# Does mass really increase with speed

PeterDonis
Mentor
2019 Award
The conclusion is that we need to take into account a force transformation with the result that indeed different forces are experienced, depending which frame we are in. For the case of charges this is explained away by introducing magnetic fields.
The *total* force experienced is the same; how it is taken to be split into "electric" and "magnetic" force is frame-dependent.

We should however be able to apply these same force transformations for the case stated in the OP, with the result that we have 2 different readouts on our balance.
No, the balance readout doesn't change. But different observers will want to explain it differently: one will say it's purely due to electric force; another will say it's partly electric and partly magnetic.

Any direct observable that tells you about gravity. For example, suppose I am "hovering" at a constant height above a gravitating body, and I want to know how much I have to accelerate--how hard I have to fire my rocket engines--to do so. The answer is simple: it's the 4-acceleration of my worldline, which is a scalar invariant.
That means this 4-acceleration is independent from the velocity of the body - even if it is so fast that we are inside our common Schwarzschild radius?

PeterDonis
Mentor
2019 Award
That means this 4-acceleration is independent from the velocity of the body - even if it is so fast that we are inside our common Schwarzschild radius?
What does "velocity so fast that we are inside our common Schwarzschild radius" mean? The Schwarzschild radius of a body does not depend on velocity.

atyy
Science Advisor
(1) There are numerical solutions for many-body systems (e.g., binary pulsars) which show the bodies orbiting each other, similar to the known Newtonian analytic solutions for the two-body problem. The difference is that GR includes the emission of gravitational waves, so the two bodies' orbits about each other are not constant; they slowly spiral inwards towards each other.

(2) For an extended system with nonzero stress-energy such as a perfect fluid, the stress-energy gravitates; e.g., an expanding FRW solution with zero cosmological constant has the expansion constantly slowing down, while a contracting FRW solution with zero cosmological constant has the contraction constantly speeding up. The expanding case corresponds to all the massive objects in the universe pulling on each other and slowing the expansion down.
Let's take the FRW case. Why doesn't one say the expansion is due to the gravity of stress-energy?

Also, this gravity is defined via a preferred set of observers, which is ok. But then would a possible answer to the OP be that "the attraction of gravity between distant objects" isn't defined for arbitrary observers?

Another idea for a possible answer is that the relative motion of distant objects is not defined. Only objects at the same event can "move" relative to each other. But we surely don't expect objects at the same event to have a gravitational attraction between themselves.

Wouldn't this be equivalent to a solution where two massive objects are at rest relative to each other? Or do you mean moving on parallel worldlines but in opposite directions?
Yes. What are these like? A charged black hole of some sort?

Last edited:
The *total* force experienced is the same; how it is taken to be split into "electric" and "magnetic" force is frame-dependent.
I quote here from the above mentioned book:
Fy=Fy’/γ
In this example the 2 forces are not the same. Have you got a copy of that book?
No, the balance readout doesn't change. But different observers will want to explain it differently: one will say it's purely due to electric force; another will say it's partly electric and partly magnetic.
The second part of my previous post referred back to the OP where we just got 2 uncharged rockets. Therefore there’s no need to mention em fields. Nevertheless we should still apply the force transformations. (Or so I believe).

What does "velocity so fast that we are inside our common Schwarzschild radius" mean?
$r < \frac{{2 \cdot \gamma }}{{c^2 }}\sqrt {\left( {\frac{{m_1 \cdot c}}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot c}}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 - \left( {\frac{{m_1 \cdot v_1 }}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot v_2 }}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 }$

The Schwarzschild radius of a body does not depend on velocity.
The common Schwarzschild radius of two bodies does (see above). In the rest frame of the center of mass it is proportional to the sum of the relativistic masses.

pervect
Staff Emeritus
Science Advisor
PeterDonis said:
Per Oni said:
The conclusion is that we need to take into account a force transformation with the result that indeed different forces are experienced, depending which frame we are in. For the case of charges this is explained away by introducing magnetic fields.
The *total* force experienced is the same; how it is taken to be split into "electric" and "magnetic" force is frame-dependent.
I suspect that the cause of this disagreement is that PeterDonis is using the four-force, the rate of change of 4-momentum with respect to proper time. Because this transforms as a tensor via the Lorentz transformations, it can be and is be considered to be independent of the frame of reference in the context of special relativity.

