- #1

- 1,681

- 3

direction of motion increases. But what about it's intertial mass

perpendicular to the direction of motion? Does it also increase?

And finally, what about the gravitational field "produced" by the

mass. Has it increased as well?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Antiphon
- Start date

- #1

- 1,681

- 3

direction of motion increases. But what about it's intertial mass

perpendicular to the direction of motion? Does it also increase?

And finally, what about the gravitational field "produced" by the

mass. Has it increased as well?

- #2

- 10,055

- 1,223

https://www.physicsforums.com/showpost.php?p=687759&postcount=71

or scrutinize the addendum to the sci.physics.faq

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

or just differentiatie the expression

p = m v / sqrt(1-(v/c)^2)

with respect to v.

Perpendicular to it's direction of motion, the "transverse mass" increases by a factor of gamma (same two references).

The gravitational field of a moving mass is not the same in all directions. It experiences a large increase when the motion is transverse, and no change when the motion is directly towards or away from the source.

I've done some calculations on the tidal force of a moving mass, which is easier to define in the strong field case. You can't measure the gravitational field of a moving object directly without an external reference frame to define what a "straight line" is, making the calculation a bit ambiguous, but you can measure the tidal force an object experiences directly with no ambiguity.

Early versions of this caclulation were flawed, but I think I'm getting the right answer this go-around. I've actually had this result for a while, but have been too busy to post it (also I thought I'd just wait until the question came up again, this is a question that comes up all the time).

The results in Schwarzschild coordinates are

For an unmoving mass, and in Schwarzschild coordinates, the tidal forces are

in the [itex]\hat{r}[/itex] direction 2m/r^3

in the [itex]\hat{\theta}[/itex] direction -m/r^3

in the [itex]\hat{\phi}[/itex] direction -m/r^3

For a mass moving directly towards or away from the source (i.e in the [itex]\hat{r}[/itex] direction), there is no change in tidal force. (This is documented in MTW's gravitation, for instance, I can find the page # if anyone is interested).

For a mass moving at right angles to the source (i.e. in the [itex]\hat{\theta}[/itex] direction), the tidal forces become

in the [itex]\hat{r}[/itex] direction (2m/r^3)*(1+v^2/2)/(1-v^2)

in the [itex]\hat{\theta}[/itex] direction -m/r^3 - no change!

in the [itex]\hat{\phi}[/itex] direction -(m/r^3)*(1+2*v^2)/(1-v^2)

Note that geometric units were used for the above caclulations, so factors of G and c have been replaced with '1'.

These results haven't been checked by another person yet, and there isn't any textbook reference for them that I've been able to find (except the MTW reference about how the tidal forces do not change when you move directly towards the source).

The good news is that the trace (sum of the digaonals, i.e. the three formula above) is zero, which is what one expects. The fact that the 'r' and 'phi' components behave differently is not surprising because Schwarzschild coordinates are not isotropic.

A rough outline of the calculation: calculate the Riemann tensor in the Schwarzschild basis (or in Schwarzschild coordinates and normalize the metric so that the basis vectors are all the same length), boost the Riemann by the velocity, and calculate the tidal forces via the geodesic deviation equation.

[sidenote: there is an induced rotation relative to the fixed stars due to the passage of a large mass, quite analogous to geodetic precession, but the centrifugal forces due to this induced rotation do not contribute to the geodesic deviation.]

One of these days I want to do a similar analysis with the energy pseudotensors, which will answer the original question a bit more directly. This should be a "gauss-law" sort of intergal of the "force-at-infinity" over a sphere. MTW's coverage of this is a bit murky, however.

- #3

- 1,681

- 3

Are you saying that a body in uniform motion (but at very high speed) would

experience static tidal forces? Or is this an effect connected with acceleration?

- #4

- 214

- 0

Pervect's analysis is flawed. Relativistic mass has no physical meaning. The definition is slightly retarded itself.Antiphon said:

Are you saying that a body in uniform motion (but at very high speed) would

experience static tidal forces? Or is this an effect connected with acceleration?

Pervect: I am willing to debate this issue with you as I am sure you disagree. However, you've yet to respond to any of my posts.

- #5

- 1,681

- 3

Relativistic mass has no physical meaning.

I'm prepared to accept this. But can you tell me this then: what happens to

the intertial mass as you approach c?

