# Special relativity - Trivial exercise with spacetime interval

• LCSphysicist
In summary: In special theory of relativity, you have to keep in mind that there are two sets of quantities:(𝑖) proper quantities are those that are moving with the observer, so no changes are seen;(𝑖𝑖) apparent quantities are those that move relative to the observer, so changes are noticed.In the length contraction formula$$\dots L = L_0 sqrt{ 1 - \frac { v^2 } { c^2 } } \to L = L_0 \text{ when } v = 0 \dots$$ \\ \dots v = 0 \to \begin{align} (1) & \text{ object whose length is being measured LCSphysicist Homework Statement . Relevant Equations . For a observer on Earth, a rocket takes Mike from Earth to Pluto with a speed of 0.82 c for 33.72 yr. Find the space-time interval for the two events such as Mike leaving the Earth and reaching Pluto considering Pluto is at rest relative to Earth for the observer on Earth. I confess that i am rather confused reading this question. See:(1) \implies \Delta S² \text{is invariant}.$$Knowing (1), i thought that the better approach to this question would be to use the framework of the traveller. In his framework, and probably here is my error, i think: (2) the time it takes, IN HIS FRAMEWORK, to travel, was 33.72 yr (PS: The reasoning i used to conclude that is basically the symmetry of the lorentz transformation. The traveller believe he is stopped and the universe is flowing by him, and he measures the time to be 33.72 yr) (PSS: I think that ##\Delta t/\gamma## is the time in his reference frame measured by us, in another words, as it were measured by our point of view, is that right, isn't?) (3) The distance between both events, to him, is 0 m.$$Ds^2 = (3*10^8*33.72*R)^2$$, where R is the conversion from years to second. Apparenttly my answer is wrong. I don't know why and where is my mistake. The problem statement says: Herculi said: For a observer on Earth, a rocket takes Mike from Earth to Pluto with a speed of 0.82 c for 33.72 yr. You calculate based on the assumption: Herculi said: (2) the time it takes, IN HIS FRAMEWORK, to travel, was 33.72 yr Time dilation is a thing. The difference in the time coordinate for two events depends on the frame of reference that one chooses. You are given figures for velocity and for time difference both relative to the Earth rest frame. You should not interpret them as though they are valid for the traveler's rest frame. So you have a velocity in the Earth frame and an elapsed time in the Earth frame. That should allow you to compute a distance in the Earth frame. Doc Al, LCSphysicist and PeroK First, I don't understand the numbers. Pluto is not 30 light years from Earth. Your mistake is to use simple time dilation inappropriately, where the full Lorentz transformation is required. However, if you are given all the data in one reference frame, why transform to another? Doc Al, LCSphysicist and jbriggs444 In special theory of relativity, you have to keep in mind that there are two sets of quantities: (𝑖) proper quantities are those that are moving with the observer, so no changes are seen; (𝑖𝑖) apparent quantities are those that move relative to the observer, so changes are noticed. In the length contraction formula$$ \dots L = L_0 sqrt{ 1 - \frac { v^2 } { c^2 } } \to L = L_0 \text{ when } v = 0 \dots $$\\$$ \dots v = 0 \to \begin{align} (1) & \text{ object whose length is being measured [COLOR=blue ]does not move[ /COLOR ] } \\ (2) & \text{ relative to the observer making the measurement } \dots \end{align} $$\\$$ \dots v \neq 0 \to \begin{align} (1) & \text{ object whose length is being measured [COLOR=blue ]does move[ /COLOR ] } \\ (2) & \text{ relative to the observer making the measurement } \dots \end{align} $$\\$$ \to \begin{align} (1) & \rm{ L_0 on the right-hand side of length contraction } \\ (2) & \text { formula is a proper length } \dots \end{align} $$\\$$ \to \begin{align} (1) & \text{ 𝐿 on the left-hand side of length contraction } \\ (2) & \text { formula is an apparent length } \dots \end{align} $$\\ The same thing is true with the time dilation formula t = \frac {t_0} {sqrt{ 1 - \frac { v^2 } { c^2 } } \dots \\ In special theory of relativity, you have to keep in mind that there are two sets of quantities: (𝑖) proper quantities are those that are moving with the observer, so no changes are seen; (𝑖𝑖) apparent quantities are those that move relative to the observer, so changes are noticed. In the length contraction formula ##$$ \dots L = L_0 sqrt{ 1 - \frac { v^2 } { c^2 } } \to L = L_0 \text{ when } v = 0 \dots ##
##$$\\$$ \dots v = 0 \to \begin{align} (1) & \text{ object whose length is being measured [COLOR=blue ]does not move[ /COLOR ] } \nonumber \\ (2) & \text{ relative to the observer making the measurement } \dots \nonumber \end{align} $$## \\ ##$$ \dots v \neq 0 \to \begin{align} (1) & \text{ object whose length is being measured [COLOR=blue ]does move[ /COLOR ] } \nonumber \\ (2) & \text{ relative to the observer making the measurement } \dots \nonumber \end{align} $$## \\ ##$$ \to \begin{align} (1) & \rm{ L_0 on the right-hand side of length contraction } \nonumber \\ (2) & \text { formula is a proper length } \dots \nonumber \end{align} $$## \\ ##$$ \to \begin{align} (1) & \text{ 𝐿 on the left-hand side of length contraction } \nonumber \\ (2) & \text { formula is an apparent length } \dots \nonumber \end{align}  ## \\
The same thing is true with the time dilation formula ## t = \frac {t_0} {sqrt{ 1 - \frac { v^2 } { c^2 } } \dots ## \\

## 1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 to explain the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion and that the speed of light is constant regardless of the observer's frame of reference.

## 2. What is a spacetime interval?

A spacetime interval is a measure of the distance between two events in the fabric of spacetime. It takes into account both the spatial distance and the time difference between the events, and is used to calculate the proper time and distance between objects in different frames of reference.

## 3. Why is special relativity considered a "trivial exercise"?

Special relativity is often considered a trivial exercise because it can be derived from just two postulates: the principle of relativity and the constancy of the speed of light. These postulates, along with basic mathematical equations, can be used to explain many of the phenomena observed in the universe.

## 4. How does special relativity impact our understanding of the universe?

Special relativity has had a significant impact on our understanding of the universe. It has helped to explain the behavior of objects moving at high speeds, such as particles in particle accelerators and spacecraft traveling near the speed of light. It has also led to the development of other important theories, such as general relativity and quantum mechanics.

## 5. Can special relativity be proven?

Special relativity has been extensively tested and has been found to be consistent with numerous experiments and observations. However, like all scientific theories, it cannot be proven definitively. It is constantly being refined and expanded upon as new evidence and technologies become available.

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