- #1
JohnnyGui
- 796
- 51
Good day to you all,
First, I want to let you all know that I'm new at this and that my question could be a bit vague so I'll try and do my best to explain what I want to know.
I read on a forum about the Hubble's value decreasing over time despite the fact that the expansion of the Universe is accelerating. Reading the explanation about this, it says that, since Hubble's value is a speed at a particular distance ((ΔD / Δt) / Dt), the fact that it's decreasing is because the increase in speed can't catch up with the increase in distance change which results in the numerator (ΔD / Δt) having a smaller factor increase than the denominator (Dt) and thus the Hubble's value decreases over time.
After thinking this a bit through, I tried to think of scenarios of the formula in which H would stay constant or even increase and I came to a (probably a wrong) conclusion.
Here's the thing; if there's, for example, an object receding at 6 m/s from a distance of 10m, thus the Hubble being 6m/s / 10m, and there's an acceleration of 1 m/s per second on this object because of expansion, this would make the Hubble's value after 1 second 7 m/s / 16m at first glance. However, this is only the case if the acceleration increase goes in discrete steps of 1 m/s2. If the acceleration was however continuous, this would mean that within that 1 second there are infinitesimally small speed increases of that object (starting from 6 m/s) until it reaches 7 m/s after a second. But this would mean that after that 1 second, the object would have reached a larger distance than 16m because it had a continuous increase in speed in within that second. Thus, the distance increase is always larger in a continuous acceleration than in a discrete one.
The conclusion that a continuous increase in speed means that it's tied to a larger factor increase of the denominator (Dt) led me to think that it's inevitable that the Hubble value must keep decreasing. I can't seem to think of a continuous accelerating expansion in which the numerator and the denominator of the Hubble's value would both be increased by the same factor to keep the Hubble constant. That would only be possible if the acceleration goes in discrete steps.
However, the fact that the Hubble value IS reaching a more or less constant value, makes me think that the accelerating expansion of the Universe does go in somewhat, probably a very very small, discrete steps of Δt or that there's a very small discrete unit of distance.
I'm probably thinking this the wrong way through and I'd appreciate some correction on this if that's the case.
First, I want to let you all know that I'm new at this and that my question could be a bit vague so I'll try and do my best to explain what I want to know.
I read on a forum about the Hubble's value decreasing over time despite the fact that the expansion of the Universe is accelerating. Reading the explanation about this, it says that, since Hubble's value is a speed at a particular distance ((ΔD / Δt) / Dt), the fact that it's decreasing is because the increase in speed can't catch up with the increase in distance change which results in the numerator (ΔD / Δt) having a smaller factor increase than the denominator (Dt) and thus the Hubble's value decreases over time.
After thinking this a bit through, I tried to think of scenarios of the formula in which H would stay constant or even increase and I came to a (probably a wrong) conclusion.
Here's the thing; if there's, for example, an object receding at 6 m/s from a distance of 10m, thus the Hubble being 6m/s / 10m, and there's an acceleration of 1 m/s per second on this object because of expansion, this would make the Hubble's value after 1 second 7 m/s / 16m at first glance. However, this is only the case if the acceleration increase goes in discrete steps of 1 m/s2. If the acceleration was however continuous, this would mean that within that 1 second there are infinitesimally small speed increases of that object (starting from 6 m/s) until it reaches 7 m/s after a second. But this would mean that after that 1 second, the object would have reached a larger distance than 16m because it had a continuous increase in speed in within that second. Thus, the distance increase is always larger in a continuous acceleration than in a discrete one.
The conclusion that a continuous increase in speed means that it's tied to a larger factor increase of the denominator (Dt) led me to think that it's inevitable that the Hubble value must keep decreasing. I can't seem to think of a continuous accelerating expansion in which the numerator and the denominator of the Hubble's value would both be increased by the same factor to keep the Hubble constant. That would only be possible if the acceleration goes in discrete steps.
However, the fact that the Hubble value IS reaching a more or less constant value, makes me think that the accelerating expansion of the Universe does go in somewhat, probably a very very small, discrete steps of Δt or that there's a very small discrete unit of distance.
I'm probably thinking this the wrong way through and I'd appreciate some correction on this if that's the case.