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Why c2 (speed of light squared)?

by neoweb
Tags: light, speed, squared
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neoweb
#1
Sep1-04, 05:45 AM
P: 24
Sorry if this sounds dumb, but in the famous equation e = mc2 why is the speed of light squared?

I recently read E=mc2: A Biography of the World's Most Famous Equation by David Bodanis... thoroughly enjoyed it... but the answer to this question eludes me.

Can anyone help in layman's terms please?
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Mike2
#2
Sep1-04, 07:31 AM
P: 1,308
Quote Quote by neoweb
Sorry if this sounds dumb, but in the famous equation e = mc2 why is the speed of light squared?

I recently read E=mc2: A Biography of the World's Most Famous Equation by David Bodanis... thoroughly enjoyed it... but the answer to this question eludes me.

Can anyone help in layman's terms please?
I think if anyone were to have time to do a dimensional analysis, he would discover that the units required a velocity squared times a mass to get an energy.
gonzo
#3
Sep1-04, 07:39 AM
P: 277
Do you also wonder why kinetic energy is .5mv^2?

neoweb
#4
Sep1-04, 08:15 AM
P: 24
Why c2 (speed of light squared)?

No. Do you?
Alkatran
#5
Sep1-04, 11:36 AM
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P: 944
Look at the units: If c wasn't squared, the units on the left side of the equation wouldn't match the ones on the right.
Nenad
#6
Sep1-04, 11:50 AM
P: 699
Quote Quote by neoweb
Sorry if this sounds dumb, but in the famous equation e = mc2 why is the speed of light squared?

I recently read E=mc2: A Biography of the World's Most Famous Equation by David Bodanis... thoroughly enjoyed it... but the answer to this question eludes me.

Can anyone help in layman's terms please?
you cant just ask a question like that. Why is it c^2, why dont we make it c^5. The reason it is c^2 is because of its derrivation. If you look athe way the eqation is derrived, the concludion statememnt equates that energy and mc^2 are the same.
And to answer you question about kinetic energy. It is simple.

[tex] F = \frac{dp}{dt} = ma [/tex]

now we can integrate: (anti-derivative)

[tex] \frac{dE}{dt} = p = mv [/tex]

[tex] E = \frac{1}{2}mv^2 [/tex]

do you get it, (my notation is a lotlle off, I dont know how to use the integration symbol in laytex)
neoweb
#7
Sep1-04, 01:24 PM
P: 24
Quote Quote by Nenad
The reason it is c^2 is because of its derrivation. If you look athe way the eqation is derrived, the concludion statememnt equates that energy and mc^2 are the same.
Could someone kindly describe/set out the equation's derivation or point me to a web page?
Nenad
#8
Sep1-04, 07:39 PM
P: 699
visit www.google.ca and do a web search on the eqaution, youll ge plenty of results.
geometer
#9
Sep1-04, 10:30 PM
P: 196
Quote Quote by neoweb
Could someone kindly describe/set out the equation's derivation or point me to a web page?
The equation falls right out of the Work-Energy Theorem if you use the 4- momentum and proper time.
da_willem
#10
Sep3-04, 05:44 AM
P: 599
In Special relativity (SR) an objects inertia increases as it approaches the speed of light, making it more and more difficult to increase the speed. You could assign this extra inertia to the mass of the object by saying the mass (this is actually called 'relativistic mass' , but I'll call it just mass) increases. Working this out it turns out the mass increases with a factor [itex]\gamma (v)[/itex]. This is a velocity dependent function increasing to infinity as the speed v approaches c thus making it impossible to acquire a speed large than c. I SR time slows down and length is shortened by the same factor!

So mathematically this means the mass ([itex]m(v)[/itex]) in terms of it's rest mass (m) will be:
[tex]m(v) = \gamma m[/tex] with [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

For low speeds this can be approximated mathematically by:
[tex]m(v)=m+\frac{1}{c^2}(\frac{1}{2}m \gamma v^2)[/tex]

But this last term is a particles low speed kinetic energy divided by c^2! So the kinetic energy of a particle contributes to its mass in a way wich is consistent with:

[tex]E=m \gamma c^2[/tex]

Or in terms of relativitsic mass: [tex]E=m(v) c^2[/tex]. Einsteins famous equation!