I believe Per Oni is using the traditional concept of force, the "three-force", which is the rate of change of 3-momentum with coordinate time rather than proper time. This is not frame independent, because coordinate time isn't frame independent, so it does not transform as a tensor in special relativity and ca not be considered to be independent of the frame of reference.

PeterDonis
Mentor
2019 Award
$r < \frac{{2 \cdot \gamma }}{{c^2 }}\sqrt {\left( {\frac{{m_1 \cdot c}}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot c}}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 - \left( {\frac{{m_1 \cdot v_1 }}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot v_2 }}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 }$

The common Schwarzschild radius of two bodies does (see above). In the rest frame of the center of mass it is proportional to the sum of the relativistic masses.
Do you have a reference for this formula?

PeterDonis
Mentor
2019 Award
I suspect that the cause of this disagreement is that PeterDonis is using the four-force, the rate of change of 4-momentum with respect to proper time. Because this transforms as a tensor via the Lorentz transformations, it can be and is be considered to be independent of the frame of reference in the context of special relativity.

I believe Per Oni is using the traditional concept of force, the "three-force", which is the rate of change of 3-momentum with coordinate time rather than proper time. This is not frame independent, because coordinate time isn't frame independent, so it does not transform as a tensor in special relativity and ca not be considered to be independent of the frame of reference.
I agree this may well be why we are saying different things, but the reason I think about this in terms of the 4-force is that the 4-force is what is directly observable as, for example, the "readout on a balance". The 3-force is *not*. So the fact that the 3-force is frame dependent does *not* mean the actual observable, such as a balance readout, is frame dependent.

It could also be that the two "forces" Per Oni is referring to are different because they refer to different observables: for example, the actual force felt by observers moving on different worldlines. If you change the conditions of the scenario, obviously you can change the predicted observations.

PeterDonis
Mentor
2019 Award
Let's take the FRW case. Why doesn't one say the expansion is due to the gravity of stress-energy?
Do you mean why *does* one say this? I am saying the (rate of change of the) expansion *is* due to the gravity of stress-energy.

Also, this gravity is defined via a preferred set of observers
No, it isn't. The redshift of Galaxy X observed at Earth is a physical observable; anyone who calculates it will calculate the same answer. And the Earth is not at rest in the "comoving" frame, so it's not a preferred observer anyway. The point is that the *change with time* of a particular galaxy's redshift, if it is far enough away not to be gravitationally bound to our local cluster of galaxies, is an observable that tells us about the gravity of the universe as a whole, i.e., the combined gravity of all the objects in it.

Another idea for a possible answer is that the relative motion of distant objects is not defined.
I don't have to define this to make the argument I'm making. I don't have to interpret the redshift as actual "relative motion"; I only have to interpret its change over time as "deceleration".

Yes. What are these like? A charged black hole of some sort?
Not sure which of my "what ifs" you were responding to here. I am not aware of any particular solutions for masses moving on parallel worldlines but in opposite directions. If you mean the Chazy-Curzon vacuum, no, the two massive objects in it are not charged. I was hoping to find a good quick reference about it online, but I haven't.

atyy
Science Advisor
Do you mean why *does* one say this? I am saying the (rate of change of the) expansion *is* due to the gravity of stress-energy.
Well, I was asking why it seemed that the rate of change of expansion, rather than the expansion itself was called "gravity".

No, it isn't. The redshift of Galaxy X observed at Earth is a physical observable; anyone who calculates it will calculate the same answer. And the Earth is not at rest in the "comoving" frame, so it's not a preferred observer anyway. The point is that the *change with time* of a particular galaxy's redshift, if it is far enough away not to be gravitationally bound to our local cluster of galaxies, is an observable that tells us about the gravity of the universe as a whole, i.e., the combined gravity of all the objects in it.

I don't have to define this to make the argument I'm making. I don't have to interpret the redshift as actual "relative motion"; I only have to interpret its change over time as "deceleration".
Anyway, I wasn't referring to the redshifts, but to the idea that expansion is "gravity". The expansion is defined relative to a set of preferred observers.

