pmb Posted May 25, 2012 Share Posted May 25, 2012 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. Link to comment Share on other sites More sharing options...

elfmotat Posted May 25, 2012 Share Posted May 25, 2012 It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m^{3} by dividing its energy density by c^{2}. In that case it would just be: [math]\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2[/math] 1 Link to comment Share on other sites More sharing options...

pmb Posted May 25, 2012 Author Share Posted May 25, 2012 (edited) It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m^{3} by dividing its energy density by c^{2}. In that case it would just be: [math]\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2[/math] 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 10^{3}. 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 May 25, 2012 by pmb Link to comment Share on other sites More sharing options...

juanrga Posted May 25, 2012 Share Posted May 25, 2012 (edited) 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. [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}[/math] Convention summation and c=1 Edited May 25, 2012 by juanrga Link to comment Share on other sites More sharing options...

elfmotat Posted May 25, 2012 Share Posted May 25, 2012 [math]\rho \equiv \sqrt{T^{00} - T^{i0}}[/math] 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. 1 Link to comment Share on other sites More sharing options...

juanrga Posted May 25, 2012 Share Posted May 25, 2012 (edited) 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 May 25, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted May 25, 2012 Author Share Posted May 25, 2012 [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}[/math] Convention summation and c=1 Hint: I'm staying out of this thread for most of it. But consider the following regarding your expression [math]\rho_i \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}[/math] Link to comment Share on other sites More sharing options...

juanrga Posted May 25, 2012 Share Posted May 25, 2012 [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}[/math] Convention summation and c=1 Hint: I'm staying out of this thread for most of it. But consider the following regarding your expression [math]\rho_i \equiv \sqrt{(T^{00})^2 - (T^{i0})^2}[/math] Thank you by the hint, but mass density is a scalar. There is no index i on scalars. Link to comment Share on other sites More sharing options...

elfmotat Posted May 26, 2012 Share Posted May 26, 2012 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: [math]\rho = \sqrt{(T^{00})^2-\eta_{ij}T^{i0}T^{j0}}[/math] Link to comment Share on other sites More sharing options...

juanrga Posted May 26, 2012 Share Posted May 26, 2012 I think the expression you're looking for is: [math]\rho = \sqrt{(T^{00})^2-\eta_{ij}T^{i0}T^{j0}}[/math] [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = (T^{i0})^2[/math] Link to comment Share on other sites More sharing options...

pmb Posted May 26, 2012 Author Share Posted May 26, 2012 It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m^{3} by dividing its energy density by c^{2}. In that case it would just be: [math]\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2[/math] Hint: What is the energy density of a magnetic field? Question: What units are you using? Link to comment Share on other sites More sharing options...

elfmotat Posted May 26, 2012 Share Posted May 26, 2012 (edited) [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = (T^{i0})^2[/math] [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \sum_i(T^{i0})^2 \neq (T^{i0})^2[/math] There is no implied summation in T^{i0} 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 [math]\frac{1}{2\mu_0}|\bold{B}|^2[/math]. Dividing by c^{2} yields what I have in my first reply. As for units, energy density (above) is in J/m^{3} while "mass" density (first post) is in kg/m^{3}. Edited May 26, 2012 by elfmotat 1 Link to comment Share on other sites More sharing options...

juanrga Posted May 26, 2012 Share Posted May 26, 2012 [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \sum_i(T^{i0})^2 \neq (T^{i0})^2[/math] There is no implied summation in T^{i0} by itself - "i" is still a free index. You either need an explicit summation, or you can do what I did with the metric. [math] \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 [/math] Using summation convention, this can be rewritten in a simpler form [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 = (T^{i0})^2[/math] Link to comment Share on other sites More sharing options...

elfmotat Posted May 26, 2012 Share Posted May 26, 2012 (edited) [math] \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 [/math] Using summation convention, this can be rewritten in a simpler form [math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 = (T^{i0})^2[/math] 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 [math](T^{i0})^2[/math]. For example: [math]A_\mu A^\mu [/math] implies a summation. [math]A_\mu A^\nu [/math] does not imply summation [math]A^\mu [/math] are the components of a vector. It clearly does not imply summation. Likewise, [math](T^{i0})^2[/math] does not imply summation. Edited May 26, 2012 by elfmotat 2 Link to comment Share on other sites More sharing options...

