# Does space resist the passage of light?

Please excuse my naive use of algebra.

$$R_{s}=c\frac{\phi_1 \phi_2}{r^2}$$

I have concocted this equation in an attempt to show that the stuff of which space is made resists the passage of light.
I wonder how more experienced members of this forum would interpret the equation?
R_s is the resistivity of space, c is the speed of light in a vacuum, ø is the flux of light in a location and r^2 is distance.

Does this make any sense at all?

It is not the velocity but the power of light which I think may encounter resistance as it moves through space. The velocity of light is constant across distance, its power is not.
It is the reduction of power that I am trying to describe.

Any better?

I doubt the above equation has any equation has meaning, either, and if it did... A lot more information would be required to show that it does.. I don't even think "impedance of free space" is a (known) real property of free space (?)

But, I think saying "the speed of light in free space is based on Maxwell's equations" isn't right either.... The actual speed of light in vacuum isn't based on anything known.. It is just measured "property" of the universe, based on something we don't really understand fully. Right?

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Doc Al
Mentor
It is not the velocity but the power of light which I think may encounter resistance as it moves through space. The velocity of light is constant across distance, its power is not.
It is the reduction of power that I am trying to describe.
Perhaps you're trying to describe the inverse square law, which shows how the intensity (but not the total power) from a point source drops off with distance? See http://hyperphysics.phy-astr.gsu.edu/HBASE/vision/isql.html#c1"

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I doubt the above equation has any equation has meaning, either, and if it did... A lot more information would be required to show that it does.. I don't even think "impedance of free space" is a real property of free space (?)

But, I think saying "the speed of light in free space is based on Maxwell's equations" isn't right either.... The actual speed of light in vacuum isn't based on anything known.. It is just measured "property" of the universe, based on something we don't really understand fully. Right?

I was going to ask a question related to this actually, why is the speed of light what it is?

Even if metres and seconds are some arbitrary human concoction that doesn't change the fact that the speed of light is a a certain thing. I have heard the weak anthropic principle, saying that 'if the speed of light were different the universe would be different and we wouldn't be here to ask that question', but i think it is a weak answer to a cool question...

But, I think saying "the speed of light in free space is based on Maxwell's equations" isn't right either.... The actual speed of light in vacuum isn't based on anything known.. It is just measured "property" of the universe, based on something we don't really understand fully. Right?

I think the speed of light as its known is based on the energy of a photon and its wavelength:

$$E = hc$$$$/$$$$\lambda$$

$$c = E\lambda/h$$

On another note wouldn't the equation above allow for super luminous (FTL) photons where E is greater. So when it comes to virtual particles c is truely a barrier that can be tunneled?

Frank

The speed of light in a vacuum is constant everywhere in any inertial system, but we do not know what specifically determines this velocity. Nevertheless, we have named two "constants", the permeability of free space u0, and the permittivity of free space e0, and have via their placement in Maxwell's equations defined them as
1/sqrt(u0e0) = 2.99792458 m/s (speed of light in vacuum), and
sqrt(u0/e0) = 376.7254 ohms (the impedance of free space).

ZapperZ
Staff Emeritus

It is not the velocity but the power of light which I think may encounter resistance as it moves through space. The velocity of light is constant across distance, its power is not.
It is the reduction of power that I am trying to describe.

Any better?

That makes even less sense.

Start this way: show how what you are proposing is actually consistent quantitatively with what we have observed now. After all, we have a huge range of laser power as it is already.

And just in case you missed it, the https://www.physicsforums.com/showthread.php?t=5374" that you agreed to already allow for very little leeway for personal unverified theory.

Zz.

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The speed of light in a vacuum is constant everywhere in any inertial system, but we do not know what specifically determines this velocity.

Shouldn't the real question be: What determines the mass of a photon? Given the realtionship of engery, wavelength and the Planck's constant leaves only one answer for c.

But as I noted earlier c could change by a photon tunneling with more mass (engery) since its wavelength is fixed.

Frank

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f95toli
Gold Member
I don't even think "impedance of free space" is a (known) real property of free space (?)

Well, the impedance of free space is just $Z_0=1/\epsilon_0 \mu_0$ and is about 377 Ohm. It is certainly a "real property" in that it has real, measurable effects. However, it is obviously just a reactive impedance, there is no power loss (meaning free space can sort of be thought of as a lossless transmission line with an impedance of 377 ohms).
A good example of this is an antenna which can be thought of as a device that matches the impedance of an electrical circuit to that of free space in order to maximize power transfer.

There are also other -more subtle- effects. One interesting example is that if you try to study quantum mechanical effects of macroscopic electrical circuits you will find that the impedance of free space is effectively shunting your circuit which in turn leads to a "leakage" and therefore decoherence. Nowadays we've learned to design our circuits to minimize this effects, but if no precautions are taken the measured shunting impedance works out to be about 100 Ohms just as one would expect(the exact value predicted by theory is Z0 divided by pi).

OmCheeto
Gold Member
Please excuse my naive use of algebra.

$$R_{s}=c\frac{\phi_1 \phi_2}{r^2}$$

I have concocted this equation in an attempt to show that the stuff of which space is made resists the passage of light.
I wonder how more experienced members of this forum would interpret the equation?
R_s is the resistivity of space, c is the speed of light in a vacuum, ø is the flux of light in a location and r^2 is distance.

Does this make any sense at all?

http://en.wikipedia.org/wiki/Tired_light#General_features_of_tired_light_models"

E(x) = E(0)e − x / R

I used to love to think about it. Then someone said it was crackpottery. I had to give it up.

Then I discovered http://en.wikipedia.org/wiki/Quantum_foam" [Broken].

I decided that light traveling through free space might be sapped by the mechanism of creating said foam, as a function of the cosmological constant; each wavelength giving up a minute amount of energy, unhindered in velocity, but forever weakening.

Supposedly, http://www.cfa.harvard.edu/~sblondin/publications/timedilation/timedilation.pdf" [Broken] written a couple of years ago put the final nail in the tired light coffin, which made me quite happy as I very seldom spend much time on my outlandish ideas of which I have very little background in the first place.

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Cyosis
Homework Helper
When you come up with an equation make sure that at least the units are the same on both sides. This doesn't seem to be the case with your equation therefore it must be wrong.

That makes even less sense.

Start this way: show how what you are proposing is actually consistent quantitatively with what we have observed now. After all, we have a huge range of laser power as it is already.

And just in case you missed it, the https://www.physicsforums.com/showthread.php?t=5374" that you agreed to already allow for very little leeway for personal unverified theory.

Zz.
Yes, sorry. I can see that I was being 'overly speculative'.

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diazona
Homework Helper
Well, the impedance of free space is just $Z_0=1/\epsilon_0 \mu_0$
Shouldn't that be $\sqrt{\mu_0/\epsilon_0}$ to get the units of an impedance?

f95toli
Shouldn't that be $\sqrt{\mu_0/\epsilon_0}$ to get the units of an impedance?
Indeed it should I guess I was thinking of $1/\epsilon_o c_0$ when I wrote down the equation above.