Challenge to all - Mass Density of a Magnetic field

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People often ask me about when proper mass can't work as good as relativistic mass or even better. I wrote an entire paper to answer this question but people don't read it carefully enough. Probably because they believe that no matter what the paper says they've already made up their mind because the already thought about it carefully alread a long time ago. And that is a good reason. I'd probably do the same thing - Too much reason with no real expectations of changing what they think in any sape or form

So to take a shot at perhaps clarifying why physicists hold on to the notion of relativist mass. So I've decided to create a challenge for everyone. I have a SR text by Hans C. Ohanian. One of the homework problems is to find the mass density of a magnetic field. That's my challenge to you all. Solve this introductory level SR problem. Let's make it as simple as possible and assume that the magnetic field be uniform. Find the mass density of a the magnetic field. Use whatever definition of mass that you see fit.

Good luck.

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It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m3 by dividing its energy density by c2. In that case it would just be:

$\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2$

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It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m3 by dividing its energy density by c2. In that case it would just be:

$\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2$

Thank's for responding. re - It's tricky to know what exactly you're asking for. - I'm asking that people take a crack at solving this problem. It's the same question asked in the SR text I mentioned, i.e. from Special Relativity: A Modern Introduction by Hans C. Ohanian, Physics Curriculum & Instruction, 2001, page 149, problem 14.

The strongest magnetic fields produced in Laboratories, by explosive compression of magnetic field lines, is 103. In such a magnetic field, what is the energy density and what is the mass density.

So basically the question I posed is to solve this problem. Let's wait until we get multiple responses before I comment on the answers.

Of course, there is a good reason why I chose this problem and why I called it challenging.

Pete

ps - If you don't want to wait then let me know and I'll PM you my comments.

Edited by pmb
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People often ask me about when proper mass can't work as good as relativistic mass or even better. I wrote an entire paper to answer this question but people don't read it carefully enough. Probably because they believe that no matter what the paper says they've already made up their mind because the already thought about it carefully alread a long time ago. And that is a good reason. I'd probably do the same thing - Too much reason with no real expectations of changing what they think in any sape or form

So to take a shot at perhaps clarifying why physicists hold on to the notion of relativist mass. So I've decided to create a challenge for everyone. I have a SR text by Hans C. Ohanian. One of the homework problems is to find the mass density of a magnetic field. That's my challenge to you all. Solve this introductory level SR problem. Let's make it as simple as possible and assume that the magnetic field be uniform. Find the mass density of a the magnetic field. Use whatever definition of mass that you see fit.

Good luck.

$\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}$

Convention summation and c=1

Edited by juanrga
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$\rho \equiv \sqrt{T^{00} - T^{i0}}$

Convention summation and c=1

That expression doesn't make any sense. It's dimensionally inconsistent (due to the square root), and your indices don't match up.

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That expression doesn't make any sense. It's dimensionally inconsistent (due to the square root), and your indices don't match up.

I forgot the squares, but I have corrected it now. Thank you. You won a vote!

Edited by juanrga
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$\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}$

Convention summation and c=1

Hint: I'm staying out of this thread for most of it. But consider the following regarding your expression

$\rho_i \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}$

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$\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}$

Convention summation and c=1

Hint: I'm staying out of this thread for most of it. But consider the following regarding your expression

$\rho_i \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}$

Thank you by the hint, but mass density is a scalar. There is no index i on scalars.

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Thank you by the hint, but mass density is a scalar. There is no index i on scalars.

I think the expression you're looking for is:

$\rho = \sqrt{(T^{00})^2-\eta_{ij}T^{i0}T^{j0}}$

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I think the expression you're looking for is:

$\rho = \sqrt{(T^{00})^2-\eta_{ij}T^{i0}T^{j0}}$

$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = (T^{i0})^2$

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It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m3 by dividing its energy density by c2. In that case it would just be:

$\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2$

Hint: What is the energy density of a magnetic field?

Question: What units are you using?

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$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = (T^{i0})^2$

$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \sum_i(T^{i0})^2 \neq (T^{i0})^2$

There is no implied summation in Ti0 by itself - "i" is still a free index. You either need an explicit summation, or you can do what I did with the metric.

Hint: What is the energy density of a magnetic field?

Question: What units are you using?

