# Quick question on "free indices" - four-vector notation

1. Feb 16, 2015

### rwooduk

We've just started with this new notation in class and I'm just trying to get a little more insight into the basics so have a few very basic questions if someone here can help?

1)

dxμ = (dt,dx,dy,dz) I understand thats it's just a four-vector with four componants but, what would
dxμ be equal to? the same thing?

2) FREE INDICES what does this term mean? free to do what? i.e.

"AμAμ is a scalar because you have no free indices"

Do the upper and lower indices pair?

similarly:

"gμvgμv is also a scalar as it has no free indices"

also

"AμBv has two free indices"

Why because they are not paired, what does all this mean?

Thanks for any points in the right direction.

2. Feb 16, 2015

### Orodruin

Staff Emeritus
No. Depending on metric convention you would get a minus sign in the time or space components. Note that it is not particularly common to see this when it comes to the differential $dx^\mu$, you will see it more with general vectors.
It essentially means that the index is not being summed over and you will have one relation for every possible combination of free indices. According to the Einstein summation convention, paired indices should be summed over and free indices typically only appear once (there are exceptions when the summation convention gets in the way, these will typically be explicitly pointed out).

In order to construct scalars, you need to pair an upper idex with a lower one. This has to do with how contravariant and covariant vectors transform under coordinate changes.

In general, the position of a free index tells you the transformation properties and an object with n free indices should be viewed as the components of a tensor of rank n within the chosen basis. Often you will see these components referred to as the tensor itself, which is ok as long as the basis chosen is clear.

3. Feb 16, 2015

### rwooduk

Thanks thats very helpful!

please could you explain the term "relation", i.e. relation of what to what?

ahh ok that makes sense!

thanks again

4. Feb 16, 2015

### Orodruin

Staff Emeritus
Generally you will have not only expressions, but relations with two things on different sides of an equal sign (or other type of relation). If this is the case, it should hold regardless of what values you put into the free indices. Also, you must have the same free indices on both sides of the relation. An example would be taking the equation $\vec v = \nabla \phi$ on component form, which is written
$$v_i = \partial_i \phi,$$
which must hold for all i, i.e., for all components.

5. Feb 16, 2015

### bcrowell

Staff Emeritus
This is only in the special case where the metric looks like (1,-1,-1,-1) or (-1,1,1,1). The OP is probably taking a course on GR, where the metric doesn't necessarily look like that. (Even in SR, it's only in special coordinates that we get a metric of this form.)

6. Feb 17, 2015

### Orodruin

Staff Emeritus
This is of course true, but I have a hard time imagining someone getting through SR and still considering 4-vector notation as something they just started with.

7. Feb 17, 2015

### rwooduk

I see, thanks again for your time

8. Feb 19, 2015

### rwooduk

one more step I need help with if anyone can oblige:

$F^{12}= \delta ^{1}A^{2} - \delta ^{2}A^{1} = -\delta _{1}A^{2} + \delta _{2}A^{1}$

but then he uses:

$B_{i} = \varepsilon _{ijk} \delta _{j}A_{k}$

to get:

$F^{12}= \delta ^{1}A^{2} - \delta ^{2}A^{1} = -\delta _{1}A^{2} + \delta _{2}A^{1}= -B_{3}$

I dont understand how he has used that identity to get it equal to -B3, and why would there be a 3?

thanks for any further help

9. Feb 19, 2015

### Orodruin

Staff Emeritus
The $\epsilon_{ijk}$ is the permutation symbol. It is totally antisymmetric and one for $ijk = 123$. Since you have 1 and 2 in the F, you only have the 3 left for a non-zero component of $\epsilon$.

10. Feb 19, 2015

### rwooduk

thanks that really helps! one more if you have time (if not it's okay)

he writes

$\Psi = R exp (-i\omega t + ik_{i}x_{i})$

as

$\Psi = R exp (-i\omega t + ik_{i}x_{i}) = R exp (-ik^{\mu }x_{\mu })$

any explanation of how he has transformed it to covarient form would be really helpful, as I just cant see it.

thanks again

11. Feb 19, 2015

### Orodruin

Staff Emeritus
$k^\mu$ is a 4-vector with the components $(\omega, k_1, k_2, k_3)$. Try taking the inner product with the 4-vector $x^\mu$, which has components $(t,x_1,x_2,x_3)$.

12. Feb 19, 2015

### rwooduk

Awesome thanks very much, will try that and keep practising! it will sink in eventually!

Thanks again!

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