Computing a term in worldlines

  • #1
help1please
167
0

Homework Statement



L is the langrangian

[tex]\dot{q}[/tex] is the velocity with time derivative


Homework Equations



[tex]\frac{\partial L}{\partial \dot{q}} = -M \frac{1}{2} \frac{1}{\sqrt{1- \dot{q} \cdot \dot{q}}} \frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

The Attempt at a Solution



A few questions here. My work is telling me to calculate the right hand term

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

First of all, what is the dot in [tex]\dot{q} \cdot \dot{q}[/tex], is this just the dot product which would be the velocity squared?

Secondly, the work demonstrates how to calculate the right hand side as

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q}) = -\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j) = -\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

This part has me totally confused. What exactly is happening here... ? How do you get, mathematically from

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

then subscripts of j appear and a negative sign shows up..

[tex]-\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j)[/tex]

Then what exactly is this

[tex]-\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

Can some one help me please, it has me stuck in my tracks... :(
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,514
2,157

Homework Statement



L is the langrangian

[tex]\dot{q}[/tex] is the velocity with time derivative


Homework Equations



[tex]\frac{\partial L}{\partial \dot{q}} = -M \frac{1}{2} \frac{1}{\sqrt{1- \dot{q} \cdot \dot{q}}} \frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]
There's something wrong with your equation. You can't have a free index on one side of the equation and not the other.
You're also missing dots all over the place in this post.

All of the derivatives in your post should be with respect to ##\dot{q}_i##, not ##q## or ##q_i##.

The Attempt at a Solution



A few questions here. My work is telling me to calculate the right hand term

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

First of all, what is the dot in [tex]\dot{q} \cdot \dot{q}[/tex], is this just the dot product which would be the velocity squared?
Yes.

Secondly, the work demonstrates how to calculate the right hand side as

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q}) = -\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j) = -\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

This part has me totally confused. What exactly is happening here... ? How do you get, mathematically from

[tex]\frac{\partial}{\partial q_i} (-\dot{q} \cdot \dot{q})[/tex]

then subscripts of j appear and a negative sign shows up..

[tex]-\frac{\partial}{\partial q_i}(\dot{q}_j \cdot \dot{q}_j)[/tex]
I'm not sure why you're confused about the negative sign. It was there right from the beginning, no?

You're using implied-summation notation. Repeated indices indicate they should be summed over. In "regular" notation, you'd write
$$\dot{\vec{q}}\cdot\dot{\vec{q}} = \sum_{j=1}^{n} \dot{q}_j \dot{q}_j.$$ Now what do you get if you differentiate this sum with respect to ##\dot{q}_i##? Keep in mind that the ##\dot{q}_i##'s are independent. Try it and compare what you get to the expression below.
Then what exactly is this

[tex]-\delta_ij \dot{q}_j - \dot{q}_j \delta_ij[/tex]

Can someone help me please? It has me stuck in my tracks... :(
If you don't understand what ##\delta_{ij}## means, look up "Kronecker delta."
 
Last edited:
  • #3
help1please
167
0
There's something wrong with your equation. You can't have a free index on one side of the equation and not the other.
You're also missing dots all over the place in this post.

All of the derivatives in your post should be with respect to ##\dot{q}_i##, not ##q## or ##q_i##.


Yes.


I'm not sure why you're confused about the negative sign. It was there right from the beginning, no?

You're using implied-summation notation. Repeated indices indicate they should be summed over. In "regular" notation, you'd write
$$\dot{\vec{q}}\cdot\dot{\vec{q}} = \sum_{j=1}^{n} \dot{q}_j \dot{q}_j.$$ Now what do you get if you differentiate this sum with respect to ##\dot{q}_i##? Keep in mind that the ##\dot{q}_i##'s are independent. Try it and compare what you get to the expression below.

If you don't understand what ##\delta_{ij}## means, look up "Kronecker delta."

Yes, I now realize that negative sign was there all along my fault.

No, I am not asking about the Kronacker delta, I am asking why are we taking it away in this form? How do we end up with this

[tex]\delta_{ij}\dot{q} - \dot{q}\delta_{ij}[/tex]

How does one end up with that mathematically? Is there a significant name for this form? How does one get from [tex]\dot{q}\cdot \dot{q}[tex] to that above?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,514
2,157
You calculate the derivative. There's nothing tricky going on here really other than the notation. Just differentiate ##\dot{\vec{q}}\cdot\dot{\vec{q}}## with respect to ##\dot{q}_i##.
 
  • #5
help1please
167
0
You calculate the derivative. There's nothing tricky going on here really other than the notation. Just differentiate ##\dot{\vec{q}}\cdot\dot{\vec{q}}## with respect to ##\dot{q}_i##.

Just differentiation? I take it I will use the product rule?
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,514
2,157
Yes, since you're differentiating a sum of products, you need to use the product rule.
 
  • #7
help1please
167
0
See when I read this and thought, ''ok product rule'' I thought everything was sorted, but I when to calculate it but I am a bit lost.

I got as far as [tex]-\dot{q} \cdot \dot{q} = -\dot{q}' \cdot \dot{q} - \dot{q} \cdot \dot{q}'[/tex]

But what am I doing now exactly? As I said before, I know what the Kronacker Delta is, but why has it been attached to both sides of this?

[tex]-\delta_{ij}\dot{q}_j - \dot{q}_j \delta_{ij}[/tex]

What is the meaning behind this? I understand why the coefficient of 2 shows up, this is because the derivative of [tex]\dot{q}^2[/tex] is [tex]2\dot{q}[/tex], but why the delta sign?
 
  • #8
help1please
167
0
Hi Vela, just in case you forgot about my thread, above I had some more questions!

Thanks!
 

Suggested for: Computing a term in worldlines

  • Last Post
Replies
3
Views
984
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
7
Views
701
Replies
1
Views
6K
  • Last Post
Replies
19
Views
1K
  • Last Post
Replies
1
Views
5K
Replies
3
Views
5K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
7K
Top