Creation of Photons

[New Science]

right. A km (kilometer) is a unit of distance. And a mpc (megaparsec) is a unit of distance. Therefore, 72 km/mpc/s has units of inverse time.

 

Period.

 

The math you posted doesn't have the right units and it therefore 100% wrong. If you can't even get the units correct when performing a calculation, any interpretation of that calculation is completely wrong and meaningless and useless.

 

Start getting the units right (on this particular calculation, and in your your thread about the "MAJOR" discovery) and then maybe we can talk slightly intelligently about the interpretation of those calculations.

 

But, without the correct units on your calculations, you are spouting nonsense.

 

I might as well say "I have to drive 47 Kelvin in my car to get to work every day, and my car gets 26 amperes per mole. I get paid 0.0812 deciliters per lumen squared. On the way home, I buy dinner at the grocery store and pay an average of 14.87 hair follicles." Do you see how ridiculous this gets? Without correct units, I might as well be talking in a foreign language.

 

I asked you how do you incorporate the mpc into a formula.

Is the BB expanding at 72 kms/s or 72kms + a mpc /second.

 

Both cannot be right. So which one is right?

 

NS

[Klaynos]

What "constant" is this, and were are you getting teh value from, link? Hubble constant I see now I was just missing where it was mentioned...

 

And if you've got the value for it in a formula when doing dimensional analysis you have to take into account all the units.

Edited July 3, 2008 by Klaynos

[D H]

I asked you how do you incorporate the mpc into a formula.

Is the BB expanding at 72 kms/s or 72kms + a mpc /second.

 

Both cannot be right. So which one is right?

Neither is right. Your lack of understand of the meaning of the Hubble constant can only mean that you don't understand much about the underlying concept of the expansion of space. How can you possibly disagree with something that you don't even understand?

 

A rubber band analogy to help you understand: Imagine a long rubber band that is being stretched so the length increases at a constant rate. The distance between any two points away from the ends will also increase at a constant rate, but that rate will be less than expansion rate of the rubber band as a whole.

 

Now consider some specific point on the rubber band. Relative to this specific point, the time derivative of the distance to some other point on the rubber band will be proportional to the distance to that point. In other words, [math]\dot d = k d[/math] Solving for the proportionality constant, [math]k = \dot d / d\ [/math] This constant has units of length/time/length, or just inverse time.

 

While the proportionality constant has units of inverse time, it is more convenient to express the proportionality constant in terms of velocity per distance because that is exactly the nature of the relation that the proportionality constant addresses. Now suppose the expansion rate [math]\dot d[/math] is measurable only at very large distances. In this case, it makes sense to express the velocity and length using different units of length.

 

This is exactly the situation with the Hubble constant. It measures the expansion of space just as the rubber band constant measured the stretching of the rubber band. While both the Hubble constant and rubber band constant have primitive units of inverse time, it is more convenient to express the constants in terms of velocity per distance because that is the purpose of these constants. The Hubble constant is expressed in kilometers/second/parsec: velocity per distance, or inverse time.

[New Science]

What "constant" is this, and were are you getting teh value from, link?

 

And if you've got the value for it in a formula when doing dimensional analysis you have to take into account all the units.

 

I am talking about the expansion of the BBT.

The latest value for this was extracted from the WMAP calculations and it is 70.1 kms/mpc/s .

 

If you include the mpc in the expansion of the BB, then the true rate of expansion would be 3.24^-23 meters/mpc/second.

This would not be a detectable expansion.

 

NS

[D H]

Show your math. Last time I checked, [math]70.1\, \text{km}/\text{sec}/\text{mpc} = 70100 \, \text{m}/\text{sec}/\text{mpc}[/math].

[Bignose]

72kms + a mpc /second.

 

 

What does this + notation even mean? I don't even know how to interpret this.

 

Anywho...

 

The constant is 72 km/mpc/s.

 

This can also be written as:

 

[math] 72 \frac{km}{mpc \cdot s}[/math]

 

Now, 1 megaparsec is 3.08*10^19 km

 

So, if you multiply the above fraction by that conversion you get:

 

[math] 72 \frac{km}{mpc \cdot s} \cdot \frac{1}{3.08 \cdot 10^{19}}\frac{mpc}{km} = 2.34 \cdot 10^{-18} \frac{1}{s}[/math]

 

the length units (the km and the mpc) and you get units of inverse time.

 

This isn't even physics 101, this is high school stuff. Haven't you ever had to convert a mile into a kilometer? Or a U.S. gallon into a Liter? I have serious, serious doubts about your claim of working as a scientist when you cannot even get simple unit conversions correct.

[New Science]

What does this + notation even mean? I don't even know how to interpret this.

 

Anywho...

 

The constant is 72 km/mpc/s.

