Relativistic Pythagorean theorem
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When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
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In physics exercises, we try to follow this pattern:
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Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
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When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
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Exercise:
Using the relativistic expressions for energy and momentum show that the following relation is fulfilled: E^ leftp c right^ + leftm c^ right^
Solution:
The relativistic energy and the relativistic momentum of a particle with mass m moving at a speed v are given by E gamma m c^ & p gamma m v where the Lorentz factor is gamma fracsqrt-fracv^c^ fracsqrt-beta^ with the abbreviation betafracvc. medskip We start from the right hand side of the relation and substitute both expressions: leftp cright^ + leftm c^right^ leftgamma m v cright^ + leftm c^right^ gamma^ m^ v^ c^ + m^ c^ m^ c^ leftgamma^ fracv^c^ + right m^ c^ leftgamma^ beta^ + right The bracket can be simplified using the definition of the Lorentz factor: gamma^ beta^ + fracbeta^-beta^ + fracbeta^ + left-beta^right-beta^ frac-beta^ gamma^ Inserting this result yields leftp cright^ + leftm c^right^ m^ c^ gamma^ leftgamma m c^right^ E^ quad square medskip Two special cases confirm the plausibility of the relation: itemize item For a particle at rest v i.e. p the relation reduces to E m c^ the rest energy. item For a massless particle m e.g. a photon it reduces to E p c. itemize Dividing the momentum by the energy furthermore gives a useful relation between the two quantities and the speed of the particle: fracp cE fracgamma m v cgamma m c^ fracvc beta
Using the relativistic expressions for energy and momentum show that the following relation is fulfilled: E^ leftp c right^ + leftm c^ right^
Solution:
The relativistic energy and the relativistic momentum of a particle with mass m moving at a speed v are given by E gamma m c^ & p gamma m v where the Lorentz factor is gamma fracsqrt-fracv^c^ fracsqrt-beta^ with the abbreviation betafracvc. medskip We start from the right hand side of the relation and substitute both expressions: leftp cright^ + leftm c^right^ leftgamma m v cright^ + leftm c^right^ gamma^ m^ v^ c^ + m^ c^ m^ c^ leftgamma^ fracv^c^ + right m^ c^ leftgamma^ beta^ + right The bracket can be simplified using the definition of the Lorentz factor: gamma^ beta^ + fracbeta^-beta^ + fracbeta^ + left-beta^right-beta^ frac-beta^ gamma^ Inserting this result yields leftp cright^ + leftm c^right^ m^ c^ gamma^ leftgamma m c^right^ E^ quad square medskip Two special cases confirm the plausibility of the relation: itemize item For a particle at rest v i.e. p the relation reduces to E m c^ the rest energy. item For a massless particle m e.g. a photon it reduces to E p c. itemize Dividing the momentum by the energy furthermore gives a useful relation between the two quantities and the speed of the particle: fracp cE fracgamma m v cgamma m c^ fracvc beta
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Exercise:
Using the relativistic expressions for energy and momentum show that the following relation is fulfilled: E^ leftp c right^ + leftm c^ right^
Solution:
The relativistic energy and the relativistic momentum of a particle with mass m moving at a speed v are given by E gamma m c^ & p gamma m v where the Lorentz factor is gamma fracsqrt-fracv^c^ fracsqrt-beta^ with the abbreviation betafracvc. medskip We start from the right hand side of the relation and substitute both expressions: leftp cright^ + leftm c^right^ leftgamma m v cright^ + leftm c^right^ gamma^ m^ v^ c^ + m^ c^ m^ c^ leftgamma^ fracv^c^ + right m^ c^ leftgamma^ beta^ + right The bracket can be simplified using the definition of the Lorentz factor: gamma^ beta^ + fracbeta^-beta^ + fracbeta^ + left-beta^right-beta^ frac-beta^ gamma^ Inserting this result yields leftp cright^ + leftm c^right^ m^ c^ gamma^ leftgamma m c^right^ E^ quad square medskip Two special cases confirm the plausibility of the relation: itemize item For a particle at rest v i.e. p the relation reduces to E m c^ the rest energy. item For a massless particle m e.g. a photon it reduces to E p c. itemize Dividing the momentum by the energy furthermore gives a useful relation between the two quantities and the speed of the particle: fracp cE fracgamma m v cgamma m c^ fracvc beta
Using the relativistic expressions for energy and momentum show that the following relation is fulfilled: E^ leftp c right^ + leftm c^ right^
Solution:
The relativistic energy and the relativistic momentum of a particle with mass m moving at a speed v are given by E gamma m c^ & p gamma m v where the Lorentz factor is gamma fracsqrt-fracv^c^ fracsqrt-beta^ with the abbreviation betafracvc. medskip We start from the right hand side of the relation and substitute both expressions: leftp cright^ + leftm c^right^ leftgamma m v cright^ + leftm c^right^ gamma^ m^ v^ c^ + m^ c^ m^ c^ leftgamma^ fracv^c^ + right m^ c^ leftgamma^ beta^ + right The bracket can be simplified using the definition of the Lorentz factor: gamma^ beta^ + fracbeta^-beta^ + fracbeta^ + left-beta^right-beta^ frac-beta^ gamma^ Inserting this result yields leftp cright^ + leftm c^right^ m^ c^ gamma^ leftgamma m c^right^ E^ quad square medskip Two special cases confirm the plausibility of the relation: itemize item For a particle at rest v i.e. p the relation reduces to E m c^ the rest energy. item For a massless particle m e.g. a photon it reduces to E p c. itemize Dividing the momentum by the energy furthermore gives a useful relation between the two quantities and the speed of the particle: fracp cE fracgamma m v cgamma m c^ fracvc beta
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Lecture notes Relativity by by
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| Massless Particle | tags |
| Finding Particle Mass | tags |