# Why does light travel at the speed it does?

ZapperZ
Staff Emeritus
E and M are the only variables, right? There's no medium that would determine the speed of light, right? So what else could be responsible for the speed of light constant C except some fixed relationship between the two variables?
But you are making a faulty assumption that THAT is the only valid equation to consider. I've given you one other. So which one do you choose? It appears that you simply chose one and ignored the rest of the equations in physics to make your deduction about the "cause" of the speed of light. This is puzzling logic.

Zz.

But you are making a faulty assumption that THAT is the only valid equation to consider. I've given you one other. So which one do you choose? It appears that you simply chose one and ignored the rest of the equations in physics to make your deduction about the "cause" of the speed of light. This is puzzling logic.

Zz.
Zapper, I chose a reasonably simple equation without any dimensionless constants for a very good reason. I'm aware that there are other equations with c in them, but E=Mc^2 was the simplest one I could think of.

I assumed it would be easier for most readers of this thread to relate to, rather than more complex equations, for example, ones including the reduced plank's constant and alpha.

I was trying to keep it simple, and to make a point at the same time.

I wasn't suggesting that the relationship between E and M was the 'cause' of the speed of light, but that the 'cause' of the speed of light is also related to the 'reason' why the relationship between E and M is what it is.

If we can explain WHY the relationship between E and M is a constant, ie c^2, then maybe this will help us understand why the speed of light in a vacuum is what it is.

Raphie, I haven't said E=(Mc)^2, you've not followed the maths.

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Well, my personal opinion is: If the universe is expanding, yet nothing travels faster than light ( to our knowledge ), then the universe is expanding at <= c. Thus this changes the fundamental question to: "why does nothing travel faster than c". And that is a philosophical question. :)

ZapperZ
Staff Emeritus
Zapper, I chose a reasonably simple equation without any dimensionless constants for a very good reason. I'm aware that there are other equations with c in them, but E=Mc^2 was the simplest one I could think of.
So just using the "simplest" equation as the criteria, you propose to use that equation as the foundation for explaining the speed of light? That's it? And you think this is a legitimate criteria without making any consideration to the actual physics? You are comfortable with doing that?

This is getting to be a bit silly. You are saying that just because F=ma is "simple", then m is "caused" by F and a, regardless of the fact that if F=a=0, m is undefined! So an object not acted upon by any force as an undefined mass! All this nonsense is obtained using your principle of using the simplest equation and then deriving profound meaning out of it as containing the "cause" of it. This makes sense to you? If not, apply the same thing to E=mc^2 to massless particles (photons) and see if you think this causes "c".

You are confusing the ability to make QUANTITATIVE measurement of something with the "cause" of something. Nothing in E=mc^2 indicates that E and m are the "cause" of the constant speed of light. Unless you think Einstein is utterly dumb, then you have to argue why HE, of all people, never made such a connection, and physics continues to consider that the constancy of the speed of light is a POSTULATE of Special Relativity.

Zz.

D H
Staff Emeritus
Well, my personal opinion is: If the universe is expanding, yet nothing travels faster than light ( to our knowledge ), then the universe is expanding at <= c.
The expansion rate of the universe is not a velocity; it doesn't even have units of velocity. Even if it did have units of velocity, that still would not mean that the expansion rate is constrained by the speed of light.

So just using the "simplest" equation as the criteria, you propose to use that equation as the foundation for explaining the speed of light? That's it? And you think this is a legitimate criteria without making any consideration to the actual physics? You are comfortable with doing that?

This is getting to be a bit silly. You are saying that just because F=ma is "simple", then m is "caused" by F and a, regardless of the fact that if F=a=0, m is undefined! So an object not acted upon by any force as an undefined mass! All this nonsense is obtained using your principle of using the simplest equation and then deriving profound meaning out of it as containing the "cause" of it. This makes sense to you? If not, apply the same thing to E=mc^2 to massless particles (photons) and see if you think this causes "c".

