General Relativity and galactic rotational speeds

[Pierre007080]

Please note that I am not even a physicist ... so please be gentle with me!
Question: When you guys work out rotational speeds of various suns in distant galaxies, do you consider General Relativity and the relative gravitational potential difference between us (the observer) and the star being measured?

 

[Jonathan Scott]

Please note that I am not even a physicist ... so please be gentle with me!
Question: When you guys work out rotational speeds of various suns in distant galaxies, do you consider General Relativity and the relative gravitational potential difference between us (the observer) and the star being measured?

As far as I know (I am not a physicist either), most calculations of that type are based on Newtonian gravity as an approximation to GR, having previously used more accurate GR approximations as a guide to the range of situations in which this approximation is valid. The calculations effectively assume that space is flat at a sufficient distance from the system being studied, and potentials are relative to that flat space. The relative potentials of ourselves (due to Earth, Sun, Milky Way galaxy etc.) compared with flat space outside our own galaxy is negligible (of the order of 1 part in 10⁶) and in any case has no effect on the structure of distant galaxies.

 

[Pierre007080]

Thanks for you reply Jonathan. Do I understand from your explanation that the difference (let's use time dilation and rod shortening of GR) between the length of a one meter (our sun potential) is negligiblely different to the length in "outer space" even though we are still well within the visible extent of our galaxy? Is there a SIMPLE approximation to determine how quickly this potential drops off? Distance square? or is it much more complicated? It seems weird that when you see a galaxy from far that the GR effects INSIDE that "conglomerate mass" are negligibly different from "outer space"

 

[Jonathan Scott]

Thanks for you reply Jonathan. Do I understand from your explanation that the difference (let's use time dilation and rod shortening of GR) between the length of a one meter (our sun potential) is negligiblely different to the length in "outer space" even though we are still well within the visible extent of our galaxy? Is there a SIMPLE approximation to determine how quickly this potential drops off? Distance square? or is it much more complicated? It seems weird that when you see a galaxy from far that the GR effects INSIDE that "conglomerate mass" are negligibly different from "outer space"

The fractional difference in scale (time rate and ruler size) due to a gravitational potential difference is simply the Newtonian gravitational potential expressed in dimensionless units, that is energy per energy rather than energy per mass. For example, the fractional difference in time rate between infinity and radius r from mass m is -Gm/rc². This approximately adds up for separate masses, so for a whole system of masses at distance r it's approximately the same as the effect of one total mass. You can plug in masses and distances to check the effect. Even if you plug in the mass of the whole galaxy and our distance from the middle to get an idea of the total effect, this is still only of the order of 10^-6.

You normally only need to use GR instead of Newtonian approximations when the square of this term becomes significant (which means this term is approaching order 1) or when handling relativistic speeds.

 

[Pierre007080]

Hi Jonathan.
I used this formula and made a table using the light years of 27000 (sun) and 100000 (outer space). I then divided all the potentials by the sun's potential to "earth adjust" the the potentials. It turns out VERY significant! The metre on Earth is only .23 metres in outer space? Please show me my mistake!

 

[Jonathan Scott]

Hi Jonathan.
I used this formula and made a table using the light years of 27000 (sun) and 100000 (outer space). I then divided all the potentials by the sun's potential to "earth adjust" the the potentials. It turns out VERY significant! The metre on Earth is only .23 metres in outer space? Please show me my mistake!

These values are very small fractions. The ratio of clock rates or ruler sizes at potentials -a and -b is not a/b but rather (1-a)/(1-b). (1-a)/(1-b) where a and b are small is approximately the same as (1-a) + b = 1 - (a-b).

 

[Jonathan Scott]

I should clarify that the Newtonian potential is the fractional difference in total energy, frequency and so on. It's the potential energy divided by the total energy, which is the difference in energy between two locations divided by the total energy at either location (where in the Newtonian approximation the difference is so small compared with the total that it doesn't matter which value you use as the total).

The approximate multiplicative factor related to the potential is (1 - Gm/rc²). If you compare the potential at two different distances r₁ and r₂ from the same mass, then the relative frequency would be (1 - Gm/r₁²)/(1 - Gm/r₂²), but given how small the potential is compared with 1 in the Newtonian approximation, this can be approximated as (1 - (Gm/r₁² - Gm/r₂²)), just using the difference of the potentials.

 

[Pierre007080]

The approximate multiplicative factor related to the potential is (1 - Gm/rc²). Gm/r₂²)), just using the difference of the potentials.

