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- Thread starter Tregg Smith
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Please read our FAQ in the General Physics forum.

Zz.

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[tex]E^2 = (pc)^2 + (m_0c^2)^2[/tex]

applies specifically to the case where momentum, p, is zero. We can't talk about the rest energy of a photon because a photon has no

[tex]E = pc.[/tex]

(On the other hand, a system of photons, not all moving in the same direction, does have a rest mass because we can choose a reference frame (spacetime coordinate system) in which there's no total velocity, e.g. if we have two photons travelling in opposite directions, we can pick a point between them in which the

But acceleration due to gravity (the rate at which the velocity of a particle changes, i.e. how much its path is bent) doesn't depend on the mass of the particle being accelerated. If, like whichever of the Apollo astronauts it was, you drop a feather and a rock on the moon (where there's no air to impede the fall of the feather) they hit the moondust at the same time. Gravity is an unusual force in this respect, and, because of this, in general relativity isn't even considered a force, but rather the "curvature" of spacetime. The mass of the sun curves spacetime, and whatever moves through space in time (regardless of its own mass) moves along a curve called a

To get an intuitive idea of what a geodesic means in general, it's best to start with a 2-d surface such as the surface of the earth. If you go in what looks at the small scale like a straight line on the earth, you're following a geodesic; these are the paths which on the surface of a sphere will--if you follow them far enough--bring you back to where you started (but that's not a general property of geodesics, it just happens to be true for geodesics on a sphere).

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Dale

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Yet it requires an object to have mass.

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To include the case where *m* = 0, I'm guessing you'd define acceleration due to gravity as the limit

[tex]\lim_{m\rightarrow0}\frac{\mathbf{F}}{m}=\lim_{m\rightarrow0}\frac{GMm}{mr^2}\hat{\mathbf{r}}=\lim_{m\rightarrow0}\frac{GM}{r^2}\hat{\mathbf{r}}=\frac{GM}{r^2}\hat{\mathbf{r}},[/tex]

where*G* is the gravitational constant, *M* the mass whose influence is being calculated, and *m* the mass of whatever it is that you're calculating *M*'s influence on, *r* being the distance between the centre of mass *M* and what is being influenced, and [itex]\hat{\mathbf{r}}[/itex] a position vector of unit length pointing from the centre of mass *M* towards the thing being influenced.

[tex]\lim_{m\rightarrow0}\frac{\mathbf{F}}{m}=\lim_{m\rightarrow0}\frac{GMm}{mr^2}\hat{\mathbf{r}}=\lim_{m\rightarrow0}\frac{GM}{r^2}\hat{\mathbf{r}}=\frac{GM}{r^2}\hat{\mathbf{r}},[/tex]

where

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Dale

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How so? The passive acceleration due to gravity is well-known to be independent of mass (e.g. the hammer and feather demo from the moon landing). If x is independent of y then why would it be "required"? I don't know why this error is so common.Yet it requires an object to have mass.

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Dale

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Exactly. Also, note that by Newton's second law: F = ma, if m=0 then f=0 regardless of a. So the fact that there is no force is consistent with a finite acceleration of a massless object.To include the case wherem= 0, I'm guessing you'd define acceleration due to gravity as the limit

[tex]\lim_{m\rightarrow0}\frac{\mathbf{F}}{m}=\lim_{m\rightarrow0}\frac{GMm}{mr^2}\hat{\mathbf{r}}=\lim_{m\rightarrow0}\frac{GM}{r^2}\hat{\mathbf{r}}=\frac{GM}{r^2}\hat{\mathbf{r}},[/tex]

whereGis the gravitational constant,Mthe mass whose influence is being calculated, andmthe mass of whatever it is that you're calculatingM's influence on,rbeing the distance between the centre of massMand what is being influenced, and [itex]\hat{\mathbf{r}}[/itex] a position vector of unit length pointing from the centre of massMtowards the thing being influenced.

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