- #6

- #7

- 10,055

- 1,223

Antiphon said:

Are you saying that a body in uniform motion (but at very high speed) would

experience static tidal forces? Or is this an effect connected with acceleration?

I'm saying that if you have an observer stationary in space, set up to measure tidal forces, and a large (non-test) mass whizzes by it in at a large fraction of 'c', that the tidal force said observer measures will be larger than the tidal force that he would measure if the large mass was stationary (rather than moving rapidly).

The situation looks like this

observer.....................large-mass^

The large mass is moving up the page.

Note that if the large mass were moving either to the left, or right (directly towards or away from the obsever), the velocity would NOT affect the tidal force.

Last edited:

- #8

- 10,055

- 1,223

Aer said:Pervect's analysis is flawed. Relativistic mass has no physical meaning. The definition is slightly retarded itself.

Pervect: I am willing to debate this issue with you as I am sure you disagree. However, you've yet to respond to any of my posts.

I've been a bit busy recently, alas. I will try and post a more detailed analysis of exactly what I calculated and how I calculated it when I get a chance (I have to go now).

This is a full-blown GR caclulation, though, and it's rather messy, even with symbolic algebra assist.

- #9

- 10,055

- 1,223

Antiphon said:I'm prepared to accept this. But can you tell me this then: what happens to

the intertial mass as you approach c?

It's unclear how you are defining inertial mass. Both

http://arxiv.org/abs/physics/0309075

and

http://online.cctt.org/physicslab/content/Phy1/labs/inertialmass/inertialgravitational.asp

seem to suggest that inertial mass is measured using an "inertial balance", i.e. one measures the oscillation frequency of a spring-mass system. This is as opposed to measuring the gravitational mass, for instance by weighing an object in a gravitational field, or finding which mass balances the pans in a pan balance (which also requires gravity).

It's unclear how an inertial balance could be used to find the mass of a relativistically moving object.

- #10

- 10,055

- 1,223

OK, here is the detail of the tidal force calculation:

First, one assumes a Schwarzschild metric, and defines an orthonormal frame of basis vectors [itex]\hat{r},\hat{\theta},\hat{\phi},\hat{t}[/itex] at some specific point in space-time.

One then finds the components of the Riemann tensor at that point - this much is done in textbooks, see for instance MTW pg 281. I'm too lazy to list all the terms of the tensor in this post, but a sample term from MTW is

[tex]

R_{\hat{r}\hat{t}\hat{r}\hat{t}} = -2M/r^3

[/tex]

using MTW's sign conventions. (As I check my worksheet, I see I used the oppposite sign conventions, using the convenient schwb metric which is part of the library of metrics supplied with GRTensorII).

It is useful, at this point, to apply the geodesic deviation equation to find the known tidal forces due to a non-moving mass.

One computes

[tex]

R_{abcd} u^{b} u^{d}

[/itex]

where u is a unit vector in the time direction (representing a stationary observer). The result is a 4x4 diagonal matrix. The three nonzero coefficients of this matrix represent the tidal forces in the [itex]\hat{r},\hat{\theta},\hat{\phi}[/itex] directions.

One finds that there is a stretching component of the tidal force of magnitude 2M/r^3 in the [itex]\hat{r}[/itex] direction, and compressive compoents of magnitude m/r^3 in the [itex]\hat{\theta},\hat{\phi}[/itex] directions.

For a web example, see for instance

http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Tidal_Forces_%20In_%20A_%20Black_%20Hole.htm [Broken]

http://www.sissa.it/~rezzolla/lnotes/virgo/node6.html [Broken]

The first reference shows the method clearly, but has some extraneous factors of two. The second reference gets the right answer, but used the Geodesic equation rather than the simpler geodesic deviation equation.

The next step is to boost this Riemann to convert it to the orthonormal set of coordinates associated with a moving observer. A boost in the [itex]\hat{r}[/itex] direction leaves the Riemann unchanged (and hence the tidal forces unchanged). However, a boost in the [itex]\hat{\theta}[/itex] direction changes the Riemann, and is the case that gives the interesting results of some increased tidal force components.

The boost is done by computing

[tex]

R_{efgh} = R_{abcd} L^a{}_e L^b{}_f L^c{}_g L^d{}_h

[/tex]

where L is the transformation matrix representing the boost.

Finally one computes the tidal forces from the boosted Riemann

[tex]

R_{abcd} u^b u^d

[/itex]

These give the tidal force in the boosted frame, and are the results I presented.