Hi Johnathan,
Thanks for your patience. Where does the 1 come into the formula. My understanding of the nature of the "multiplactive factor" seems to be the problem. May I give you the way I am reasoning, because although I see what you are doing, I obviously still don't understand the reasoning. Can we confine the measurement to one potential ... let's say our sun at about 27000 light years. This returns the "warp factor" as about 1 x10^-6. Surely if we on Earth (sun) define the metre as 1 metre, the we have to divide the 1 x10^-6 by 1 x10^-6 to adjust our reference to our standard one metre?

 

[Jonathan Scott]

Hi Johnathan,
Thanks for your patience. Where does the 1 come into the formula. My understanding of the nature of the "multiplactive factor" seems to be the problem. May I give you the way I am reasoning, because although I see what you are doing, I obviously still don't understand the reasoning. Can we confine the measurement to one potential ... let's say our sun at about 27000 light years. This returns the "warp factor" as about 1 x10^-6. Surely if we on Earth (sun) define the metre as 1 metre, the we have to divide the 1 x10^-6 by 1 x10^-6 to adjust our reference to our standard one metre?

The ratio of a clock rate or (assuming an appropriate coordinate system) ruler size at one place relative to another is the same as the ratio of the TOTAL energy of identical objects at those two locations. Those two total energy values only differ by the potential energy, which is a very tiny fraction of the total. In the Newtonian approximation, that tiny fraction is -Gm/rc². We can therefore represent the total energy of an object in a given potential at distance r from mass m as being multiplied by a factor (1-Gm/rc²) compared with what the total energy would be at infinity.

 

[Pierre007080]

The ratio of a clock rate or (assuming an appropriate coordinate system) ruler size at one place relative to another is the same as the ratio of the TOTAL energy of identical objects at those two locations. Those two total energy values only differ by the potential energy, which is a very tiny fraction of the total. In the Newtonian approximation, that tiny fraction is -Gm/rc². We can therefore represent the total energy of an object in a given potential at distance r from mass m as being multiplied by a factor (1-Gm/rc²) compared with what the total energy would be at infinity.

Jonathan, You must be exasperated by my persistence. I would love to just say that I understand, but please hear me out for the last time, then I will wallow in my ignorance. When Einstein compared gravitation to a spinning disc, he used the speeds v0 at the centre of the disc and v further out on the radius and used the difference of potential of the centrifugal force to define the potential energy between the centre (v0) and the point spinning at speed v. He then used the difference (v0 - v)/v0 to determine the RATIO of the potential from the CENTRE. The GM/rc^2 he derived from this RATIO has nothing to do with the fact that we defined the metre as one or the second as one. We are sitting at a potential of 1.0 x 10^-6. Is this not the RATIO of the difference (V0 - V) to the V0 potential at infinity radius of the galaxy? This is why I ask where the 1-GM/rc^2 comes from when GM/rc^2 is already a ratio?

 

[Jonathan Scott]

Jonathan, You must be exasperated by my persistence. I would love to just say that I understand, but please hear me out for the last time, then I will wallow in my ignorance. When Einstein compared gravitation to a spinning disc, he used the speeds v0 at the centre of the disc and v further out on the radius and used the difference of potential of the centrifugal force to define the potential energy between the centre (v0) and the point spinning at speed v. He then used the difference (v0 - v)/v0 to determine the RATIO of the potential from the CENTRE. The GM/rc^2 he derived from this RATIO has nothing to do with the fact that we defined the metre as one or the second as one. We are sitting at a potential of 1.0 x 10^-6. Is this not the RATIO of the difference (V0 - V) to the V0 potential at infinity radius of the galaxy? This is why I ask where the 1-GM/rc^2 comes from when GM/rc^2 is already a ratio?

Gm/rc² is the ratio of potential energy (extra energy compared with being at some other specific location) to total energy, primarily consisting of rest energy due to rest mass. This ratio (which is typically extremely small) is useful in some contexts.

However, if you want the ratio of clock rates and ruler sizes etc, you need the ratio of the total energy at different locations. (1-Gm/rc²) is the ratio of the total energy compared at infinity and at distance r from the central mass. If you want the ratio of total energy between two places where neither of them is at infinity, you take the ratio of 1-Gm/rc² values at the two locations, but since the potential part is tiny, that ratio is approximately 1-(difference in Gm/rc²).

 

[Pierre007080]

Jonathan, Thanks for your time. I am going to read up some more along the lines you have indicated and come back when I am more informed.Regards.
Pierre