Do gravitational fields depend on relativistic mass, or just rest mass?

[Sobeita]

I was reading the Wikipedia page on "Mass in Special Relativity" (http://en.wikipedia.org/wiki/Mass_in_special_relativity) and I came across two equations:

M = m/sqrt(1-v²/c²)
and
p = mv/sqrt(1-v²/c²)

along with the following quote:

It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.

Einstein's comment seems to suggest that the perceived increase in mass is due to the increase in momentum. If mass itself were increasing, you would expect to see an increase in the gravitational field around particles moving at relativistic speeds. But that increase might be undetectable. Can anyone here tell me how scientists might be able to discern between the two possibilities, or better yet, if it's already been done?

 

[PeterDonis]

If mass itself were increasing, you would expect to see an increase in the gravitational field around particles moving at relativistic speeds.

The short answer is no, you don't, because the object's gravitational field depends on its rest mass, not its relativistic mass. Put another way, you can't change an object's gravitational field by changing your state of motion relative to the object.

The somewhat longer answer is that an object's gravitational field actually depends on something called the "stress-energy tensor", which is a more complicated mathematical object that takes into account the object's rest mass (more precisely its rest energy density), its momentum, and other things like pressure and internal stresses inside the object. The stress-energy tensor transforms when you change reference frames in such a way that the observed gravitational field of the object stays the same. So again, you can't change the object's observed gravitational field by changing your state of motion relative to the object.

The really long answer brings in a number of complications that I ignored in the shorter answers above. I don't know if you want to go into that level of detail.

 

[pervect]

We get this question a lot, it probably deserves a FAQ but certain aspects are tricky.

Let's start with trying to link the question to some actual measurements (even if they are thought measurement), rather than talking purely the abstract. Specifically, let's talk about how we'd actually measure the gravitational field of a moving object.

In the literature, we have doi:10.1119/1.14280, by Olson and Guarino, which suggests one way to do this, not necessarily unique.

f a heavy object with rest mass M moves past you with a velocity comparable to the speed of light, you will be attracted gravitationally towards its path as though it had an increased mass. If the relativistic increase in active gravitational mass is measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path, then we find, with this definition, that Mrel=γ(1+β2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not γM but is approximately 2γM.

Now, O & G have offered one definition of "active gravitational mass" that we can measure but it's not quite clear what theory this definition applies to. In Newtonian theory, "mass" causes gravity. In General relativity, as has been pointed out by previous posts, it's not mass (any sort of mass), but the stress energy tensor that causes gravity.

But rather that worry about the words at the moment, let's look at the meaning, as explained by the results of the thought experiment.

We can first ask - "is the disturbance of the test particles motion independent of the flyby velocity" And the answer is clear - it's not.

We can then ask "is the disturbance of the velocity better related to the energy of the flyby object". Note that energy is another name for "relativistic mass".

And the answer is, "Yes, it's better, but not perfect. Ther's still a factor of 1 + \beta^2 in there that we can't explain in this way.

So where this leaves us is that it's better to think of gravity as being caused by "energy" rather than mass, and thinking in these simple terms gets us with a factor of 2:1 of the actual answer predicted by General Relativity. But it doesn't get us ALL the way there. The full answer requires something that's more complex than just a single number to describe "the source of gravity" - that "more complex" something is the stress energy tensor. You can think of the Stress Energy Tensor as being composed of energy, momentum, and pressure. But to even start doing this, you need to abandon the notion that there's "something represented by one number" that "causes gravity", and start thinking "a bunch of numbers, representing energy and momentum and pressure cause gravity", which is the point of view of General Relativity.

Now that we've talked about the meaning, let's go back to some semantic issues (words). Olson's paper is informative, but his use of the term "active gravitational mass" never caught on, and for a good reason. It's still trying to create a single number out of something that really requires more than one number to represent.

Another semantic issue here is the one of "energy" vs "relativistic mass". I view the later as very outdated, in large part because it invites the erroneous idea that we can plug "relativistic mass" into Newton's law of gravity and get General Relativity.

