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When discussing alternatives to General Relativity, it's often stated that scalar gravity theories fail because they violate the equivalence principle. The problem is that rest mass is used as the source of gravity and that of course leads to violation of the equivalence principle.

Now, I'm wondering, why can we not just use the relativistic mass $\gamma m_0$ where $\gamma = 1/\sqrt{1-\frac{v^2}{c^2}}$ as the source of gravity in those cases?

Qmechanic
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AngBar
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4 Answers4

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Now, I'm wondering, why can we not just use the inertial mass $\gamma m_0$ where $\gamma = \sqrt{1-\frac{v^2}{c^2}}$ as the source of gravity in those cases?

Because inertial mass is not frame-invariant. There’s a frame in which the density of the Earth (due to γ) is high enough to be a black hole; that doesn’t mean that it becomes a black hole.

The Schwarzschild metric is only valid in the rest frame of the mass. As it turns out that’s okay, because the ideas of special relativity (“everything has to be frame-independent!”) get thrown out when there is actually a preferred frame. Special relativity, after all, is an edge case of general relativity where gravity goes to zero, and where those simplifying ideas actually do work.

controlgroup
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This extends an existing answer.

To formulate a physical theory one has to ask oneself at the outset what aspects of known physics it will have to respect if it is to have any chance of being correct. One of the guiding principles is symmetry, and that includes invariance of the equations under change of reference frame (or of coordinates). You can of course throw away that constraint, and argue that the cosmos has a preferred frame. That will produce one kind of theory. But if you don't want to build a preferred frame into the very equations, then you have to avoid doing that. Mathematics offers ways to guarantee such an invariance. It is done by requiring that everything in the equations is written in tensor language. That is, scalar quantities are not just scalar but invariant scalars. First rank quantities are 4-vectors; 2nd rank quantities are tensors, etc. So your equations can involve an invariant such as rest mass, and a 4-vector such as energy-momentum, and a tensor such as stress, but they cannot directly involve a frame-dependent quantity such as the energy alone, or the Lorentz factor, or the speed of a particle.

Andrew Steane
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Because using inertial mass (which includes velocity-dependent $\gamma$) as the source of gravity leads to a huge problem: it makes gravity depend on how fast something is moving — and relative to what? That breaks the principle that physics should look the same in all inertial frames. In other words, if gravity changes depending on velocity, then different observers would see different gravitational effects, violating both special relativity and the equivalence principle.

General Relativity avoids this by saying it's not "mass" but the stress-energy tensor (which includes energy, momentum, pressure, etc.) that curves spacetime — and it does so in a frame-independent way.

Trying to use $\gamma m_o$ as a gravitational source just reintroduces a preferred frame and messes with causality. So... clever idea, but it collapses under its own weight.

safesphere
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Chetti
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This could work if there was an absolute rest frame, but there is no absolute rest frame.

You can have any $\gamma$ you want right now: no rockets, no ion-drive, no di-lithium crystals, it's just depends on how you choose your coordinates.

So I choose $\gamma=10^{32}$, for you, in my coordinates.

Do you feel any different?

JEB
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