Dimensionless and dimensioned fundamental constants

[Lapidus]

There are 25 or so dimensionless constants in the standard models, such as the masses of the fundamental particles (that can be divided by Planck mass or some other mass to become dimensionless).

And there are the three dimensionful constants c, h, G (speed of light, Planck's constant, Newton gravitational constant).

I understand that the latter can have different units, given different conventions. BUT they are still constants after all! Would not our universe look very different if these three constants were different? Put another way, the Planck units of time, length and mass that we get from using c, h and g would be different when c, h, G were different. Would different Planck units not affect the nature of our Universe?

So why do people say that the unexplained parameters are only the dimensionless? In which sense are c, h and G already explained? Why does it not matter if c, h, G were different?

thanks

 

[phyzguy]

Suppose I re-defined the second so that there were 12 hours in a day, with each hour 3600 seconds long. Then c, h, and G would all have different values ( c, for example, would be twice as big), and yet the universe would look no different than it does today. This is what is meant.

 

[Lapidus]

Suppose I re-defined the second so that there were 12 hours in a day, with each hour 3600 seconds long. Then c, h, and G would all have different values ( c, for example, would be twice as big), and yet the universe would look no different than it does today. This is what is meant.

Suppose the speed of light were only a tenth of what it is and Planck's constant were 100-fold bigger. Would not that mean that relativistic effects and quantum effects were bigger?

Does not c, h, G tell us how relativistic, quantum-physically and spacetime-curved the universe is?

 

[phyzguy]

OK, let's re-define the second so there are 24 hours in a day and 36000 seconds in an hour. Then c is only 1/10 of what it is now. Have relativity effects gotten any bigger? The point is that you need to separate changes that are due to changes in your units from real changes in the physics. This is why people focus on the dimensionless constants. If the fine structure constant is changed by 10X, then things are really different.

 

[PeterDonis]

Would not our universe look very different if these three constants were different?

This question can't be answered as it stands, because there is no well-defined meaning to a dimensionful constant being "different". What you should be asking is, how would the universe look different if particular dimensionless constants were different?

For example, asking what the universe would look like if $c$ were different is not well-defined; you have to ask what the universe would look like if the fine structure constant $\alpha$ (i.e., the dimensionless coupling constant for the electromagnetic interaction) were different.

Does not c, h, G tell us how relativistic, quantum-physically and spacetime-curved the universe is?

No.

First of all, you're mixing up different kinds of things. The universe being "relativistic" and "quantum-physical" are binary properties. Since the universe is Lorentz invariant, it's relativistic; that's a binary property. In terms of $c$, it equates to $c$ being finite, or more precisely, to there being a finite invariant speed; the numerical value of that speed has no direct physical meaning, the only physical fact is that it's finite. Similarly, since Planck's constant $h$ is finite, the universe is quantum-physical; the numerical value of $h$ has no direct physical meaning, it depends on our choice of units.

How curved spacetime is, however, is not a binary property: it depends on how much stress-energy there is, and that can vary continuously from place to place (or time to time), which means spacetime curvature can vary continuously from place to place (or time to time). The constant $G$, like $c$ and $h$, depends on your choice of units, so its numerical value has no direct physical meaning; but the property of the universe that it represents is not how curved spacetime is, but how strongly spacetime curvature is coupled to stress-energy density. One way of describing the issue with constructing a theory of quantum gravity is that we don't know how to express this coupling in a dimensionless way, unlike with the Standard Model interactions.

 

[Dale]

Would not our universe look very different if these three constants were different?

Not in the slightest. All that would look different is our units. This is actually an enjoyable exercise to go through.

In which sense are c, h and G already explained?

They are explained in the sense that our choice of units fully determines the values of those constants. There is no physical experiment which can be done to determine their values independently of the choice of unit.

 

[Dale]

See

https://www.physicsforums.com/showpost.php?p=2011753&postcount=55

https://www.physicsforums.com/showpost.php?p=2015734&postcount=68

http://math.ucr.edu/home/baez/constants.html

 

[ftr]

c, h and G already explained?

where did you see that statement, it certainly looks wrong.

 

[Lapidus]

OK, let's re-define the second so there are 24 hours in a day and 36000 seconds in an hour. Then c is only 1/10 of what it is now. Have relativity effects gotten any bigger? The point is that you need to separate changes that are due to changes in your units from real changes in the physics. This is why people focus on the dimensionless constants. If the fine structure constant is changed by 10X, then things are really different.

I do not want to redefine anything. I want to measure the speed of light and I want to use the metres and seconds as commonly defined. Are you saying if the speed of light turns out be 792458 m / s or 458 m / s instead of 299792458 m / s we would not notice any changes?

 

[jbriggs444]

I do not want to redefine anything. I want to measure the speed of light and I want to use the metres and seconds as commonly defined. Are you saying if the speed of light turns out be 792458 m / s or 458 m / s instead of 299792458 m / s we would not notice any changes?

The speed of light [in meters per second] is a defined constant. The meter is defined in terms of the second and the speed of light. Given those definitions, it is impossible to measure the speed of light.

However, it is possible to measure the length of your foot [in meters] using those definitions.

In days gone by it would have been impossible to measure the average length of the left feet of 16 people exiting from the church on Sunday -- it would have been one foot by legal definition:

"Stand at the door of a church on a Sunday and bid 16 men to stop, tall ones and small ones, as they happen to pass out when the service is finished; then make them put their left feet one behind the other, and the length thus obtained shall be a right and lawful rood to measure and survey the land with, and the 16th part of it shall be the right and lawful foot."

 

[Lapidus]

This question can't be answered as it stands, because there is no well-defined meaning to a dimensionful constant being "different". What you should be asking is, how would the universe look different if particular dimensionless constants were different?