You are confusing the ability to make QUANTITATIVE measurement of something with the "cause" of something. Nothing in E=mc^2 indicates that E and m are the "cause" of the constant speed of light. Unless you think Einstein is utterly dumb, then you have to argue why HE, of all people, never made such a connection, and physics continues to consider that the constancy of the speed of light is a POSTULATE of Special Relativity.

Zz.
No, Zapper, you've misunderstood.

I've just re-pr=hrased the question, not provided an answer.

There are other ways to re-phrase the question as well, as you have pointed out.

(by the way, photons are energy, they have no mass, but they do travel at the speed of light.If you want me to put this mathematically, E/c^2=0)

The expansion rate of the universe is not a velocity; it doesn't even have units of velocity.
The boundary of an FINITE universe can be expanding and expressed in terms of a velocity.

You could also express it as a change in volume per time. Its a rate so time must be there - there are a few different quantities you can use on top of time

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ZapperZ
Staff Emeritus
No, Zapper, you've misunderstood.

I've just re-pr=hrased the question, not provided an answer.
Really? Let's review, shall we?

Yes. I've just re-phrased the question. Surely this is the reason why speed of light in a vacuum is constant?
The energy constituting the particle internally.

Surely in order to understand why c is a constant we need to explain the relationship between E and M?
You've just claimed that the relationship between E and M is the reason why the speed of light is a constant in vacuum. Where did I misunderstood you here?

(by the way, photons are energy, they have no mass, but they do travel at the speed of light.If you want me to put this mathematically, E/c^2=0)
This is nonsense. E is not zero for a photon, and c isn't zero for a photon. How in the world were you able to convince yourself that non-zero/non-zero = 0? What kind of math are you using?

I think you still have a lot to learn about basic physics before making such outlandish speculation.

Zz.

D H
Staff Emeritus
The boundary of an FINITE universe can be expanding and expressed in terms of a velocity.
Assuming a finite universe, yes you could express the expansion of the universe in terms of a velocity. The diameter, for example. The speed of light would not be a constraint on that velocity.

(by the way, photons are energy, they have no mass, but they do travel at the speed of light.If you want me to put this mathematically, E/c^2=0)
Ash Small, whether correct or incorrect, is this the logic you were getting at?

A) If mass of photon = 0, then e=mc^2 = e = 0*c^2
B) Anything multiplied by 0 = 0, therefore e = 0 for energy of photon
C) If E = MC^2, then by substitution e = mc^2, therefore 0 = 0*c^2
D) By rearrangement of terms, 0/c^2 = 0

If the energy of a photon is not 0 Joules, but its mass is 0 kilograms, then there does seem to be a contradiction here, not a flaw in your ability to perform basic mathematics. Two obvious "kludges" to reconcile that contradiction, it seems to me, again whether wrong or right, is to assume either 1) that photon mass is not equal to 0, or b) that the M term should be considered in terms of n!/|n-1|!

1) That a photon has mass, however small, has been considered by physicists and is an issue of open debate.
2) The factorial argument is no more than a mathematically, but not physically, valid "workaround" to handle 0 that leaves one with the following logical (in isolation), but absurd (not in isolation), statement:

A) If mass of photon = 0, then e=mc^2 = e = (0!/|0-1|!) kg.*c^2 = c^2
B) "Anything" multiplied by 1 = "Anything", therefore, e = c^2 for energy of photon
C) If E = MC^2, then by substitution e = 1 kg. *c^2 = mc^2
D) By rearrangement of terms, e/c^2 = 1 kg.
Which is absurd because it suggests that the mass of a photon and the mass of 1 kilogram are equivalencies in terms of total energies...

