# Does relativistic mass exhibit gravitiational effects?

Groundhog Day Update, 2014

The simple and dumb way to ask my main question is this: If something like a neutron start goes sailing by at very close to the speed of light, say fast enough to double its total mass-energy, do you "feel" the energy it is carrying as gravity as it passes by?

I'm about 99% sure the answer is yes, but I'd sure appreciate some confirmation of that. I was surprised (and appreciative) at the complexity of the first two attempts at answers, but must confess I thought it was a simpler problem than that.

For one thing, frame equivalence should not an issue, since in terms of gravitational effects it doesn't matter whether it was you or the star that doubled in mass. And since the mass used to speed up the star cannot simply disappear, its gravitational pull has to go somewhere, so why not in the star itself?

I now gather that such a simple argument gets very, very messy when expressed in tensor form. Ah... oops, but again, thanks. And I think that's what John Rennie answered...

Contingent on relativistic mass actually having gravitational effects, I think I can now answer my second question myself.

Imagine a self-contained cluster of masses within a compact region of space, all initially motionless relative to each other. At a large distances the gravitational behavior of the cluster will asymptotically approach that of a single large gravitational mass, one equal in magnitude to a simple sum of the individual masses in the cluster.

Next, have the cluster go crazy, all by itself and without any external stimulus. Huge parts of it are annihilated into pure energy, which in turn drives other parts outward at relativistic velocities. (If that sounds a bit outrageous, look up black hole jets sometime.) Since there are no external influences, the outbound parts must have momenta that sum to zero, with the simplest case being two equal-mass objects moving in opposite directions at the same velocities.

Now from a great enough distance, the cluster will continue to look asymptotically close to a single mass that is still equal to the sum of the original cluster masses. To that distant observer, the conversion of huge chunks of the cluster into pure momentum makes not a whit of difference: The cluster still has exactly the same mass-energy, with the only difference being that it is getting harder to approximate as a point.

So of course the few chunks of the original cluster that were never accelerated during the explosion will look a bit unique to the distant observer. In particular they will appear to have the lowest masses, since they did not acquire any of the converted mass energy during the explosion. Nothing profound, that... yet still interesting, especially as you broaden the argument to include larger and larger assemblages of mass-energy.

The black holes at the center of most (all?) galaxies would be examples of mass-gravity minimum frames, and their dual jets examples of entities with excess gravitation.

Oh, and one other point: Will stars or mass (large jets?) moving at relativistic speed exhibit higher levels of gravitational lensing? I would assume so...

My original more specific and inadvertently homework-like version of the above thought experiment is below.

Start with five large objects $\{m_a,m_b,m_c,m_d,m_e\}$ of equal mass $m$. Oddly, $m_b$ is made of antimatter.

$m_a$ remains unchanged at the center of mass of the group.

$m_b$ and $m_c$ are mutually annihilated. Their energy is used to launch $m_d$ and $m_e$ along ${\pm}x$ paths. The energy imparts to each a velocity as measured from $m_a$ of $(\sqrt{\frac{3}{4}})c$, which in turn gives $m_d$ and $m_e$ each a relativistic mass of $2m$, again as measured from $m_a$. The annihilated masses of $m_b$ and $m_c$ have in effect been "added" in the form of momentum to the rest masses of $m_d$ and $m_e$.

1. In terms of gravity, where in the resulting three-body system do the masses of $m_b$ and $m_c$ reside?

2. If you answered "$m_d$ and $m_e$", what happened to frame invariance?

Notes

A related (but definitely different) question is:

Does the increase of (relativistic) mass, while flying near speed of light, has any impact on astronauts?

I could not find any exact matches, but I also cheerfully acknowledge that my question search skills are not as good as some on this group.

• This looks like a homework problem. Please reword the problem so as to ask a conceptual question rather than an answer to the HW problem. If required, show us what work you have done to answer this problem and where you are stuck. Jan 7, 2014 at 21:35
• It kind of does read like a homework problem. Terry, perhaps you could edit it to highlight the underlying conceptual issue you're trying to get at? Jan 7, 2014 at 22:35
• That is delightful! It never even crossed my mind that this one might be interpreted as a homework question, but I guess I did get a bit specific in all the terms. I'll edit the question to see if I can clarify my intent. Feb 2, 2014 at 5:44

I'm guessing at the underlying conceptual question, so ignore this if I've missed the target.

