
#1
Nov411, 08:16 AM

P: 1,011

The relativistic relativistic energy–momentum relationship is:
[tex]E^2 = \left(pc\right)^2 + \left(m_0c^2\right)^2[/tex] The relativistic energy (for a mass) is: [tex]E = \gamma m_0c^2[/tex] It has been said that the curvature of space is contributed to by photons, not just mass. However:
If it is still true that photons themselves produce some amount of curvature, then the invariant mass of system [itex]m_0[/itex] cannot account for this curvature. However, neither can the following:
B.T.W. In SR, [itex]m_0[/itex] is not a timeinvariant mass, but a frameinvariant mass. Photons colliding headon have been shown to produce additional [itex]m_0[/itex]. So I ask, "What energy term determines gravity in GR?" My basic premise of course is that this quantity should not be dependent on the observer. But, maybe velocity time dilation relative to each mass has something to do with the apparent difference in curvature (more velocity time dilation would mean greater perceived acceleration of bodies in orbit around the mass). Does [itex]\sqrt{\left(pc\right)^2 + \left(m_0c^2\right)^2}[/itex] give us the amount of energy involved in this curvature? 



#2
Nov411, 08:36 AM

Sci Advisor
Thanks
P: 3,850

Please remember that Maxwell's Equations have not been repealed, and it is still perfectly permissible to describe electromagnetic waves in terms of ... electromagnetic fields! Photons are appropriate for very low intensity or very high frequency, but otherwise they are not the best way to go.
T^{μν} = F^{μα} F^{ν}_{α}  1/4 g^{μν} F^{αβ} F_{αβ} and this is the source of the gravity. 



#3
Nov411, 08:37 AM

Sci Advisor
P: 8,004

In special relativity to conserve energy for fields it is not sufficient to assign fields energy and momentum, but also "pressure" via a stressenergymomentum tensor (which may be called the "energymomentum tensor" or "stressenergy tensor" for convenience).
In general relativity, the energymomentum tensor is the generalization of Newtonian gravitational mass, and is the source of the gravitational field (the metric field). 



#4
Nov411, 08:51 AM

P: 1,011

What energy term determines gravity in GR? 



#5
Nov511, 01:10 PM

P: 476





#6
Nov511, 01:23 PM

Mentor
P: 16,470

http://en.wikipedia.org/wiki/Electro...energy_tensor And some further details about the components of stressenergy tensors in general: http://en.wikipedia.org/wiki/Stressenergy_tensor 



#7
Nov511, 03:56 PM

P: 647

Since we are on the topic of light waves, I asked a question a year or so ago and never got any answers... ( i figured it was kind of a dumb question) but now a year later, it still bugs me if it still applies.... can I get a hand with this?
One way to "build" the SEM tensor is to associate a 3volume with the energymomentum 4vector [tex] \Delta p^{\alpha\beta}=T^{\alpha\beta}n_\beta \Delta V [/tex] This can get one the energy density, momentum density, and pressure by studying the different spacelike/timelike surfaces that comprise the volume. So my question is: is it appropriate to substitute [itex]p^\alpha =\hbar k^\alpha [/itex] for the momentum 4vector? This would give the SEM tensor in terms of wave quantities. I find this interesting since it phrases SEM in terms of k and ω and the fluxes of these quantities. Thanks, 



#8
Nov511, 04:42 PM

Sci Advisor
P: 8,004

You could check it by computing the SEM tensor the usual way (via the Lagrangian), then substituting the plave wave solution. 


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