Note that this was more or less assumed by the OP , without even thinking about it. This is one of several reasons I have for disliking the term "relativistic mass".

 

[Bill_K]

It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.

Einstein's comment seems to suggest that the perceived increase in mass is due to the increase in momentum.

No, he's just saying that they both increase. The energy of a relativistic particle is γmc², while its momentum is γmv.

The stress-energy tensor transforms when you change reference frames in such a way that the observed gravitational field of the object stays the same. So again, you can't change the object's observed gravitational field by changing your state of motion relative to the object.

Sorry, if the short, short answer is "no, it doesn't change", that's incorrect.

First, you have to recognize that "the" gravitational field is not a single number, it's a tensor with ten independent components, and the components transform differently when you change reference frames.

Secondly, a useful guide in answering this question is to ask the same question about the field of a moving charge. The electromagnetic field is a tensor F_μν, and the electric part of it is the three components F_i0. Under a Lorentz transformation, F_μν' = Λ_μaΛ_νbF_ab. If Λ is a boost along the z-axis,

\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&γ&γβ\\0&0&γβ&γ\end{array}\right)

Then E_x' = γE_x, E_y' = γE_y and E_z' = E_z. The transverse components of E get larger by a factor γ, while the longitudinal component does not.

In exactly the same way, the gravitational field is described by a tensor R_μνστ. For a stationary particle the relevant components are the "electric" part, R_i0j0. And under a Lorentz tranformation, it's easy to see that the transverse components grow: R_x0x0' = γ² R_x0x0, while the longitudinal component R_z0z0does not.

 

[PeterDonis]

Sorry, if the short, short answer is "no, it doesn't change", that's incorrect.

Well, I was unclear on exactly which question "no" is the short answer to; the question I was giving the short answer "no" to was something like "if you go too fast, do you become a black hole?". That is, if I have a spacetime containing a gravitating body, can I change the spacetime geometry generated by that body by changing my state of motion relative to it? The answer to that question is "no" (at least, it is assuming that I am a test object).

You and pervect are answering a different question, which I agree is a better way of addressing the issue the OP was probably trying to get at. The question you're answering is "can I change my *response* to the field of a gravitating body by changing my state of motion relative to it?" The answer to that question is "yes".

First, you have to recognize that "the" gravitational field is not a single number, it's a tensor with ten independent components, and the components transform differently when you change reference frames.

This is part of the complications I referred to, that I wasn't sure if the OP wanted to get into.

Secondly, a useful guide in answering this question is to ask the same question about the field of a moving charge. The electromagnetic field is a tensor F_μν, and the electric part of it is the three components F_i0. Under a Lorentz transformation, F_μν' = Λ_μaΛ_νbF_ab. If Λ is a boost along the z-axis,

\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&γ&γβ\\0&0&γβ&γ\end{array}\right)

Then E_x' = γE_x, E_y' = γE_y and E_z' = E_z. The transverse components of E get larger by a factor γ, while the longitudinal component does not.

Yes, but the boost also makes the magnetic part of the field nonzero (assuming it was zero in the original frame), and the effect of the magnetic part on a test charge opposes the effect of the increased transverse electric part, correct? Similar remarks would apply to boosting a gravitational field that is purely "electric" in the original frame.

 

[Bill_K]

Yes, but the boost also makes the magnetic part of the field nonzero (assuming it was zero in the original frame), and the effect of the magnetic part on a test charge opposes the effect of the increased transverse electric part, correct? Similar remarks would apply to boosting a gravitational field that is purely "electric" in the original frame.

That's the correct answer, but to an entirely different FAQ.

Question: Two charges side by side at rest have a certain Coulomb attraction. But viewed from a moving frame, their electric fields are greater. How can this be? Answer: In the moving frame they also produce magnetic fields which cancel the increased attraction.

In the present case, we're talking about the field of one moving particle. A stationary observer is not affected by the magnetic field produced by the particle, all he feels is the increased electric field.