This doesn't seem to be working? Relativity

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In summary, the derivative of \vec{F} is equal to the acceleration \vec{a} of the particle, which is given by the formula \frac{dv}{dt}=\frac{\vec{a}\cdot \vec{v}}{v}. The incorrect answers were obtained by treating m and the denominator as constants and using the quotient rule. However, the correct approach is to use the above formula to find the derivative.
  • #1
IntegrateMe
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Evaluate the derivative of [tex]\vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}}[/tex] to find the acceleration [tex] a = \frac {dv}{dt} [/tex] of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

[tex]\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}[/tex]
[tex]\frac {F}{m} (1 + \frac {v^2}{3c^2})[/tex]
 
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  • #2
IntegrateMe said:
Evaluate the derivative of [tex]\vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}}[/tex] to find the acceleration [tex] a = \frac {dv}{dt} [/tex] of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

[tex]\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}[/tex]
[tex]\frac {F}{m} (1 + \frac {v^2}{3c^2})[/tex]
How can you treat v2 as a constant?
 
  • #3
Remember the fact that:

[tex]\frac{dv}{dt}=\frac{\vec{a}\cdot \vec{v}}{v}[/tex]
 
Last edited:

1. How does the theory of relativity explain the concept of time dilation?

The theory of relativity explains time dilation as a result of the relative motion between two observers. According to this theory, time appears to pass slower for an observer who is moving at a high velocity compared to an observer who is at rest. This is because the faster an object moves, the more it warps the fabric of spacetime, causing time to slow down.

2. What is the difference between general relativity and special relativity?

Special relativity deals with the laws of physics in a uniform and non-accelerating frame of reference, while general relativity extends these laws to include the effects of gravity. In other words, special relativity explains the behavior of objects in the absence of gravity, while general relativity takes into account the effects of gravity on the fabric of spacetime.

3. How does the theory of relativity challenge our understanding of space and time?

The theory of relativity challenges our understanding of space and time by showing that they are not absolute, but rather they are relative and can change based on the observer's frame of reference. This means that the laws of physics may appear different to different observers, depending on their relative motion and the strength of gravity in their respective frames of reference.

4. Can the theory of relativity be proven?

The theory of relativity has been extensively tested and confirmed through various experiments and observations. However, like all scientific theories, it is open to further testing and refinement. Many of its predictions, such as the bending of light by gravity and the existence of black holes, have been observed and confirmed, providing strong evidence for its validity.

5. How does the theory of relativity impact our daily lives?

The theory of relativity has had a significant impact on our daily lives. It has led to the development of technologies such as GPS, which relies on precise timing and the effects of relativity on the satellites in orbit. It has also revolutionized our understanding of the universe, from the behavior of subatomic particles to the behavior of massive objects like galaxies. In addition, it has challenged our fundamental concepts of space and time, leading to new ways of thinking about the world around us.

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