Suggestions
Use up and down arrows to review and enter to select.Please wait while we process your payment
If you don't see it, please check your spam folder. Sometimes it can end up there.
If you don't see it, please check your spam folder. Sometimes it can end up there.
Please wait while we process your payment
By signing up you agree to our terms and privacy policy.
Don’t have an account? Subscribe now
Create Your Account
Sign up for your FREE 7-day trial
Already have an account? Log in
Your Email
Choose Your Plan
Individual
Group Discount
Save over 50% with a SparkNotes PLUS Annual Plan!
Purchasing SparkNotes PLUS for a group?
Get Annual Plans at a discount when you buy 2 or more!
Price
$24.99 $18.74 /subscription + tax
Subtotal $37.48 + tax
Save 25% on 2-49 accounts
Save 30% on 50-99 accounts
Want 100 or more? Contact us for a customized plan.
Your Plan
Payment Details
Payment Summary
SparkNotes Plus
You'll be billed after your free trial ends.
7-Day Free Trial
Not Applicable
Renews May 3, 2024 April 26, 2024
Discounts (applied to next billing)
DUE NOW
US $0.00
SNPLUSROCKS20 | 20% Discount
This is not a valid promo code.
Discount Code (one code per order)
SparkNotes PLUS Annual Plan - Group Discount
Qty: 00
SparkNotes Plus subscription is $4.99/month or $24.99/year as selected above. The free trial period is the first 7 days of your subscription. TO CANCEL YOUR SUBSCRIPTION AND AVOID BEING CHARGED, YOU MUST CANCEL BEFORE THE END OF THE FREE TRIAL PERIOD. You may cancel your subscription on your Subscription and Billing page or contact Customer Support at custserv@bn.com. Your subscription will continue automatically once the free trial period is over. Free trial is available to new customers only.
Choose Your Plan
For the next 7 days, you'll have access to awesome PLUS stuff like AP English test prep, No Fear Shakespeare translations and audio, a note-taking tool, personalized dashboard, & much more!
You’ve successfully purchased a group discount. Your group members can use the joining link below to redeem their group membership. You'll also receive an email with the link.
Members will be prompted to log in or create an account to redeem their group membership.
Thanks for creating a SparkNotes account! Continue to start your free trial.
We're sorry, we could not create your account. SparkNotes PLUS is not available in your country. See what countries we’re in.
There was an error creating your account. Please check your payment details and try again.
Please wait while we process your payment
Your PLUS subscription has expired
Please wait while we process your payment
Please wait while we process your payment
This section is really an extension of 4-vectors which introduced the energy-momentum 4-vector. Here we see how the concept of a 4-vector, in particular the fact that the inner product is invariant between frames, can be applied to solve problems involving collisions and decays. Many such particle-particle collisions occur on the atomic or sub-atomic level; such small particles require little (by macroscopic standards) energy to accelerate them to speeds near the speed of light. Thus, Special Relativity is necessary for describing many of these interactions.
Recall that the energy-momentum 4-vector or 4-momentum is given by:
PâÉá(E/c, |
P2âÉáP.P = E2/c2 - | |
Now let us tackle an example of first a collision problem and then a decay problem. Consider a particle with energy E and mass m. This particle moves towards another identical particle at rest. The particles collide elastically and both scatter at an angle θ with respect to the incident direction. This is illustrated in . We want to find θ in terms of E and m. We can write down the 4-momenta of the two particles. The moving particle has P1 = (E/c, p, 0, 0) and the stationary particle P2 = (mc, 0, 0, 0), where p = . The 4-mometa after the collision are: P1' = (E'/c, p'cosθ, p'sinθ, 0) and P2' = (E'/c, p'cosθ, - p'sinθ, 0), where p' = . We know from the symmetry of the situation that the energy and momentum of the two particles must be equal after the collision. Conserving energy gives E' = . Conserving momentum (only the x- direction is significant since the y components cancel) gives: p'cosθ = p/2. Thus:
P1' = ,,, 0 |
m2c2 | = | - (1 + tan2θ) | |
âá4m2c4 | = | (E + mc2)2 - | |
âáE2 + m2c4 +2Emc2 -4m2c4 | = | ||
âácos2θ | = | = |
Decay problems can be solved in a similar manner; that is, by conserving energy and momentum. The situation in which a particle of mass M and energy E decays into two identical particles is also shown in . As shown, one particle heads off in the y-direction, and the other at an angle θ. Our problem is to calculate the energies of these particles resulting from the decay. Again, we begin by writing down the 4-momenta before and after the collision. Before the decay P = (E/c,, 0, 0) and after P1 = (E1/c, 0, p1, 0) and P2 = (E2/c, p2cosθ, - p2sinθ, 0); if the created particles have mass m, then, p1 = and p2 = . This problem becomes quite algebraically messy if we proceed in the same manner as we did above, conserving energy and momentum. Instead let us exploit the invariance of the inner product to solve the problem. Conservation of energy and momentum tells us that P = P1 + P2 which implies P2 = P - P1. Taking inner products we have:
(P - P1).(P - P1) = P2.P2 | |||
âáP2 -2P.P1 + P12 = P22 | |||
âáM2c2 -2EE1/c2 + m2c2 = m2c2 | |||
âáE1 = |
Please wait while we process your payment