- #1

- 24

- 0

Thank you for reading and if it has already been said please just post a link or a few keywords I can search for. :)

-Coughlan

- Thread starter Coughlan
- Start date

- #1

- 24

- 0

Thank you for reading and if it has already been said please just post a link or a few keywords I can search for. :)

-Coughlan

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

I specified "in special relativity" because, of course, general relativity has a completely different way of computing gravity.

- #3

- 5,428

- 291

Clocks are affected by gravity. Obviously the gravitational field at a certain place may vary if masses are moving nearby.Is gravity affected by time or is time affected by gravity?

- #4

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

I think what you're asking is:

Gravitational force depends on mass and distance.

Is that mass the ordinary mass (rest-mass) m, or, for something moving at a very fast speed v, is it the increased mass [tex]\frac{m}{\sqrt{1\,-\,v^2/c^2}}[/tex]?

Erm … I don't actually know the answer to that!

- #5

- 24

- 0

- #6

- 28

- 0

- #7

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Sorry, but this is just wrong. Momentum in each direction is rest-mass times velocity-in-that-direction times theI'm going to take a guess here. From my understanding, at relativistic speeds, an object doesn't gain mass, per se {snip} different masses in different directions of the same object doesn't make much sense {snip}

The increased mass, rest-mass times [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], is the same for all directions.

Yes, but the question is whetherSince object A is not traveling at relativistic speeds in this direction, it's normal mass would be used.

- #8

- 6

- 0

At the moment I have no idea about how doing that, but on one point I am quite sure. Newton's gravity has as framework the Euclidean space and three-vectors, while Special Relativity resides in the space-time (Minkowski space) and uses four-vectors.

So, Newton's law cannot be used as it stands, but needs to be generalized to the four-dimensional case, as one usually does for Electromagnetism, introducing the four-potential A^{\mu}.

However, the generalization of Newton's law to Minkowski space (if it is possible) has not to be trivial!

If we assume it possible, then one uses four-momenta and four-forces and should be able to deal properly with newtonian gravity.

I think I have been right!

- #9

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

You just have to be careful!

Usually, it just involves multiplying the mass by [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex].

In electromagnetism, for example, E and B are parts of a six-component tensor in four-dimensional space, and would be different for an observer with a different velocity, but for any particular observer the usual three-dimensional rules for E and B apply.

- #10

- 1,548

- 0

You will be much better off applying the accepted standard of only using rest mass.Yes, but the question is whetherB'snormal mass would be used. I believe the answer is no, because of the equality of inertial mass and gravitational mass, but that's general relativity ....

Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

Energy is defined by applying gamma to momentum no more than that is needed.

No need to assume that the momentum value change is the result of a change in any actual mass inside the equation for momentum. Physicist today find it makes much better sense to view the momentum itself as being altered relativisticly by time and distance changes defined in the gamma factor. Those effects are what is important not the idea that it “looks as if the mass changed”. That is all that has happened, the relativistic effects are real. Not what only seems like a change in mass from our point of view.

- #11

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

We apply the "gamma factor", [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], because the Lorentz equations tell us to. That then tells us what effects will be apparent, not the other way round!You will be much better off applying the accepted standard of only using rest mass.

Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

What?Energy is defined by applying gamma to momentum no more than that is needed.

I don't really understand what you're saying here.Physicist today find it makes much better sense to view the momentum itself as being altered relativisticly by time and distance changes defined in the gamma factor. Those effects are what is important not the idea that it “looks as if the mass changed”. That is all that has happened, the relativistic effects are real. Not what only seems like a change in mass from our point of view.

I'm quite happy to accept that there's more than one way of looking at the same mathematics, and which you choose is largely a matter of taste.

My taste is to say that the mass increases. That's perfectly consistent with the mathematics.

If you want to state clearly what your taste is, I'll probably be happy to agree with you that that's fine too.

I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass!

- #12

- 6

- 0

He also said that Special Relativity was already contained in Maxwell equations, for which Galileo transformations give problems.

The same problems are absent for newtonian gravity, so one has to investigate some sort of "gravitodynamics among bodies"! :-)

Namely, which are the relativistic effects in newtonian gravity? And how do you build the gravitational four-potential?

Anyway, in my opinion the rest mass, being a scalar, should be used as the source of the gravitational force. The gamma factor would enter through Lorentz transformations!

- #13

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

So far as I know, all forces are affected in the same way by the Lorentz transformation.Electromagnetism is a very special case!

