Why did I make this error?What is the error in my equation?

AI Thread Summary
The discussion centers on a misunderstanding of momentum and energy conservation in a scenario involving a photon striking a solar sail. The original calculations incorrectly used the difference in photon frequencies for momentum conservation, while the correct approach requires summing the changes in momentum due to the reflection. Additionally, the momentum of the photon was misrepresented; it should be calculated as h/λ rather than h/c*ν. The final speed of the solar sail cannot equal the speed of light, highlighting the need for proper application of mass and momentum in the equations. The participants clarified that the change in momentum must be divided by the mass of the sail to determine its velocity change accurately.
JoAuSc
Messages
197
Reaction score
1
The problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense.

Let's say a photon travels from the left and hits a stationary solar sail which is perpendicular to the light. There is perfect reflection and a change in momenta for both objects. For momentum we have:

p_iphoton + p_iSS = p_fphoton + p_fSS =>
h/c*nu1 + m*(0)^2 = h/c*nu2 + mv
h/c*(nu1-nu2) = mv

energy:

KE_iphoton + KE_iSS = KE_fphoton + KE_fSS
h*nu1 + 0 = h*nu2 + mv^2
h*(nu1-nu2) = mv^2 = c*mv => v = c

(KE_iphoton is the photon's initial kinetic energy, p_iphoton is its initial momentum, KE_fphoton is the final KE, etc.; h = Planck's constant, c = speed of light, m = mass of solar sail, v = final velocity of solar sail, nu1 = initial frequency of photon, nu2 = final frequency of photon.)

Obviously, the final speed of the solar sail isn't the speed of light. What did I do wrong?
 
Physics news on Phys.org
You did not take direction into account with your momentum conservation calculation.
 
JoAuSc said:
The problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense.

Let's say a photon travels from the left and hits a stationary solar sail which is perpendicular to the light. There is perfect reflection and a change in momenta for both objects. For momentum we have:

p_iphoton + p_iSS = p_fphoton + p_fSS =>
h/c*nu1 + m*(0)^2 = h/c*nu2 + mv
h/c*(nu1-nu2) = mv
As Andrevh pointed out, the change in momentum is the sum of, not difference of, the magnitudes of the momentum before and after due to the 180 degree change in direction.

Also, you are using h/c\nu for momentum of the photon which is not correct. Momentum of the photon is E/c = h\nu/c = h/\lambda.

energy:

KE_iphoton + KE_iSS = KE_fphoton + KE_fSS
h*nu1 + 0 = h*nu2 + mv^2
h*(nu1-nu2) = mv^2 = c*mv => v = c

(KE_iphoton is the photon's initial kinetic energy, p_iphoton is its initial momentum, KE_fphoton is the final KE, etc.; h = Planck's constant, c = speed of light, m = mass of solar sail, v = final velocity of solar sail, nu1 = initial frequency of photon, nu2 = final frequency of photon.)

Obviously, the final speed of the solar sail isn't the speed of light. What did I do wrong?
I don't follow your math here. You have to divide the change in momentum of the photon by the sail's mass to find the speed change in the sail. The m in your equation is the sail's mass not the mass of the photon.

\Delta P_{ph}/m_s = v_s = P_s/m_s

KE_s = \frac{1}{2}m_sv_s^2 = P_s^2/2m_s

AM
 
Andrew Mason said:
As Andrevh pointed out, the change in momentum is the sum of, not difference of, the magnitudes of the momentum before and after due to the 180 degree change in direction.
Yes, I see it now. Thank you and Andrevh. I'm guessing you'd have a difference of frequencies in the energy equation, but a sum of frequencies in the momentum equation, so you can't substitute a change in photon energy with a change in photon momentum, at least not like I did.

Andrew Mason said:
Also, you are using h/c\nu for momentum of the photon which is not correct. Momentum of the photon is E/c = h\nu/c = h/\lambda.
That's what I had, except that rather than clarify things with parentheses like I should have, I relied on order of operations. I used h/c*\nu = (h/c)*\nu = (h*\nu)/c.

Andrew Mason said:
I don't follow your math here. You have to divide the change in momentum of the photon by the sail's mass to find the speed change in the sail. The m in your equation is the sail's mass not the mass of the photon.

\Delta P_{ph}/m_s = v_s = P_s/m_s

KE_s = \frac{1}{2}m_sv_s^2 = P_s^2/2m_s

AM
Let me clarify. In the last line of equations, I had
h*(\nu_1-\nu_2) = mv^2 = c*mv => v = c
I believe the first two terms are, in fact, equal to each other, right? Now, based on my erroneous assumption that
\Delta P_{ph} = (h/c) * (\nu_1 - \nu_2)
I got
\Delta P_{ph} = c * \Delta E_{ph}[/itex]<br /> which implies that mv^2 = cmv.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top