- #1

villiami

- 27

- 0

I know this is basic, but how is E=mc^c derived.

Thanks Heaps

Thanks Heaps

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter villiami
- Start date

- #1

villiami

- 27

- 0

I know this is basic, but how is E=mc^c derived.

Thanks Heaps

Thanks Heaps

- #2

- 13,288

- 1,740

[tex] W=\vec{v}\cdot\frac{\partial L}{\partial\vec{v}}-L [/tex]

Daniel.

- #3

da_willem

- 599

- 1

https://www.physicsforums.com/showthread.php?t=16754

https://www.physicsforums.com/showthread.php?t=42098

https://www.physicsforums.com/showthread.php?t=41354

By using the search function you might find several more threads about the subject.

- #4

cepheid

Staff Emeritus

Science Advisor

Gold Member

- 5,196

- 38

- #5

villiami

- 27

- 0

What do you mean?

- #6

villiami

- 27

- 0

Sorry, now I see

- #7

- 13,288

- 1,740

Daniel.

- #8

scilover89

- 78

- 0

Can you explain in detail? I found myself difficult to understand the statement.dextercioby said:

Daniel.

- #9

- 13,288

- 1,740

I think this stuff is described in electrodynamics books.In order to discuss relativistic particle in interaction with a classical EM field,u need to analyze the free particle first.I don't remember seing this issue in Jackson,nor in Goldstein,but I'm sure it has to be somewhere,just have the interest to look for.

Daniel.

- #10

- 13,288

- 1,740

Okay,here's an elementary proof,i'd call it HS level,using differential calculus.

Assume for simplicity only one space-component.So the whole discussion would involve scalars.

Newton's second law:

[tex] \frac{d(m_{rel}v)}{dt}=F [/tex](1)

Multiply with dt:

[tex] d(m_{rel}v)=Fdt [/tex] (2)

Substitute dt with:

[tex] dt=\frac{dx}{v} [/tex](3)

[tex] d(m_{rel}v)=F\frac{dx}{v} [/tex](4)

Define differential work:

[tex] \delta L=Fdx [/tex](5)

Use the Leibniz theorem in differential form:

[tex] dE=\delta L\Rightarrow dE=Fdx [/tex](6)

Rewrite (4) in terms of the differential of energy:

[tex] v d(m_{rel}v)=dE [/tex] (7)

Everybody knows that:

[tex] m_{rel}=\gamma m_{0}=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} [/tex] (8)

Expand (7):

[tex] v^{2}dm_{rel}+m_{rel}vdv=dW [/tex](9)

Differentiate (8):

[tex] dm_{rel}=\frac{m_{0}\gamma v dv}{c^{2}-v^{2}} [/tex](10)

Plug (10) in (9) and factor:

[tex] m_{0}\gamma v dv(\frac{v^{2}}{c^{2}-v^{2}}+1)=dE [/tex](11)

Therefore:

[tex] m_{0}\gamma v dv \frac{c^{2}}{c^{2}-v^{2}}=dE [/tex](12)

Or:

[tex] c^{2}(\frac{m_{0}\gamma v dv}{c^{2}-v^{2}})=dE [/tex](13)

Taking into account (10),one finally finds the diferential form of Einstein's formula:

[tex] c^{2}dm_{rel}=dE [/tex](14)

Integrating with corresponding limits (zero relativistic mass,zero energy),one finds:

[tex] m_{rel}c^{2}=E [/tex]

Daniel.

EDIT:**THAT** is a proof... It took me 10 minute to cook. :yuck: Though the Lagrangian approach is simply PERFECT.

Assume for simplicity only one space-component.So the whole discussion would involve scalars.