No, they're there, but gravity is many many orders of magnitude weaker than electromagnetism (think how fast an electron can be accelerated form rest across a cathode ray tune, and how long gravity would take to pull it the same distance!), so nobody noticed them.The same problems are absent for newtonian gravity

- #14

- 24

- 0

- #15

DrGreg

Science Advisor

Gold Member

- 2,294

- 918

I stand to be corrected, but I believe you have no choice but to use general relativity. I'm no expert, but I recall other answers to similar questions saying that, in GR, gravity relates to energy-momentum rather than to (rest) mass. So even photons have some (tiny) gravitational influence.

- #16

- 24

- 0

- #17

DrGreg

Science Advisor

Gold Member

- 2,294

- 918

I agree.I'm quite happy to accept that there's more than one way of looking at the same mathematics, and which you choose is largely a matter of taste.

(...and mine is not!)My taste is to say that the mass increases.

I would suggest to everyone that when we talk about mass, we should make it clear whether we mean "relativistic mass" (proportional to energy, including kinetic energy) or "rest mass" (="invariant mass"). The modern fashion is for "mass" to mean "invariant mass". In these forums, many readers are struggling to work out what mass is, so we should avoid ambiguity.

Yes, it has increased invariant mass. A rotating flywheel has more invariant mass than an identical non-rotating wheel (one reason why "rest mass" isn't such a good name).I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass!

For the benefit of anyone who doesn't know why, the invariant mass

[tex](\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2 [/tex].

which reduces to

[tex]\Sigma E = m_{tot} c^2 [/tex]

in the box's frame, where the total momentum is zero. For each individual particle of invariant mass

[tex]E^2 = m^2 c^4 + ||\textbf{p}||^2 c^2[/tex], so [tex]E > m c^2[/tex], and [tex]m_{tot} > \Sigma m[/tex].

- #18

jtbell

Mentor

- 15,614

- 3,638

https://www.physicsforums.com/showthread.php?t=205073

https://www.physicsforums.com/showthread.php?t=153173

https://www.physicsforums.com/showthread.php?t=150342

- #19

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

As you say, invariant mass = rest mass.Yes, it has increased invariant mass. A rotating flywheel has more invariant mass than an identical non-rotating wheel (one reason why "rest mass" isn't such a good name).

And you agree something can have "increased invariant mass".

Which is the same as "increased rest mass"!

Which is why my taste is to regard mass as increased in all situations!

(And don't forget, there's nothing unusual about rotational motion contributing to invariant/rest mass: the electrons in you and me have very fast rotational motion which does contribute to increasing our "rest mass"! )

No. If a wheel is rotating so that its rim is moving with speed U relative to the centre of the wheel, and if the whole wheel is moving parallel to its axis with speed V relative to us, then the Lorentz-increased mass (energy) of the rim is the rest-mass multiplied by:For the benefit of anyone who doesn't know why, the invariant massmof a collection of particles can be defined by_{tot}

[tex](\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2 [/tex].

which reduces to [tex]\Sigma E = m_{tot} c^2 [/tex] in the box's frame, where the total momentum is zero.

[tex]\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}[/tex]

Even in the box's frame, the Lorentz-increased mass (energy) of the rim is still multiplied by [tex]\Large\frac{1}{\sqrt{1 - U^2}}[/tex].

Last edited:

- #20

- 24

- 0

thank you jtbell. I will look when my boss isn't here!

- #21

- 1,548

- 0

Lorentz equations do not require they be applied to mass.We apply the "gamma factor", [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], because the Lorentz equations tell us to.

If you want to state clearly what your taste is, I'll probably be happy to agree with you that that's fine too.

I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass!

I did state it – I use gamma on momentum and NOT on mass as most modern science does.

In the rotational case you are comparing the effects of inertia.

Inertia is not defined by changing mass but by Energy, Momentum, and invariant Mass. You are simply attempting to justify changes in Energy and Momentum following relativistic rather than Classical rules by attributing the changes in Energy and Momentum to actual changes in the physical mass of an object.

Most modern practicing science does not find it necessary to use that way of thinking as a crutch to understanding what is going on. They are able to think in terms of Energy and Momentum as being independent things that need to be treated relativisticly, that’s all.

True you can treat Energy and Momentum Classically by correctly applying relativistic effects only to mass, most of modern science just does not do that.

I suppose either way might be OK, with the possible exception that applying it to mass might require using a preferred reference frame- but I’m not sure on that point. Probably not that important a distinction since a preferred frame is desirable to some anyway.

If you want to survey how most scientists think, I think you find I’m correct on how the majority view the issue.

- #22

- 28

- 0

I'll have to rethink my reasoning--thanks for pointing out the troubles. I found a different thread on here that explains what I wanted to say in a much better way. Here is the conclusion, and I'll give a link to the mathematics:Sorry, but this is just wrong. Momentum in each direction is rest-mass times velocity-in-that-direction times thesamefactor, [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], where v is the total velocity.