The energy density of a magnetic field is $\frac{1}{2\mu_0}|\bold{B}|^2$. Dividing by c2 yields what I have in my first reply.

As for units, energy density (above) is in J/m3 while "mass" density (first post) is in kg/m3.

Edited by elfmotat
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$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \sum_i(T^{i0})^2 \neq (T^{i0})^2$

There is no implied summation in Ti0 by itself - "i" is still a free index. You either need an explicit summation, or you can do what I did with the metric.

$\sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 = \sum_i (T^{i0})^2$

Using summation convention, this can be rewritten in a simpler form

$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 = (T^{i0})^2$

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$\sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 = \sum_i (T^{i0})^2$

Using summation convention, this can be rewritten in a simpler form

$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 = (T^{i0})^2$

You seem to have a misunderstanding about what the summation convention is. The summation convention implies a sum over a repeated upper+lower index. There is no summation implied by $(T^{i0})^2$.

For example:

$A_\mu A^\mu$ implies a summation.

$A_\mu A^\nu$ does not imply summation

$A^\mu$ are the components of a vector. It clearly does not imply summation.

Likewise, $(T^{i0})^2$ does not imply summation.

Edited by elfmotat
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You seem to have a misunderstanding about what the summation convention is. The summation convention implies a sum over a repeated upper+lower index. There is no summation implied by $(T^{i0})^2$.

You seem do not understand what $\delta_{ii}$ is. Start with

$\sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2$

Now, using $\delta_{ii} = 1$

$\sum_{i} \delta_{ii}(T^{i0})^2 = \sum_i 1 (T^{i0})^2 = \sum_i (T^{i0})^2$

If you prefer to work with the summation convention the starting point is

$\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2$

Using again $\delta_{ii} = 1$ you obtain what I wrote

$\delta_{ii}(T^{i0})^2 = 1 (T^{i0})^2 = (T^{i0})^2$

But if you do not like my notation you can use your favourite notation, including your original

$\sum_{ij} \eta_{ij}T^{i0}T^{j0}$

Although, the mine is less ambiguous. The yours only gives the correct value when using the trace -2 convention. Whereas my notation is valid with independence of the trace convention.

My notation is also simpler and you would understand that the summation convention was invented to write less stuff and simplify the notation.

Edited by juanrga
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You seem do not understand what $\delta_{ii}$ is. Start with

$\sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2$

You can't sum with $\delta_{ii}$ in a valid tensor equation. An index can appear no more than twice in any term, or else it becomes unclear which index you're summing over.

Although, the mine is less ambiguous.

Your notation is far more ambiguous. For example, how would you express the i0'th component squared, but not summed over (i is a free index)?

The yours only gives the correct value when using the trace -2 convention.

The what convention?

My notation is also simpler and you would understand that the summation convention was invented to write less stuff and simplify the notation.

You sacrifice clarity for brevity.

This topic has become sufficiently derailed at this point, so I won't be responding (in this thread) to any more points about notation. Feel free to PM me if you'd like to continue this conversation.

Anyway, the SET for an electromagnetic field is:

$T^{\mu\nu} = \frac{1}{\mu_0} \left[ F^{\mu \alpha}F^\nu_{\ \alpha} - \frac{1}{4} \eta^{\mu\nu}F_{\alpha\beta} F^{\alpha\beta}\right]=\left [\begin{matrix} \frac{1}{2}\left(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\right) & S_x/c & S_y/c & S_z/c \\S_x/c & -\sigma_{xx} & -\sigma_{xy} & -\sigma_{xz} \\S_y/c & -\sigma_{yx} & -\sigma_{yy} & -\sigma_{yz} \\S_z/c & -\sigma_{zx} & -\sigma_{zy} & -\sigma_{zz} \end{matrix} \right ]$

where F is the electromagnetic tensor, S is the Poynting vector, and:

$\sigma_{ij} = \epsilon_0 E_i E_j + \frac{1}{{\mu _0 }}B_i B_j - \frac{1}{2} \left( \epsilon_0 E^2 + \frac{1}{\mu _0}B^2 \right)\delta _{ij}$

The energy density is given by T00 (no other components contribute to the energy density - they represent energy and momentum flux), which in this case is:

$T^{00}=\frac{1}{2}\left(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\right)$

Setting E=0 (it's a purely magnetic field) and dividing by c2 to get it in units of mass density gives:

$\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2$

which is what I had in my original post. Of course, should one choose to boost to another frame then things would get a lot messier and other components of T (from the original frame) would contribute.

