I think my Prof did this wrong, Need

[RagincajunLA]

Hey guys, I just had a quiz on relativistic energy and there was a question on there that I was sure I got right but got marked wrong and I think my prof did it wrong. Please tell me what you guys think of this...

The relativistic momentum of a particle traveling with velocity u in frame S is p= γ_u mu where γ_u = (1-(u^2)/(c^2))^-.5. Recalling that the relativistic force is F = dp/dt. What is the best expression for the force in terms of acceleration, the mass, and the lorentz factor γ_u?

OK, so this is how i did it. I said that F = dp/dt = (dp/du)*(du/dt). du/dt is just acceleration, and dp/du is (γ_u)*m, from the above equation. so this means that F = dp/dt = (γ_u)*m*a. Its basically the regular F = ma equation but multiplied by gamma.

My prof said that the answer is F = (d/dγt)*mu + (dm/dt)*γu + (du/dt)*mγ. He said he used chain rule but this doesn't make sense to me at all. I am 100% sure I did it the correct way. Please give me some feed back, I can get an extra 15 pts on my quiz if i can get this right!

 

[JesseM]

and dp/du is (γ_u)*m, from the above equation

How did you get that? If p = mu * (1 - u^2/c^2)^-.5, then by the product rule dp/du = mu * d/du[(1 - u^2/c^2)^-.5] + m * (1 - u^2/c^2)^-.5, and by the chain rule d/du[(1 - u^2/c^2)^-.5] = -0.5 * (1 - u^2/c^2)^-1.5 * -2u/c^2 = (u/c^2) / (1 - u^2/c^2)^1.5. So plugging that back into the product rule equation:

dp/du = [m*(u^2/c^2) / (1 - u^2/c^2)^1.5 ] + [m / (1 - u^2/c^2)^.5 ]
= [m*(u^2/c^2) / (1 - u^2/c^2)^1.5 ] + [m*(1 - u^2/c^2) / (1 - u^2/c^2)^1.5 ]
= m / (1 - u^2/c^2)^1.5

So this would not be equal to m*(γ_u), but rather to m*(γ_u)^3

 

[JesseM]

My prof said that the answer is F = (d/dγt)*mu + (dm/dt)*γu + (du/dt)*mγ. He said he used chain rule but this doesn't make sense to me at all.

Incidentally it seems like your prof was using the product rule, not the chain rule (see the part in the opening of the wiki article that says "The derivative of the product of three functions is:")--I assume there's a typo in that first term, and that the first term should actually be (dγ/dt)*mu.

 

[Meir Achuz]

$$\frac{d\bf p}{dt}= m\frac{d}{dt}\left[\frac{\bf v}<br /> {\sqrt{1-{\bf v}^2}}\right]<br /> = m[\gamma{\bf a}+\gamma^3{\bf v(v\cdot a)}]<br /> =m\gamma^3[{\bf a}+{\bf v\times(v\times a)}]$$

 

[RagincajunLA]

but why use the product rule in the first place? dp/dt should equal (dp/du)*(du/dt). why isn't dp/du equal to γm? isn't it the same thing as taking dy/dx of y=2x? in that case the derivative is just 2. so why isn't the derivative of γmu equal to γm? then you just multiply that by du/dt which is a. that part, ^^^, is what is confusing me...

 

[PAllen]

but why use the product rule in the first place? dp/dt should equal (dp/du)*(du/dt). why isn't dp/du equal to γm? isn't it the same thing as taking dy/dx of y=2x? in that case the derivative is just 2. so why isn't the derivative of γmu equal to γm? then you just multiply that by du/dt which is a. that part, ^^^, is what is confusing me...

Because gamma is a function of u.