Special relativity and acceleration

[Ich]

Would the concept of ‘gravitational time dilation’ even be necessary if what was really happening were changes in the value of c caused by a variable refractive index.

Of course. The idea of an effective refraction index is not new, a quick search gave me http://homepages.ecs.vuw.ac.nz/~visser/Seminars/Conferences/refractive-index-2.pdf" .
GR can't be reduced to a scalar theory, as the refractive index would be. So that's not an intrpretation, it's a new theory which contradicts some experiments (I don't know which at the moment, you can look it up in MTW's Gravitation).
Even in cases where a scalar quantitiy is enough, it is not evident how a refractive index should produce real time dilatation, the one accumulating over measurement time.

 

[Quentin]

Thanks for the quick reply and the reference. My comment on the 'gravitational time dilation' was only in the context of the Harvard Tower experiment. Just to clarify the gist of my argument:

Suppose that a rocket containing an atomic clock was accelerated on an outward journey and was then decelerated to return to the starting point. If the refractive index was affected as suggested by the gravitational fields generated in the acceleration/deceleration periods, then the clock frequency would be reduced for both those periods and would of course show a corresponding cumulative time dilation.

 

[Ich]

Suppose that a rocket containing an atomic clock was accelerated on an outward journey and was then decelerated to return to the starting point. If the refractive index was affected as suggested by the gravitational fields generated in the acceleration/deceleration periods, then the clock frequency would be reduced for both those periods and would of course show a corresponding cumulative time dilation

But time dilatation is proportional rather to the total time you spend on the journey, not to the acceleration phases. You can't model this "SR" time dilatation with a local refraction index.
You can express time dilatation in static spacetimes with a single parameter, but I don't see how this would be due to a different refraction index. But as I said, the theory can't replace GR anyway.

 

[Austin0]

It does: and this is what I describe above in the previous post.

Note that you cannot simply compare an accelerated observer with an unaccelerated observer, because in that case you also get an increasing velocity difference, and that dominates any time dilation.

However, if you have a long spaceship experiencing a constant acceleration at all points, it turns out that the front has slightly less acceleration than the back, and there is a time dilation between the front and the back of the ship... but no change in the distance between them, as measured by anyone on the ship. THIS is what turns out to be exactly analogous to the time difference of two observers at different altitudes in a gravitational field.

Cheers -- sylas

Hi sylas I am still thinking over this topic.
I did a work up of a hypothetical case but my math is rusty so I thought I would run it by you.
Inertial frame F
Accelerating System S'
rest L'= 1 km
a= 1000g= 10km /s^{2}

Range .6c ===> .7c
.7c-.6c =.1c = 3 x 10^{4}km/s

Time dt= (3 x 10^{4}km/s)/(10km/s) =3000 s

Contraction v_{i}=.6c ------- \gamma=1.25 --- = L'_{0}=.8 km
v_{f} =.7 -------- \gamma= 1.4 --- =L'_{1} =.71km

Difference in length over course of acceleration = .09 km
.09km/ 3000s = 3 x 10 ^{-5} km /s

relative velocity between front and back v_{fb}= (3 x 10 ^{-5} km /s) /(3 x 10 ^{5}km /s ) = 10^{-10}c
Additive average relative velocity between front and back = (.65+10^{-10})+ .65c = .65+ (1.7316 e^{-10} )

average velocity difference v_{d}= 1.7316 e^{-10} c
avg \gamma= 1 +( 2.9484 x 10 ^{-20} ) between front and back

Relative to inertial frame F ,,, S' avg v=.65c \gamma= 1.32
dt/1.32 = 3000/1.32 = 2,273 s = overall elapsed time on S'

2,273 / 1 +( 2.9484 x 10 ^{-20} ) = 6.782 x 10 ^{-17} s
elapsed time difference between back and front.

As I said I am rusty and could have easily dropped an exponent or counted all the zeros or 9's on the calculator screen wrong but does this seem in the ballpark?
Or is there some other fundamentally different way to calculate?
I assumed constant acceleration as observed in the inertial frame , of course the calculated (a) factor wouldn't neccessarily be healthy for humans but it made for smaller numbers
Thanks cia0

 

[Quentin]

Silas:
I was notified that your reply to AustinO (the quote in message #64) pertained to a question that I raised in message #63 and indeed it does. I get the point about the acceleration on a long rocket being different at each end, because of time dilation over the length of the thing, due to the gravitational field which it generates being different from end to end.

Can you give me some idea of what the polar diagram for the field looks like? Surely it cannot have anything to do with the position of the propulsive force. Suppose for example the source of this force was situated at the centre of the rocket, rather than at the rear end, would that change the polar diagram of the gravitational field? I can't imagine that it would.

For example what would the field look like if the accelerating object was not a rocket but a spherical mass with the propulsive force being a point source at the centre of the sphere? Don't ask me how such an arrangement could be devised without a radial hole to the centre for the gas or the ions or whatever to escape and provide thrust, but let's assume that it is in the realms of mythical frictionless surfaces and perfectly reflecting ones along with the other mythical assumptions which are made to illustrate a basic principle.