Edited by elfmotat
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You can't sum with $\delta_{ii}$ in a valid tensor equation. An index can appear no more than twice in any term, or else it becomes unclear which index you're summing over.

$\sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2$

The expression is well-defined and unambiguous. $\delta_{ii}$ has only one index, because the second i is the same than the first i. You seem to believe that there are two i, but there is only one. Moreover, if you use $\delta_{ii} = 1$ and sum you will get that the right hand side is equivalent to the left hand side.

Your notation is far more ambiguous. For example, how would you express the i0'th component squared, but not summed over (i is a free index)?

$\sum_{ij} \eta_{ij}T^{i0}T^{j0}$ can mean both the sum over i and j or the ij element. You know what it means from context.

The same happens with $(T^{i0})^2$. It can mean both the sum over i or the i element. From context, it is evident that $(T^{i0})^2$ did mean the sum over 1...3, because otherwise what i you choose? The 1? The 2? The 3?

The what convention?

Your expression only agrees with mine when using the trace -2 convention. Mine works for both conventions.

You sacrifice clarity for brevity.

Both are subjective terms. But as said before if you do not like my notation use your favourite one!

Setting E=0 (it's a purely magnetic field) and dividing by c2 to get it in units of mass density gives:

$\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2$

which is what I had in my original post. Of course, should one choose to boost to another frame then things would get a lot messier and other components of T (from the original frame) would contribute.

I prefer the mine which in c=1 units and convention summation (i.e. sum over index i) is

$\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = \sqrt{(T^{00})^2 - \sum_{ij} \delta_{ij}T^{i0}T^{j0}}$

This also gives the energy in the general case when the electric field is not zero and coincides with the fundamental definition used in particle physics when the above is applied to matter in a quantum context.

Edited by juanrga
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$\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2$

which is what I had in my original post. Of course, should one choose to boost to another frame then things would get a lot messier and other components of T (from the original frame) would contribute.

Quite interesting! I've never see it written that way and didn't recognize it. Thanks!

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It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m3 by dividing its energy density by c2. In that case it would just be: $\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2$

...$\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2$ which is what I had in my original post. Of course, should one choose to boost to another frame then things would get a lot messier and other components of T (from the original frame) would contribute.

There seems to be a subtle difference here...

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There seems to be a subtle difference here...

I don't see it. Is there something obvious I'm missing? $B=|\bold{B}|$ if that's what you're worried about.

Edited by elfmotat
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I don't see it. Is there something obvious I'm missing? $B=|\bold{B}|$ if that's what you're worried about.

That's it. I was just clarifying that for others following the thread since it's had over 240 views thus far...

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I don't see it. Is there something obvious I'm missing? $B=|\bold{B}|$ if that's what you're worried about.

I was thinking the same thing.

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This could go on forever. I don't see a day that where everyone would decide on when they've arrived at a solutio. I'm not sure when to explain the solution.

Perhaps I could point to a relativity text which would give you a hint to the solution if you'd like. Here is a list of the ones I have.

http://home.comcast.net/~peter.m.brown/ref/physics_textbooks.htm

I could also point to an article in the American Journal of physcs which has a similar problem with the electric field. They won't give the answer but it should head you off to the right direction.

Do any of you have the text Gravity From The Ground Up by Bernard F. Schutz?

When would you like to see the answer, if at all?

Edited by pmb
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http://en.wikipedia.org/wiki/Magnetar

Magnetars are primarily characterized by their extremely powerful magnetic field, which can often reach the order of ten gigateslas.

A 10 gigatesla field...has an energy density of 4.0×10²⁵ J/m³, with an E/c² mass density > 10,000 times that of lead.

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I prefer the mine which in c=1 units and convention summation (i.e. sum over index i) is

$\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = \sqrt{(T^{00})^2 - \sum_{ij} \delta_{ij}T^{i0}T^{j0}}$

This also gives the energy in the general case when the electric field is not zero and coincides with the fundamental definition used in particle physics when the above is applied to matter in a quantum context.

I forgot to ask before, why do you think that energy flux would contribute to the energy density?

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