# Special Relativity GRE question (Where is my method flawed?)

• PsychonautQQ
In summary, the conversation discusses the non-relativistic approximations for Total Energy and Kinetic Energy, and how they are derived from the full relativistic equation by expanding the square root and keeping only the leading term. The conversation also highlights the importance of having common approximations at hand in order to solve physics problems.

## Homework Statement

http://grephysics.net/ans/9277/70

I'm not interested in how you get the right answer, I'm really just curious if somebody could show me where my thought process was flawed.

Total Energy = E = 100mc^2
Rest Energy = Mc^2
Total Energy = Kinetic Energy + Rest Energy
100mc^2 = p^2/2m + mc^2
99mc^2 = p^2/2m
198(mc)^2 = p^2
sqrt(198)mc = P
and yet the correct answer is
10^4(mc)

where is my method flawed?

The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:
$$E^2 = p^2 c^2 + m^2 c^4$$
$$E = \sqrt{p^2 c^2 + m^2 c^4} = mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4}) = mc^2 + \frac{p^2}{2m}$$

phyzguy said:
The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

$$mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})$$

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?

PsychonautQQ said:
phyzguy said:
The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

$$mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})$$

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?
Phyzguy is just showing you where your non-relativistic approximation came from. He used the binomial expansion to approximate the square root, and retained only the first two terms in the expansion. Please notice the "approximate" symbol in his equation.

PsychonautQQ said:
phyzguy said:
The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

$$mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}$$
$$E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})$$

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?

When I expand the square root of (1+ε) in a Taylor series and keep only the first order term, I get:

$$\sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}$$

For example: $$\sqrt{1.01} = 1.004987 \ldots \approx 1.005$$

One more comment. You need to have some of these more common approximations at your fingertips in order to do physics. Some of the most important are:
$$\sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}$$
$$\frac{1}{1+\epsilon} \approx 1-\epsilon$$
$$sin(\epsilon) \approx \epsilon$$
$$tan(\epsilon) \approx \epsilon$$
$$cos(\epsilon) \approx 1+\frac{\epsilon^2}{2}$$
$$ln(1+\epsilon) \approx \epsilon$$

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## 1) What is special relativity and how does it differ from general relativity?

Special relativity is a theory proposed by Albert Einstein that explains the relationship between space and time in objects moving at high speeds. It differs from general relativity in that it only applies to objects moving at constant speeds in a straight line, while general relativity applies to all objects and takes into account the effects of gravity.

## 2) What is the main concept behind special relativity?

The main concept behind special relativity is the idea that the laws of physics are the same for all observers in uniform motion, regardless of their speed or direction. This means that the laws of physics are independent of the observer's frame of reference.

## 3) How does special relativity affect the perception of time and space?

Special relativity states that time and space are relative concepts and can be affected by the speed of an observer. Time can appear to pass slower for objects moving at high speeds, and distances can appear to be shorter in the direction of motion. This is known as time dilation and length contraction.

## 4) Can special relativity be applied to everyday situations?

Yes, special relativity has been experimentally proven and is used in various technologies, such as GPS systems. However, the effects of special relativity are only noticeable at speeds close to the speed of light, so it is not typically noticeable in everyday situations.

## 5) What are some common misconceptions about special relativity?

Some common misconceptions about special relativity include the idea that it only applies to objects moving at the speed of light, or that it only applies to very large objects. In reality, special relativity applies to any object moving at high speeds, regardless of its size, and the effects of special relativity can be observed at speeds much slower than the speed of light.