
6) ## (k \vec u) \times \vec v = \vec u \times (k\vec v) = k(\vec u\times \vec v) ## where ##k## is a scalar
7) ##\vec u \times (\vec v \times \vec w) \neq (\vec u \times \vec v ) \times \vec w ## (the cross product is non-associative)
8) ## \vec u \cdot (\vec u \times \vec v) = 0 ## and ##\vec v \cdot (\vec u \times \vec v) = 0 ## (the vector produced by the cross product is orthogonal to both of its input-vectors)
9) ##| \vec u \times \vec v| = |\vec u | |\vec v | \sin \theta##
10) ##\vec u \times (\vec v \times \vec w) + \vec v \times (\vec w \times \vec u) + \vec w \times (\vec u \times \vec v) = 0## (Jacobi identity)

You should be able to prove these with just basic high-school algebra such as factorization and some basic trigonometry. I.e. no fancy math required. When you can do these on your own, everything else should be piece of cake.
 
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  • #19
malawi_glenn said:
You basically only need this for vectors in ##\mathbb{R}^3##.

##\vec u = (u_x, u_y, u_z)##

Definition of scalar product ## \vec u \cdot \vec v = u_xv_y + u_yv_y + u_zv_z = |\vec u | |\vec v | \cos \theta ##, where ##\theta## is the angle bewteen the two vectors and ##|\vec u | = \sqrt{u_x^2 + u_y^2 + u_z^2}## is the norm (lenght) of a vector in R^3.

Definition of cross product ##\vec u \times \vec v = \vec w ## where ##w_x=u_y v_z - u_zv_y##, ##w_y = u_z v_x - u_xv_z## and ##w_z = u_x v_y - u_yv_x##.

As a first set of excersices, show that
1) ## \vec u \cdot \vec u = |\vec u |^2##
2) ## \vec u \cdot \vec v = |\vec u | |\vec v | \cos \theta ##, the law of cosines is helpful here.
3) ##\vec u \times \vec v = - (\vec v \times \vec u)## (the cross product is anti-commutative)
4) ##\vec u \times \vec u = \vec 0## where ##\vec 0 = (0, 0, 0)## is the zero-vector in ##\mathbb{R}^3##.
5) ## \vec u \times (\vec v + \vec w ) = \vec u \times\vec v + \vec u \times \vec w ##
6) ## (k \vec u) \times \vec v = \vec u \times (k\vec v) = k(\vec u\times \vec v) ## where ##k## is a scalar
7) ##\vec u \times (\vec v \times \vec w) \neq (\vec u \times \vec v ) \times \vec w ## (the cross product is non-associative)
8) ## \vec u \cdot (\vec u \times \vec v) = 0 ## and ##\vec v \cdot (\vec u \times \vec v) = 0 ## (the vector produced by the cross product is orthogonal to both of its input-vectors)
9) ##| \vec u \times \vec v| = |\vec u | |\vec v | \sin \theta##
10) ##\vec u \times (\vec v \times \vec w) + \vec v \times (\vec w \times \vec u) + \vec w \times (\vec u \times \vec v) = 0## (Jacobi identity)

You should be able to prove these with just basic high-school algebra such as factorization and some basic trigonometry. I.e. no fancy math required. When you can do these on your own, everything else should be piece of cake.
Thank you very much for this! I shall do that!
 
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  • #20
Thanks @malawi_glenn . It has been so long since I did anything with vectors, I think I will do these myself just for the mental exercise!
 
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  • #21
gmax137 said:
Thanks @malawi_glenn . It has been so long since I did anything with vectors, I think I will do these myself just for the mental exercise!

Slimy0233 said:
Thank you very much for this! I shall do that!

Work slowly and be meticulous. A sloppy error early on is futile. In particular number 9 and 10...

Good luck and enjoy the ride.

as excerise 11, consider proving Lagrange identity ## | \vec u \times \vec v |^2 = | \vec u| ^2 | \vec v|^2 - ( \vec u \cdot \vec v )^2 ## which is a special case of the first equation in that red box in the first post of this thread.
 

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