How to Get E=mc2: Exploring Einstein's Equations

  • Thread starter fawk3s
  • Start date
  • Tags
    E=mc2
In summary: E=0.1*m*c^2?I'm going to read the other link now and post back after I've read it.In summary, the conversation discusses the derivation of the formula E=mc2 and different arguments and proofs for it. The conversation also mentions a video and a website where the formula is explained. There is a debate about the validity of certain derivations and equations, including p=mv and p=E/c for electromagnetic waves. Ultimately, the conversation ends with the request for a link showing how to derive p=mv from Maxwell's equations.
  • #1
fawk3s
342
1
Hi

Somewhat usual question for you guys I guess. How do we get the formula E=mc2?

I saw in a video where it said Einstein combined these equations

Mv=E/c
t=L/c
x=vt
Mx=mL

and got

EL/c2=mL => E/c2=m

But I don't quite follow. What is Mv? What is Mx? And what time does L/c represent? How exactly are they combined?
Always interested me.

Thanks in advance,
fawk3s
 
Physics news on Phys.org
  • #2
For those who haven't seen it:
http://images1.wikia.nocookie.net/uncyclopedia/images/0/0a/Kuntry%27stein_by_adj.gif
:D
 
  • #5
Feldoh said:
It looks like they assumed some of the results in their derivation which is circular logic.
Now as you say it, I agree. That derivation already assumes the result from the beginning.

In fact after assuming
[tex]p=\frac{E}{c}[/tex]
one could directly use [itex]p=mc[/itex] to get the result.

However, I don't see a derivation in Wikipedia either?!
 
  • #6
I've written up a derivation that I think is pretty concise.

go to www.shadycrypt.com

Click on the E=mc2 link at the top.
 
  • #7
Here is the argument that Einstein originally published: http://fourmilab.ch/etexts/einstein/E_mc2/www/

Here is a different argument: http://www.lightandmatter.com/html_books/6mr/ch01/ch01.html#Section1.3 [Broken]

Feldoh said:
It looks like they assumed some of the results in their derivation which is circular logic.

No, the derivation on the fotonowy.pl page is not circular. This is another well known proof, originating with Einstein. One delicate issue in it is that in the original form of this thought experiment, the box is implicitly assumed to be perfectly rigid. This is a flaw, but it can be fixed: http://galileo.phys.virginia.edu/classes/252/mass_and_energy.html

Feldoh said:
That isn't a derivation. They point out that it's a special case of the relativistic energy-momentum relation, but they haven't proved the energy-momentum relation.
 
Last edited by a moderator:
  • #8
bcrowell said:
No, the derivation on the fotonowy.pl page is not circular.
It's not even circular. It already assumes E=mc^2 from the very beginning. After writing p=E/c you do not need a lengthy derivation, but just use p=mc to derive the final result. So the two equations are one step apart from being equivalent.
 
  • #9
Gerenuk said:
It's not even circular. It already assumes E=mc^2 from the very beginning. After writing p=E/c you do not need a lengthy derivation

No, p=E/c for electromagnetic waves follows directly from Maxwell's equations, so that had been known for 30 or 40 years before Einstein published SR in 1905. Here is an explanation: http://www.lightandmatter.com/html_books/genrel/ch01/ch01.html [Broken] (see subsection 1.5.7).

Gerenuk said:
but just use p=mc to derive the final result. So the two equations are one step apart from being equivalent.
No, this is incorrect. You can't just plug v=c into p=mv and expect it to be correct for a photon. p=mv is a nonrelativistic equation, which can't be expected to hold for light.

Considering that the argument given at fotonowy.pl is due to Einstein, I really don't think you're going to find obvious logical fallacies in it.
 
Last edited by a moderator:
  • #10
I need to check the links you proposed.

Can you point me to a link where they show the prove how to derive p=mv from Maxwell?
(I'm not surprised it works, since Maxwell is already relativistic?!)

However:
bcrowell said:
No, this is incorrect. You can't just plug v=c into p=mv and expect it to be correct for a photon. p=mv is a nonrelativistic equation, which can't be expected to hold for light.
p=mv is always true. Also in relativity, since judging by the proof they use the relativistic mass. They use E=p/c, so E is the total energy. They derive E=mc^2 with the same variable E, so m must be the relativistic mass. In that case p=mv in both classical and relativistic theory.
 
