# In E=mc^2, why is the speed of light squared?

Discussion in 'General Science & Technology' started by strategicman, Jun 17, 2003.

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1. ### strategicmanRegistered Senior Member

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Hey, I'm just wondering why in E=mc^2, the speed of light is squared. Could somebody just try to explain it all? Thanks.

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3. ### MystechAdult Supervision RequiredRegistered Senior Member

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It's squared because you're supposed to multiply it by itself in the equation ^.^ *Capitan obvious zooms off into the clouds*

That is a good question, though why are you supposed to multiply the speed of light by itself? And why represent the speed of light with a variable, when it is a constant? Guess that's just so you can decide what units you want to work in *shrug*

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5. ### strategicmanRegistered Senior Member

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lol, thanks captian obvious! But how could Einstein have thought of something like that? how would he have just thought "hey, it's not working when I use E=mc so I should square "c"". I just don't see how. ::shrugs::

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7. ### NasorValued Senior Member

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Usually in mathematics letters at the beginning of the alphabet (like C) are used to denote constants, while letters at the end of the alphabet (X,Y,Z) are used for variables.

8. ### tempusmeRegistered Senior Member

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Energy = Mass * Speed of light ^ 2

J = kg * m^2/s^2

J = (kg * m/s^2)*m

J = N * m

J = J

But this might not be what you're looking for...

9. ### James RJust this guy, you know?Staff Member

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tempusme is right. If you're going to relate energy to speed and mass, the only way to do it is to have and equation of the form E=kmc<sup>2</sup>, where k is a dimensionless constant.

That's necessary for the units (dimensions) on both sides of the equation to match.

If we choose consistent units to define E, m and c, then k turns out to be equal to 1.

10. ### one_ravenGod is a Chinese WhisperValued Senior Member

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I think it has to do with the inverse square law.
I am not a mathematician (if you want an answer from a mathematician, you should post this in the Physics and Math forum), however...
When light propagates, it does so in much the same way that sound does.
It does not go in a straight line, it does so omnidirectionally.
It spreads out in all directions at once, and at a constant speed.
Therefore, at any given time, if you follow light propagating from a single point, the light would spread evenly in all diorections, so the "front" of the light wave will be in the shape of a sphere.
(see attachment. NOTE: Not my image, I stole it from somewhere, but I don't want to steal credit)

All energy (includiong light) travels via electromagnetic waves and the Inverse Square Law applies to all energy propagation.

Since the formula for the surface of a sphere is 4(pi)r^2, and light propagates in the shape of the face of a sphere I think that is where the ^2 comes from.
I don't yet have the ability to break it down for you matematically (still learning the math behind it), but I think this is where the connection lies.

Wish I could be of more help than that, but I am still trying to figure it all out myself.
As I said, you would likely have a much better response from the Physics and Math forum.

Good luck.

11. ### (Q)Encephaloid MartiniValued Senior Member

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I'm just wondering why in E=mc^2, the speed of light is squared.

Because at that speed (c^2) mass and energy are indistinguishable from one another.

12. ### James RJust this guy, you know?Staff Member

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Actually, the c<sup>2</sup> has nothing to do with the inverse square law. In the equation E=mc<sup>2</sup> it also has nothing to do with the speed of anything. It is merely a numerical conversion factor.

13. ### rapid transitRegistered Member

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http://www.tardyon.de/luxon.htm

This guy presents an interesting theory.

In his view it works out to e = pc where p = mv (momentum).

It's hard to explain. Just read the site.

Please Register or Log in to view the hidden image!

14. ### MyqRegistered Member

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Due to momentum conservation arguments in special relativity theory, you can get a formula for how the mass increases with velocity of an object. If you assume that the Rest Energy of mass can only vary with the first power of M (makes sense, double the mass double the mass-energy), you can then use a low velocity limit to show that the constant in front of mass should be C^2 and only C^2.

It works out in units, it works out in relativity, and in the low velocity limit, you get what you should expect:

E=1/2mV^2 + mC^2

15. ### tempusmeRegistered Senior Member

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Also, E=hf.