As I mentioned before, it is true that the boosted frame is also a rotating frame. This may initially give one pause as to applying the equation of geodesic deviation to compute the tidal forces. However, in another series of posts I have calculated the Riemann of a rotating Minkowski frame and shown that it is zero at the origin of the rotation, so rotation does not affect the value of the Riemann at the origin. A close study of the diagrams of the the computation of the geodesic devaition equation in MTW also cause one to reach the same conclusion.

One may ask - why go to all the effort of boosting the Riemann? Why not just use a 4-velocity that represents the moving observer directly in the geodesic deviation equation?

This is _almost_ OK, but it has the problem that it does not take into account the fact that the orthonormal basis vectors of the moving observer are different than the orthonormal basis vectors of the non-moving observer. In this problem, one really wants the former, not the later.

Specifically, the [itex]\hat{\theta}[/itex] vector is different when the boost is in the [itex]\theta[/itex] direction.

First, one assumes a Schwarzschild metric, and defines an orthonormal frame of basis vectors [itex]\hat{r},\hat{\theta},\hat{\phi},\hat{t}[/itex] at some specific point in space-time.

One then finds the components of the Riemann tensor at that point - this much is done in textbooks, see for instance MTW pg 281. I'm too lazy to list all the terms of the tensor in this post, but a sample term from MTW is

[tex]

R_{\hat{r}\hat{t}\hat{r}\hat{t}} = -2M/r^3

[/tex]

using MTW's sign conventions. (As I check my worksheet, I see I used the oppposite sign conventions, using the convenient schwb metric which is part of the library of metrics supplied with GRTensorII).

It is useful, at this point, to apply the geodesic deviation equation to find the known tidal forces due to a non-moving mass.

One computes

[tex]

R_{abcd} u^{b} u^{d}

[/itex]

where u is a unit vector in the time direction (representing a stationary observer). The result is a 4x4 diagonal matrix. The three nonzero coefficients of this matrix represent the tidal forces in the [itex]\hat{r},\hat{\theta},\hat{\phi}[/itex] directions.

One finds that there is a stretching component of the tidal force of magnitude 2M/r^3 in the [itex]\hat{r}[/itex] direction, and compressive compoents of magnitude m/r^3 in the [itex]\hat{\theta},\hat{\phi}[/itex] directions.

For a web example, see for instance

http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Tidal_Forces_%20In_%20A_%20Black_%20Hole.htm [Broken]

http://www.sissa.it/~rezzolla/lnotes/virgo/node6.html [Broken]

The first reference shows the method clearly, but has some extraneous factors of two. The second reference gets the right answer, but used the Geodesic equation rather than the simpler geodesic deviation equation.

The next step is to boost this Riemann to convert it to the orthonormal set of coordinates associated with a moving observer. A boost in the [itex]\hat{r}[/itex] direction leaves the Riemann unchanged (and hence the tidal forces unchanged). However, a boost in the [itex]\hat{\theta}[/itex] direction changes the Riemann, and is the case that gives the interesting results of some increased tidal force components.

The boost is done by computing

[tex]

R_{efgh} = R_{abcd} L^a{}_e L^b{}_f L^c{}_g L^d{}_h

[/tex]

where L is the transformation matrix representing the boost.

Finally one computes the tidal forces from the boosted Riemann

[tex]

R_{abcd} u^b u^d

[/itex]

These give the tidal force in the boosted frame, and are the results I presented.

As I mentioned before, it is true that the boosted frame is also a rotating frame. This may initially give one pause as to applying the equation of geodesic deviation to compute the tidal forces. However, in another series of posts I have calculated the Riemann of a rotating Minkowski frame and shown that it is zero at the origin of the rotation, so rotation does not affect the value of the Riemann at the origin. A close study of the diagrams of the the computation of the geodesic devaition equation in MTW also cause one to reach the same conclusion.

One may ask - why go to all the effort of boosting the Riemann? Why not just use a 4-velocity that represents the moving observer directly in the geodesic deviation equation?

This is _almost_ OK, but it has the problem that it does not take into account the fact that the orthonormal basis vectors of the moving observer are different than the orthonormal basis vectors of the non-moving observer. In this problem, one really wants the former, not the later.

Specifically, the [itex]\hat{\theta}[/itex] vector is different when the boost is in the [itex]\theta[/itex] direction.

Last edited by a moderator:

- #11

- 10,055

- 1,223

I realize that there is one thing I need to slightly change in my description.