Not well-defined? Does light reach the moon in one second or in two minutes? I still fail to see why this is a not well-defined question.

 

[ftr]

Not well-defined? Does light reach the moon in one second or in two minutes? I still fail to see why this is a not well-defined question.

That is because we only care how they show up( affect) in the equations, that is why we can define many such unit systems. The distance to the moon is not fundamental.
https://en.wikipedia.org/wiki/Natural_units
https://arxiv.org/pdf/1412.2040.pdf

 

[Lapidus]

Not in the slightest. All that would look different is our units. This is actually an enjoyable exercise to go through.

They are explained in the sense that our choice of units fully determines the values of those constants. There is no physical experiment which can be done to determine their values independently of the choice of unit.

If light would reach the moon in two minutes instead of one second, my units would change? How?

 

[Dale]

If light would reach the moon in two minutes instead of one second, my units would change? How?

That would mean that your unit of time is approximately 2 "minutes" = 1 second.

 

[Nugatory]

I want to measure the speed of light and I want to use the metres and seconds as commonly defined.

The meter is defined to be the distance that light travels in $\frac{1}{299792458}$ seconds. Thus, there's no such thing as "measuring the speed of light" - what you're really doing is checking the accuracy of your meter stick (assuming that you trust your clock).

Are you saying if the speed of light turns out be 792458 m / s or 458 m / s instead of 299792458 m / s we would not notice any changes?

We'd be throwing out a lot of meter sticks and redesigning a lot of machine tool readouts... But as far as any real physics changing, it's no different from switching an economy over from inches and feet to the metric system.

 

[PeterDonis]

Does light reach the moon in one second or in two minutes?

It depends on how you define "seconds" and "minutes". There is nothing in the laws of physics that says we must define our units in a particular way--for example, defining a "second" such that it takes about 1.25 seconds for light to go from the Earth to the Moon (to be somewhat pedantic, this is in a frame in which the Earth is at rest).

I still fail to see why this is a not well-defined question.

The issue is not that it isn't a well-defined question. The issue is that it is a question about how we define our units, not about physics.

 

[Lapidus]

I'm convinced now.

If I choose Planck units for measuring, it becomes super-obvious that whenever I measure a different c,h,G my Planck units change as well.

Thanks everybody!

 

[Lapidus]

First of all, you're mixing up different kinds of things. The universe being "relativistic" and "quantum-physical" are binary properties. Since the universe is Lorentz invariant, it's relativistic; that's a binary property. In terms of $c$, it equates to $c$ being finite, or more precisely, to there being a finite invariant speed; the numerical value of that speed has no direct physical meaning, the only physical fact is that it's finite. Similarly, since Planck's constant $h$ is finite, the universe is quantum-physical; the numerical value of $h$ has no direct physical meaning, it depends on our choice of units.

How curved spacetime is, however, is not a binary property: it depends on how much stress-energy there is, and that can vary continuously from place to place (or time to time), which means spacetime curvature can vary continuously from place to place (or time to time). The constant $G$, like $c$ and $h$, depends on your choice of units, so its numerical value has no direct physical meaning; but the property of the universe that it represents is not how curved spacetime is, but how strongly spacetime curvature is coupled to stress-energy density. One way of describing the issue with constructing a theory of quantum gravity is that we don't know how to express this coupling in a dimensionless way, unlike with the Standard Model interactions.

That's good stuff. What got me confused, among other things, was the Bronstein cube, that also appears in the opening chapter of Zee's "Einstein gravity", which makes you think that by increasing c, h or G gets you deeper in the domain of the respective theories.

 

[Lapidus]

Perhaps, one final question. What physical meaningful observation/ conclusion can we make about the Planck units? They are special units after all, ratios of three (dimensional) constants. The fact that the Planck mass is many orders higher than Planck time and length, would be one, as I read many time before. Anything else?

 

[jbriggs444]

Perhaps, one final question. What physical meaningful observation/ conclusion can we make about the Planck units? They are special units after all, ratios of three (dimensional) constants. The fact that the Planck mass is many orders higher than Planck time and length, would be one, as I read many time before. Anything else?

The Planck mass is a unit of mass. It can be neither greater than nor less than a unit of time. The two are not commensurable.

The fact that the Planck time (in seconds) is numerically less than the Planck mass (in kilograms) is as much a factoid about the size of the second and the kilogram as it is a factoid about the size of the Planck time and the Planck mass.

 

[Lapidus]

The Planck mass is a unit of mass. It can be neither greater than nor less than a unit of time. The two are not commensurable.

The fact that the Planck time (in seconds) is numerically less than the Planck mass (in kilograms) is as much a factoid about the size of the second and the kilogram as it is a factoid about the size of the Planck time and the Planck mass.

That makes sense.

But when we compare Planck mass and Planck time with other mass and time scales. Like described here

 

[Dale]

The Planck mass is a unit of mass. It can be neither greater than nor less than a unit of time. The two are not commensurable.

Well, that also depends on your system of units. In geometrized units mass and time both have units of length. So in geometrized units it does make sense to compare the Planck mass to the Planck time. In those units they are equal.

 

[Lapidus]

Some last words before this thread disappears into oblivion.

It is always amazing to see how much physics and deep insight someone can gain from dimensional analysis!

 

[Meir Achuz]

I haven't read all the other answers, but my answer is that a number with dimensions, like cm/sec, cannot be a fundamental physical quantity. Its value is whatever the unitarians pick for their units.

 

[Dale]

whatever the unitarians pick for their units.

Hahaha!

 

[sophiecentaur]

Some last words before this thread disappears into oblivion.

It is always amazing to see how much physics and deep insight someone can gain from dimensional analysis!

Runner has it that Mr Stokes produced his theorem when solving an Exam question that was set as being insoluble. He did it using D.A..