If you are willing to adopt the observation that a photon has a momentum [itex]p=\frac{h}{\lambda}[/itex] and energy [itex]E=hf[/itex] with (h Planck's constant) it's much easier:

[tex]E=hf=\frac{hc}{\lambda}=pc=mc^2[/tex]

With p=mv=mc for a photon.

NOTE: These are only two out of many derivations of the relation, and I'm not sure they can be understood by 'a layman'. I'm also not certain a derivation of the equation is in place. It will always need other assumptions to be derived and in the end it's just an observation, wich is the only true 'proof' of the equation!
da_willem
#11
Sep3-04, 06:11 AM
P: 599
And about why kinetic energy is (1/2)mv^2? You need Newtons second law that states the motion of an object changes according to: F=ma. And that the energy of an object changes under influence of a force when moving a distance x parallel to the force (work!) by an amount [itex]W= \int F dx[/itex]

[tex]W=\int F dx = \int ma dx= m\int \frac{dv}{dt} dx = m\int \frac{dv}{dt} \frac{dx}{dt} dt = m\int \frac{dv}{dt} v dt = m\int v dv = \frac{1}{2}mv^2 [/tex]

[You could also derive it using the general case (not requiring the movement is parallel to the force) but you would need to treat Force, speed, acceleration as a vector and use the dot-product, but this is essentially the same, and makes things a little bit more complicated]
Recher
#12
Dec18-10, 05:19 PM
P: 1
What NEOWEB is getting at is though there are all these incredible numbers vis a vis the fine tuning of the universe (Just Six Numbers by Martin Rees a great read) is C2 is on its own in import.

Another way of writing the formula is E/C2 = M

We know what C2 means in terms of maths but the what and why....We have failed to focus...There is much more to this C2 than we have ever thought of
WonderWatcher
#13
Dec22-10, 12:12 PM
P: 2
I think recher is the only one who even came close to answering neoweb’s question.

I think the questions: “Why the speed of light and why the heck does it needs to be squared?” are answered like this:
E=mC^2, or m=E/C^2, C^2=E/m, or however you want to write it...what it’s saying is that it is observed that there is a constant proportion between Energy and Mass. Apparently this constant proportion is equivalent to the speed of light multiplied by the speed of light. The speed of light is a constant, regardless of the mass involved, or time dilation, or any of the other weird things that Einstein talked about. It can’t be the speed of sound, or the number of stars in a galaxy, or the number of farts in an elevator, or the speed of a falling turtle, because none of those things survive as a UNIVERSAL constant. Based on what is observed, only the speed of light is a UNIVERSAL constant. Therefore, it’s a really good number to use.
So, the REAL questions is, what the heck is light, and mainly why does it travel at ~186 thousand miles per second all the time?? I believe that if we could answer this question, we could replace C^2 with that answer in some way. Like, the speed of light is ~186k m/s because the Ether makes it so. Therefore, E=m(c^2) could be written E=m(because the Ether makes it so), or more logically (because the Ether makes it so) = E/m.

What do you all think….is this on the right physics-osophical path?
ZirkMan
#14
Dec22-10, 01:48 PM
P: 136
Quote Quote by WonderWatcher View Post
What do you all think….is this on the right physics-osophical path?
The layman's view path is probably right. But I think it's wrong to blame the light for the strangeness of the observed phenomena as c is not only the speed of light in vacuo but a speed limit in general for any form of energy capable of carrying information and light just happens not to cross this limit too. The speed limit itself doesn't have to have anything in common with the physical nature of light and probably has deeper roots.

