Understanding Mass-Energy Equivalence to Fdx and dm in E=mc^2

Click For Summary
SUMMARY

The discussion centers on the equation Fdx = dm c^2 and its relationship to mass-energy equivalence expressed in E=mc^2. Participants clarify that Fdx represents work done, while dm indicates a change in mass. The integration of both sides leads to the conclusion that work equals mc^2, aligning with the fundamental theorem of calculus. The conversation emphasizes the importance of understanding derivatives and integration in physics, particularly in the context of well-behaved functions.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and integrals.
  • Familiarity with the concept of work in physics.
  • Knowledge of mass-energy equivalence and the equation E=mc^2.
  • Basic understanding of the fundamental theorem of calculus.
NEXT STEPS
  • Study the process of integrating expressions in physics, focusing on work and energy.
  • Learn about the fundamental theorem of calculus and its applications in physics.
  • Explore the implications of mass-energy equivalence in modern physics.
  • Review calculus concepts related to derivatives and their physical interpretations.
USEFUL FOR

Students of physics, educators teaching calculus and physics, and anyone interested in the mathematical foundations of mass-energy equivalence.

rrrright
Messages
5
Reaction score
0
Hi I was wondering if anyone could help me with this equation.

Fdx &= dm c^2

First of all, excuse me for my limited knowledge of calculus, but how exactly can you just use the numerator of a derivative? What do Fdx and dm mean if they are not in respect to anything? Do they simply mean a change in x and a change in m?

Secondly, I have seen this equation used to get to E=mc^2 through integration. How exactly do we integrate either side of this equation? What is the step by step process for doing this? Again, please excuse my limited math knowledge.
 
Physics news on Phys.org
Splitting the derivatives is technically not kosher in math; however, physicists do it all the time. I think there are probably weird pathological functions (such as discontinuous everywhere functions or some such) where splitting the derivative will lead to the wrong answer, but physics usually only deals with well behaved functions. Usually something like dx just means a small displacement in x.

You can immediately integrate that expression to obtain W=mc^2. Integrateing Fdx gives you the work and integrating d(mc^2) just gives you mc^2 back (sorta like the fundamental theorem of calculus).
 
rrrright said:
Hi I was wondering if anyone could help me with this equation.

Fdx &= dm c^2

First of all, excuse me for my limited knowledge of calculus, but how exactly can you just use the numerator of a derivative? What do Fdx and dm mean if they are not in respect to anything? Do they simply mean a change in x and a change in m?

Secondly, I have seen this equation used to get to E=mc^2 through integration. How exactly do we integrate either side of this equation? What is the step by step process for doing this? Again, please excuse my limited math knowledge.

Hi, welcome on board!

For the answers, look at the first file in my blog. It explains the differentiation as well as the physics you are asking about.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 55 ·
2
Replies
55
Views
6K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K