juanrga Posted May 26, 2012 Share Posted May 26, 2012 (edited) 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 [math](T^{i0})^2[/math]. You seem do not understand what [math] \delta_{ii}[/math] is. Start with [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 [/math] Now, using [math] \delta_{ii} = 1 [/math] [math] \sum_{i} \delta_{ii}(T^{i0})^2 = \sum_i 1 (T^{i0})^2 = \sum_i (T^{i0})^2 [/math] If you prefer to work with the summation convention the starting point is [math] \eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \delta_{ii}(T^{i0})^2 [/math] Using again [math] \delta_{ii} = 1 [/math] you obtain what I wrote [math] \delta_{ii}(T^{i0})^2 = 1 (T^{i0})^2 = (T^{i0})^2 [/math] But if you do not like my notation you can use your favourite notation, including your original [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} [/math] 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 May 26, 2012 by juanrga Link to comment Share on other sites More sharing options...

elfmotat Posted May 27, 2012 Share Posted May 27, 2012 (edited) You seem do not understand what [math] \delta_{ii}[/math] is. Start with [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 [/math] You can't sum with [math]\delta_{ii}[/math] 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: [math]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 ][/math] where F is the electromagnetic tensor, S is the Poynting vector, and: [math]\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}[/math] The energy density is given by T^{00 }(no other components contribute to the energy density - they represent energy and momentum flux), which in this case is: [math]T^{00}=\frac{1}{2}\left(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\right)[/math] Setting E=0 (it's a purely magnetic field) and dividing by c^{2} to get it in units of mass density gives: [math]\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2[/math] 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 May 27, 2012 by elfmotat 1 Link to comment Share on other sites More sharing options...

juanrga Posted May 27, 2012 Share Posted May 27, 2012 (edited) You can't sum with [math]\delta_{ii}[/math] 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. [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0} = \sum_{ij} \delta_{ij}T^{i0}T^{j0} = \sum_{i} \delta_{ii}(T^{i0})^2 [/math] The expression is well-defined and unambiguous. [math] \delta_{ii}[/math] 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 [math] \delta_{ii} = 1 [/math] 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)? [math] \sum_{ij} \eta_{ij}T^{i0}T^{j0}[/math] can mean both the sum over i and j or the ij element. You know what it means from context. The same happens with [math](T^{i0})^2[/math]. It can mean both the sum over i or the i element. From context, it is evident that [math](T^{i0})^2[/math] 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 c^{2} to get it in units of mass density gives: [math]\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2[/math] 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 [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = \sqrt{(T^{00})^2 - \sum_{ij} \delta_{ij}T^{i0}T^{j0}}[/math] 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 May 27, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted May 27, 2012 Author Share Posted May 27, 2012 [math]\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2[/math] 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! Link to comment Share on other sites More sharing options...

doG Posted May 27, 2012 Share Posted May 27, 2012 It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m^{3} by dividing its energy density by c^{2}. In that case it would just be: [math]\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2[/math] ...[math]\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2[/math] 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... Link to comment Share on other sites More sharing options...

elfmotat Posted May 27, 2012 Share Posted May 27, 2012 (edited) There seems to be a subtle difference here... I don't see it. Is there something obvious I'm missing? [math]B=|\bold{B}|[/math] if that's what you're worried about. Edited May 27, 2012 by elfmotat Link to comment Share on other sites More sharing options...

doG Posted May 27, 2012 Share Posted May 27, 2012 I don't see it. Is there something obvious I'm missing? [math]B=|\bold{B}|[/math] 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... Link to comment Share on other sites More sharing options...

ydoaPs Posted May 27, 2012 Share Posted May 27, 2012 I don't see it. Is there something obvious I'm missing? [math]B=|\bold{B}|[/math] if that's what you're worried about. I was thinking the same thing. Link to comment Share on other sites More sharing options...

pmb Posted May 27, 2012 Author Share Posted May 27, 2012 (edited) 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 May 27, 2012 by pmb Link to comment Share on other sites More sharing options...

granpa Posted May 28, 2012 Share Posted May 28, 2012 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. Link to comment Share on other sites More sharing options...

elfmotat Posted May 28, 2012 Share Posted May 28, 2012 I prefer the mine which in c=1 units and convention summation (i.e. sum over index i) is [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = \sqrt{(T^{00})^2 - \sum_{ij} \delta_{ij}T^{i0}T^{j0}}[/math] 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? Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now