Consider the following metric which is 5 dimensional (2-parameter) spherically symmetric Kaluza-Klein solution


where $x_5$ is the periodic fifth coordinate. In this paper (page 15) I read

the inertial mass of the star can be determined by the asymptotic behavior of this metric, assuming no interior singularities, and is equal to $M_{in}=\frac{(1+\beta)m}{\alpha}$

What should I understand as "inertial mass" of a star here? How can I see that this formula represents inertial mass?

In the same page, just a little below it says

the gravitational mass of the stars can be determined by the asymptotic form of $g_{00}$ and is given by $M_g=m/\alpha$

In this case, what should I understand as "gravitational mass" here? Again, how can I see that this formula represents gravitational mass?

  • $\begingroup$ Understanding the distinction between "inertial mass" and "gravitational mass" is quite literally the first thing you should learn before you even attempt to look at anything in general relativity. So, are you asking what those terms mean, or why they are equal to what is written in the paper? $\endgroup$ – Jim Jul 7 '14 at 20:22
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    $\begingroup$ @Jim no, this is a perfectly legitimate rather advanced technical question and not homework at all. $\endgroup$ – Dilaton Jul 7 '14 at 23:34
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    $\begingroup$ @Jim In lieu of putting this on hold, I've edited the question to remove the homework-like parts. I think that should take care of the issue? $\endgroup$ – David Z Jul 8 '14 at 0:15
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    $\begingroup$ @Jim: coming up with a clear distinction between the two in full general relativitiy is hardly trivial, considering that the two body problem in full generality is generally solved via supercomputer simulation. $\endgroup$ – Jerry Schirmer Jul 8 '14 at 23:00
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    $\begingroup$ There is an interesting discussion on these definitions of the inertial mass and gravitational mass in this paper $\endgroup$ – Trimok Jul 12 '14 at 12:19

Are they identifying inertial mass with the ADM mass?

There is an old style way of writing $g_{00}$ of a stationary metric, far from the matter source, as $-\left(1 - \frac{1}{2}\phi({\vec x})\right)$, where $\phi$ is the potential function for the metric, and then you can monopole expand this and identify the numerator of the $\frac{1}{r}$ term of the expansion with the "mass" of the body.${}^{1}$. This interpretation would explain the rest of their statements (as well as the dependence of the "gravitational mass" on $\alpha$ but not $\beta$). This is difficult to answer without more context, though.

${}^{1}$Myself, this metric is simple enough that I"d just look for geodesics with $\dot x_{5} = 0$, and do ordinary physics to find the "gravitational mass" of the object.

  • $\begingroup$ It would seem to me that looking at approximate form of $g_{00}$ would give $M = 4 m/\alpha$ with your formula, but again the relativistic formula would still give an extra factor of 2 as pointed out by Kyle. $\endgroup$ – Void Jul 7 '14 at 20:36
  • $\begingroup$ @Void: yeah, I'm much happier looking for circular orbits and matching periods to classical periods and using that method for describing the gravitational mass, which doesn't match the method described in the question, though. They likely are doing what Kyle's answer is -- comparing to schwarzschild in isotropic coordinates. That's ugly and unphysical, though. $\endgroup$ – Jerry Schirmer Jul 7 '14 at 20:39
  • $\begingroup$ Also, I think I should have written $1-\frac{1}{2}\phi(r)$, since it otherwise doesn't even work for Schwarzschild. The newtonian potential is, of course, $\frac{GM}{r}$. $\endgroup$ – Jerry Schirmer Jul 7 '14 at 20:45
  • $\begingroup$ Hi Jerry, thanks for the answer. I do not know if they identify it with ADM mass. They are not overly explicit in that. $\endgroup$ – Yossarian Jul 9 '14 at 12:31

Disclaimer: while I have a good grasp of GR fundamentals, it is not my area of expertise.

The gravitational mass is distinct from the inertial mass as follows. The gravitational mass defines how strongly the body curves space-time, qualitatively it answers the question "how strong is the gravitational force from this object?". The inertial mass defines how the body reacts to forces, qualitatively it answers the question "how will the body react to the application of a given force".

The metric you give is approximated to first order near $r\rightarrow\infty$ by:

$${\rm d}s^2 \approx -\left(1-\frac{4m}{\alpha r}\right){\rm d}t^2 + \left(1+\frac{4(\beta+1)m}{\alpha r}\right)\left({\rm d}r^2+r^2{\rm d}\Omega^2\right) + \left(1-\frac{4\beta m}{\alpha r}\right){\rm d}{x_5}^2$$

For the gravitational mass, there is a similarity with the Schwarzschild metric. Begin with the approximate (at $r\rightarrow\infty$) $g_{00}$:

$$g_{00} \approx -\left(1-\frac{4m}{\alpha r}\right)$$

Compare this to the Schwarzschild solution in isotropic coordinates, also expanded at $r\rightarrow\infty$:

$$ds^2_{\rm Scwarzschild} \approx -\left(1-\frac{2M}{r}\right){\rm d}t^2 + \left(1+\frac{2M}{r}\right)\left({\rm d}r^2+r^2{\rm d}\Omega^2\right)$$

$$g_{00,\rm Schwarzschild} = -\left(1-\frac{2M}{r}\right)$$

Notice that $M\leftrightarrow 2m/\alpha$. I'm not sure why there's a difference of a factor of $2$, but I think it may have to do with what Jerry Schirmer mentioned in his answer, with $g_{00} = -\left(1-\frac{1}{2}\phi(\vec{x})\right)$; they do mention (about the Schwarzschild metric):

[...] and the gravitational potential is $\frac{1}{2}(g_{00}+1)=\frac{M}{R}$.

For the inertial mass, there's a $4(\beta+1)m/\alpha r$ (the inertial mass they quote, and again a factor of $2$) in the approximation to the metric that I gave where you'd expect to find a $2M$ in the Schwarzschild case, this time in the spatial coordinate term.

If anyone can explain that factor of $2$, I'm curious...

  • $\begingroup$ This metric is not in an analogue of Schwarzschild coordinates. It's in an analogue of isotropic coordinates. Note that $g_{rr} \neq \frac{1}{-g_{tt}}$. That will explain your factor of two. $\endgroup$ – Jerry Schirmer Jul 7 '14 at 20:33
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Jerry Schirmer Jul 7 '14 at 20:34
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    $\begingroup$ @JerrySchirmer ah, I noticed that the Scwarzschild metric is usually given in Schwarzschild coordinates, and that this was probably important, but stopped there. Thanks, I think this clears things up a bit... going to see if I can sort out an edit. $\endgroup$ – Kyle Oman Jul 7 '14 at 20:36

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