E=pc would be the same thing as e=mc^2, if v=c.

16. ### (Q)Encephaloid MartiniValued Senior Member

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James

In the equation E=mc2 it also has nothing to do with the speed of anything.

I was watching an interview with Einstein (black and white) some time ago when someone in the crowd posed this question. That was the reply he gave. It caught me completely off gaurd. If I find the interview, I'll post a reference.

17. ### GiftedWorld WandererRegistered Senior Member

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Becuase it's easier to write/say C (one digit) rather than 186,000 mi/sec (thirteen digits).

18. ### MystechAdult Supervision RequiredRegistered Senior Member

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Ahh, thanks for that. I myself am something of a mathematical moron, so I appreciate that. Every little bit helps, seeing how as I am required to take more math credits than I can handle to get the degree I'm going for, heh.

19. ### errandirRegistered Senior Member

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The paradigm for a contravariant tensor is the differential displacement 4-vector. The first term is the time term. The speed of light is the relationship between time and space. The energy-strees tensor is a symetric second rank tensor. Since the paradigm contains a c in the first term, and since the stress tensor is symetric, then it follows that there should be a c^2 in this term in the stress tensor, which is basically the E = mc^2.

To be more specific, the stress tensor is the product of the current density 4-vector with the proper velocity 4-vector. The 0th component of the current density is pho*c. The 0th component of the velocity is just c. Therefore, in the product, the 00 component is just rho*c*c = rho*c^2. This is just the mc^2 per unit mass. I don't remember the exact consequence of the top of my head, but I know that, if you just use mass instead of density, you will wind up with something that has ugly transformation properties.

20. ### myhrRegistered Member

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Einstein's theory of relativity isn't the formula E = m*c^2 , though it is the most known consequence of it. He actually set out to think about how would different observers (with different velocities) observe (or measure) the same mechanical and electromagnetical phenomena. From this he discovered formulas for addition of velocities, time dilation, length contraction, etc. In the end he also noted that in the limit of zero velocity the inertia (that is: mass) of a body is E/c^2. If I remember correctly the formula is never stated in the form E = m*c^2 in the original paper (which can by the way be found at http://fourmilab.to/etexts/einstein/E_mc2/www/)

21. ### apoloRegistered Senior Member

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STRATIGIGMAN is posing a good question a real good question

Why is energy equal to the mas multiplied by the speed of light multiplyed by itself ??
I have always asumed this formula to be corect,-the bedrock of relativity theory.- It basically says, that if you convert mas into pure energy, you get a hell of a lot of enenergy. (as in the atomic bomb) or in the fueson reaction in the sun-H to He.
As far as I know Einstein did not arive at this formula by experimentation or observation. It was a postulate !
My question is; Has anybody ever coroborated this by actual experiments

PS. I'll read your answer JAMES R and I wont argue.(At least not on this subject)

22. ### James RJust this guy, you know?Staff Member

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<i>Has anybody ever coroborated this by actual experiments</i>

Yep. It is verified in nuclear reactions and in accelerator experiments.

23. ### Janus58Valued Senior Member

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No, it is not a postulate in itself, but a natural conculsion of the postulates of Relativity, That the physical laws in all frames are the same, and that the speed of light in a vacuum is a constant.

These two postulates lead naturally to the Lorentz contractions,

t = t'/sqrt(1-v²/c²)
x = x sqrt(1-v²/c²)

If you re-evalulate the classical formula E= mv²/2 taking these transformations into effect you get:

E= mc²/sqrt(1-v²/c²)

Expanding this equation gives:

E = mc² + mv²/2 + 3mv^4/8c² + ....

Note that if v=0, we are left with the famous

E=mc²

Which is the potenetial energy content of the mass at rest.

Also, if 0 < v <<c, everything from the third term on becomes insignificant and can be ignored, leaving

E= mc²+ mv²/2

In normal circumstances, we are only interested in that part of the energy which is due to motion, so we get:

E= mv²/2

Which is why we can still use the classical formula for kinetic energy in most common situations where v<<c.

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