The increased mass, rest-mass times [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], is the same for all directions.

Yes, but the question is whetherB'snormal mass would be used. I believe the answer is no, because of the equality of inertial mass and gravitational mass, but that's general relativity, so I'd rather let someone else comment on it.

"Given this form of the dynamics equation for special relativity it becomes evident that a mass "changing with speed" is not the correct explanation for why a massive object can not be accelerated up to the speed of light. The mass m here does not change with speed. The correct explanation is that given this law of motion an arbitrary amount of ordinary force produces a diminishing coordinate acceleration as the coordinate velocity approaches c due to the time dilations involved in relating coordinate to four-vector expressions. These time dilations are in turn due to the Lorentzian structure of spacetime(the Lorentzian structure being why we defined the momentum four vector in terms of a Lorentz transform in the first place). So the reason really that nothing with mass can be accelerated up to c speeds in special relativity is ultimately that spacetime has a Lorentzian structure."

https://www.physicsforums.com/showthread.php?t=11072

- #23

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

You've quoted me out of context (and without giving the post's reference).Lorentz equations do not require they be applied to mass.

I was simply denying that I was getting the gamma factor from what you called the "apparent effect", as in:

We apply the "gamma factor", [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], because the Lorentz equations tell us to. That then tells us what effects will be apparent, not the other way round!You will be much better off applying the accepted standard of only using rest mass.

Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

Sorry … that's too unclear for me to either agree or disagree with.I did state it – I use gamma on momentum and NOT on mass as most modern science does.

I'm really not following most of this.In the rotational case you are comparing the effects of inertia … {saving space} … preferred frame is desirable to some anyway.

For example, you say "They are able to think in terms of Energy and Momentum as being independent things that need to be treated relativisticly" - but Energy and Momentum are most certainly

The simple fact is that "Most modern practicing science"

- #24

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Hi AB! (btw, I

Yes, in that the rest-mass or inertial mass is the same for all observers.{snip} The mass m here {of something accelerating towards the speed of light} does not change with speed. {snip}

But that doesn't help with the case of something with moving parts, which has an increased inertial mass even when the things as a whole is stationary!

I can't see any

Ultimately, I think the difference here is between physicists and mathematicians.

For example, mathematicians will say that:

But physicists are entitled to say "we want to relate the maths to things which we, as experimental physicists, can actuallythe reason really that nothing with mass can be accelerated up to c speeds in special relativity is ultimately that spacetime has a Lorentzian structure.

Now, no experimental physicists in their right minds are going to measure the rest-mass of the individual bits of me, or of a vehicle containing a rotating flywheel. They'll measure the mass as a whole, either by weighing or by pushing.

It might be better if they called it the "resistance-to-movement" or "inertia" - but:

it is the same for all observers (no gamma factor); and

it can be used as [tex]m_0[/tex] in all equations I know of

Some call it "rest-mass", but that's slightly misleading because it takes account of some parts which aren't at rest.

Some call it "inertial mass", which is probably best because it is a measure of its resistance to movement (which is, of course, the way it is usually measured).

A mathematician may say "you are missing the fundamental nature of the group structure of space-time."

But a physicist should reply "you are missing the simplicity of your equations, and the necessity of applying them to observable measurements in the actual universe!"

This thing - whatever we call it - is a perfectly valid measurement, and we are entitled to put brackets in the equations round the terms that describe it, and to defend those brackets against all attacks!

Last edited:

- #25

- 9,815

- 1,033

Thank you for mentioning this. Yes, it is a frequently asked question, and many of the answers given here have been rather speculative.

https://www.physicsforums.com/showthread.php?t=205073

https://www.physicsforums.com/showthread.php?t=153173

https://www.physicsforums.com/showthread.php?t=150342

There are some fine technical points here which make the gravitational field of a moving mass hard to even talk about rigorously.

If you think of the gravitational field of a moving mass as being like the electric field of a moving charge, i.e. concentrated in the transverse direction, however, while you will be wrong, you won't be far wrong. By "not far wrong", I mean that you'll probably get the right order of magnitude when you translate the not-well-defined concept of "gravitational field" into experiment so you can actually measure it.

I have the feeling from skimming the threads that most posters here aren't even familiar with how relativity treats the electric field of a relativistically moving particle (a much simpler question. See for instance

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

and look and numbers above and below 14 in the url to get a fuller treatment.

Last edited by a moderator:

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 26

- Views
- 4K

- Replies
- 74

- Views
- 803

- Last Post

- Replies
- 8

- Views
- 4K

- Last Post

- Replies
- 23

- Views
- 3K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 10

- Views
- 2K

- Replies
- 61

- Views
- 13K

- Replies
- 4

- Views
- 616

- Replies
- 3

- Views
- 2K