Extracting Maxwell's equations

In summary, Maxwell's equations can be obtained from the given equations, and from these equations, it can also be shown that ##\partial^{\mu}T_{\mu \nu} = 0##, where ##T_{\mu \nu} = F_{\mu \rho}F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}F^{\alpha \beta}##.
  • #1

CAF123

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Homework Statement


Show one can obtain Maxwell's equations from $$\partial^{\mu} F_{\mu \nu} = 0\,\,\,; \,\, \partial_{\mu}F_{\nu \rho} + \partial_{\nu} F_{\rho \mu} + \partial_{\rho}F_{\mu \nu} = 0,$$ where ##F_{ij} = \epsilon_{ijk}B_k## and ##F_{i0} = E_i## with ##F_{\mu \nu} = - F_{\nu \mu}##.
From these two equations, show that one also has ##\partial^{\mu}T_{\mu \nu} = 0##, where ##T_{\mu \nu} = F_{\mu \rho}F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}F^{\alpha \beta}##

Homework Equations



All given in problem statement

The Attempt at a Solution


[/B]
I've done the first part of the question.

My working for the second part is shown below. Insert ##T_{\mu \nu}## and contract with ##\partial^{\mu}##. This gives

$$(\partial^{\mu}F_{\mu \rho}) F_{\nu \sigma} \eta^{\rho \sigma} + F_{\mu \rho}(\partial^{\mu} F_{\nu \sigma}) \eta^{\rho \sigma} - \frac{1}{4}\eta_{\mu \nu}(\partial^{\mu}F_{\alpha \beta})F^{\alpha \beta} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}(\partial^{\mu}F^{\alpha \beta})$$
The first term is zero by Maxwell eqn. In the last term, lower the index on the ##\partial^{\mu}## term and then use the fact that ##\eta_{\mu \nu} \eta^{\mu \gamma} = \delta^{\gamma}_{\nu}## to reduce the last term to ##\frac{1}{4} F_{\alpha \beta} \partial_{\nu} F^{\alpha \beta}##. The same analysis can be applied on third term. Then use the fact that ##F^{\alpha \beta} = F_{\rho \sigma} \eta^{\alpha \rho} \eta^{\beta \sigma}##. The third and fourth terms are then identical and the whole expression reduces to $$F_{\mu \rho} \eta^{\mu \alpha} \partial_{\alpha} F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta} = F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$. I'm unsure of a next step. Possibly rearrange ##\partial_{\alpha}F_{\nu \sigma}## using the second Maxwell eqn above but not sure if this would help.

Many thanks.
 
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  • #2
CAF123 said:
$$ F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$. I'm unsure of a next step. Possibly rearrange ##\partial_{\alpha}F_{\nu \sigma}## using the second Maxwell eqn above but not sure if this would help.

Yes, that would be a good idea. Also, note that in the first term, the index ##\sigma## is a dummy index. Think about renaming this index so that the term looks more similar to the last term.
 
  • #3
TSny said:
Yes, that would be a good idea. Also, note that in the first term, the index ##\sigma## is a dummy index. Think about renaming this index so that the term looks more similar to the last term.
Hi TSny, yes that would give $$\frac{1}{2 } ( \partial_{\nu} F_{\alpha \beta })F^{\alpha \beta} - (\partial_{\beta}F_{\alpha \nu})F^{\alpha \beta}$$ Still a bit unclear as to how to progress, thanks.
 
  • #4
CAF123 said:
Hi TSny, yes that would give $$\frac{1}{2 } ( \partial_{\nu} F_{\alpha \beta })F^{\alpha \beta} - (\partial_{\beta}F_{\alpha \nu})F^{\alpha \beta}$$ Still a bit unclear as to how to progress, thanks.

Not sure of all the things you did here. Especially the sign changes. Let's see. You had $$ F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$
And, for now, you just want to rename the dummy index ##\sigma## in the first term to something more appropriate. If that's all you do, then what do you get?
 
  • #5
TSny said:
Not sure of all the things you did here.
I used the second Maxwell equation to rearrange the ##\partial_{\alpha}F_{\nu \sigma}## term and collected terms by setting ##\sigma \rightarrow \beta##.
And, for now, you just want to rename the dummy index ##\sigma## in the first term to something more appropriate. If that's all you do, then what do you get?
This would give $$F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} - \frac{1}{2}(\partial_{\nu} F_{\alpha \beta})F^{\alpha \beta}$$ where I relabelled ##\sigma## to ##\beta##.
 