To ZapperZ et al.: What do the physicists have to say regarding energy of a photon beyond, for instance:

The minimum energy required to eject an electron from the surface is called the photoelectric work function. The threshold for this element corresponds to a wavelength of 683 nm. Using this wavelength in the Planck relationship gives a photon energy of 1.82 eV.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html

A) Is photon mass 0 or not 0?
B.i) If not 0, then why do text books repeatedly propagate this meme?
B.ii) If actually 0, then why do physicists repeatedly tell those who point out the contradiction that Ash Small has pointed out that it is their logic, basic ability to perform mathematics, and/or knowledge of physics that is faulty rather than the equations they are relying upon?

Best,
Raphie

P.S. I don't mean this post in a confrontational way, and, straight-up, I am not contending that the energy of a photon is either 0 or c^2 Joules, but I do desire to defend Ash Small's basic reasoning capabilities here. Any reasonable person following the logic through, using not "made-up" math, but elementary algebra, might well ask similar questions. Intended as a response to:

This is nonsense. E is not zero for a photon, and c isn't zero for a photon. How in the world were you able to convince yourself that non-zero/non-zero = 0? What kind of math are you using?

I think you still have a lot to learn about basic physics before making such outlandish speculation.

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ZapperZ
Staff Emeritus
To ZapperZ et al.: What do the physicists have to say regarding energy of a photon beyond, for instance:

The minimum energy required to eject an electron from the surface is called the photoelectric work function. The threshold for this element corresponds to a wavelength of 683 nm. Using this wavelength in the Planck relationship gives a photon energy of 1.82 eV.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html

A) Is photon mass 0 or not 0?
B.i) If not 0, then why do text books repeatedly propagate this meme?
B.ii) If actually 0, then why do physicists repeatedly tell those who point out the contradiction that Ash Small has pointed out that it is their logic, basic ability to perform mathematics, and/or knowledge of physics that is faulty rather than the equations they are relying upon?

Best,
Raphie

P.S. I don't mean this post in a confrontational way, and, straight-up, I am not contending that the energy of a photon is either 0 or c^2 Joules, but I do desire to defend Ash Small's basic reasoning capabilities here. Any reasonable person following the logic through, using not "made-up" math, but elementary algebra, might well ask similar questions. Intended as a response to:

Zz.

Really? Let's review, shall we?

You've just claimed that the relationship between E and M is the reason why the speed of light is a constant in vacuum. Where did I misunderstood you here?

This is nonsense. E is not zero for a photon, and c isn't zero for a photon. How in the world were you able to convince yourself that non-zero/non-zero = 0? What kind of math are you using?

I think you still have a lot to learn about basic physics before making such outlandish speculation.

Zz.
Zapper, Einstein said his theories were only approximations. I seem to remember this is why he gave us general relativity and special relativity.

a very small number (~0),divided by a very large number (~infinity) equals approximately zero.

I think you'll find that a lot of Einstein's equations worked like this.

(I admit that the other point you raised wasn't worded very well. The reason why E/M=c^2 is the same as (or related to) the reason why the speed of light is a constant, I accept that I didn't phrase it well. I was trying to make a point as simply as possible.)

D H
Staff Emeritus
Ash Small, whether correct or incorrect, is this the logic you were getting at?

A) If mass of photon = 0, then e=mc^2 = e = 0*c^2
B) Anything multiplied by 0 = 0, therefore e = 0 for energy of photon
C) If E = MC^2, then by substitution e = mc^2, therefore 0 = 0*c^2
D) By rearrangement of terms, 0/c^2 = 0

If the energy of a photon is not 0 Joules, but its mass is 0 kilograms, then there does seem to be a contradiction here, not a flaw in your ability to perform basic mathematics.
The only flaw is using the wrong equation. The correct equation is

$$E^2 = (m_0c^2)^2 + p^2c^2$$

from which it should be clear that a particle with zero mass but non-zero momentum will have non-zero energy.