We start with a moving large mass, with an associated gravitational field, and a co-moving observer. The co-moving observer has Schwarzschild coordinates r, theta, and phi.

What also have a stationary observer.

What we are doing is comparing the two observers measurement of tidal force when they meet, i.e for transverse motion, it looks like this:

CASE "T" - transverse motion

state at t<0 (#1 comoving observer, #2 stationary observer)

#2

#1...............mass (observer #1 has Schwarzschild coords [itex]r, \theta, \phi[/itex]

Note that observer 2 starts out above observer 1

state at t=0

#1+2.............mass

observer #1 and #2 are at the same spot and compare their readings.

Observer 1 measures a radial stretching tidal force of 2M/r^3 (in geometric units), observer 2 measured a radial stretching tidal force of

[tex]\left(\frac{2m}{r^3}\right)\left(\frac{1+v^2/2}{1-v^2}\right)[/tex] (in geometric units).

**Note that the r-coordinate used in both expressions above is the Schwarzschild r-coordiantes of observer #1**

This matters more when the velocity is towards the mass

CASE "I" - INLINE MOTION

t<0

#2............#1..........mass

t=0

#1#2..........mass

Observers #1 and #2 both measure the same tidal forces, so I have said that the tidal force is not affected by the velocity. However, this could be misleading, because the distance that observer #2 measures to the mass is not the same as the distance that observer #1 measures to the mass.

If we were in flat space-time, we would say that the distance from observer #1 to the mass was Lorentz-contracted by a factor of gamma. In curved space-time this would be a questionable statement.

An instructive comparison can be made with the much simpler analysis of the electric field of a point charge. Here we can measure the electric field directly.

In case "T", the observer sees the electric field increase by a factor of [itex]\gamma = 1/\sqrt{1-v^2}[/itex] (in geometric units).

In case "I", the observer sees no change in the electric field.

see for instance

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_13.pdf

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

some of the early PHY2061 lecture notes are also useful, the base page for these notes is at

http://www.phys.ufl.edu/~rfield/PHY2061/2061_notes.html

We start with a moving large mass, with an associated gravitational field, and a co-moving observer. The co-moving observer has Schwarzschild coordinates r, theta, and phi.

What also have a stationary observer.

What we are doing is comparing the two observers measurement of tidal force when they meet, i.e for transverse motion, it looks like this:

CASE "T" - transverse motion

state at t<0 (#1 comoving observer, #2 stationary observer)

#2

#1...............mass (observer #1 has Schwarzschild coords [itex]r, \theta, \phi[/itex]

Note that observer 2 starts out above observer 1

state at t=0

#1+2.............mass

observer #1 and #2 are at the same spot and compare their readings.

Observer 1 measures a radial stretching tidal force of 2M/r^3 (in geometric units), observer 2 measured a radial stretching tidal force of

[tex]\left(\frac{2m}{r^3}\right)\left(\frac{1+v^2/2}{1-v^2}\right)[/tex] (in geometric units).

This matters more when the velocity is towards the mass

CASE "I" - INLINE MOTION

t<0

#2............#1..........mass

t=0

#1#2..........mass

Observers #1 and #2 both measure the same tidal forces, so I have said that the tidal force is not affected by the velocity. However, this could be misleading, because the distance that observer #2 measures to the mass is not the same as the distance that observer #1 measures to the mass.

If we were in flat space-time, we would say that the distance from observer #1 to the mass was Lorentz-contracted by a factor of gamma. In curved space-time this would be a questionable statement.

An instructive comparison can be made with the much simpler analysis of the electric field of a point charge. Here we can measure the electric field directly.

In case "T", the observer sees the electric field increase by a factor of [itex]\gamma = 1/\sqrt{1-v^2}[/itex] (in geometric units).

In case "I", the observer sees no change in the electric field.

see for instance

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_13.pdf

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

some of the early PHY2061 lecture notes are also useful, the base page for these notes is at

http://www.phys.ufl.edu/~rfield/PHY2061/2061_notes.html

Last edited by a moderator:

- #12

- 214

- 0

Ok, before I go into any discussion on this topic (I've not studied this particular topic, thus I would have to do some reading up to be able to converse on an intellectual basis) I just want to point out, that I would much prefer to discuss experiments that have been shown to prove "relativistic mass" has any physical meaning. Because, otherwise "relativistic mass" is just a definition and as such may appear to have significance in mathematical equations when in fact is has no physical significance.pervect said:OK, here is the detail of the tidal force calculation:

I hope you understand my argument. Anyway, if you

- #13

- 10,055

- 1,223

m_r = m/sqrt(1-v^2) = E. There are apparently some differences for non-isolated distributed systems - assuming one is convinced that the concept of mass applies to systems that aren not essentially isolated. (This is a question that I have not yet fully resolved to my satisfaction).