  • #6
CAF123 said:
I used the second Maxwell equation to rearrange the ##\partial_{\alpha}F_{\nu \sigma}## term and collected terms by setting ##\sigma \rightarrow \beta##.

OK. For me, it seems to work out quicker if I use the 2nd Maxwell equation on the term with the 1/2 factor.

This would give $$F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} - \frac{1}{2}(\partial_{\nu} F_{\alpha \beta})F^{\alpha \beta}$$ where I relabelled ##\sigma## to ##\beta##.

Now try the 2nd ME on the term with the 1/2.
 
  • #7
TSny said:
Now try the 2nd ME on the term with the 1/2.
Ok, so reexpress the second term as ##-\frac{1}{2} (-\partial_{\alpha}F_{\beta \nu} - \partial_{\beta} F_{\nu \alpha})## and altogether this gives $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\}.$$ Then interchange ##\alpha## and ##\beta## to give $$\frac{1}{2}F^{\beta \alpha} ( \partial_{\beta} F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta}) = - \frac{1}{2}F^{\alpha \beta} (\partial_{\beta}F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta})$$ So we have something like ##X=-X## and this conclude ##X=0## as required. Right?
 
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  • #8
CAF123 said:
Ok, so reexpress the second term as ##-\frac{1}{2} (-\partial_{\alpha}F_{\beta \nu} - \partial_{\beta} F_{\nu \alpha})## and altogether this gives $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\}.$$ Then interchange ##\alpha## and ##\beta## to give $$\frac{1}{2}F^{\beta \alpha} ( \partial_{\beta} F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta}) = - \frac{1}{2}F^{\alpha \beta} (\partial_{\beta}F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta})$$ So we have something like ##X=-X## and this conclude ##X=0## as required. Right?

Yes, that will get it. Good.

Or, note that if you contract ##F^{\alpha \beta}## with any tensor ##B_{\mu \nu \sigma}##, say, to get the expression ##F^{\alpha \beta}B_{\alpha \beta \nu}## , then due to the antisymmetry of ##F^{\alpha \beta}##, you get the identity $$F^{\alpha \beta}B_{\alpha \beta \nu} = -F^{\beta \alpha}B_{\alpha \beta \nu} = -F^{\alpha \beta}B_{\beta \alpha \nu}$$ where the last equality is just renaming dummy indices.

Thus interchanging the contracted indices on B in the expression ##F^{\alpha \beta}B_{\alpha \beta \nu}## just changes the overall sign of the expression.

You have $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\} = \frac{1}{2}F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}F^{\alpha \beta}\partial_{\beta}F_{\nu \alpha} $$ How does the second term compare to the first term?
 
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  • #9
TSny said:
You have $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\} = \frac{1}{2}F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}F^{\alpha \beta}\partial_{\beta}F_{\nu \alpha} $$ How does the second term compare to the first term?

The second term is minus the first term. ##\frac{1}{2}F^{\alpha \beta}\partial_{\beta}F_{\nu \alpha} = \frac{1}{2}F^{\beta \alpha}\partial_{\alpha}F_{\nu \beta} = -\frac{1}{2}F^{\alpha \beta}\partial_{\alpha}F_{\nu \beta}## and so the whole expression reduces to 0.
 
  • #10
Yes, that's right.
 

1. What are Maxwell's equations?

Maxwell's equations are a set of four fundamental equations that describe the behavior of electromagnetic fields. They were developed by physicist James Clerk Maxwell in the 19th century and are considered one of the most important achievements in the field of electromagnetism.

2. How are Maxwell's equations derived?

Maxwell's equations are derived from a combination of experimental observations, mathematical principles, and theoretical models. They were initially proposed by Maxwell based on his interpretation of the laws of electricity and magnetism, and have since been refined and expanded upon by other scientists.

3. What is the significance of Maxwell's equations?

Maxwell's equations have had a profound impact on our understanding and application of electricity and magnetism. They have been used to develop technologies such as radio, television, and wireless communication, and are the basis for many modern theories and experiments in physics.

4. Can Maxwell's equations be simplified?

While Maxwell's equations may seem complex, they can be simplified and applied to different situations depending on the specific variables and conditions involved. There are also simplified versions of the equations, such as the time-independent Maxwell's equations, which are commonly used in certain fields of study.

5. How are Maxwell's equations used in practical applications?

Maxwell's equations are used in a wide range of practical applications, including electronics, telecommunications, and electromechanical systems. They are also used in fields such as optics, acoustics, and quantum mechanics. Understanding and applying Maxwell's equations is essential for the development of new technologies and advancements in science.

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