1) That a photon has mass, however small, has been considered by physicists and is an issue of open debate.
There is no contention amongst physicists that the photon has zero mass. None. Experimental physicists have attempted to measure the photon's rest mass because one of the main jobs of experimentalists is to verify the assumptions and predictions of theoreticians. Experimentalists don't believe anything told to them by theoreticians unless they see first-hand evidence in their experiments that the theoreticians got it right this time. That certainly is the case here. Every experimental attempt to measure the rest mass of the photon has come up with zero, to within experimental error.

2) The factorial argument is no more than a mathematically, but not physically, valid "workaround" to handle 0 that leaves one with the following logical (in isolation), but absurd (not in isolation), statement: ...
What factorial argument?

Your entire post is based on extrapolations from the overly simplistic E=mc2.

ZapperZ
Staff Emeritus
Zapper, Einstein said his theories were only approximations. I seem to remember this is why he gave us general relativity and special relativity.

a very small number (~0),divided by a very large number (~infinity) equals approximately zero.
What does this have anything to do with that you did? You were dividing 0 with 0!

I think you'll find that a lot of Einstein's equations worked like this.

(I admit that the other point you raised wasn't worded very well. The reason why E/M=c^2 is the same as (or related to) the reason why the speed of light is a constant, I accept that I didn't phrase it well. I was trying to make a point as simply as possible.)
Er.. I don't think you know Einstein very well, or for that matter, understand what is meant by "theory".

There has been a lot of nonsense posted in this thread.

Zz.

@ZapperZ: Thank you for the FAQ reference.

@DH:
Your entire post is based on extrapolations from the overly simplistic E=mc2.
You are correct. And that is exactly my point. e=mc^2 is what most regular people (the layman) are taught. I am saying they are taught wrong (or rather, over-simplistically) and therefore extrapolate incorrectly.

As such, I was not arguing that a photon has no energy, nor that a photon has mass, but that such suppositions follow from e=mc^2 using elementary algebra. As far as that goes, thank you for including the correct equation in your response. It's very helpful in regards to my personal understanding and hopefully to other's as well.

As for the issue of a photon having mass or no, I fully understand that the general consensus is definitely "no," but I also know that this question has been debated over and over again (although it would seem I overestimated the seriousness with which that debate is viewed...) and I also know that photons can be considered to have mass in a "relativistic sense," unless I am being "taught wrong" by Steven S. Zumdahl.

Chemical principles, Page 528 by Steven S. Zumdahl
"A photon has mass only in a relativistic sense -- it has no rest mass"

Best,
Raphie

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What does this have anything to do with that you did? You were dividing 0 with 0!.
Zapper, I was actually dividing E by c^2 (within the context of E=Mc^2, which we all know is flawed)

In this case, E is a very small number and c^2 is a very large number, so the answer is ~0.

On the subject of dividing 0 by 0, however, zero divided by anything is zero, and anything divided by zero is infinity, however, in this case, zero is nothing, not 'anything', so I assume that this follows the case where anything divided by itself is one. (0/0=1). I'm not sure of this, just doing my best to answer your question.

ZapperZ
Staff Emeritus
Zapper, I was actually dividing E by c^2 (within the context of E=Mc^2, which we all know is flawed)

In this case, E is a very small number and c^2 is a very large number, so the answer is ~0.
This is wrong. What makes you think "E" is always "very small number"? For example, what is E for an electron? Do you get this to be "~0" as well?

On the subject of dividing 0 by 0, however, zero divided by anything is zero, and anything divided by zero is infinity, however, in this case, zero is nothing, not 'anything', so I assume that this follows the case where anything divided by itself is one. (0/0=1). I'm not sure of this, just doing my best to answer your question.
You need to learn calculus, especially complex algebra. What you said here is mathematically incorrect.

In case this is something you don't realize already, this forum is populated with professional physics, mathematicians, engineers, etc. Considering that from my perspective, you are barely understanding basic physics, I would strongly suggest that you do not take it upon yourself to try and "teach" basic physics and math on here, something which I don't think you understand already.