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Is this derivative in terms of tensors correct?

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1. Homework Statement
Solve this, $$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}$$
where q is a constant vector.
2. Homework Equations


3. The Attempt at a Solution
$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}[g^{\mu \nu}[q_{\mu,\nu}x_{\nu}+q_{\mu}x_{\nu,\nu}]] \\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}[g^{\mu \nu}[0+q_{\mu}x_{\nu,\nu}]] \\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}[[q_{\mu}\delta^\mu _\nu]]\\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}q_{\nu}$$
Is this correct? Besides is there an easier or faster way to solve this? Or can it be reduce further to lesser terms?
 
Last edited:

PeroK

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Did you think of checking with an example? When you say reduced further, you have left the answer with the expanded form of the inner product.

Note that you have ##\nu## as the free index, so you can't use this for a dummy index in your inner product. You should see that's wrong as you have three ##\nu##'s in the same term.
 
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Did you think of checking with an example? When you say reduced further, you have left the answer with the expanded form of the inner product
Sorry about the inner product. I just want confirm the validity of my answer.

$$\frac{-9}{(q.x)^4}q_{\nu}$$
 

PeroK

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It's not right.
I don't get why my answer is not right. Even when I reduced it to a normal derivative of x, I will get the same result.

$$\frac{\partial}{\partial x}\frac{3}{(qx)^3}=\frac{-9}{(q x)^4}q$$
 

PeroK

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I don't get why my answer is not right. Even when I reduced it to a normal derivative of x, I will get the same result.

$$\frac{\partial}{\partial x}\frac{3}{(qx)^3}=\frac{-9}{(q x)^4}q$$
Do you know any examples where the metric tensor is not ##\eta^{\mu \nu}##?
 
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Do you know any examples where the metric tensor is not ##\eta^{\mu \nu}##?
No unless is curved spacetime. Correcting for the ημν and v,

$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\kappa)^4}[\eta^{\mu \kappa}[q_{\mu,\nu}x_{\kappa}+q_{\mu}x_{\kappa,\nu}]] \\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\kappa)^4}[\eta^{\mu \kappa}[0+q_{\mu}x_{\kappa,\nu}]] \\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\kappa)^4}[[q_{\mu}\delta^\mu _\nu]]\\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\nu)^4}q_{\nu}\\=\frac{-9}{(q.x)^4}q_{\nu}$$
 

PeroK

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No unless is curved spacetime. Correcting for the ημν and v,

$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\kappa)^4}[\eta^{\mu \kappa}[q_{\mu,\nu}x_{\kappa}+q_{\mu}x_{\kappa,\nu}]] \\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\kappa)^4}[\eta^{\mu \kappa}[0+q_{\mu}x_{\kappa,\nu}]] \\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\kappa)^4}[[q^{\nu}\delta^\mu _\nu]]\\ =\frac{-9}{(\eta^{\mu \kappa}q_\mu x_\nu)^4}q_{\nu}\\=\frac{-9}{(q.x)^4}q_{\nu}$$
The metric tensor (in flat spacetime) is ##\eta^{\mu \nu}## when you have Cartesian coordinates. You could try polar coordinates in 2D (flat) space, where the metric tensor is not ##\eta^{\mu \nu}##.

In general, 2D polar coordinates are the simplest example to use when you want to check something like this. Note, however, that they still have a diagonal metric. So, your other option is just to make one up. For example:

##a \cdot b = a_1b_1 + 2a_2b_2 + 3a_1b_2 + 3a_2b_1##

Also, your problem hinges on the derivative of the inner product wrt ##x^{\nu}## so you can just focus on that:
 
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The metric tensor (in flat spacetime) is ##\eta^{\mu \nu}## when you have Cartesian coordinates. You could try polar coordinates in 2D (flat) space, where the metric tensor is not ##\eta^{\mu \nu}##.

In general, 2D polar coordinates are the simplest example to use when you want to check something like this. Note, however, that they still have a diagonal metric. So, your other option is just to make one up. For example:

##a \cdot b = a_1b_1 + 2a_2b_2 + 3a_1b_2 + 3a_2b_1##

Also, your problem hinges on the derivative of the inner product wrt ##x^{\nu}## so you can just focus on that:
When you mean check answer, do you mean something like this?
E.g.
$$q=(a\ \ \ \ \ b)\\ x=(x_1\ \ \ \ \ x_2)$$
$$\partial_\nu \frac{3}{(ax_1+bx_2)^{3}}=\begin{pmatrix}-9a(ax_1+bx_2)^{-4}\\-9b(ax_1+bx_2)^{-4}\end{pmatrix} \\=\frac{-9}{(ax_1+bx_2)^{4}}q_\nu \\=\frac{-9}{(q.x)^{4}}q_\nu$$
And how does the metric tensor fit into the checking here? Is it used to check the validity of the intermediate steps here?
 

PeroK

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When you mean check answer, do you mean something like this?
E.g.
$$q=(a\ \ \ \ \ b)\\ x=(x_1\ \ \ \ \ x_2)$$
$$\partial_\nu \frac{3}{(ax_1+bx_2)^{3}}=\begin{pmatrix}-9a(ax_1+bx_2)^{-4}\\-9b(ax_1+bx_2)^{-4}\end{pmatrix} \\=\frac{-9}{(ax_1+bx_2)^{4}}q_\nu \\=\frac{-9}{(q.x)^{4}}q_\nu$$
And how does the metric tensor fit into the checking here? Is it used to check the validity of the intermediate steps here?
I think you have several conceptual problems here. First, I better check. You used a general form of the inner product in your OP that used a general metric tensor ##g^{\mu \nu}##. Did you really mean that? Or, are you assuming that ##g^{\mu \nu} = diag(1, 1, 1)##?