In most cases, one can assume that m_r is just a synonym for energy, E, however.

Energy has definite physical consequences. Gravity couples directly to energy, (actually both energy and pressure), so energy has definite physical consequences. Since relativistic mass is essentially energy, it has the same physical consequences. People who are fond of relativistic mass are also fond of saying that gravity couples to it (which is true as far as it goes, though it's also important to note that gravity couples to pressure, too, IMO).

- #14

- 214

- 0

pervect said:Energy has definite physical consequences. Gravity couples directly to energy, (actually both energy and pressure), so energy has definite physical consequences. Since relativistic mass is essentially energy, it has the same physical consequences. People who are fond of relativistic mass are also fond of saying that gravity couples to it (which is true as far as it goes, though it's also important to note that gravity couples to pressure, too, IMO).

Well, I will certainly look into this in the next few upcoming days.

- #15

- 10,055

- 1,223

Aer said:Well, I will certainly look into this in the next few upcoming days.

You might find the following helpful

http://arxiv.org/abs/gr-qc/9909014

This does not go into all the subtle points, but it's a starting point for discussing why a hot gas has a higher mass than a cold one.

https://www.physicsforums.com/showpost.php?p=671119&postcount=25

might also be helpful, but note that a lot of the comments here are restricted to calculating the mass of an object in a

- #16

- 2,952

- 0

For a rather complete discussion on this topic you can see thye write up I posted hereAntiphon said:

direction of motion increases. But what about it's intertial mass

perpendicular to the direction of motion? Does it also increase?

And finally, what about the gravitational field "produced" by the

mass. Has it increased as well?

http://www.geocities.com/physics_world/sr/inertial_mass.htm

The gravitational field, defined by the components of the metric tensor (gravitational potentials) are a function of speed so the answer is yes. The components of the Riemann tensor (aka tidal force tensor) has componenents which are also functions of speed so the tidal force is also a function of speed (components change upon change of frame of referance).

Pete

- #17

- 10,055

- 1,223

I've posted links to the fact that the E-field of a moving charge is not symmetrical. Neither is the gravitational field of a moving mass.

For the electric charge, when we integrate the electric field over a surface, we can find the total enclosed charge by Gauss's law. This serves more or less as the defintion of charge.

So the electric field of a moving charge is not symmetrical, but if we integrate the field over the surface of a surrounding sphere, we will get a constant number, and it doesn't matter if the charge is moving or not moving.

With gravity, things are a bit more complicated, and the results are different. We do not get a conserved quantity (intergal) for the total energy at all, *unless* we perform the integration in flat space-time. And when we do get a conserved quantity, it's value depends on how fast the enclosed mass is moving.

The sci.physics.faq on "is energy conserved in GR" might be a useful reference for this point. I'll let the interested reader track down the link, it's easy to find.

When we require the system under consideration to be "flat" far away from the central mass (asymptotic flatness"), we can perform the integration for energy and get a conserved number. We can do even more - we get not only one conserved intergal for the energy, but three more for the momentum (it takes 3 quantities to describe the momentum vector).

The simplest way of performing this calculation that I'm aware of are the energy psuedo-tensors. They are called psuedo-tensors because they work properly only in asymptotically flat space-times. (A general tensor has no such limitation).

There's a writeup of energy pseudo-tensors in MTW's "Gravitation", but I must say I'm finding it a bit unclear.

The important point about the resulting numbers for enclosed energy and momentum is how they transform. The resulting numbers transform just like a standard energy-momentum 4-vector in special relativity. This means that in the center of mass frame, the energy is the lowest, and is equal to the invariant mass of the system, and the momentum is zero. E^2 - p^2 = m^2 in systems where the center of mass is not stationary.

When we chose a reference frame other than the center-of-mass frame, the energy increases, which means that gauss's law intergal is increasing, which means basically that the "average" gravitational field (expressed as an acceleration relative to infinity) increases in that frame.

Share: