According to special theory of relativity, we know that any equation that contains mass is corrected by $$ m\to \frac{m_0}{\sqrt{1-v^2/c^2}},$$

where $m_0$ is the rest mass. In the case of momentum, the classical momentum i.e. $mv$ we modify it by writing $$ p = \frac{m_0v}{\sqrt{1-v^2/c^2}}.$$

And since Newton's second law says "Force is the rate of change of momentum", we can write it as $$ F = m_0 \frac{d}{dt}\left(\frac{v}{\sqrt{1-v^2/c^2}}\right).$$

So far so good, now let's turn to Newton's law of universal gravitation which says that gravitational field produced my mass $m$ is $$ g= \frac{Gm}{r^2}.$$

Now, why can't we modify it by writing $m$ out like $$ g = \frac{Gm_0}{r^2\sqrt{1-v^2/c^2}}?$$ Is there any physical violation in writing this out or it is just because of general relativity which doesn't consider gravity as force but as consequence of curved space-time?

EDIT : @Ruslan answer gives the drawbacks of putting the $\gamma$ factor in the equation of Gravitoelectromagentism and it is very fascinating that this concept was introduced by Oliver Heaviside much before Einstein's GR. But I want to know the drawbacks of putting $\gamma$ factor in Newtonian Gravity,the gravity which is governed totally by mass. Why Newtonian Gravitational Equation can't be modified by introducing $\gamma$ factor? Is the non-invariance the only problem (I know it's a big problem but is it the only one)?

A Side Note:- Due to some reason people are thinking that my question is about the concept of relativistic mass, the concept to which many links are already present. I want to clarify that my question is not: How to interpret $m=\frac{m_0}{\sqrt{1-v^2/c^2}}$ ?. But my question is why a direct substitution of this formula in Newtonian Gravity equation (as @QMechanics has said in his answer) would lead to contradictions and what are those contradictions?

@Ruslan answers about the drawbacks of my insertion of $\gamma$ factor in GEM, his answer focuses as to why my insertion needs to be modified even further in GEM . My question is why did Newtonian Gravity failed in relativistic mechanics? Why GEM was introduced, was it only for invariance principle or were the predictions of introducing $\gamma$ factor in Newtonian Gravity very inconsistent?. I'm talking about introducing $\gamma$ factor because of it's advent in the momentum equation, my question is not just a random thought. To restate my question, I would write :

What are the illogical predictions of Newtonian Gravity if we introduce the $\gamma$ factor with mass?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Chris Dec 28 '19 at 21:21

OP is essentially asking if Newton's gravitational formula $F=Gm_1m_2/r^2$ could work in GR if we interpret the masses $m_1$ and $m_2$ as relativistic masses (rather than rest-masses), say for the deflection of a relativistic test-mass in a Schwarzschild geometry? Knowing that gravity couples to energy (rather than rest-mass) that's a natural question to ask. However, the actual GR calculation shows that OP's proposal yields incorrect results. E.g. in the ultra-relativistic case, there is a famous factor 2 in difference for small deflection angles. Taken more broadly, OP's proposal has various other shortcomings.

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  • $\begingroup$ Thank you for understanding me. Even if we consider gravity as a force then, can we replace $m_0$ by $m_0\gamma$ ? $\endgroup$ – user240696 Dec 26 '19 at 8:57

Consider for a moment electric field of a charge $q$ moving with speed $V^\dagger$. If we denote the angle between direction of motion and position $\vec R$ of test charge as $\theta$, we'll have (in Gaussian units)

$$\vec E=\frac{q\vec R}{R^3}\frac{1-\frac{V^2}{c^2}}{\left(1-\frac{V^2}{c^2}\sin^2\theta\right)^{^3/_2}}.$$

We can see that, if the test charge is on the line of motion of the moving charge, so that $\theta=0$ or $\theta=180°$, the electric field "felt" by it will be


while for the test charge to the side of the moving charge, i.e. with $\theta=\pm90°$, the electric field is


In weak-field approximation, relativistic gravity can be treated in the framework of gravitoelectromagnetism (GEM). In this framework, particle mass is the gravitational charge, and classical gravitational acceleration corresponds to electric (electrostatic) field.