In GR we don't use mass but rather the energy density, though in most cases the two are simply related by Einstein's famous equation $E = mc^2$. More precisely the metric is a function of an object called the stress-energy tensor, $T$.

We write this tensor as a 4x4 matrix, and as usually written the top left element of this tensor, $T_{00}$, is the energy density. For a stationary object $T_{00}$ is therefore related to the rest mass and the other entries will be zero - nice and simple. For a moving object $T_{00}$ includes the energy of motion, and now the other entries in the matrix are non-zero.

The stress-energy tensor is an invariant, i.e. it's the same for all observers. However it's representation, i.e. the individual values of the entries in the matrix, are dependant on the coordinates your using. So for a static object only $T_{00}$ is non-zero, but if we choose a coordinate system where that object is moving then $T_{00}$ changes, but so do the other entries and overall the tensor remains the same.

The point of all this rambling is that when you ask where do the masses reside? this really means where in the stress-energy tensor do we write those masses?. And the answer is that it will depend on the coordinates you choose. The obvious choice for your problem is the rest frame of $m_a$, so the energy densities of $m_a$, $m_d$ and $m_e$ go in $T_{00}$ (in this frame $m_d$ and $m_e$ have a higher energy than $m_a$) and the momenta of $m_d$ and $m_e$ go in $T_{i0}$ and $T_{0j}$. If you choose for example the rest frame of $m_d$ then $T_{00}$ would still contain the energy densities of the three objects but now $m_a$ and $m_e$ have a higher energy than $m_d$, and the momenta entries contain the momenta of $m_a$ and $m_e$. The entries in the tensor would be different but it's still the same tensor - just written differently - so it doesn't violate frame invariance.

A quick footnote ...

... because it confused me at first: the stress-energy tensor is a function of position in spacetime. So when I casually talk about putting the energy density in as $T_{00}$ this means $T_{00}$ is zero at all spacetime points outside the object and equal to $\rho c^2$ for spacetime points inside the object. For many of the examples we study when learning GR the system is time invarient, so $T_{00}$ is not a function of time. In your example $T_{00}$ would be a function of time as well as space.

• Shouldn't the field actually increase (at least change) as seen by observer who sees the relativistic particle moving with constant speed? $T_{\mu \nu}=T_{0\mu}=cp_{\mu}$ for a relativistic particle. Since the energy-momentum tensor is the 'source' term of Einsteins equation and since $p_{\mu}=(\gamma m c,\gamma \overline{v}m)$ the gravity field as seen by an observer with respect to whom the particle moves at speed $v$ should be increasing or at least should be different form the rest frame of the particle where only $T_{00}≠0$? Jul 31, 2016 at 8:51
• No, and it's easy to understand why. We simply switch reference frames to one in which the mass is stationary and it's the observer who is moving. But in this frame the observer just orbits (presumably a hyperbolic orbit) the mass. But if you're orbiting a mass then the gravitational field of that mass doesn't increase as your orbital velocity gets faster. Jul 31, 2016 at 9:03
• Ok but take for instance the motion of an electron in a cyclotron. In this case the magnetic field should be increased in order to keep the electron from flying away. So the magnetic field, which is stationary sees some increase? So if I understand you correctly, if an ultra-relativistic cannonball flew near me, I would not feel and increased attraction. I don't really see that to be obvious. I understand that due to $g_{\mu\nu}p^{\mu}p^{\nu}=m^2$ being invariant the mass can't change, but I can't reconcile this with what I learned for Cyclotron resonance for instance. Aug 1, 2016 at 21:14
• @JohnRennie In this paper, which analyzes the orbit of Mercury, each mass in Newton's law of gravitation is multiplied by $\gamma^3$, thus apparently increasing the sun's gravitational field. Does this contradict that last sentence of your comment? Mar 17 at 9:15
• No, they are using the fact that in special relativity if you observe an accelerating object the acceleration you observe, $a'$, is related to the acceleration in the object's frame, $a$, by $a' = a/\gamma^3$. It is an interesting experiment to use this in Newton's law of gravitation and see how it compares with the result from GR, but that's all it is. Mar 17 at 9:25