Newton's second law:

[tex] \frac{d(m_{rel}v)}{dt}=F [/tex](1)

Multiply with dt:

[tex] d(m_{rel}v)=Fdt [/tex] (2)

Substitute dt with:

[tex] dt=\frac{dx}{v} [/tex](3)

[tex] d(m_{rel}v)=F\frac{dx}{v} [/tex](4)

Define differential work:

[tex] \delta L=Fdx [/tex](5)

Use the Leibniz theorem in differential form:

[tex] dE=\delta L\Rightarrow dE=Fdx [/tex](6)

Rewrite (4) in terms of the differential of energy:

[tex] v d(m_{rel}v)=dE [/tex] (7)

Everybody knows that:

[tex] m_{rel}=\gamma m_{0}=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} [/tex] (8)

Expand (7):

[tex] v^{2}dm_{rel}+m_{rel}vdv=dW [/tex](9)

Differentiate (8):

[tex] dm_{rel}=\frac{m_{0}\gamma v dv}{c^{2}-v^{2}} [/tex](10)

Plug (10) in (9) and factor:

[tex] m_{0}\gamma v dv(\frac{v^{2}}{c^{2}-v^{2}}+1)=dE [/tex](11)

Therefore:

[tex] m_{0}\gamma v dv \frac{c^{2}}{c^{2}-v^{2}}=dE [/tex](12)

Or:

[tex] c^{2}(\frac{m_{0}\gamma v dv}{c^{2}-v^{2}})=dE [/tex](13)

Taking into account (10),one finally finds the diferential form of Einstein's formula:

[tex] c^{2}dm_{rel}=dE [/tex](14)

Integrating with corresponding limits (zero relativistic mass,zero energy),one finds:

[tex] m_{rel}c^{2}=E [/tex]

Daniel.

EDIT:

Last edited by a moderator:

- #11

Andrew Mason

Science Advisor

Homework Helper

- 7,740

- 439

The box part really is not needed. All you need are two masses. Conceptually, it goes like this:

Two equal masses, m1 and m2 (=m) at co-ordinates -d,0 and d,0 (origin at centre of mass). Since light has energy E = hf and momentum E/c (= hf/c), when a photon leaves m1, m1 recoils with momentum E/c. When the photon is stopped by m2, m2 takes on momentum E/c = mv (so v=E/cm). In time t=2d/c, m1 moves distance s=vt = E2d/mc^2. At time t after m1 begins moving, m2 receives momentum E/c.

Now you can see where there is a problem. Unless some mass is transferred, the centre of mass has moved! Newton's third law takes care of this where the masses are not separated by a distance as the changes in motion occur at the same instant. But when the momentum change is provided by light, there will be a shift in the centre of mass unless light carries mass with it.

How much mass does it have to carry with it? Work it out: In order to conserve the centre of mass at time t, m1(-d) + m2d = 0 = m1'(-d1') + m2'd. Since d1'=d+s = d+E2d/mc^2, we have:

[tex](m + \Delta m)(d) = (m-\Delta m)(d + E2d/mc^2)[/tex]

[tex]md + \Delta md = md - \Delta md + mE2d/mc^2 - \Delta mE2d/mc^2)[/tex] or:

[tex]2\Delta md + \Delta mE2d/mc^2 = 2Ed/c^2[/tex]

Ignoring the negligible 1/mc^2 term (this actually disappears if you take into account the [itex]\gamma[/itex] factor but we can avoid the math because we can see that [itex]m>>\Delta m[/itex]):

[tex]\Delta m = E/c^2[/tex] or:

[tex]E = \Delta mc^2[/tex]

The explanation from Einstein's "m = L/c^2" paper in 1905 is also quite understandable. The original translation is here:

http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

AM

Share:

- Last Post

- Replies
- 12

- Views
- 560

- Last Post

- Replies
- 3

- Views
- 249

- Last Post

- Replies
- 3

- Views
- 354

- Last Post

- Replies
- 7

- Views
- 1K

- Replies
- 1

- Views
- 349

- Last Post

- Replies
- 26

- Views
- 555

- Last Post

- Replies
- 0

- Views
- 453

- Replies
- 10

- Views
- 1K

- Last Post

- Replies
- 15

- Views
- 689

- Replies
- 17

- Views
- 933