Cox & Forshaw: Explaining E=mc^2's "Negative" Option

  • I
  • Thread starter EveningLight
  • Start date
  • Tags
    E=mc^2
In summary, the conversation discusses a book that is being read for the second time. The book presents "jumps" in reasoning that are easier to follow but also leave the reader questioning certain assumptions and statements. On page 77, the authors present two options for the length of the hypotenuse in the spacetime model, but the negative version is not thoroughly explained. It is later shown that this is the only other option based on the assumptions of special relativity. Other sources, such as a physics textbook, may provide a more in-depth understanding of this concept.
  • #1
EveningLight
3
0
Great book. Reading it for the second time. There are several "jumps" that are made in their lines of reasoning that certainly make the explanations easier to follow, but also left me wondering "Why or where did that assumption or statement come from?"

On page 77 (in my 2009 Da Capo Press version) when working towards setting up the spacetime model, they state: "The length of the hypotenuse must be either s^2 = (ct)^2+x^2 or s^2 = (ct)^2-x^2. There is no other option."

So why is the negative version even considered as a option? They try to address this in the paragraph following this, but really don't. They just say it is the only other option.

Can anyone explain to me why that is the only other option? Why is not s = ct + x not considered an option? Or any other s = function(ct,x)?
 
Physics news on Phys.org
  • #2
EveningLight said:
Great book. Reading it for the second time. There are several "jumps" that are made in their lines of reasoning that certainly make the explanations easier to follow, but also left me wondering "Why or where did that assumption or statement come from?"

On page 77 (in my 2009 Da Capo Press version) when working towards setting up the spacetime model, they state: "The length of the hypotenuse must be either s^2 = (ct)^2+x^2 or s^2 = (ct)^2-x^2. There is no other option."

So why is the negative version even considered as a option? They try to address this in the paragraph following this, but really don't. They just say it is the only other option.

Can anyone explain to me why that is the only other option? Why is not s = ct + x not considered an option? Or any other s = function(ct,x)?
Welcome to PF EveningLight! I can't really comment on the book without seeing it. Can you post the relevant pages?

From the assumptions of special relativity, it can be shown that for two events separated by x and t in a "rest frame" and x' and t' in a frame moving at speed v relative to the rest frame, ##c^2t^2 - x^2 = c^2t'^2 - x'^2##. This is easier to see where x' = 0 (the two events occur at the same place in the moving frame) i.e.: ##c^2t^2 - x^2 = c^2t'^2##. Since this equality ##c^2t^2 - x^2 = c^2t'^2 - x'^2## does not depend on v, the relative speed of the two observers, the relationship must be true for all observers. The spacetime interval between two events is defined as ##s^2 = x^2 - c^2t^2## precisely because it is invariant for observers in all frames.

AM
 
Last edited:
  • #3
Thanks.

"From the assumptions of special relativity, it can be shown that for two events separated by x and t in a "rest frame" and x' and t' in a frame moving at speed v relative to the rest frame, c 2 t 2 − x 2 = c 2 t ′ 2 − x ′ 2"

So I guess I need the bit about "It can be shown that". I probably need someone to look at the book and understand the authors' progression of their presentation to help me out here.

Reference https://www.physicsforums.com/threads/cox-and-forshaw-why-e-mc2.877248/
 
  • #4
Would probably been good for me to also read another source about this topic. Can anyone else suggest other laymens' sources to this topic?

Thanks.
 
  • #5
EveningLight said:
Would probably been good for me to also read another source about this topic. Can anyone else suggest other laymens' sources to this topic?

Thanks.
There are any number of good physics texts on SR. My old https://www.amazon.com/dp/0393097935/?tag=pfamazon01-20 is easy to follow.

EveningLight said:
"From the assumptions of special relativity, it can be shown that for two events separated by x and t in a "rest frame" and x' and t' in a frame moving at speed v relative to the rest frame, c 2 t 2 − x 2 = c 2 t ′ 2 − x ′ 2"

So I guess I need the bit about "It can be shown that". I probably need someone to look at the book and understand the authors' progression of their presentation to help me out here.
It is simply a matter of applying the Lorentz transformations to ##c^2t'^2 - x'^2##. Texts usually skip the detailed algebra steps but I have put them in for you, one at a time:

##c^2t'^2 - x'^2 = c^2(\gamma(t-vx/c^2))^2 - (\gamma(x -vt))^2##
## = c^2\gamma^2(t^2 - 2vtx/c^2 + v^2x^2/c^4) - \gamma^2(x^2 - 2vtx + v^2t^2)##
## = \gamma^2(c^2t^2 - 2vtx + v^2x^2/c^2 - x^2 + 2vtx - v^2t^2)##
## = \gamma^2(c^2t^2 + v^2x^2/c^2 - x^2 - v^2t^2)##
## = \gamma^2(c^2t^2 - x^2 + (v^2x^2/c^2 - v^2t^2) ##
## = \gamma^2(c^2t^2 - x^2 + v^2(x^2 - c^2t^2)/c^2 ##
## = \gamma^2(c^2t^2 - x^2 - v^2(c^2t^2 - x^2)/c^2 ##
## = \gamma^2(c^2t^2 - x^2)(1 - v^2/c^2) ##
## = \frac{1}{(1 - v^2/c^2)}(c^2t^2 - x^2)(1 - v^2/c^2)##
## = c^2t^2 - x^2##

AM
 
Last edited by a moderator:
  • Like
Likes Battlemage!, SiennaTheGr8 and m4r35n357

1. What is the significance of the "negative" option in Cox & Forshaw's explanation of E=mc^2?

The "negative" option in Cox & Forshaw's explanation refers to the fact that in certain situations, the energy and momentum of a particle can have negative values. This is counterintuitive, as we typically think of energy and momentum as positive quantities. However, this concept is crucial in understanding how E=mc^2 works and its implications in physics.

2. How does the "negative" option affect our understanding of E=mc^2?

The "negative" option allows us to better understand the relationship between mass and energy. In traditional understandings of E=mc^2, mass is seen as a constant and energy as a variable. However, with the inclusion of the "negative" option, we can see that mass and energy are two sides of the same coin and can be interchanged under certain circumstances.

3. Can you provide an example of how the "negative" option works in practice?

One example is the annihilation of a particle and its antiparticle. When a particle and its corresponding antiparticle collide, they can annihilate each other and release energy in the form of photons. In this scenario, the total energy of the system before and after the annihilation is zero, as the positive energy of the particles is cancelled out by the negative energy of the photons.

4. Does the inclusion of the "negative" option change the overall equation of E=mc^2?

No, the inclusion of the "negative" option does not change the equation itself. It simply expands our understanding of how the equation works and its implications in different scenarios. The equation remains the same as E=mc^2, with the option of including a negative sign for certain values of energy and momentum.

5. How does the concept of the "negative" option impact other areas of physics?

The concept of the "negative" option has far-reaching implications in various fields of physics, including particle physics, quantum mechanics, and cosmology. It allows for a better understanding of fundamental particles and their interactions, as well as the behavior of energy and matter in extreme conditions such as black holes or the early universe.

Similar threads

  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
22
Views
2K
  • Special and General Relativity
3
Replies
75
Views
3K
  • Special and General Relativity
Replies
1
Views
1K
Replies
4
Views
1K
  • Special and General Relativity
Replies
6
Views
2K
  • Special and General Relativity
Replies
20
Views
4K
  • Special and General Relativity
Replies
2
Views
2K
  • Special and General Relativity
Replies
9
Views
14K
  • Special and General Relativity
Replies
4
Views
2K
Back
Top