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The theory of relativity shows that the inertial mass of a body increases with the energy it contains; if the increase of energy amounts to $E$, the increase in inertial mass is equal to $E/c^2$, where $c$ denotes the velocity of light. Now, is there an increase of gravitational mass corresponding to this increase of inertial mass? If not, then a body would fall in the same gravitational field with varying acceleration according to the energy it contained. And then the highly satisfactory result of the theory of relativity, by which the law of the conservation of mass leads to the law of conservation of energy, could not be maintained, because it would compel us to abandon the law of the conservation of mass in its old form for inertial mass, but maintain it for gravitational mass.

This must be regarded as a very bad model. On the other hand, the usual theory of relativity does not provide us with any argument from which to infer that the weight of a body depends on the energy contained in it. However Einstein has shown that gravitation of energy is a necessary consequence.

However, some very serious issues are raised when these arguments for maintaining the law of the conservation of both the inertial and the gravitational mass are applied to the red shift of a photon influenced by the gravitational field of a body M. When doing so we arrive at the following solution when resorting to certain well-known physical expressions:

$$ E = h\nu $$

$$ E = mc^2 $$

$$ m = \frac{h\nu}{c^2} $$

$$ F = \frac{GMm}{r^2} $$

$$ F = \frac{GMh\nu}{r^2 c^2} $$

From the work done against the gravity force in bringing a photon in from infinity where the potential energy is assigned the value zero, the expression for gravitational potential energy $U$ is

$$ U = -\frac{GMh\nu}{rc^2} $$

If the original frequency and energy of a photon “launched” from a body $M$ were equal in magnitude to the potential energy, then when would it have enough energy to escape from $M$?

This escape energy would be the original energy $h\nu$ of a photon which is totally red shifted to total extinguishment by the gravitational field of the body $M$.

That would mean that any photon with an original energy less than the escape energy would be extinguished before it leaves the gravitational field of the body $M$ and that photons with higher original frequencies, and thus higher original energies, would escape that field less red shifted the more the original frequency of the photon is increased.

However, as far as I know this kind of redshift has never been observed.

My question to Red Bunn is: what is then the reason for using the metaphor: "In Newtonian language, if you imagine the source of the light at the centre of a spherically symmetric expanding spacetime, then the light travels 'uphill' in a gravitational potential all the way to the observer. The observed redshift is partially due to this redshift and partially due to the observer's motion. Conversely, if you put the observer at the centre, the light 'falls downhill' all the way to the observer. This gives a different breakdown of the observed redshift into gravitational and Doppler contributions."

I would rather ask whether it's proven or not that the photon could have no such gravitational mass which interacts with a gravitational field in the way it would be expected if the photon would act as a mass particle?

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  • $\begingroup$ The photon in the standard model has 0 mass, no adjectives attached. en.wikipedia.org/wiki/Standard_Model . This question belongs to science fiction. I reacts to to gravity through its energy, by special and general relativity rules that have been validated over and over. Mass is defined as the length of the four vector describing a particle and it is an invariant. en.wikipedia.org/wiki/Invariant_mass $\endgroup$ – anna v Jul 19 '14 at 4:58
  • $\begingroup$ No its not sci-fi, it is just a little confused. $\endgroup$ – Kevin Kostlan Jul 19 '14 at 5:14
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"Now, is there an increase of gravitational mass corresponding to this increase of inertial mass?"

Yes, both increase by the same amount. For any object inertial = gravitational. Even antimatter will fall down at the same speed.

"If the original frequency and energy of a photon “launched” from a body M were equal in magnitude to the potential energy, then when would it have enough energy to escape from M?"

As long as your launch point is not inside the event horizon of a black hole, "disappearing photons" are impossible. The photon will always have more energy than the escape energy. If you cut the energy in half the escape energy will also be cut by 1/2 since the photon is lighter and easier to lift. However, if the object is a black hole and you launch the photon from below the horizon, then it immediately gets pulled inward, along with you and anything else that got unlucky enough to fall in.

"I would rather ask whether it's proven or not that the photon could have no such gravitational mass"

The photon does exert gravity, a beam of light will attract objects by a tiny bit. However, the motion will double the gravitational effect on stationary objects. This strange effect applies for any near-light speed object. This effect is different from binding enery adding to mass! Binding energy adds to inertial mass as well, but this "high-speed double gravity effect"does not add to the inertial mass.

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  • $\begingroup$ Well, the empirical results indicate that the FREQUENCY of a photon is NOT related to the CHANGE of its FREQUENCY caused by gravitational effects. The CHANGE of its FREQUENCY is caused by time dilation. When trying to apply the gravitational potential (GP) to the problem of solving red shift issues, you inevitably land in science fiction. Thus, isn't it contra productive to insist on using the difference in GP as an explanation of the red shift? In GR, the GP is replaced by the metric tensor. But when the GP is weak, GR reduces to Newtonian gravity and it's relevant to apply time dilation. $\endgroup$ – Carl Johan Soderquist Jul 19 '14 at 7:38
  • $\begingroup$ Which results are you talking about? The frequency IS related to the change in frequency. Time dilation is a relative number, i.e "twice as slow". If the original frequency is 1000Hz, the new frequency will be 500Hz. For 10000Hz it will be 5000 Hz. $\endgroup$ – Kevin Kostlan Jul 19 '14 at 14:58
  • $\begingroup$ Dear Kevin Kostlan, I just don't know what you base your claim on when you assert that the FREQUENCY of a photon IS indeed related to such CHANGE of its FREQUENCY that is indirectly caused by gravitational effects. If you were correct, any redshift z of a photon in a gravitational field should depend upon which the original frequency of the red-shifted photon was. However no such dependency exists. The expression of z doesn't contain the frequency or wavelength. Alas! Could you prove me to be wrong, or you think I should just trust your judgment? However, I think I need some evidence. $\endgroup$ – Carl Johan Soderquist Jul 19 '14 at 16:00
  • $\begingroup$ I would emphasize my standpoint by saying that the expression of the redshift or blueshift ν1 – ν2 = Δν of a photon moving 'uphill' or 'downhill' in a gravitational field, is NOT mathematically or physically related to ν1. $\endgroup$ – Carl Johan Soderquist Jul 19 '14 at 16:13
  • $\begingroup$ Kevin Kostlan, in the GR no fundamental, gravitational force originates from interaction between masses, i.e. that force is fictitious. Gravitational time dilation is an actual difference of elapsed time between two events as measured by observers in regions of different gravitational potential. A radiation source may serve as a clock. The lower the gravitational potential, i.e. the closer the time measuring radiation source is to the source of gravitation, the slower will the measured time pass and the lower will the measured frequency be. This effect has been confirmed by many tests of GR. $\endgroup$ – Carl Johan Soderquist Jul 20 '14 at 4:13

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