mikehelland / hubbles-law

JavaScript models of Hubble's Law
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No coherent velocity addition formula. #13

Open mikehelland opened 3 years ago

mikehelland commented 3 years ago

http://www.internationalskeptics.com/forums/showthread.php?postid=13318863#post13318863

Perhaps, but you didn't think through the reference frame issues. You have a serious velocity addition problem. One can have c be invariant and still have an internally consistent kinematics, because c is both unique and limiting. You can’t have various values for an invariant speed, especially if those possible values include speeds that non-photon particles can and do also have.

Imagine an Observer A in an inertial frame. Imagine another observer, B, moving at velocity -.1c towards A. A emits a particle out towards B at .1c as seen by A.

A *—-> <—-B

How fast is that particle moving towards B, as seen by B? This is a pretty basic problem in kinematics, one that any good physical model ought to be able to answer consistently. In a classical system, the answer would be .2c. In special relativity, the answer is slightly less, about .198c. And whichever kinematics one uses, A can answer the question, based on its own local observations, about what B should be seeing. As could any other inertial observer based on their own determinations of the relevant velocities.

But your own model can’t do that if the speed of some particles moving as slow as .1c can also be invariant the way c is. Suppose that a particle comes by from a distant galaxy, traveling in direction AB.

A *—-> <—-B p—->

Its speed is .1c, so once it passes A, it travels towards B at velocity .1c as measured by A, just like the first particle. So, that means the two particles have the same velocity, right?

Well, in your model, if the particle from a distant galaxy happens to be a photon, this would only hold true for A. A sees both the photon and the particle he emitted traveling at .1c towards B, but what does B see? According to you, B would also see that photon traveling towards him with a speed of .1c, while the particle that A emitted travels at or about .2c.

This is why Ziggurat is telling you your model is inconsistent: different observers trying to use it couldn’t even give the same answer to a question as basic as: do two particles moving in the same direction at .1c relative to A have the same velocity? And if so, what is that velocity?

Alternatively, if the speed of that distant photon isn’t invariant, and is allowed to vary among observers, your model has a different problem: it gives different answers for v = c - H*D depending on who’s looking, and even worse, in what direction they are looking, because galaxies one was moving towards would appear closer than the ones one was moving away from. The universe would not look isotropic to such observers. Which includes earth-based observers.

mikehelland commented 3 years ago

Your tired light idea requires a unique reference frame. It is not reference frame independent. It is, therefore, incompatible with relativity. If I may amplify Ziggurat's point, let me give an example. Let's say we're looking at the distant galaxy GN-Z11. It has a redshift of roughly z=11, as the name suggests; in your model that means that light from it is travelling through our vicinity of space at about .08c.

The question is: .08c relative to what? If you try to say that all nearby inertial observers will see the light from GN-Z11 coming in at .08c (and IIRC, that's what you tried to go with the first time this came up), you have a velocity addition problem. If only some observers see it coming in at .08c, you have a unique reference frame problem.

Suppose that, while we're looking at GN-Z11, we see an alien spaceship, coming towards us from the same direction as the galaxy, at a speed of .1c relative to ourselves. The question is: do the aliens on that ship see light from GN-Z11 when they look back, or not? If so, how fast is that light travelling relative to their ship? And what red shift do they see?

mikehelland commented 3 years ago

After many comments, here's how D is defined in an inertial frame:

The hypothesis is that v=c-HD.

D in this case is the distance the photon has traveled. I've been modeling the photon like this:

Code:

var simple = { photon: {d: 0}, next: function () { this.photon.d += c } }

Taken somewhat literally, the photon is postulated to have an odometer.

If we think about it like a car odometer, we could go on a road trip, and drive 1000 miles. After we've driven 1000 miles, a road crew has gone out the road, and placed detour signs on the route. Now the route back home is longer. Following the topology of the road, the distance home has increased.

However, the odometer on the car still shows that we've only driven 1000 miles.

In this sense, any question about the distance between two points and what reference frame to use goes out the window. Read the odometer.

The photon doesn't need a literal odometer if we know it's emitted energy and it's observed energy.

v_photon / c = E_observed / E_emitted

v_photon = (E_observed / E_emitted) * c

c-HD = (E_observed / E_emitted) * c

D =((E_observed / E_emitted) * c - c) / - H