PS note that ##\eta^{\mu \nu} = diag(-1, 1, 1, 1)##. So, even with that as the metric the answer is not correct.
 
Last edited:

Ray Vickson

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1. Homework Statement
Solve this, $$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}$$
where q is a constant vector.
2. Homework Equations


3. The Attempt at a Solution
$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}[g^{\mu \nu}[q_{\mu,\nu}x_{\nu}+q_{\mu}x_{\nu,\nu}]] \\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}[g^{\mu \nu}[0+q_{\mu}x_{\nu,\nu}]] \\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}[[q_{\mu}\delta^\mu _\nu]]\\ =\frac{-9}{(g^{\mu \nu}q_\mu x_\nu)^4}q_{\nu}$$
Is this correct? Besides is there an easier or faster way to solve this? Or can it be reduce further to lesser terms?
Is ##q\cdot x = q_\mu x^\mu##? If so, an elementary differentiation process gives
$$\frac{\partial}{\partial x^\nu} \frac{3}{(q \cdot x)^3} = -\frac{9}{(q \cdot x)^4} q_\nu$$ with essentially no work.
 
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I think you have several conceptual problems here. First, I better check. You used a general form of the inner product in your OP that used a general metric tensor ##g^{\mu \nu}##. Did you really mean that? Or, are you assuming that ##g^{\mu \nu} = diag(1, 1, 1)##?
My mistake because I did assume that ##g^{\mu \nu} = diag(1, 1, 1)## which shouldn't be the case.
$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[q^{\mu}_{,\nu}x^{\kappa}+q^{\mu}x^{\kappa}_{,\nu}]] \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[0+q^{\mu}x^{\kappa}_{,\nu}]] \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[q^{\mu}\delta^\kappa _\nu]]\\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \nu}q^{\mu}]\\=\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}q_{\nu}$$
where qv=[-q1 q2 q3 ...]

PS note that ##\eta^{\mu \nu} = diag(-1, 1, 1, 1)##. So, even with that as the metric the answer is not correct.
However, what do you mean by this? Why won't that metric work?
 

Ray Vickson

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My mistake because I did assume that ##g^{\mu \nu} = diag(1, 1, 1)## which shouldn't be the case.
$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[q^{\mu}_{,\nu}x^{\kappa}+q^{\mu}x^{\kappa}_{,\nu}]] \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[0+q^{\mu}x^{\kappa}_{,\nu}]] \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[q^{\mu}\delta^\kappa _\nu]]\\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \nu}q^{\mu}]\\=\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}q_{\nu}$$
where qv=[-q1 q2 q3 ...]


However, what do you mean by this? Why won't that metric work?
In #11, if ##q \cdot x = \eta_{\mu \nu} q^\mu x^\nu## we can write it as ##q \cdot x = q_\nu x^\nu,## where ##q_\nu = \eta_{\mu \nu} q^\mu.## So we just have ##q \cdot x = q_1 x^1 + q_2 x^2 + \cdots + q_n x^n##, irrespective of the exact form of the metric ##\eta_{\mu \nu}.##
 
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In #11, if ##q \cdot x = \eta_{\mu \nu} q^\mu x^\nu## we can write it as ##q \cdot x = q_\nu x^\nu,## where ##q_\nu = \eta_{\mu \nu} q^\mu.## So we just have ##q \cdot x = q_1 x^1 + q_2 x^2 + \cdots + q_n x^n##, irrespective of the exact form of the metric ##\eta_{\mu \nu}.##
How do I then ensure that the property of metric tensor is expressed in the equation then? Do I leave the metric tensor and the contravariant vector as separate then?
 

PeroK

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My mistake because I did assume that ##g^{\mu \nu} = diag(1, 1, 1)## which shouldn't be the case.
$$\frac{\partial}{\partial x^{\nu}}\frac{3}{(q.x)^3}=3\frac{\partial(q.x)^{-3}}{\partial (q.x)}*\frac{\partial (q.x)}{\partial x^{\nu}} \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[q^{\mu}_{,\nu}x^{\kappa}+q^{\mu}x^{\kappa}_{,\nu}]] \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[0+q^{\mu}x^{\kappa}_{,\nu}]] \\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \kappa}[q^{\mu}\delta^\kappa _\nu]]\\ =\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}[\eta_{\mu \nu}q^{\mu}]\\=\frac{-9}{(\eta_{\mu \kappa}q^\mu x^\kappa)^4}q_{\nu}$$
where qv=[-q1 q2 q3 ...]


However, what do you mean by this? Why won't that metric work?
Sorry, I think I misled you. I didn't understand why you were not differentiating the tensor components and how you got rid of them. I didn't notice, I must admit, that you had the correct answer all along. Sorry for that.

I thought that if you used an example metric that depended on the coordinates (like polar coordinates) you would see that you have to differentiate the metric components as well.

In any case, @Ray Vickson has given you the best approach.
 
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Sorry, I think I misled you. I didn't understand why you were not differentiating the tensor components and how you got rid of them. I didn't notice, I must admit, that you had the correct answer all along. Sorry for that.

I thought that if you used an example metric that depended on the coordinates (like polar coordinates) you would see that you have to differentiate the metric components as well.

In any case, @Ray Vickson has given you the best approach.
Okay thanks for the help!
 

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