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Earlier we noted that even though spatial and temporal separations, $\Delta x$ and $\Delta t$, change from one reference frame to another, the combination $(-c^2 (\Delta t)^2 + (\Delta x)^2)$ does not change. As discussed, this is an example of an “invariant”.
We have now encountered two new quantities, $E=m_0c^2\gamma$. Because $\gamma$ depends directly on $v$, both $E$ and $p$ change values from one frame to another. However, as with $\Delta x$ and $\Delta t$, there is a combination of $E$ and $p$ that does not change, which is the subject of this problem.
Module 43 – $E=mc^2$ Problem 2
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Question 1 of 6
1. Question
Consider the combination $E^2 – p^2 c^2$. What does this equal?
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Question 2 of 6
2. Question
Since $E^2 – p^2 c^2$ does not contain any dependence on velocity, we learn immediately that its value does not change upon changing your frame of reference. It is an invariant. Let’s see this directly by considering how $E$ and $p$ themselves transform under a change of reference frame. First, to align with the notation we’ve been using in the course, consider an object moving along the positive $x$-axis with speed $w$, according to the ground frame. Now, change to a frame moving along the $x$-axis with speed $v$.
What is the energy $E^\prime$ of the object from this new frame of reference.
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Question 3 of 6
3. Question
What is the momentum $p^\prime$ of the object from this new frame of reference?
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Question 4 of 6
4. Question
Express $E^\prime$ in terms of $E$ and $p$.
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Question 5 of 6
5. Question
Express $p^\prime$ in terms of $E$ and $p$.
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Question 6 of 6
6. Question
Calculate $E^{\prime 2}-c^2 p^{\prime 2}$, in terms of $E$ and $p$:
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