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43.4 E=mc2E=mc^2
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Module 43 - E=mc2E=mc^2 Problem 2

Earlier we noted that even though spatial and temporal separations, Δx\Delta x and Δt\Delta t, change from one reference frame to another, the combination (−c2(Δt)2+(Δx)2)(-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=m0c2γE=m_0c^2\gamma. Because γ\gamma depends directly on vv, both EE and pp change values from one frame to another. However, as with Δx\Delta x and Δt\Delta t, there is a combination of EE and pp that does not change, which is the subject of this problem.

  1. 1. Question

    Consider the combination E2−p2c2E^2 - p^2 c^2. What does this equal?

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  2. 2. Question

    Since E2−p2c2E^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 EE and pp 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 xx-axis with speed ww, according to the ground frame. Now, change to a frame moving along the xx-axis with speed vv.

    What is the energy E′E^\prime of the object from this new frame of reference.

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  3. 3. Question

    What is the momentum p′p^\prime of the object from this new frame of reference?

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  4. 4. Question

    Express E′E^\prime in terms of EE and pp.

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  5. 5. Question

    Express p′p^\prime in terms of EE and pp.

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  6. 6. Question

    Calculate E′2−c2p′2E^{\prime 2}-c^2 p^{\prime 2}, in terms of EE and pp:

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