 

This can also be written as:

 

[math] 72 \frac{km}{mpc \cdot s}[/math]

 

Now, 1 megaparsec is 3.08*10^19 km

 

So, if you multiply the above fraction by that conversion you get:

 

 

[math] 72 \frac{km}{mpc \cdot s} \cdot \frac{1}{3.08 \cdot 10^{19}}\frac{mpc}{km} = 2.34 \cdot 10^{-18} \frac{1}{s}[/math]

 

the length units (the km and the mpc) and you get units of inverse time.

 

This isn't even physics 101, this is high school stuff. Haven't you ever had to convert a mile into a kilometer? Or a U.S. gallon into a Liter? I have serious, serious doubts about your claim of working as a scientist when you cannot even get simple unit conversions correct.

 

You do not use kms in your calculations because the SI unit is the 'meter'.

Secondly, you do not include seconds in the calculation because the calculations are based on one second.

 

So to include the megaparsec in the expansion of space, you just divide

72^6 meters by a mpc in meters +72^6 to get the correct rate of expansion.

 

This equals 2.33^-15 meters per second.

 

The point here is that the Hubble expantion is spread out over a mpc.

 

NS

[ajb]

You do not use kms in your calculations because the SI unit is the 'meter'.

Secondly, you do not include seconds in the calculation because the calculations are based on one second.

 

So to include the megaparsec in the expansion of space, you just divide

72^6 meters by a mpc in meters +72^6 to get the correct rate of expansion.

 

This equals 2.33^-15 meters per second.

 

The point here is that the Hubble expantion is spread out over a mpc.

 

NS

 

(Why am I doing this?)

 

You are free to use any set of units you want. As long as you are consistent there is no problem.

 

Seconds in a unit of time. It must be included. Unless you mean something like "second per second" which is indeed unitless. But you have to say this.

[Bignose]

 

The point here is that the Hubble expantion is spread out over a mpc.

 

 

There is nothing sacrosanct about any unit of length. They are just conversion factors from each other. If Hubble expansion is spread out over megaparsecs, then it is also spread out over microns. A different number of micros -- determined by the conversion factor -- but one length is just as good as any other length.

 

If Hubble expansion is spread out over megaparsecs, then it is also spread out over Angstroms, astronomical units, cubits, chains, fathoms, feet, furlongs, inches, leagues, light-days, microns, miles, nautical miles, parsecs, and yards, not to mention any of the even more esoteric units of length like finger or palm or cable length or telegraph mile.

 

And when dealing with a constant, you can't just "not include" of a unit if you don't want it there. If you want something to last a second, you have to multiply by that second. You have to explicitly state that, and keep track of the units. In this case, if you multiply Hubble's constant by a period of time, then you're going to be left with a dimensionless number.

 

Look, if you're trying to calculate something, you have to keep track of the units. There is no ifs, ands, or buts about this. You know this, whether you've consciously thought about it or not. If someone gave you a rope with a length of 22 inches and tied another rope that had a length of 2 meters to the end of the first one, you'd convert one or the other to get the total length. You wouldn't just sit there confused wondering what good a rope of 2 m and 22 inches is. You'd figure out that it is 100 3/4 inches long or 2.55 m long.

 

As another example, if you had a 1000 US dollar bill, a 1000 Canadian dollar bill, a 1000 Euro bill, and 100 000 yen bill, you traveled to the UK, you would find it very hard to spend that money in its current form. You'd convert it to British pounds.

 

You know what conversion factors are. You wouldn't be able to live daily life without them. This calculation is the same thing. When a constant has units of inverse time, you can't just "not include" the unit of time. Anymore than I can describe the length of rope using units of brightness like the lumen or anymore than I can describe how much money you have in gallons. You can't just change the units. The units are what give the number meaning. If you were trying to build a table and you asked how long the legs are supposed to be, and i just said "1", you'd have to ask "1 what?". It is supposed to be 1 foot tall like a coffee table or is it supposed to be 1 yard tall more like a dining table? Or 1 meter? Just saying "1" is completely meaningless.

 

Just like not including the units on Hubble's constant, or the speed of light, or Planck's constant or the gas constant or the gravitational constant or any other physical constant is meaningless. You don't get to change or "not include" a unit because you don't like it in there. You can use the constants in a calculation -- where you multiply, divide, add and subtract other physical qualities also with units -- and then try to interpret the results. But if the units don't match in the calculations, you have nonsense again. You have to observe very detailed bookkeeping with the units to ensure that your calculations are correct. Just like you check your paycheck to ensure you were paid the correct amount, and you check your bank statement to ensure that no errors were made. You have to check the units to make sure that they come out correctly. There is no other choice.

 

NS, I have tried to be very patient, but this is some very basic stuff. Like I wrote above, you know how important conversion factors are -- you wouldn't survive day to day life without some knowledge of them. Conversion factors are just as important in physics calculations.