After we accept this analogy and apply it to an initially stationary test mass, we can directly compare its predictions with what you suggest. Namely, your equations, where you replace gravitational charge with one multiplied by $\gamma$, result in the gravitational force increasing isotropically with the increase in speed $V$ of the field source. But in gravitoelectromagnetism, gravitostatic field only increases this way in the direction perpendicular to motion of the field source, while in the direction parallel to its motion the field decreases.

Moreover, your suggestion completely omits the gravitomagnetic effects, which actually become more pronounced with increasing $V$.

Since gravitoelectromagnetism is an approximation to Einstein field equations, it's valid in certain scenarios — in particular, in weak fields. This means that in these scenarios your model, which disagrees with gravitoelectromagnetism, is wrong.

$^\dagger$ Reference: L.D. Landau, E.M. Lifshitz, The Classical Theory of Fields, $\S38$ "The field of a uniformly moving charge".

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  • $\begingroup$ Wow, I never knew that there existed a gravitatonal charge, how is it different from the mass that produces the gravity. $\endgroup$ – user240696 Dec 26 '19 at 10:49
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    $\begingroup$ Note that it's still an approximation. It's not that there exists some separate charge. In the context of gravitoelectromagnetism, gravitational charge is identical to invariant mass. In the GR context gravity is created not by mass, but the stress-energy tensor. $\endgroup$ – Ruslan Dec 26 '19 at 10:52

Is there any physical violation in writing this out or it is just because of general relativity which doesn't consider gravity as force but as consequence of curved space-time?

Yes, there indeed is one right away: You started with a result from Special Relativity. But from that framework you already know that there's a finite speed of causality. Newtonian gravity however still assumes an instantaneous force. So this is already a hint that you are combining two things that may not be compatible.

Your resulting gravitational law also has an observer dependent part. But this again goes against another principle of Special Relativity, the principle of covariance. Every observer should "see" the same physics, but in your case, an observer resting with repsect to the mass creating the gravitational field and an observer moving with respect to it would see different fields.

Another one is that any non-uniform gravitational field (which we know exist) is not compatible with a global inertial frame - which SR has. People back in the beginning of the 1900's tried hard to come up with theories of gravity that reconciled graity with Special Relativity, and they ran into several problems like these, and it led to many different theories, of which General Relativity finally succeeded.

In Misner, Thorne and Wheeler's Gravitation there's a long chapter devoted to this.

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  • $\begingroup$ I think in the last second para there is a typo, graity is written instead of gravity. $\endgroup$ – user240696 Dec 26 '19 at 10:53
  • $\begingroup$ Your problem in the first para can be easily overcome just by writing a retarded factor , i.e. $ (t- r/c)$ . But this doesn't affect the replacement of $m_0$ by $m_0 \gamma$. $\endgroup$ – user240696 Dec 26 '19 at 10:56
  • $\begingroup$ Yes, there are ways of fixing those problems, and those were attempted. But people quickly ran into other inconsistencies by doing that. In fact, Newtonian gravity with retarded potentials is a worse approximation than the weak field limit of GR. In the end, trying go reconcile special relativity and Newtonian gravity usually leads to inconsistencies and you need metric theories of gravity. $\endgroup$ – Xeno Dec 26 '19 at 12:59

According to Special Theory of Relativity, we know that any equation that contains mass gonna be corrected as ...

This is incorrect. This correction allows some Newtonian formulae to work, but not all of them, and Newton's gravitational equation is one of the equations that don't work.

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  • $\begingroup$ Well my question is that why it doesn’t work? Please re-read it. $\endgroup$ – user240696 Dec 26 '19 at 8:20
  • $\begingroup$ @Knight I don't understand your question. Given that Newtonian mechancis cannot be modified by replacing $m$ -> $\gamma m$, what made you think it should work? $\endgroup$ – Allure Dec 26 '19 at 8:25
  • $\begingroup$ My question is why we can’t replace $m$ in Newtonian gravity by $m_0 \gamma$ ? $\endgroup$ – user240696 Dec 26 '19 at 8:34
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    $\begingroup$ @Knight why can't we replace $m$ in Newtonian gravity with $2214m^4 - m^2$? Again, the real question is: what makes you think it should work? $\endgroup$ – Allure Dec 26 '19 at 8:38
  • $\begingroup$ Well I think I was very clear in my question. Thank you for your answer, I will wait for other answers. $\endgroup$ – user240696 Dec 26 '19 at 8:40

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