My suggestion about relative motion not being defined was not related to any of those. It was trying to think of a completely different way of answering the OP question.

Not sure which of my "what ifs" you were responding to here. I am not aware of any particular solutions for masses moving on parallel worldlines but in opposite directions. If you mean the Chazy-Curzon vacuum, no, the two massive objects in it are not charged. I was hoping to find a good quick reference about it online, but I haven't.
I was asking about the one with two massive objects stationary relative to each other.

PeterDonis
Mentor
2019 Award
Well, I was asking why it seemed that the rate of change of expansion, rather than the expansion itself was called "gravity".
Because the expansion itself is just due to an initial impulse that got applied to all the matter in the universe. (We're talking about the standard, zero cosmological constant FRW model, so we're leaving out stuff like inflation and dark energy.) Without gravity, the expansion would just keep on at the same rate forever; the limiting case of the FRW model with zero stress-energy tensor does exactly this. It's the *change* in the expansion, the fact that it slows down because there is nonzero stress-energy present, that indicates gravity.

The expansion is defined relative to a set of preferred observers.
This is true in the sense that the usual meaning of the phrase "the universe is expanding" is that the "comoving" observers are all moving away from each other (they all see each other as having a redshift). But those observers are not picked out arbitrarily: they are picked out by the fact that they see the universe as homogeneous and isotropic. In other words, their worldlines, and the 3-surfaces orthogonal to them, match up with a particular symmetry of the spacetime. So actually, the fact that the universe is expanding is a property of the spacetime, not just of the observers.

I was asking about the one with two massive objects stationary relative to each other.
Ah, ok. If I find any good online source on the Chazy-Curzon vacuum I'll post a link.

atyy
Science Advisor
Because the expansion itself is just due to an initial impulse that got applied to all the matter in the universe. (We're talking about the standard, zero cosmological constant FRW model, so we're leaving out stuff like inflation and dark energy.) Without gravity, the expansion would just keep on at the same rate forever; the limiting case of the FRW model with zero stress-energy tensor does exactly this. It's the *change* in the expansion, the fact that it slows down because there is nonzero stress-energy present, that indicates gravity.
I see, so you count even the zero stress-energy tensor FRW case as expanding. But isn't that spacetime flat, so although it could be described as expanding, those observers are no longer reflecting such a particular symmetry of the spacetime, are they?

This is true in the sense that the usual meaning of the phrase "the universe is expanding" is that the "comoving" observers are all moving away from each other (they all see each other as having a redshift). But those observers are not picked out arbitrarily: they are picked out by the fact that they see the universe as homogeneous and isotropic. In other words, their worldlines, and the 3-surfaces orthogonal to them, match up with a particular symmetry of the spacetime. So actually, the fact that the universe is expanding is a property of the spacetime, not just of the observers.
Agreed. It's basically a matter of convention whether one says the expansion is observer-dependent or not. I was picking the former.

Ah, ok. If I find any good online source on the Chazy-Curzon vacuum I'll post a link.
Thanks. At least now I know what to google.

I guess the odd thing in GR is that one could try to say all questions are meaningless unless they talk about gauge invariant quantities. OTOH, we can make any gauge variant quantities gauge invariant by invariantly specifying events and worldlines and frames. The latter could be a bit restricted by saying that we only allow worldlines and frames that are not test particles, ie. their stress-energy must be accounted for in the stress-energy tensor in the Einstein field equations. But that's probably too strong, because even if we could do that, those observers would be left without light beams and test particles to probe their spacetime.

PeterDonis
Mentor
2019 Award
I see, so you count even the zero stress-energy tensor FRW case as expanding. But isn't that spacetime flat, so although it could be described as expanding, those observers are no longer reflecting such a particular symmetry of the spacetime, are they?
Yes, you're right, I was speaking loosely. The case where the stress-energy tensor is exactly zero is just Minkowski spacetime with a weird coordinate chart. But if we look at a case with non-zero stress-energy tensor, and gradually let the density approach zero, we find that the deceleration goes to zero. So viewed as a limiting case, the "zero energy, zero deceleration" case just shows that it's the deceleration that indicates gravity, not the expansion itself.