As for the E=mc2 this can be literally interpreted as that the total energy of a system is mathematically equivalent to that many of light squares (c2) as there are units of rest mass we choose. There must be a deeper meaning why this translation is mediated through multiples of areas of the maximum speed limit in two dimensions but I do not think anybody will be able to give you a satisfying answer just now other than mathematical derivation.
It's because we don't have a final theory of space&energy yet. I think the SR&GR is an important step to such a theory and gives some important clues but on this level it still just provides answers of HOW things as energy-space relate to each other rather than why.
AnotherDave
#15
Jun27-12, 05:38 PM
P: 7
Ok, let me rephrase the context of the original question.

Grandpa has been a mailman all of his life. But he likes all that "sciencey stuff" on the discovery channel. He knows his smart Grandson is a physics student and he's always wondered about that whole E=MC2 thing.

So Grandpa comes over and says..."hey, maybe you can explain something to me. That crazy guy, Michio Kaku was on Fox News talking about Einstein and how nothing can travel faster than the speed of light. So why do we have that whole C2 thing?"

The smart physics student respects his grandfather and doesn't want to tell him that he's too old or not smart enough to understand. But he also won't understand an explanation that goes something like:

m(v)=m+\frac{1}{c^2}(\frac{1}{2}m \gamma v^2)

What's the best way to explain "why C2" in non-mathematical terms that the retired mailman can understand?
Muphrid
#16
Jun27-12, 06:15 PM
P: 834
First, remember that every object travels through space and time at the speed of light. In mathematical language, that says that every object as a four-velocity [itex]u[/itex] such that [itex]|u \cdot u| = c^2[/itex].

From there, remember that any object also has four-momentum [itex]p = mu[/itex]. When the object has no three-momentum, this equation reduces to [itex]E/c = mc[/itex]. That just follows from the idea that energy and momentum come together to form a single object.

So what do you have?

1) Every object travels through space and time at the speed of light
2) Momentum = mass x velocity
3) Energy and momentum are part of the same object, the four-momentum
4) A little algebra

That's all it is.
AnotherDave
#17
Jun27-12, 06:44 PM
P: 7
Quote Quote by Muphrid View Post
First, remember that every object travels through space and time at the speed of light. In mathematical language, that says that every object as a four-velocity [itex]u[/itex] such that [itex]|u \cdot u| = c^2[/itex].

From there, remember that any object also has four-momentum [itex]p = mu[/itex]. When the object has no three-momentum, this equation reduces to [itex]E/c = mc[/itex]. That just follows from the idea that energy and momentum come together to form a single object.

So what do you have?

1) Every object travels through space and time at the speed of light
2) Momentum = mass x velocity
3) Energy and momentum are part of the same object, the four-momentum
4) A little algebra

That's all it is.
Quote Quote by Muphrid View Post
First, remember that every object travels through space and time at the speed of light.
I'm quite sure this was a typo. But you said it twice, so I think you mean to say something else. Everything does not travel through space and time at C. Everything travels at a percentage of C...but for oversimplified purposes, only C travels at C.

Grandpa isn't going to understand four-momentum. Grandpa also isn't going to understand "mathematical language" and his poor old eyes will glaze over if you try :-) Although even Grandpa has some math education and we may be able to remind him about balanced equations. Maybe not Algebra, but certainly fractions. What you do to one side you have to do to the other.

He understands the E=M part. He just doesn't get C2. How can you have the square of something that can't be exceeded? Can you see my problem in explaining this? LOL
Muphrid
#18
Jun27-12, 06:50 PM
P: 834
Quote Quote by AnotherDave View Post
I'm quite sure this was a typo. But you said it twice, so I think you mean to say something else. Everything does not travel through space and time at C. Everything travels at a percentage of C...but for oversimplified purposes, only C travels at C.
I meant what I said. Every non-massless object has a four-velocity with magnitude [itex]c[/itex]. Hence, it travels through spacetime (space and time together, not separately) at the speed of light. I included the "in mathematical language" part to make that statement precise; I wouldn't expect it to be used among laymen. Still, I feel that understanding this point and how all four-velocities for massive objects lie on a hyperbola (and hence, all of them have magnitude [itex]c[/itex]) is critical in the chain of reasoning here.


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