If the universe is truly expanding, why aren't any stars getting dimmer?

[dcorbett]

The intensity of light observed from a given source decreases as the square of the distance between the observer and the source. Hence, if stars are truly receding from the earth, the intensity of light observed from any such star should be getting dimmer as time goes by. But the intensity of light from distant stars is not decreasing. How can this be explained if the universe is truly expanding and stars are really receding?

 

[marcus]

The rate distances are increasing is 1/140 of one percent per million years. Go figure

The expansion of distances is very easy to detect by the redshift it causes. Measuring wavelengths of light can be done very accurately. So this is a sensitive way to measure.

By contrast, waiting around for for a million years to see if a galaxy gets a tiny fraction of a percent dimmer would be a ridiculously crude way to detect expasion. You wouldn't see anything. It's too small a change to measure.

Aren't we lucky to have the redshift method for detecting expansion?

 

[Chalnoth]

The intensity of light observed from a given source decreases as the square of the distance between the observer and the source. Hence, if stars are truly receding from the earth, the intensity of light observed from any such star should be getting dimmer as time goes by. But the intensity of light from distant stars is not decreasing. How can this be explained if the universe is truly expanding and stars are really receding?

In addition to what Marcus said, nearly everything in the sky that we can see with the naked eye is within our own galaxy. Our galaxy is not expanding. The expansion is an average thing that happens to the universe as a whole, but there are little pockets of the universe that are not expanding, pockets where the mass of the local matter has overcome the expansion.

It is only the far-away galaxies that are receding from us on average. And as Marcus notes, their distances are changing too slowly to matter on a scale of a few decades (except, perhaps, with extremely accurate observations).

 

[yogi]

In addition to what Marcus said, nearly everything in the sky that we can see with the naked eye is within our own galaxy. Our galaxy is not expanding. The expansion is an average thing that happens to the universe as a whole, but there are little pockets of the universe that are not expanding, pockets where the mass of the local matter has overcome the expansion.

It is only the far-away galaxies that are receding from us on average. And as Marcus notes, their distances are changing too slowly to matter on a scale of a few decades (except, perhaps, with extremely accurate observations).

If you include the globular clusters, everything we see with the naked eye is part of the milky way -----except one item - M31

 

[Nabeshin]

If you include the globular clusters, everything we see with the naked eye is part of the milky way -----except one item - M31

Which, to top it all off, is getting closer!

 

[Chronos]

The are a few, keyword few, extragalactic bodies visible to the unaided eye.

 

[dcorbett]

If the universe is expanding, then stars must be moving away from us. And as the red shift of a star is proportional to its distance from us, then the red shift of all stars must also be increasing. Is the red shift of any star increasing?

If stars are moving away from us, then they must be getting dimmer. No observable stars are getting dimmer. To say that we would not notice the dimming effect for millions of years is absurd unless the original light intensity of stars is infinite. If we move from the Earth's orbit to that of Mars, the intensity of the light from the sun is significantly less.

If the red shift of stars is proportional to their velocity of recession from us.
This velocity of recession could be given by the equation

v = Hd

v.....is the velocity of recession
H.....is the Hubble constant
d.....is the distance of the star from the Earth

If we put Hubble’s equation into differential equation form, we get:

dx/dt (= x ̇ ) = Hx

dx/dt...is the velocity of recession
H … is the Hubble constant
x … is the distance of the star from the Earth

The solution to this simple first-order differential equation is:

x = Ae^Ht + c

x …….is the distance of the star from the Earth
A and c ……are constants
H …………… is the Hubble constant
t …………… is time

Differentiating x above, we get x ̇= HAe^Ht = Hx.

But Ae^Ht is continuous and differentiable over all time, so we can differentiate x ̇ again and get:

x ̈ = Hx ̇ = H^2 x

Indeed, we can continue differentiating ad infinitum to get implied accelerations and super-accelerations until the cows come home. Clearly, the energy required for such things to be possible is infinite. As it is impossible for new energy to be continuously generated out of nothing, the initial postulate that the red shift of distant stars is a Doppler effect is not borne out by the mathematics.

 

[dcorbett]

Let us take a simple and generally accepted law such as the law of conservation of momentum. Let us say we are dealing with a closed system and are ignoring extraneous factors such as friction and energy dissipation due to heat during the following collision:
We have two masses traveling in the same straight line with the smaller and faster mass following the larger mass. Let mass M1 traveling at a velocity of V1 be following mass M2 traveling at a velocity of V2. After collision, the combined mass (M1 + M2) would then travel at a velocity of V3.
The law of conservation of momentum would state that:
M1V1 + M2V2 = ( M1 + M2)V3

Now let us plug some simple values into this equation – e.g.
M1 = 2 V1 = 7
M2 = 5 V2 = 3
(M1 + M2) = 7

This would give us a calculated value of V3 = 4.14 (to 2 decimal places)

Now let us consider the energy in the system before and after the collision.

Surely 0.5M1 V1^2 + 0.5M2V2^2 = 0.5 (M1 + M2) V3^2.
Thus 49 + 22.5 = 59.99

But the left side of the equation sums to 71.5 while the right sums to 59.99.

So, if the law of conservation of momentum is correct, 11.51 units of energy have magically disappeared as a result of this collision. Thus, either the law of conservation of momentum is wrong or the law of conservation of energy is wrong. They can’t both be right (unless my ignorance has led me into some error in calculation).

By the way, if a black hole is simply a point, how can we have massive black holes and microscopic black holes?

 

[Drakkith]

Black holes are not points. The event horizon is three dimensional and the diameter is larger when the mass of the black hole is larger. What lies behind is unknown.
Also, I don't believe you are using the right math to account for the kinetic energy. You would need to figure out the amount of acceleration provided by the smaller faster object impacting the larger object and figure the final velocity of both objects combined. Assuming there are no losses the two sides should be equal.

 

[Dadface]

dcorbett the mistake you made with your momentum example is that you assumed that the collision should be perfectly elastic.Perfectly elastic collisions are ones where there is no loss of kinetic energy and such collisions are rare.They can happen with microscopic objects for example with gas atoms under certain conditions but never with macroscopic objects such as in your example.When macroscopic objects collide it is total energy that is conserved,the lost kinetic being converted to other forms such as heat energy.