Agreed. It's basically a matter of convention whether one says the expansion is observer-dependent or not. I was picking the former.
No real argument, but it's worth noting that the fact that the set of comoving observers exists and that they all see the universe as expanding--i.e., the fact that it is possible to find a frame in which the universe is expanding everywhere--places limits on what observers in other states of motion can see as well. For instance, there is no observer who will see the universe contracting everywhere (assuming that all observers have the same direction of time). In fact, I believe there's no observer who will even see a "preponderance" of contraction over expansion (but I'll have to think some more to formulate that intuitive guess more precisely).

I guess the odd thing in GR is that one could try to say all questions are meaningless unless they talk about gauge invariant quantities. OTOH, we can make any gauge variant quantities gauge invariant by invariantly specifying events and worldlines and frames.
I agree, in the sense that specifying worldlines and frames specifies *which* particular invariants you are talking about. However, I would *not* say that this counts as making a variant quantity into an invariant quantity. Or at least, I would not word it that way. This may be a matter of choice of words rather than physics, but I think it's important. I would say that what a particular observer actually observes, in the sense of observable numbers (such as redshifts) can always be specified in an invariant way--that is, it can always be specified in terms of *only* invariant quantities, i.e., only scalars (which may be formed by contracting covariant objects like vectors and tensors). What we sometimes call "frame-dependent" quantities can actually be specified in frame-independent terms; for example, the energy you observe an object as having is the contraction of its 4-momentum with your 4-velocity.

The latter could be a bit restricted by saying that we only allow worldlines and frames that are not test particles, ie. their stress-energy must be accounted for in the stress-energy tensor in the Einstein field equations. But that's probably too strong, because even if we could do that, those observers would be left without light beams and test particles to probe their spacetime.
Yes, I agree, this restriction would be way too strong.

PeterDonis
Mentor
2019 Award
That means this 4-acceleration is independent from the velocity of the body - even if it is so fast that we are inside our common Schwarzschild radius?
On re-reading I realized I may not have made clear the scenario I was describing, with an observer "hovering" above a gravitating body. By "hovering" I mean maintaining a constant height above the gravitating body--in other words, the relative velocity of the "hoverer" and the body is zero, and stays that way. So there is no "velocity" involved.

atyy
Science Advisor
I agree, in the sense that specifying worldlines and frames specifies *which* particular invariants you are talking about. However, I would *not* say that this counts as making a variant quantity into an invariant quantity. Or at least, I would not word it that way. This may be a matter of choice of words rather than physics, but I think it's important. I would say that what a particular observer actually observes, in the sense of observable numbers (such as redshifts) can always be specified in an invariant way--that is, it can always be specified in terms of *only* invariant quantities, i.e., only scalars (which may be formed by contracting covariant objects like vectors and tensors). What we sometimes call "frame-dependent" quantities can actually be specified in frame-independent terms; for example, the energy you observe an object as having is the contraction of its 4-momentum with your 4-velocity.
So is there such a thing as a gauge variant quantity in GR if we are always allowed to put test events in spacetime?

The only thing that comes to mind is Shapiro delay, but that's not really even gauge variant since it's a delay compared to a non-existent Newtonian trajectory. As I understand, the real observable in Shapiro delay is the logarithmic form.

PeterDonis
Mentor
2019 Award
So is there such a thing as a gauge variant quantity in GR if we are always allowed to put test events in spacetime?
Not sure what you mean here. I was trying to say that you don't even need the concept of "gauge variant" quantities at all; you can express everything in terms of invariants, even things that are often taken to be "frame dependent".

atyy
Science Advisor
Not sure what you mean here. I was trying to say that you don't even need the concept of "gauge variant" quantities at all; you can express everything in terms of invariants, even things that are often taken to be "frame dependent".
Yes. What I'm asking is whether we can even define "gauge variant" as something distinct from "gauge invariant".

PeterDonis
Mentor
2019 Award
Yes. What I'm asking is whether we can even define "gauge variant" as something distinct from "gauge invariant".
Hmm. We do talk about the independence of physical observables from the choice of coordinates in GR as being a kind of gauge invariance, by analogy with electromagnetism, but thinking about it I'm not sure if the analogy fully holds.