 

Now,let's this simple question: before doing any calculations with it, before using it at all -- what are the units of Hubble's constant? Not just the specific units in any one version of them, but what does each specific unit represent, i.e. a yard is a length. So is a km. So is a furlong. A min is a time. So is an hour, or a day. A kg is a mass. etc.

What are the fundamental units of Hubble's constant?

 

I'll give a few examples to build up to answer my question.

 

60 miles per hour is the speed of a car. What are the fundamental units of 60 mph? A mile is a length, and an hour is a time. So 60 mph is a length per time.

 

Here's a more complicated example. I fill a tank from a hose at a rate of 3 gallons per minute, what are the fundamental units of gallons per minute? Well, a gallon is a volume, and volume is a length cubed. And a minute is a unit of time. So, 3 gallons per minute is a length cubed (length^3) per time.

 

Now let's look at an even more complicated example. Let's say I own a farm and I am using a sprayer to put insecticide on my crops. Based on the speed I drive the sprayer and the speed I set the flow rate of the sprayer, I find how many gallons of insecticide I out in the crops per acre. Let's say it 5 gallons per acre. Now, let's look at the fundamental units. We saw above that a gallon is unit of length cubed. An area is a unit of area, which is a length squared. A length cubed divided by a length squared is just a length. The fundamental unit of 5 gallons per acre is a length. Now, you may ask, well why don't we just express all things in terms of fundamental units? Well, if your neighbor farmer and you were talking and you neighbor asked how much insecticide you put down and you said "oh, about 4.6 microns" the other guy would look at you very strangely. But, if you told him "oh, about 5 gallons per acre" then he'd know exactly what you are talking about. Sometimes keeping the numbers in non-fundamental units carries a lot more meaning.

 

In the same way, you should be able to find that the gravitational acceleration at sea level, 9.8 m/s/s had fundamental units of length per time squared.

 

Now, finally, let's look at Hubble's constant. 72 km/mps/s. A km is a length. A megaparsec is also a length. A s is a unit of time. Just like the above example of the sprayer, you have a length in the numerator and in the the denominator, so they cancel. And you are left with just time in the denominator. Therefore, in fundamental units, Hubble's constant has fundamental units of inverse time. It doesn't matter if you express it in inverse seconds or inverse millenia. It is inverse time no matter what unit of time you pick. It also doesn't matter if you express Hubble's constant in terms of meters, or yards or miles, or furlongs or cubits. There is a length on the top and length on the bottom so they cancel out. The reason Hubble's constant is still expressed as km/mpc/s is because, just like the sprayer example, boiling it down to the fundamental units loses some of it's intuitive meaning.

 

But, finally, you can't change what fundamental units a number has, just because you want to. A speed is a length per time. It is wasn't length per time, it isn't a speed. If it was a length per time squared, then it is an acceleration. If a number doesn't have fundamental units of inverse time, then it cannot be an expression of Hubble's constant. Hubble's constant is inverse time. There is no other choice.

 

I hope that you'll take this long post under advisement. I don't have much hope that you will, but I do have some small kernel of hope. I hope that maybe, this time, it'll get through why your calculation here where you just change the units haphazardly is wrong, and your calculations in your other threads where you change units is just wrong. The units are a necessary part of any physical calculation, you cannot just change them however you like. You end up with completely meaningless nonsense unless you keep the units.

[swansont]

The units are a necessary part of any physical calculation, you cannot just change them however you like. You end up with completely meaningless nonsense unless you keep the units.

 

Agreed.

 

If there's any more "new math" or "new" dimensional analysis, then I'll take that as an indication that there's no point in continuing the discussion.

[New Science]

Agreed.

 

If there's any more "new math" or "new" dimensional analysis, then I'll take that as an indication that there's no point in continuing the discussion.

 

Iwould like to revise that expansion above as an added expansion that adds up to the Hubble constant of 72^6 meters per second.

So when you multiply the 72^6 meters x the lifespan of the BB in seconds, it gives the BBT a size of 3.4^9 light years in extent.

This is still a miniscule size when the HDFN can see more than 25 billion light years deep.

 

So this is another example of the unrealistic probability of the BB concept.

 

NS

[Bignose]

Iwould like to revise that expansion above as an added expansion that adds up to the Hubble constant of 72^6 meters per second.

 

 

I am absolutely crying.

 

Hubble's constant does NOT, NOT, NOT have units of meters per second which is a length per time. I don't know how much clearer anyone can make that. Hubble's constant has units of velocity per length, or length per (length * time) or inverse time.

 

If you have an expression with units of m/s, because you have multiplied Hubble's constant by something else, then it is no longer Hubble's constant! I can't multiply [math]\pi[/math] by 2 and still call it [math]\pi[/math]!

[swansont]

I had hoped that my last post would have made this clear. I guess we're done discussing this topic.