  • #11
Gerenuk said:
I need to check the links you proposed.

Can you point me to a link where they show the prove how to derive p=mv from Maxwell?
No, because it's not true. But I did provide a link that shows p=E/c for an electromagnetic wave.

Gerenuk said:
p=mv is always true. Also in relativity, since judging by the proof they use the relativistic mass. They use E=p/c, so E is the total energy. They derive E=mc^2 with the same variable E, so m must be the relativistic mass. In that case p=mv in both classical and relativistic theory.
You've misunderstood the content of E=mc2.

[EDIT] Actually the link I gave above only proves that p is nonzero for an electromagnetic wave (which is inconsistent with the classical relation p=mv, since m=0 for light). By linearity and units we must have [itex]p=kE/c[/itex], where k is a unitless constant. For the proof that k=1, see this link: http://www.lightandmatter.com/html_books/0sn/ch11/ch11.html#Section11.6 [Broken] (subsection 11.6.2).
 
Last edited by a moderator:
  • #12
bcrowell said:
No, because it's not true. But I did provide a link that shows p=E/c for an electromagnetic wave.
That link states this equation follows from Maxwell's equation, but there isn't even a single Maxwell equation. There might be a vague hint in the text, but there is no derivation.

Oh, but if you mean the link you posted later... I still have to go through it...

bcrowell said:
You've misunderstood the content of E=mc2.
You have to be more specific and tell where my argumentation is wrong, if you believe you know it.
 
Last edited:
  • #13
bcrowell said:
By linearity and units we must have [itex]p=kE/c[/itex], where k is a unitless constant.
Where is the proof for linearity then? Why are normal particles not linear?
 
  • #14
Gerenuk said:
Where is the proof for linearity then? Why are normal particles not linear?

Take the Maxwell's equations in vacuum. Transform them to a wave equation for B. Try the solution in the form [tex]B=B_ocos(kx-\omega t)[/tex].

The normal particles have nonzero rest masses, so their energy is proportional to [tex]p^2[/tex] rather then to [tex]p[/tex].
 

1. What does E=mc2 mean?

E=mc2 is a famous equation introduced by Albert Einstein in his theory of special relativity. It states that energy (E) is equal to mass (m) multiplied by the speed of light (c) squared. In other words, it shows the relationship between mass and energy.

2. How do I understand E=mc2?

E=mc2 can be understood by breaking down each component of the equation. The "E" stands for energy, which is measured in joules. The "m" stands for mass, which is measured in kilograms. The "c" stands for the speed of light, which is approximately 299,792,458 meters per second. When these units are plugged into the equation, we can see that energy is equal to mass multiplied by the speed of light squared.

3. Why is E=mc2 important?

E=mc2 is important because it revolutionized our understanding of the relationship between mass and energy. It showed that mass and energy are not separate entities, but are instead interchangeable. This led to a better understanding of nuclear energy and the development of nuclear weapons and nuclear power.

4. How can I apply E=mc2 in everyday life?

While E=mc2 is often associated with complex scientific concepts, it can also be applied in everyday life. For example, the equation can be used to understand the amount of energy released in nuclear reactions or to calculate the amount of energy stored in an object's mass. It can also be used to understand the amount of energy needed to travel at the speed of light.

5. Is E=mc2 still relevant today?

Yes, E=mc2 is still relevant today and is widely used in various fields of science, including nuclear physics, astrophysics, and particle physics. It is also the basis for many technological advancements, such as nuclear power and medical imaging. The equation has also been confirmed by numerous experiments and continues to play a crucial role in our understanding of the universe.

Similar threads

  • Special and General Relativity
2
Replies
41
Views
4K
  • Special and General Relativity
Replies
19
Views
3K
  • Special and General Relativity
Replies
5
Views
7K
  • Special and General Relativity
Replies
7
Views
2K
  • Special and General Relativity
Replies
23
Views
8K
  • Special and General Relativity
Replies
1
Views
2K
  • Special and General Relativity
Replies
2
Views
1K
Replies
7
Views
2K
  • Special and General Relativity
3
Replies
75
Views
3K
  • Special and General Relativity
Replies
1
Views
1K
Back
Top