In electromagnetism there are certainly "gauge variant" quantities as distinct from "gauge invariant" ones: the potential $A_{u}$ is gauge variant, but the field tensor $F_{uv}$ is gauge invariant. The reason we say this is that we can change the potential by the gradient of a scalar, $A'_{u} = A_{u} + \partial_{u} \phi$, without changing any physical predictions; but the reason it doesn't change any physical predictions is that it doesn't change the field tensor $F_{uv} = \partial_{u} A_{v} - \partial_{v} A_{u}$ (because mixed partial derivatives commute), and the field tensor is what determines the physical predictions.

In the case of gravity, the analogue of a "gauge transformation", a change of coordinates, can change the components of the things that actually determine physical predictions, such as the electromagnetic field tensor $F_{uv}$; the reason it doesn't change the actual physical predictions is that those predictions are expressed as scalars, i.e., contractions of vectors and tensors, and those do not change with a coordinate transformation, even though the individual components do. So in this case, we could say that the vectors and tensors themselves are the "gauge variant" quantities, and only the scalars are "gauge invariant"; but "gauge variant" here has a different meaning than it did in the electromagnetism case, because the things we are calling "gauge variant" are directly used to make physical predictions, not indirectly as in the electromagnetism case.

timmdeeg
Gold Member
Yes, you're right, I was speaking loosely. The case where the stress-energy tensor is exactly zero is just Minkowski spacetime with a weird coordinate chart.
Talking about the empty universe there are two cases, the empty FRW universe (hyperbolic) and the Milne universe (flat Minkowski spacetime), which can be converted to each other by coordinate transformation. Interestingly the kinematic explosion model (Milne) is equivalent to the expansion model (FRW) in this case.

PeterDonis
Mentor
2019 Award
timmdeeg, I see this is your first post. Welcome to PhysicsForums!

which can be converted to each other by coordinate transformation.
This is another way of saying they are the *same* spacetime, just described by different coordinate charts. That is, they both describe exactly the same physics.

Interestingly the kinematic explosion model (Milne) is equivalent to the expansion model (FRW) in this case.
Yes, because they both describe the same spacetime and the same physics.

timmdeeg
Gold Member
Thanks for your welcome, Peter.

As to the question "Does mass really increase with speed?" my idea is, that if true, then the creation of black holes would be observer dependent. Therefore it can't be true. Perhaps this was already mentioned in one of the posts, I didn' read all so far.

$r < \frac{{2 \cdot \gamma }}{{c^2 }}\sqrt {\left( {\frac{{m_1 \cdot c}}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot c}}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 - \left( {\frac{{m_1 \cdot v_1 }}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot v_2 }}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 }$
Do you have a reference for this formula?
The distance of the bodies shall be below their common Schwarzschild radius:

$r < r_s$

The Schwarzschildradius depends on the rest mass according to

$r_s = \frac{{2 \cdot \gamma \cdot m}}{{c^2 }}$

The rest mass is related to total energy and total momentum according to

$E^2 = m^2 \cdot c^4 + p^2 \cdot c^2$

and total energy and total momentum depends on the rest masses and velocities of the two bodies according to

$E = \frac{{m_1 \cdot c^2 }}{{\sqrt {1 - \frac{{v_1^2 }}{{c^2 }}} }} + \frac{{m_2 \cdot c^2 }}{{\sqrt {1 - \frac{{v_2^2 }}{{c^2 }}} }}$

and

$p = \frac{{m_1 \cdot v_1 }}{{\sqrt {1 - \frac{{v_1^2 }}{{c^2 }}} }} + \frac{{m_2 \cdot v_2 }}{{\sqrt {1 - \frac{{v_2^2 }}{{c^2 }}} }}$

All together results in the formula above.

PeterDonis
Mentor
2019 Award
The Schwarzschild radius depends on the rest mass according to

$r_s = \frac{{2 \cdot \gamma \cdot m}}{{c^2 }}$
This is incorrect; there is no relativistic $\gamma$ factor in the formula for the Schwarzschild radius. The correct formula is

$$r_{s} = \frac{2 G m}{c^{2}}$$

where $G$ is Newton's gravitational constant.

As to the question "Does mass really increase with speed?" my idea is, that if true, then the creation of black holes would be observer dependent. Therefore it can't be true.
You are right, but that doesn't mean that gravitational mass does not increase with speed. It might depend on the circumstances.